{
  "schema": "cie-topical-past-papers/v1",
  "subject": "9700",
  "subject_name": "Biology",
  "syllabus_url": "https://www.cambridgeinternational.org/Images/664560-2025-2027-syllabus.pdf",
  "generated": "2026-10-21",
  "coverage": {
    "years": [
      2021,
      2022,
      2023,
      2024,
      2025,
      2026
    ],
    "sessions": [
      "March",
      "May/June",
      "Oct/Nov"
    ],
    "papers": [
      "41",
      "42",
      "43",
      "44",
      "51",
      "52",
      "53",
      "54"
    ],
    "available_papers": 84,
    "missing_papers": [
      {
        "year": 2021,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_41.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_43.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_51.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_53.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "May/June",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "May/June",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "Oct/Nov",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2021,
        "session": "Oct/Nov",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_41.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_43.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_51.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_53.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "May/June",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "May/June",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "Oct/Nov",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2022,
        "session": "Oct/Nov",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_41.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_43.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_51.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_53.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "May/June",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "May/June",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "Oct/Nov",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2023,
        "session": "Oct/Nov",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_41.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_43.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_51.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_53.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "May/June",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "May/June",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "Oct/Nov",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2024,
        "session": "Oct/Nov",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_41.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_43.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_44.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_51.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_53.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2025,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_54.pdf?download=true",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-41.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-43.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-44.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-51.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-53.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "March",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-54.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "41",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-41.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "42",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-42.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "43",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-43.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "44",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-44.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "51",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-51.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "52",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-52.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "53",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-53.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      },
      {
        "year": 2026,
        "session": "Oct/Nov",
        "variant": "54",
        "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-oct-nov/9700-biology-november-2026-question-paper-54.pdf?download=1",
        "reason": "not available from the download source when checked on 2026-10-21"
      }
    ]
  },
  "topics": [
    {
      "number": 12,
      "label": "Energy and respiration",
      "slug": "9700-topic-12-energy-and-respiration",
      "questions_html": "9700-topic-12-energy-and-respiration/questions.html",
      "answers_html": "9700-topic-12-energy-and-respiration/answers.html"
    },
    {
      "number": 13,
      "label": "Photosynthesis",
      "slug": "9700-topic-13-photosynthesis",
      "questions_html": "9700-topic-13-photosynthesis/questions.html",
      "answers_html": "9700-topic-13-photosynthesis/answers.html"
    },
    {
      "number": 14,
      "label": "Homeostasis",
      "slug": "9700-topic-14-homeostasis",
      "questions_html": "9700-topic-14-homeostasis/questions.html",
      "answers_html": "9700-topic-14-homeostasis/answers.html"
    },
    {
      "number": 15,
      "label": "Control and coordination",
      "slug": "9700-topic-15-control-and-coordination",
      "questions_html": "9700-topic-15-control-and-coordination/questions.html",
      "answers_html": "9700-topic-15-control-and-coordination/answers.html"
    },
    {
      "number": 16,
      "label": "Inheritance",
      "slug": "9700-topic-16-inheritance",
      "questions_html": "9700-topic-16-inheritance/questions.html",
      "answers_html": "9700-topic-16-inheritance/answers.html"
    },
    {
      "number": 17,
      "label": "Selection and evolution",
      "slug": "9700-topic-17-selection-and-evolution",
      "questions_html": "9700-topic-17-selection-and-evolution/questions.html",
      "answers_html": "9700-topic-17-selection-and-evolution/answers.html"
    },
    {
      "number": 18,
      "label": "Classification, biodiversity and conservation",
      "slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "questions_html": "9700-topic-18-classification-biodiversity-and-conservation/questions.html",
      "answers_html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
    },
    {
      "number": 19,
      "label": "Genetic technology",
      "slug": "9700-topic-19-genetic-technology",
      "questions_html": "9700-topic-19-genetic-technology/questions.html",
      "answers_html": "9700-topic-19-genetic-technology/answers.html"
    },
    {
      "number": null,
      "label": "Planning, analysis and evaluation",
      "slug": "9700-experimental-skills",
      "questions_html": "9700-experimental-skills/questions.html",
      "answers_html": "9700-experimental-skills/answers.html"
    }
  ],
  "papers": [
    {
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf"
    },
    {
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2021-March/qp/9700_m21_qp_52.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_51.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_52.pdf"
    },
    {
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2021-May-June/qp/9700_s21_qp_53.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_51.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_52.pdf"
    },
    {
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_53.pdf"
    },
    {
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf"
    },
    {
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2022-March/qp/9700_m22_qp_52.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_51.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_52.pdf"
    },
    {
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2022-May-June/qp/9700_s22_qp_53.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_51.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_52.pdf"
    },
    {
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_53.pdf"
    },
    {
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf"
    },
    {
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2023-March/qp/9700_m23_qp_52.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_51.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_52.pdf"
    },
    {
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2023-May-June/qp/9700_s23_qp_53.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_51.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_52.pdf"
    },
    {
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_53.pdf"
    },
    {
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf"
    },
    {
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2024-March/qp/9700_m24_qp_52.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_51.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_52.pdf"
    },
    {
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2024-May-June/qp/9700_s24_qp_53.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_51.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_52.pdf"
    },
    {
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_53.pdf"
    },
    {
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf"
    },
    {
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2025-March/qp/9700_m25_qp_52.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "44",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_51.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_52.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_53.pdf"
    },
    {
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "variant": "54",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_54.pdf?download=true",
      "local_path": "_source-pdfs/2025-May-June/qp/9700_s25_qp_54.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "44",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_51.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_51.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "52",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_52.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_52.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_53.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_53.pdf"
    },
    {
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "variant": "54",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_54.pdf?download=true",
      "local_path": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_54.pdf"
    },
    {
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "local_path": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf"
    },
    {
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "variant": "52",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-52.pdf?download=1",
      "local_path": "_source-pdfs/2026-March/qp/9700_m26_qp_52.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "41",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "42",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "43",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "44",
      "paper": 4,
      "question_count": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "51",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-51.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_51.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "52",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-52.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_52.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "53",
      "paper": 5,
      "question_count": 2,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-53.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_53.pdf"
    },
    {
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "variant": "54",
      "paper": 5,
      "question_count": 3,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-54.pdf?download=1",
      "local_path": "_source-pdfs/2026-May-June/qp/9700_s26_qp_54.pdf"
    }
  ],
  "records": [
    {
      "id": "9700-2021-m-42-q01",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-m-42-q01-p01-603c73cd70.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-m-42-q01-p02-804b5fc8d6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q01",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The European eel, Anguilla anguilla, is a fish. The sizes of eel populations tend to remain 1 (a)\nrelatively stable despite eels producing large numbers of offspring.\nSuggest reasons why the population sizes of eels tend to remain relatively stable. two\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nExplain what is meant by the general theory of evolution. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe generation time of a species is the mean (average) time from one generation (parents) (c)\nto the next generation (offspring). For example, the generation time of humans is about\n25 years.\nFig. 1.1 shows a graph of the relationship between the rate of evolution and the generation\ntime for a wide range of different species.\nrate of\nevolution\ngeneration time\nFig. 1.1\nDescribe explain the relationship shown in Fig. 1.1. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n........................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-m-42-q01-8093c4e5bc.pdf"
    },
    {
      "id": "9700-2021-m-42-q02",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-m-42-q02-p01-91fe1c2e55.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-m-42-q02-p02-43aa4044f7.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-m-42-q02-p03-cff35d7050.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-m-42-q02-p04-79a6f8a17f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q02",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The grey wolf, Canis lupus, is a large predator. During the 20th century, the grey wolf in 2 (a)\nsouth-west Europe was hunted almost to extinction.\nFig. 2.1 shows a grey wolf.\nFig. 2.1\nState the genus of the grey wolf. (i)\n..................................................................................................................................... [1]\nSuggest explain the effects on the biodiversity of south-west Europe if the grey wolf (ii) and\nbecomes extinct.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nkm2. Grey wolves can have territories of up to 950 Grey wolves can travel up to 1000 km to (b)\nstart a new population.\nState reasons why the mark-release-recapture method is suitable for estimating the size not\nof a grey wolf population.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nIn 1992, new laws were introduced across south-west Europe to protect the grey wolf. (c)\nFig. 2.2 shows the distribution of grey wolf populations in south-west Europe in 1970 and\n2012. No grey wolves from captive-breeding populations were released into the wild in\nsouth-west Europe during the period from 1970 to 2012.\n1970 2012\n4\n1 1\n3 3\n2 2\nkey to populations\nkm 1000\n1 Iberian\n2 Sierra Morena\n3 Italian\n4 Alpine\nFig. 2.2\nTable 2.1 shows the sizes of the populations of grey wolf shown in Fig. 2.2.\nTable 2.1\nsize of size of\npercentage\npopulation population population\nc",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-m-42-q02-94c198c2d4.pdf"
    },
    {
      "id": "9700-2021-m-42-q03",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q03-p01-4897e31c87.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q03-p02-ed4a66c8ee.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Vitamin A deficiency is a widespread public health problem. 3\nOne source of vitamin A is carotene in the diet. Carotene can be converted to vitamin A in the\nbody.\nWhite rice grains are not a dietary source of vitamin A because they do not contain carotene.\nScientists have genetically modified a variety of rice to improve the diet of people who are vitamin A\ndeficient. The grains of this genetically modified rice contain carotene.\nDescribe how genetic engineering could be used to modify a variety of rice so that the grains (a)\ncontain carotene.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIn a 2018 report, the United Nations Children’s Fund (UNICEF) estimated that:\n• more than 140 million children are at high risk of vitamin A deficiency\n• 1.15 million child deaths are caused by vitamin A deficiency each year.\nVitamin A deficiency is also a leading cause of childhood blindness.\nμg The recommended dietary allowance (RDA) of vitamin A for a 5-year-old child is 400 per day.\nμg Genetically modified (GM) rice contains 12 of carotene per gram of rice. In 2018, four countries\napproved GM rice as safe to eat.\nThe development of GM rice has been partly paid for by governments. It has been agreed that\nseeds of GM rice will be made available to farmers in countri",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-m-42-q03-1cfd768bc5.pdf"
    },
    {
      "id": "9700-2021-m-42-q04",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-m-42-q04-p01-7f67bb9713.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-m-42-q04-p02-099ab807ef.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q04",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Red algae are multicellular photosynthetic protoctists that contain phycoerythrin. Phycoerythrin is 4\na photosynthetic pigment.\nFig. 4.1 shows: (a)\n• the absorption spectrum of phycoerythrin\n• the action spectrum of red algae.\n75 100\n60 80\npercentage rate of 45 60\nabsorption of photosynthesis\nlight / arbitrary units 30 40\n15 20\n0 0\n600 650 700 400 450 500 550\n/ nm wavelength\nkey\nabsorption spectrum of phycoerythrin\naction spectrum of red algae\nFig. 4.1\nWith reference to Fig. 4.1, state the wavelength of peak absorption by phycoerythrin. (i)\n..................................................................................................................................... [1]\nExplain how the data in Fig. 4.1 show that phycoerythrin is the only photosynthetic (ii) not\npigment in red algae.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nPhycoerythrin is the primary pigment (reaction centre pigment) for photosynthesis in (iii) not\nred algae.\nSuggest the role of phycoerythrin in photosynthesis in red algae.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nPhycoerythrin is rarely found in plants that have green leaves. (b)\nState the name of a technique that can be used to separate and identify photosynthetic (i)\npigments.\n..................................................................................................................................... [1]\nExplain how the results of this technique would be used to confirm that phycoerythrin is (ii)\npresent in red algae and present in a plant with green leaves. not\n...........................................................................................................................................\n........................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-m-42-q04-78f32e7cde.pdf"
    },
    {
      "id": "9700-2021-m-42-q05",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-m-42-q05-p01-313c8d619c.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-m-42-q05-p02-a64bc3a2ac.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Many processes and actions in plants and animals are due to the movement of ions. 5 (a)\nTable 5.1 lists several ions, the direction of movement of each ion and the action resulting\nfrom that movement.\nComplete Table 5.1.\nTable 5.1\naction resulting from movement\nion direction of movement of ion\nof ion\n.......................................................\nfrom lumen of proximal\nNa+ convoluted tubule into proximal\n.......................................................\nconvoluted tubule cell\n.......................................................\ninto guard cell opening of stoma(ta)\n.............\ndepolarisation of presynaptic\ninto presynaptic knob\nmembrane\n.............\nrelease of acetylcholine\ninto presynaptic knob\ninto synapse\n.............\n.......................................................\nfrom intermembrane space into\nH+\nmatrix of mitochondrion .......................................................\n.......................................................\n.......................................................\nfrom sarcoplasmic reticulum to\n2+ Ca\ncytoplasm of muscle fibre .......................................................\n.......................................................\n[6]\nThe Venus fly trap is a plant that is able to capture and digest insects. It has modified leaves, (b)\nwhich have sensory hairs that respond to touch. When an insect comes into contact with the\nhairs, receptor potentials are generated.\nName the ion that moves into the cells at the base of the sensory hairs to generate (i)\nreceptor potentials.\n..................................................................................................................................... [1]\nIf two or more of these hairs are stimulated within a period of 20–35 seconds, action (ii)\npotentials are generated, causing the leaf to close quickly and trap the insect.\nSuggest why it is beneficial to the plant for stimulation of two or more hairs to be\nnecessary before the leaf will close.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe trapped insect is digested by enzymes released from the leaf cells. (iii)\nName the mechanism by which the enzymes are released.\n..................................................................................................................................... [1]\nSuggest why Venus fly trap plants need to capture insects. (iv)\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-m-42-q05-9bf3105143.pdf"
    },
    {
      "id": "9700-2021-m-42-q06",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        14
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-m-42-q06-p01-c099e7f27d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q06",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Dogs have an exceptionally good sense of smell. They can detect molecules in the air with a 6\n(1012). concentration of one part per trillion\nOlfactory receptor cells are the sensory receptors in the nasal cavity that respond to chemicals (a)\nin the air. They function in a similar way to the chemoreceptor cells in the taste buds of the\ntongue.\nDescribe how olfactory receptor cells in the nasal cavity of dogs respond to chemicals to\ngenerate an action potential.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nDogs can use their sense of smell to detect differences between the breath of people with (b)\nlung cancer and the breath of people without lung cancer. These differences are due to the\npresence of particular chemicals at very low concentrations in the breath of people with\nlung cancer. Dogs can be trained to sit when a person’s breath indicates that lung cancer is\npresent.\nScreening people for lung cancer is important for early detection and treatment.\nSuggest advantages of using in this way to screen people for lung cancer. two dogs\n...................................................................................................................................................\n.........................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-m-42-q06-abada16b55.pdf"
    },
    {
      "id": "9700-2021-m-42-q07",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-m-42-q07-p01-f4a0b5ff2f.png",
        "9700-topic-16-inheritance/assets/9700-2021-m-42-q07-p02-d1c4b17035.png",
        "9700-topic-16-inheritance/assets/9700-2021-m-42-q07-p03-7589b478ad.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Fig. 7.1 shows two chromosomes in one stage of meiosis. 7 (a)\nThe letters G to M represent the dominant alleles of seven genes and the letters g to m\nrepresent the recessive alleles of the same seven genes.\ng g G G\nh h H H\nR\nI I i i\nj j J J\nS S\nk k K K\nl l L L\nm m M M\nFig. 7.1\nName the structures labelled and on Fig. 7.1. (i) R S\n........................................................... R\n........................................................... S\n[2]\nState features visible on Fig. 7.1 that identify the chromosomes as a homologous (ii) three\npair.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nFig. 7.2 shows the same two chromosomes a little later in the same stage of meiosis. (iii)\nCrossing over is beginning to occur at point T.\ng G G g\nh H H h\nI I i i\nT\nj J j J\nk K k K\nl L l L\nm M m M\nFig. 7.2 Fig. 7.3\nFig. 7.3 shows an outline of the same two chromosomes after crossing over has\noccurred.\nComplete Fig. 7.3 by writing in the letters of the alleles along both chromosomes. Take\ncare to clearly show the difference between letters representing dominant alleles and\nletters representing recessive alleles.\n[2]\nState the stage in meiosis in which crossing over occurs. (b)\n............................................................................................................................................. [1]\nCrossing over results in genetic variation. (c)\nExplain how random assortment of homologous chromosomes also results in genetic\nvariation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-m-42-q07-713189af25.pdf"
    },
    {
      "id": "9700-2021-m-42-q08",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19,
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-m-42-q08-p01-0ae8ccf0b1.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-m-42-q08-p02-20be284f99.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-m-42-q08-p03-87f7e358cf.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-m-42-q08-p04-c4bb1cf491.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-m-42-q08-p05-774b3682cf.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q08",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Fig. 8.1 shows a transmission electron micrograph of a mitochondrion. 8 (a)\nB\nFig. 8.1\nOn Fig. 8.1, use the letter with a label line to show a location where the Krebs cycle (i) A\noccurs. [1]\nName the structure labelled on Fig. 8.1 that forms part of the inner mitochondrial (ii) B\nmembrane.\n..................................................................................................................................... [1]\nThe inner mitochondrial membrane is the site of oxidative phosphorylation. (iii)\nExplain how the structure of the inner mitochondrial membrane is linked to its function.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [5]\nThe activity of succinate dehydrogenase, an enzyme in the Krebs cycle, is used as a measure (b)\nof the rate of respiration in the mitochondria. The redox reaction catalysed by succinate\ndehydrogenase is shown in Fig. 8.2.\nsuccinate\ndehydrogenase\nsuccinate fumarate\ne–\nFig. 8.2\nThe activity of succinate dehydrogenase can be measured using a redox indicator.\nName a redox indicator explain why this indicator can be used to measure the activity of and\nsuccinate dehydrogenase.\n...................................................................................................................................................\n...................................................................................................................................................\n..........................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-m-42-q08-9964eb222c.pdf"
    },
    {
      "id": "9700-2021-m-42-q09",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-m-42-q09-p01-19af3ee360.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Compare the characteristic features of the domains Eukarya and Bacteria. [8] 9 (a)\nDescribe the methods used to conserve endangered species. [7] (b) plant\n[Total: 15]",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-m-42-q09-3ea37eb748.pdf"
    },
    {
      "id": "9700-2021-m-42-q10",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23,
        24,
        25,
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p01-4bc60f01ac.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p02-e874c07125.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p03-1ddc4eb48d.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p04-e874c07125.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p05-1ddc4eb48d.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-m-42-q10-p06-1ddc4eb48d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-42-q10",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Explain what is meant by bioinformatics outline the role of bioinformatics following the 10 (a) and\nsequencing of genomes of humans and parasites. [6]\nExplain how a microarray can be used to analyse gene expression in a tissue sample. [9] (b)\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n.........................................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-m-42-q10-4aa98f0f8c.pdf"
    },
    {
      "id": "9700-2021-m-52-q01",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 22,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        5,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p01-7fb5ac4b4c.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p02-787e412984.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p03-089e94c4ca.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p04-a83f128902.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p05-bcae2ad811.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p06-3f8d43fecb.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q01-p07-2fe451d004.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Students at a college often took a shortcut between the library and the tennis courts across an 1 (a)\narea of open land. The area of open land was covered by a large number of plants of a variety\nof different species. Walking over this area many times damaged the plants by trampling. As\na result, a shortcut path formed across the area of open land between the library and tennis\ncourts. This path was approximately one metre wide.\nFig. 1.1 shows a map of the area.\ntennis\ncourts\narea of\nopen land\nlibrary\nshortcut path\n5 m\nfootpath linking library to\ntennis courts\nFig. 1.1\nSome students decided to investigate the effect of trampling on the biodiversity of plant\nspecies growing in the area of open land.\nState the data the students need to collect to assess the biodiversity of plant species. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState the independent variable in this investigation. (ii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe a method the students could use to compare the biodiversity of plant species (b)\ngrowing in the area of untrampled open land with the biodiversity of plant species growing on\nthe shortcut path.\nYour method should be set out in a logical order and be detailed enough to let another person\nfollow it.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-m-52-q01-d4273f2557.pdf"
    },
    {
      "id": "9700-2021-m-52-q02",
      "subject": "9700",
      "year": 2021,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-march/9700_m21_qp_52.pdf?download=true",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2021-March/qp/9700_m21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-m-52-q02-p01-22317c0962.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q02-p02-902441e053.png",
        "9700-experimental-skills/assets/9700-2021-m-52-q02-p03-389b524db3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Indigo carmine is a water-soluble indicator. When oxygen is bubbled through a solution of this 2 (a)\nindicator, the colour changes from colourless to blue.\nA student used indigo carmine solution to investigate some of the factors that may limit the\nrate of photosynthesis.\nFig. 2.1 shows the apparatus used by the student.\ncm3 15 colourless indigo carmine\nsolution\nand\ncm3 5 sodium hydrogencarbonate\nsolution\nlarge test-tube\nalginate beads containing\nimmobilised single-celled\naquatic photosynthetic\nprotoctists\nFig. 2.1\nThe student prepared four sets of this apparatus. For each set of apparatus, the student\ncm3 sodium hydrogencarbonate solution: 0.25%, 0.50%, used a different concentration of 5\n0.75% and 1.00%. Sodium hydrogencarbonate solution is a source of carbon dioxide for\nphotosynthesis.\nAfter adding indigo carmine solution and sodium hydrogencarbonate solution to the alginate\nbeads, the apparatus was illuminated with a bench lamp from a distance of 20 cm. To minimise\nheating effects, the bench lamp used an LED bulb.\nFor each concentration of sodium hydrogencarbonate solution, the time for the indigo carmine\nsolution to turn from colourless to blue was recorded.\nThe investigation was repeated with four new sets of apparatus illuminated with the same\nbench lamp at a distance of 10 cm, instead of 20 cm. The same concentrations of sodium\nhydrogencarbonate solution were used as previously.\nAll other variables in this investigation were standardised.\nState the dependent variable in this investigation. (i)\n..................................................................................................................................... [1]\nState variable that needed to be standardised in this investigation, than (ii) one other\ntemperature or those variables clearly standardised in Fig. 2.1.\n..................................................................................................................................... [1]\nThe time taken for each indigo carmine solution to turn blue was converted to a rate (b)\nof photosynthesis in arbitrary units. This was done in such a way that the slowest rate of\nphotosynthesis in this investigation was given a value of 1.00 arbitrary units. All other rates of\nphotosynthesis were therefore greater than 1.00 arbitrary units.\nTable 2.1 shows the results of this investigation and some of the student’s calculated values.\nTable 2.1\ndistance of apparatus from\n20 10\nbench lamp / cm\npercentage concentration of\nsodium hydrogencarbonate 0.25 0.50 0.75 1.00 0.25 0.50 0.75 1.00\nsolution\ntime taken for indicator to\n600 310 250 250 600 310 235 195\nturn blue / s\nrate of photosynthesis\n1.00 1.94 2.40 2.40 1.00 1.94 3.08\n/ arbitrary units\nComplete Table 2.1 by calculating the missing rate of photosynthesis in arbitrary units. (i)\n[1]\nState conclusions that can be made from the results shown in Table 2.1 about (ii) three\nfactors that affect the rate of photosynthesis when these factors become limiting. and\n.....................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-m-52-q02-312faef661.pdf"
    },
    {
      "id": "9700-2021-mj-41-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3,
        4
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-41-q01-p01-602188bd05.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-41-q01-p02-5d81823e90.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-41-q01-p03-0e9635f350.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q01",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is a complex process involving the transfer of light energy into chemical 1 (a)\nenergy.\nDescribe the role of photosynthetic pigments. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nName the precise location in a chloroplast of photosynthetic pigments. (ii)\n..................................................................................................................................... [1]\nName a practical technique to separate photosynthetic pigments. (iii)\n..................................................................................................................................... [1]\nThe rate of photosynthesis is affected by many environmental factors. (b)\nA student carried out an experiment to investigate the relationship between the concentration\nof carbon dioxide available to an aquatic plant, Cabomba caroliniana, and its rate of\nphotosynthesis.\nFig. 1.1 shows the experimental set-up for this investigation.\nLED lamp at fixed distance\nsealed container\n15 cm length of C. caroliniana\nsodium\nhydrogencarbonate\n) solution (NaHCO\n3\nFig. 1.1\n• Sodium hydrogencarbonate solution was used as a source of carbon dioxide.\n• The concentration of carbon dioxide was varied using six different concentrations of\nsodium hydrogencarbonate solution.\n• All C. caroliniana plants were kept in the dark before the light was switched on at\nthe start of the experiment.\n• Five replicates were carried out at each concentration.\n• The rate of photosynthesis was obtained by calculating the percentage change in\ndissolved oxygen concentration in the solution over five minutes.\nFig. 1.2 shows the results of the investigation.\n35\n30\n25\n20\n15\npercentage\nchange in\n10\ndissolved\noxygen\n5\n0\n0.00 0.25 0.50 0.75 1.00 1.25\n–5 concentration of\nsodium hydrogencarbonate\ndm–3 / mol solution –10\n–15\nFig. 1.2\ndm–3 With reference to Fig. 1.2, explain the pattern of results obtained between 0.25 mol (i)\ndm–3 of sodium hydrogencarbonate solution. and 1.25 mol\n...........................................................................................................................................\n...........................................................................................................................................\n...............................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-mj-41-q01-237450264e.pdf"
    },
    {
      "id": "9700-2021-mj-41-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-41-q02-p01-30e182f8bd.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-41-q02-p02-06cbbd5b91.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q02",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "All organisms respire to release energy from energy-rich molecules such as glucose. 2 (a)\nAnabolic reactions energy. (i) use\nState examples of anabolic reactions occurring in living organisms. two\n...........................................................................................................................................\n..................................................................................................................................... [2]\nDescribe explain the use of ATP in glycolysis. (ii) and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nSome factory workers in the early 20th century were exposed to chemical X and experienced (b)\nserious side-effects.\nSome of the effects of exposure to chemical X are:\n• decreased production of ATP\n• increased lipid metabolism, with weight loss\n• increased production of pyruvate and lactate\n• excess heat energy release, causing an increase in body temperature, which can\nbe fatal.\nChemical X increases the permeability of the inner mitochondrial membrane to protons\n(hydrogen ions), causing some protons to leak out into the matrix.\nExplain why people exposed to chemical X show: (i)\n• decreased production of ATP\n• increased lipid metabolism.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-mj-41-q02-73b8cd460c.pdf"
    },
    {
      "id": "9700-2021-mj-41-q03",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-41-q03-p01-8a99c0b989.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-41-q03-p02-ebf0038d65.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "In 2014, scientists published a phylogenetic (evolutionary) tree diagram showing how and when 3\ndifferent groups of insects developed from ancestral groups.\nTo construct the diagram, the scientists compared nucleotide amino acid sequence data for and\n1478 genes from each of 103 different insect species. The 103 species represented all 32 orders\nof the class Insecta.\nThe data were used to arrange the species in related groups according to their degree of molecular\nsimilarity and to track how sequences changed over time.\nOutline ways in which the members of a species are similar. (a) (i) three\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nOrganisms are classified in a hierarchical system of taxonomic units. (ii)\nName the domain and kingdom to which insects belong.\ndomain ............................................\nkingdom ..........................................\n[2]\nExplain why a phylogenetic tree diagram is useful for classifying insect species. (iii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nFor this study, new software for processing large quantities of data was developed. (b)\nCalculate the total number of sequ",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-mj-41-q03-2cf8854b9b.pdf"
    },
    {
      "id": "9700-2021-mj-41-q04",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-41-q04-p01-d9a8d1d18d.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-41-q04-p02-d96171dd6b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genetic engineering involves the manipulation of naturally occurring enzymes and processes. 4 (a)\nList the enzymes used in genetic engineering (genetic modification) outline their roles in and\nnatural processes.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nGenetically modified organisms (GMOs) include crop plants. Genetically modified (GM) crop (b)\nplants have been grown in North America in increasing quantities since 1996. They are now\ngrown in many areas of the world and are eaten by millions of people and farm animals.\nTable 4.1 compares the area of land used to grow GM crop plants on six continents in 2013.\nTable 4.1\narea of land used to\ncontinent grow GM crop plants edible GM crop plants grown\n/ million hectares\nAfrica 3.5 maize, soybean\nAsia 19.1 maize, papaya, tomato\nAustralasia 0.6 canola\nEurope 0.2 maize\ncanola, maize, papaya,\nNorth America 81.1\nsoybean, squash, sugar beet\nSouth America 70.9 canola, maize, soybean\nSuggest how North America and South America benefit from growing large areas of GM (i)\ncrop plants rather than non-GM crop plants.\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-mj-41-q04-32190baeb0.pdf"
    },
    {
      "id": "9700-2021-mj-41-q05",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-41-q05-p01-356866fc85.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-41-q05-p02-dd0cedcdff.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-41-q05-p03-0d29900e7b.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-41-q05-p04-56f9d48dd9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Nerves consist of bundles of different types of neurones. 5 (a)\nFig. 5.1 shows the mean transmission speed of impulses for three types of sensory neurone,\nδ β and . C, A A\n60\nβ A\n50\n40\nmean\ntransmission\nspeed of 30\nimpulse\nms–1 /\n20 δ A\n10\nC\n0\n1 3 9\nμm mean axon diameter /\nFig. 5.1\nδ • Impulses from pain receptors in the skin are sent along sensory neurones and . C A\nδ • Sensory neurones and synapse in the spinal cord with relay neurones known C A\nas projection neurones.\n• Projection neurones send impulses to the part of the brain that perceives pain.\nβ • Impulses from touch receptors in the skin pass along sensory neurones , which A\ncan also synapse with the projection neurones in the spinal cord.\nδ β • Sensory neurones and are myelinated. A A\n• Sensory neurones are unmyelinated. C\nDescribe the results shown in Fig. 5.1 suggest explanations for these results. (i) and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nGently rubbing a damaged area of skin can reduce the perception of pain. (ii)\nSuggest an explanation for this.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nBack pain in humans can be reduced by using a transcutaneous electrical nerve stimulation (b)\n(TENS) machine.\nA TENS machine uses electrical impulses to stimulate nerve endings near the site of the\npain.\nIt is thought that TENS triggers the release of natural painkillers called endorphins at\nsynapses.\nFig. 5.2 shows synapses in a pain pathway.\npresynaptic\nknob\npostsynaptic endorphin receptor\nneurone\nimpulses from area of to pain centre\npainful back muscles in brain\nvesicle\ncontaining\nACh acetylcholine (ACh)\nreceptor\nvesicle\ncontaining\nendorphins\nimpulse from area\nstimulated by TENS\nmachine\nFig. 5.2\nSugges",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-mj-41-q05-655f781316.pdf"
    },
    {
      "id": "9700-2021-mj-41-q06",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-mj-41-q06-p01-8f44b9903b.png",
        "9700-topic-14-homeostasis/assets/9700-2021-mj-41-q06-p02-2dfab7fdc1.png",
        "9700-topic-14-homeostasis/assets/9700-2021-mj-41-q06-p03-99b3afdce2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Thermoregulation is the control of the core temperature of the body. 6 (a)\nThermoreceptors send information about a change in core temperature to the hypothalamus.\nThe hypothalamus controls the core temperature by sending impulses to activate several\nphysiological responses.\nTable 6.1 lists physiological responses to a in core temperature. decrease\nComplete Table 6.1 to explain how each response contributes to maintaining a core\ntemperature within narrow limits.\nTable 6.1\nresponse explanation\n.........................................................................\nblood vessels near\nthe surface of the skin\n.........................................................................\nconstrict (narrow)\n.........................................................................\n.........................................................................\nshivering occurs\n.........................................................................\n.........................................................................\n.........................................................................\nhairs rise\n.........................................................................\n.........................................................................\n.........................................................................\nreduction in the\nproduction of sweat .........................................................................\n.........................................................................\n[4]\nAnimals can also use behaviour to adapt to temperature changes. (b)\nSuggest why some animals curl up their bodies in cold weather.\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe Sitatunga deer, Tragelaphus spekii, and the Nile monitor lizard, Varanus niloticus, live in (c)\nAfrica.\nFig. 6.1 shows a Sitatunga and Fig. 6.2 shows a Nile monitor.\nFig. 6.1 Fig. 6.2\nAn investigation was carried out to measure the core temperature of the Sitatunga and the\nNile monitor as the environmental temperature was increased.\nThe results are shown in Fig. 6.3.\n40\nSitatunga\n30\nNile monitor\ncore\ntemperature 20\n/ °C\n10\n0\n0 10 20 40 30\nenvironmental temperature / °C\nFig. 6.3\nDescribe the results shown in Fig. 6.3. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-mj-41-q06-ba0b384b3e.pdf"
    },
    {
      "id": "9700-2021-mj-41-q07",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-mj-41-q07-p01-c203df332b.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-41-q07-p02-fae64dba5a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "People vary in which molecules are present on the surface of their red blood cells. For example, in 7\nthe ABO blood group system, people may be categorised into one of four blood groups: A, B, AB\nand O.\nI, I with three alleles, determines these blood groups. Gene is an example of a A single gene,\ngene with multiple alleles.\nIA IB • Alleles and are codominant.\nIo IA IB. • Allele is recessive to alleles and\nExplain what is meant by codominant and recessive. (a)\ncodominant ................................................................................................................................\n...................................................................................................................................................\nrecessive ...................................................................................................................................\n...................................................................................................................................................\n[3]\nTable 7.1 shows a cross between a person with blood group A and a person with blood (b)\ngroup B.\nComplete Table 7.1.\nTable 7.1\nparental phenotype\nA B\n(blood group)\nIA Io IB Io parental genotype\ngametes\n........................ ........................ ........................ ........................\noffspring genotype\n........................ ........................ ........................ ........................\noffspring phenotype\n(blood group)\n........................ ........................ ........................ ........................\n[3]\nHaemophilia is a blood condition where bleeding continues for a long period of time after a (c)\nblood vessel has been damaged.\nA person with haemophilia has a mutation in a gene on the X chromosome, which results in\nthe lack of a blood clotting factor.\nExplain why a man who has haemophilia cannot pass the condition to his male offspring.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 9]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-mj-41-q07-5700e3bf1e.pdf"
    },
    {
      "id": "9700-2021-mj-41-q08",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-41-q08-p01-3e2152339d.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-41-q08-p02-5fb0be917c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Humans use selective breeding to improve the features of crop plants and domestic 8 (a)\nanimals.\nState another term for selective breeding. (i)\n..................................................................................................................................... [1]\nDescribe how selective breeding can be used to improve the milk yield of dairy cattle. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [5]\nExplain what is meant by disruptive selection. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[To",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-mj-41-q08-8a6cdc70ab.pdf"
    },
    {
      "id": "9700-2021-mj-41-q09",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-41-q09-p01-3d5b514866.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Describe how you would carry out an investigation to calculate the effect of substrate 9 (a)\nconcentration on the rate of respiration of yeast in anaerobic conditions using a redox\nindicator, such as DCPIP or methylene blue. [8]\nCompare respiration in anaerobic conditions in mammalian tissue and in yeast cells. [7] (b)\n[Total: 15]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-mj-41-q09-2f6e505b3d.pdf"
    },
    {
      "id": "9700-2021-mj-41-q10",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-41-q10-p01-20d4ed88df.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-41-q10-p02-eaf0753114.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-41-q10-p03-dfc93b82ba.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-41-q10",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Explain how speciation can occur due to geographical separation. [8] 10 (a)\nUsing examples, explain why species may become extinct. [7] (b)\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n.............................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-mj-41-q10-d6bde3ed9f.pdf"
    },
    {
      "id": "9700-2021-mj-42-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-42-q01-p01-16c08d28a9.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-42-q01-p02-205253de26.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a diagram of part of a neurone membrane at resting potential. 1 (a)\nB\ntissue\nfluid\nA phospholipid\nbilayer\naxoplasm\nD\nC\nFig. 1.1\nWith reference to Fig. 1.1, name A, B, and D. (i)\n........................................................................................... A\n........................................................................................... B\n........................................................................................... [3] D\nSubstance is required to make structure function. (ii) C A\nName substance C.\n............................................................................................ [1]\nSome drugs can affect the functioning of neuromuscular junctions or cholinergic synapses. (b)\nTable 1.1 lists three drugs and describes their action on neuromuscular junctions or cholinergic\nsynapses.\nTable 1.1\ndrug action of drug\ncurare blocks muscle cell membrane receptors at neuromuscular junctions\nnerve gas inhibits acetylcholinesterase function in synapses\nalcohol inhibits exocytosis of neurotransmitters in synapses\nSuggest describe the immediate consequence of the action of each drug on a and\nneuromuscular junction or cholinergic synapse.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\n[Total: 9]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-mj-42-q01-acd354c792.pdf"
    },
    {
      "id": "9700-2021-mj-42-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-42-q02-p01-925399cfee.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-42-q02-p02-cc7c882ff8.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-42-q02-p03-b9b36f15b3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q02",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "All organisms respire. The ATP produced as a result of respiration is used as the energy 2 (a)\ncurrency of the cell.\nOutline examples of movement in cells that use ATP. (i) two\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nATP cannot be stored in cells so it has to be continually re-synthesised to meet the (ii)\ndemands of an organism.\nA person with a total quantity of 0.2 moles of ATP needs to hydrolyse 150 moles of ATP\nper day.\nCalculate how many times the total quantity of 0.2 moles of ATP has to be re-synthesised\nper hour to meet the demand of 150 moles per day.\nShow your working and give your answer to the nearest whole number.\nanswer = ......................................................... [2]\nName the stages in which chemiosmosis occurs in respiration and in photosynthesis. (iii)\nrespiration .........................................................................................................................\nphotosynthesis ..................................................................................................................\n[2]\nFur seals are mammals that are adapted to live in cold temperatures. Fur seals have large (b)\nquantities of a type of fat tissue known as brown adipose tissue. Brown adipose cells contain\nmany mitochondria. These mitochondria contain a transport protein called thermogenin.\nFig. 2.1 shows the role of thermogenin in a mitochondrion of a brown adipose cell when\nexternal temperatures are cold.\nH+ H+ H+ H+\nH+ H+ H+\nH+ H+\nH+ H+ H+\nintermembrane\nH+ H+ H+\nspace\ninner electron\nmitochondrial transport thermogenin\nmembrane chain\nATP\nsynthase\nheat H+ H+ H+\nmatrix\nH+\nADP + Pi ATP\nH+\nFig. 2.1\nWith reference to Fig. 2.1, describe explain the effect of thermogenin on ATP (i) and\nsynthesis.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.....................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-mj-42-q02-d749d0b98a.pdf"
    },
    {
      "id": "9700-2021-mj-42-q03",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-42-q03-p01-28aed0ae1d.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-42-q03-p02-aaf09b8678.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-42-q03-p03-28a096bb8f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "One way to measure global biodiversity is to count the number of species of organisms. 3 (a)\nTable 3.1 shows estimates for 2009 of the number of species in some taxa of animals.\nThe numbers in brackets are the numbers that were updated in 2019 from the International\nUnion for Conservation of Nature and Natural Resources (IUCN) for three intensively studied\ntaxa.\nTable 3.1\nnumber of species in each taxon\nkingdom phylum class\narachnids\n102 248\ncrustaceans\n47 000\narthropods myriapods\n1 191 770 16 072\ninsects\n1 024 945\nother arthropod classes\n1505\nfish\n31 269\nanimals\n1 438 805\namphibians\n6515 (6722)\nchordates birds\n63 543 9990 (11 126)\nreptiles\n8734\nmammals\n5487 (5692)\nmolluscs\n85 000\nother animal phyla\n98 492\nList features shared by species. (i) three animal\n1 ........................................................................................................................................\n2 ........................................................................................................................................\n3 ........................................................................................................................................\n[3]\nExplain how the classification of species into a taxonomic hierarchy assists the work of (ii)\nconservation bodies such as the IUCN.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nTable 3.1 shows that the number of species of amphibians, birds and mammals has (iii)\nincreased between 2009 and 2019.\nDiscuss whether the increase in numbers means that these classes of chordate are\nbeing successfully conserved.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.....................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-mj-42-q03-d760a3d8c4.pdf"
    },
    {
      "id": "9700-2021-mj-42-q04",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-42-q04-p01-f78d1542f9.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-42-q04-p02-63278f3eec.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-42-q04-p03-2a270021a4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genetic technology involving the creation of recombinant DNA can be used to treat different 4 (a)\nhuman diseases. These include diseases such as diabetes that may have multiple causes\nand inherited disorders that are caused by a single gene.\nOutline ways of using recombinant DNA technology to treat these diseases. two different\ndiabetes ....................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nsingle gene disorder .................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n[2]\nA new application of recombinant DNA technology uses the genetic modification of a plant to (b)\nprevent disease. It aims to prevent the most common source of food-borne disease, which is\ncaused by eating food contaminated with pathogenic Escherichia coli bacteria.\nScientists genetically modified edible spinach plants to produce colicins. Colicins are\nantimicrobial proteins that can kill pathogenic E.coli. Colicins are normally made by other\nbacteria.\nAn experiment was carried out to test this application by spraying an extract of genetically\nmodified (GM) spinach in buffer solution onto raw meat contaminated with pathogenic E.coli.\nThe meat was stored at 10 °C for varying lengths of time before the numbers of viable (living)\npathogenic E.coli bacteria were counted. Fig. 4.1 shows the results.\n4\n3\nKey:\nlog number of\nno treatment\nviable bacterial cells\n2\nbuffer solution only\nper gram of meat\nbuffer solution plus\nGM spinach extract\n1\n0\n1 24 72\ntime stored at 10 °C / hours\nFig. 4.1\nWith reference to Fig. 4.1, evaluate the effectiveness of using GM spinach spray on raw (i)\nmeat to prevent food-borne disease.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n....................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-mj-42-q04-37f12505d0.pdf"
    },
    {
      "id": "9700-2021-mj-42-q05",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-42-q05-p01-f28dd1565f.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-42-q05-p02-f146b6401d.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-42-q05-p03-449eecd9d9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q05",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is a complex process involving a light dependent stage and a light independent 5 (a)\nstage.\nName the products of the light dependent stage that are needed in the light independent (i)\nstage.\n...........................................................................................................................................\n..................................................................................................................................... [2]\nDescribe the role of chlorophyll b in photosynthesis. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nA student carried out an experiment to investigate the effect of light intensity and light wavelength\non the rate of photosynthesis.\n• An aquatic plant, Elodea canadensis, was put into a beaker containing sodium\nhydrogencarbonate solution as a source of carbon dioxide.\n• To minimise changes in temperature, an LED lamp was used as a source of light.\n• The lamp was switched on and the number of bubbles released by the aquatic plant in\n1 minute was counted.\n• The lamp was placed at seven different distances from the beaker to change light intensity.\n• Five replicates were carried out at each lamp distance.\n• All other variables were controlled.\n1\n. The student calculated the light intensity for each distance (d) using (b)\n2 d\nTable 5.1 shows the calculated light intensities for each distance.\nTable 5.1\ndistance between plant 1\n/ light intensity\n2 / and lamp m d\n0.025 1600\n0.050 400\n0.100\n0.150 44\n0.200 25\n0.250 16\n0.300 11\nComplete Table 5.1 by calculating the light intensity for distance 0.100 m. [1]\nAt each distance from the lamp, the experiment was repeated using a red filter in front of the (c)\nlamp to give a different wavelength of light. The experiment was repeated using a blue filter\nand then using a green filter. Each filter transmitted the same light intensity.\nThe student calculated the mean rate of bubble production as a measure of the rate of\nphotosynthesis.\nFig. 5.1 shows a graph of the results.\n160\n140\n120\n100\nKey: rate of\nphotosynthesis white light\n80\nmin–1 / bubbles\nred light\n60 blue light\ngreen light\n40\n20\n0\n200 400 600 800 1000 1200 1400 1600 0\nlight intensity\nFig. 5.1\nWith reference to Fig. 5.1: (i)\n• state the range over which light intensity is the limiting factor\n• explain why light intensity above this range is not limiting the rate of\nphotosynthesis.\n...........................................................................................................................................\n.................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-mj-42-q05-e0c9a064c9.pdf"
    },
    {
      "id": "9700-2021-mj-42-q06",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-mj-42-q06-p01-dba39e5dfc.png",
        "9700-topic-14-homeostasis/assets/9700-2021-mj-42-q06-p02-ad3b21b11c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Glucagon-like peptide-1 (GLP-1) is a hormone consisting of 31 amino acids. It is secreted by 6\nspecialised intestinal epithelial cells known as L cells.\nFig. 6.1 outlines the secretion of GLP-1 and its effects on the pancreas, brain and stomach.\nfood\ningested\nbrain\nstomach\ndecrease\ninhibition\nin\nstomach\nappetite\nempties\nnutrients\nstimulate\npancreas L cells in\nintestine\ndecrease increase\nin in GLP-1\nglucagon insulin secreted\nsecretion secretion\nα β cells cells\nFig. 6.1\nβ GLP-1 binds to receptors on cell surface membranes of cells in the pancreas. These cells (a)\nsecrete insulin.\nName the type of membrane component that forms a receptor in the cell surface (i)\nmembrane.\n..................................................................................................................................... [1]\nState how GLP-1 is transported to the pancreas. (ii)\n..................................................................................................................................... [1]\nα GLP-1 inhibits the secretion of glucagon by cells in the pancreas. (b)\nDescribe processes in the liver that are stimulated by glucagon. two\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nSuggest reason why GLP-1 causes a decrease in appetite. (c) one\n............................................................................................................................................. [1]\nGLP-1 inhibits the emptying of the stomach. This is an example of negative feedback. (d)\nExplain what is meant by negative feedback.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...............................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-mj-42-q06-a06f8578b9.pdf"
    },
    {
      "id": "9700-2021-mj-42-q07",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q07-p01-86f0adba99.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q07-p02-d9d86137e6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. 7 (a)\nA fruit fly with normal features is described as wild type. It has red eyes and its wings are\nlonger than its abdomen. There are mutant variations such as purple eyes or short (vestigial)\nwings.\nFig. 7.1 shows a wild type fruit fly and a mutant fruit fly with purple eyes and vestigial wings.\nwild type mutant\nred eyes purple eyes\nlong wing vestigial wing\nFig. 7.1\n• The genes coding for eye colour and wing length are located on the same\nchromosome.\n• Allele for red eyes is dominant to allele for purple eyes. R r\n• Allele for long wings is dominant to allele for vestigial wings. N n\nExplain what is meant by the terms allele and dominant.\nallele .........................................................................................................................................\ndominant ...................................................................................................................................\n...................................................................................................................................................\n[2]\nA wild type fruit fly, heterozygous for both genes, was crossed with a fruit fly that was (b)\nhomozygous recessive for both genes.\nTable 7.1 is a summary of the cross. (i)\nComplete Table 7.1.\nTable 7.1\nwild type parent double homozygous\nrecessive parent\nparental\n............................................... ............................................... phenotype\nx\n............................................... ...............................................\nparental\ngenotype ............................................... ............................................... x\noffspring\noffspring\n...................... ...................... ...................... ...................... phenotype\n...................... ...................... ...................... ......................\noffspring\ngenotype ...................... ...................... ...................... ......................\nnumber of\n1339 1195 151 154\noffspring\n[5]\nExplain why the four offspring phenotypes are not present in a 1:1:1:1 ratio. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-mj-42-q07-6f0b23a726.pdf"
    },
    {
      "id": "9700-2021-mj-42-q08",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-42-q08-p01-79b0515ac4.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-42-q08-p02-5fb0be917c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "In continuous variation, a population shows a range of phenotypes between two extremes with 8 (a)\nno distinct groups. Height and mass are examples of phenotypic traits that show continuous\nvariation.\nDescribe the genetic basis for continuous variation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nEnvironmental factors can contribute to continuous variation. (b)\nSuggest environmental factors that may affect the body mass of an animal. two\n...................................................................................................................................................\n............................................................................................................................................. [2]\nHumans have used selective breeding (artificial selection) for thousands of years to improve (c)\nthe quality of livestock.\nOutline the principles of selective breeding in livestock.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.........................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-mj-42-q08-087d8cfbd2.pdf"
    },
    {
      "id": "9700-2021-mj-42-q09",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        21
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-42-q09-p01-89fc095b6d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Describe how the structure of a mitochondrion is related to its function. [9] 9 (a)\nExplain how rice is adapted to grow with its roots submerged in water. [6] (b)\n[Total: 15]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-mj-42-q09-7e91431c2f.pdf"
    },
    {
      "id": "9700-2021-mj-42-q10",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_42.pdf?download=true",
      "source_pages": [
        21,
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q10-p01-71ec86a214.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q10-p02-1ddc4eb48d.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q10-p03-e874c07125.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-42-q10-p04-dfc93b82ba.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-42-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Explain how different types of gene mutation can affect the phenotype outline the effects 10 (a) and\nof the mutant alleles that cause Huntington’s disease on the phenotype of a person. [9]\nExplain how gibberellin acts on DELLA proteins to stimulate the production of amylase in a (b)\ngerminating seed. [6]\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..........................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-mj-42-q10-22d1403e96.pdf"
    },
    {
      "id": "9700-2021-mj-43-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-43-q01-p01-4ac31c92d5.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-43-q01-p02-116c13d547.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a diagram of a part of a sarcomere in striated muscle. 1 (a)\nB\nC\nA\nactin\nD\nmyosin\nFig. 1.1\nWith reference to Fig. 1.1, name A, B, and D. C\n........................................................................................... A\n........................................................................................... B\n........................................................................................... C\n........................................................................................... D\n[4]\nWhen a muscle cell is stimulated, calcium ions are released from the sarcoplasmic reticulum. (b)\nWith reference to Fig. 1.1, describe the role of calcium ions in the contraction of the sarcomere.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nWhen a mammal dies, aerobic respiration stops. The striated muscles contract and remain (c)\ncontracted for a few hours after death.\nSuggest why the muscles remain contracted for a few hours.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\n[Total: 10]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-mj-43-q01-0f3b911a61.pdf"
    },
    {
      "id": "9700-2021-mj-43-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-43-q02-p01-8a2b2c4a4f.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-43-q02-p02-ba9e4b9dc5.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-mj-43-q02-p03-8f35de9972.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q02",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "During respiration, respiratory substrates such as glucose are used as a source of energy to 2 (a)\nproduce ATP molecules.\nCells can maintain a different concentration of ions within the cell compared with the (i)\nextracellular environment.\nIt is estimated that animals may use 50% of ATP made from respiration to maintain\nhigher concentrations of some ions inside cells.\nName the mechanism that can maintain a higher concentration of ions inside a cell.\n..................................................................................................................................... [1]\nOnly a proportion of energy released by the respiration of glucose is used to make ATP. (ii)\n• The energy yield from the respiration of glucose in aerobic conditions is\nmol–1. 2870 kJ\nmol–1 • When ATP is hydrolysed to ADP, it releases 30.5 kJ of energy.\n• It is estimated that 31 moles of ATP are made per mole of glucose respired.\nCalculate the percentage efficiency of glucose respiration.\nShow your working.\n...................................................... % [2]\nThe energy that is converted to ATP during respiration is released as heat energy. (iii) not\nState homeostatic use of this heat energy in mammals. one\n..................................................................................................................................... [1]\nState the term used to describe the reaction that results in the production of ATP during (iv)\nthe Krebs cycle.\n..................................................................................................................................... [1]\nWhen a person exercises, power is generated. (b)\ns–1). Power is measured in joules per second (J\nThe power generated when a person exercises will vary depending on the type and intensity\nof exercise. More power is generated when the intensity of exercise increases.\nAn experiment was carried out to determine whether increasing the intensity of exercise in a\nhealthy human male caused a change in the value of the respiratory quotient (RQ).\n• The man had a balanced diet.\n• The RQ was calculated at rest.\n• He carried out exercise for the same length of time on four separate days.\n• The intensity of exercise was increased over the four days, generating different\npowers.\n• The RQ was calculated for the four different powers generated.\nThe results are shown in Fig. 2.1.\n1.2\n1.0\nRQ\n0.8\n0.6\n0.4\n0.2\n0\n0 50 100 150 200\ns–1 power generated / J\nat rest\nFig. 2.1\nWith reference to Fig. 2.1, suggest what can be deduced from:\nthe RQ at rest\n...................................................................................................................................................\n...................................................................................................................................................\n–1 the RQ when the power generated is 110 J s\n......................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-mj-43-q02-02a0580c14.pdf"
    },
    {
      "id": "9700-2021-mj-43-q03",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-43-q03-p01-e7fc9d20ee.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-mj-43-q03-p02-16d2a562b8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Fruit flies of the genus Bactrocera are pests that damage fruit crops. 3\nTable 3.1 gives the names and geographical ranges of Bactrocera fruit flies that were classified as\nfour separate species.\nTable 3.1\nname geographical range\nB. dorsalis China, India, Thailand\nB. invadens Africa, India\nB. papayae Indonesia, Malaysia\nB. philippinensis Philippine Islands, Borneo\nIn 2014, the classification of these flies was changed. All four species were recognised as\nbelonging to a single species, B. dorsalis.\nSuggest reasons why the four species were originally classified as separate species. (a) (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSome students decided to investigate whether the flies were members of one species or (ii)\nfour separate species.\nSuggest a simple investigation that the students could carry out.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nFemale B. dorsalis lay their eggs in different types of fruits, such as avocados, bananas, (b)\nmangos and papayas. The eggs hatch into larvae that eat the fruit. The actions of the female\nflies and larvae allow microorganisms to enter the fruit. The microorganisms feed by secreting\nextracellular enzymes, causing the fruit to rot.\nName kingdoms that include organisms that could spoil fruit by secreting extracellular two\nenzymes.\n1 ....................................................................\n2 ....................................................................\n[2]\nBefore the four Bactrocera species were reclassified as a single species in 2014, some\ngovernments in Asia banned fruit imports from African countries to avoid introducing B. invadens\nas an alien species.\nExplain why the introduction of alien species",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-mj-43-q03-87bd55103f.pdf"
    },
    {
      "id": "9700-2021-mj-43-q04",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-43-q04-p01-eab08a2e11.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-43-q04-p02-4daf74d180.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Bt maize has been genetically engineered to resist insect attack. In 2019, Bt maize formed 4 (a)\n83% of the maize grown in the USA.\nOutline how the genetic engineering process produced Bt maize plants with resistance to\ninsect attack.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nA proposed conservation benefit of Bt maize is that, unlike chemical insecticide sprays, it (b)\nshould not harm non‑target insects.\nWhen Bt maize was first grown in the USA, scientists were concerned that maize pollen\ncontaining the Bt toxin might be deposited by the wind on wild plants such as milkweeds,\nAsclepias spp. Milkweed leaves are the food source of caterpillars (larvae) of the monarch\nbutterfly, Danaus plexippus.\nIn 1999, a laboratory experiment tested the effects of feeding milkweed leaves treated in\ndifferent ways to the caterpillars of monarch butterflies. Each group of caterpillars was provided\nwith milkweed leaves to eat throughout the experiment. The three different treatments were:\n• leaves with no pollen\n• leaves dusted with pollen from non‑Bt maize\n• leaves dusted with pollen from Bt maize.\nFig. 4.1 shows the res",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-mj-43-q04-7c96247447.pdf"
    },
    {
      "id": "9700-2021-mj-43-q05",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-43-q05-p01-dad0fbf8a9.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-43-q05-p02-facde80168.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-mj-43-q05-p03-e2aeaf5bb1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q05",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is a process that results in the synthesis of complex organic molecules. 5\nRespiration is a process that involves the breakdown of complex organic molecules.\nThe light dependent stage of photosynthesis involves cyclic and non‑cyclic (a)\nphotophosphorylation.\nDescribe differences between cyclic and non‑cyclic photophosphorylation. two\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nBiochemical processes involving carbon dioxide change the external environment of the (b)\naquatic plant, Elodea canadensis, and the pond snail, Lymnaea stagnalis. Carbon dioxide\ndissolves in water to form carbonic acid.\n3 of distilled water containing a A student set up 8 test‑tubes. Each test‑tube contained 30 cm\npH indicator, bromothymol blue.\nFig. 5.1 shows the experimental set up for these 8 test‑tubes.\nThe test‑tubes were left for 12 hours.\naquatic plant pond snail Key:\nhigh light\nintensity\n1 2 3 4\ndark\n5 6 7 8\nFig. 5.1\nFig. 5.2 shows how the colour of bromothymol blue changes with pH.\nblue increasing pH\ngreen neutral pH\nyellow decreasing pH\nFig. 5.2\nAll tubes were green at the start of the experiment.\nThe results of the experiment are shown in Table 5.1.\nTable 5.1\ntest‑tube colour after 12 hours\n1 green\n2 blue\n3 yellow\n4 green\n5 green\n6 yellow‑green\n7 yellow\n8 yellow\nExplain the purpose of test‑tube 1 and test‑tube 5. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nExplain the results of test‑tubes 2, 3 and 4. (ii)\ntest‑tube 2 .........................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\ntest‑tube 3 .........................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\ntest‑tube 4 ................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-mj-43-q05-5ab59c689d.pdf"
    },
    {
      "id": "9700-2021-mj-43-q06",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-mj-43-q06-p01-ba5dbc4924.png",
        "9700-topic-14-homeostasis/assets/9700-2021-mj-43-q06-p02-1c34ead7dd.png",
        "9700-topic-14-homeostasis/assets/9700-2021-mj-43-q06-p03-2b4c2779ef.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The bottlenose dolphin, Tursiops truncatus, is an aquatic mammal. It has adaptations to 6 (a)\nconserve heat when swimming in cold water.\nFig. 6.1 shows a bottlenose dolphin.\nflipper\nFig. 6.1\nThe arteries in the bottlenose dolphin that carry blood to the flippers are surrounded by (i)\nveins bringing blood back to the rest of the body.\nOutline how this arrangement of blood vessels is an adaptation to conserve body heat.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nWhen a dolphin has exercised for a long time, it may need to lose excess heat. Unlike (ii)\nhumans, dolphins do produce sweat. not\nSuggest how a dolphin may lose excess heat.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe control of blood glucose concentration in dolphins is the same as in most mammals. (b)\nA study was carried out to investigate the concentration of insulin in the blood of dolphins that\nwere provided with a diet of fish, supplemented with glucose.\n• One group of dolphins ate fish with 3 g of glucose for every kg of fish.\n• Another group ate fish with 11 g of glucose for every kg of fish.\nThe concentration of insulin in the blood was measured 60 minutes before being fed, at the\ntime of being fed and at regular intervals afterwards.\nMeasurements were also taken for a third group of dolphins, at the same time intervals, that\nwere not fed any fish (fasting).\nThe results of the study are shown in Fig. 6.2.\n80\nKey:\n70\nfasting\ng / kg 3\n60\ng / kg 11\n50\nmean\nconcentration\n40 of insulin\nin blood\ncm–3 30 pmol /\n20\n10\n0\n–60 0 60 120 180 240 300 360 420 480\ntime / min\nfood given\nFig. 6.2\nDescribe the trends shown in Fig. 6.2. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..........................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-mj-43-q06-60f17cf894.pdf"
    },
    {
      "id": "9700-2021-mj-43-q07",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-mj-43-q07-p01-a8ab2f3733.png",
        "9700-topic-16-inheritance/assets/9700-2021-mj-43-q07-p02-b9dd8f6b72.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "α‑globin β‑globin Haemoglobin is made of two chains and two chains. 7 (a)\nβ‑globin. A person may have a mutation in the gene coding for This is due to a base substitution\nβ‑globin and leads to the production of abnormal and therefore abnormal haemoglobin.\nA person who is homozygous for the mutant allele will have a condition called sickle cell\nanaemia.\nOutline the phenotypic effects of having abnormal haemoglobin in a person with sickle cell\nanaemia.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nA β‑globin. is the allele coding for normal (b) Hb\nS β‑globin. is the allele coding for abnormal Hb\nThe alleles are codominant.\nExplain what is meant by the alleles being codominant. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nA person who is heterozygous will have sickle cell trait but may not experience the (ii)\nsymptoms of sickle cell anaemia.\nSuggest why a person with sickle cell trait may not show the symptoms of sickle cell\nanaemia.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nConstruct a genetic diagram to show the possible offspring for a man and a woman who (iii)\nboth have sickle cell trait.\nparental sickle cell trait × sickle cell trait\nphenotype\nparental\ngenotype\ngametes\noffspring\ngenotypes\noffspring\nphenotypes\n[3]\n[Total: ",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-mj-43-q07-4c82137881.pdf"
    },
    {
      "id": "9700-2021-mj-43-q08",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-43-q08-p01-365e02cb50.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-43-q08-p02-b0e4a11da6.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-mj-43-q08-p03-5fb0be917c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Explain what is meant by stabilising selection. 8 (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nFig. 8.1 shows the distribution of fur colour in a population of mice. (b)\nThe colour of the soil where the mice lived was mid‑brown.\nnumbers\nin\npopulation\nwhite light mid- dark black\nbrown brown brown\nfur colour\nFig. 8.1\nSuggest why mice with white fur and mice with black fur were present in the lowest (i)\nnumbers in Fig. 8.1.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nAn environmental event caused the colour of the soil to change to dark brown. (ii)\nOn Fig. 8.1, sketch a curve to show the distribution of fur colour in the mouse population\nafter many generations. [2]\nExplain what is meant by genetic drift. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\n[Total: 9]\nSection B\nAnswer question. on",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-mj-43-q08-62b2cad892.pdf"
    },
    {
      "id": "9700-2021-mj-43-q09",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-mj-43-q09-p01-61cfb93490.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q09",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Explain the use of genes for fluorescence as markers in gene technology. [6] 9 (a)\nDiscuss the implications of using genetically modified organisms in food production. (b) social\n[9]\n[Total: 15]",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-mj-43-q09-f74526c7b0.pdf"
    },
    {
      "id": "9700-2021-mj-43-q10",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_43.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-43-q10-p01-f498ef307d.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-43-q10-p02-e874c07125.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-mj-43-q10-p03-dfc93b82ba.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-43-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Describe the roles of sodium ions in selective reabsorption in the nephron calcium ions 10 (a) and\nin the functioning of a cholinergic synapse. [7]\nCompare the endocrine and nervous systems in control and co‑ordination in mammals. [8] (b)\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n.............................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-mj-43-q10-4df4960b19.pdf"
    },
    {
      "id": "9700-2021-mj-51-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 21,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p01-fa9e58e7b3.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p02-76267a4cf8.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p03-9059c5aab3.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p04-c0d6cd31bb.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p05-8cc0c23025.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q01-p06-a6709f2658.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "β-galactosidase The enzyme catalyses the breakdown of the compound ONPG (o-nitrophenyl 1\n1-D-galactopyranoside) to the compound ONP (o-nitrophenyl), as shown in Fig. 1.1.\nβ-galactosidase\n+ ONPG ONP galactose\n(colourless substrate) (yellow product)\nFig. 1.1\nAs ONP is produced, the colour of the reaction mixture changes to yellow. The intensity of the\nyellow colour produced is proportional to the concentration of ONP.\nA colorimeter is used to measure the absorbance of the reaction mixture. Absorbance is a measure\nof the light absorbed by a coloured solution. In the reaction shown in Fig. 1.1, the more intense the\nyellow colour, the higher the absorbance.\nβ-galactosidase. A student was provided with a stock solution of the enzyme The student (a)\ndiluted this by a factor of 20 using a buffer solution of pH 8. The student made a final volume\n3 β-galactosidase of dilute solution. of 10 cm\ncm3 β-galactosidase of diluted solution. Describe how the student prepared the 10\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe student investigated the effect of substrate concentration on the enzyme-catalysed reaction\nshown in Fig. 1.1.\nThe student was provided with: (b)\n• β-galactosidase the diluted solution prepared in step (a), which was kept cold until\nneeded\n• a stock solution of 1.0% ONPG made up in a buffered solution of pH 8.0\n• a buffer solution of pH 8.0.\nThe procedure used by the student is outlined in step 1 to step 3.\nβ-galactosidase 1. The diluted solution was mixed with 1.0% ONPG solution.\n2. After 2 minutes a colorimeter was used to measure the absorbance of this mixture.\n3. Steps 1 and 2 were repeated using different concentrations of ONPG solution.\nSuggest why the student used a colorimeter to measure the absorbance rather than (i)\njudging the intensity of the colour by eye.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nIdentify the independent variable and the dependent variable in this inves",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-51-q01-51f7460357.pdf"
    },
    {
      "id": "9700-2021-mj-51-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_51.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-51-q02-p01-4fd3fa2516.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q02-p02-91dda7eada.png",
        "9700-experimental-skills/assets/9700-2021-mj-51-q02-p03-044f387afe.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Inheritance of flower colour and flower position in pea plants are controlled by two genes. 2\n• Gene P/p controls flower colour. Allele for purple flowers is dominant to allele for P p\nwhite flowers.\n• Gene A/a controls flower position. Allele for flowers growing from the side of the shoot A\n(axial position) is dominant to allele for flowers growing at the end of the shoot (terminal a\nposition), as shown in Fig. 2.1.\nflower position\naxial terminal\nFig. 2.1\nA biologist predicted that, if the genes are on chromosomes, the ratio of the phenotypes different\nof the F2 generation would be 9:3:3:1.\nThe biologist carried out a breeding experiment.\n• Plants homozygous for white flowers and axial position were crossed with plants\nhomozygous for purple flowers and axial position.\n• All the F1 plants had purple, axial flowers.\n• The F1 plants were crossed with each other.\nTable 2.1 shows the results for the F2 generation.\nTable 2.1\nF2 phenotype frequency\npurple, axial flowers 1756\npurple, terminal flowers 653\nwhite, axial flowers 702\nwhite, terminal flowers 234\ntotal 3345\n2 (χ The chi-squared test test) was used to analyse the data in Table 2.1. (a)\n2 (χ test) was used. State reason why the chi-squared test (i) one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState the null hypothesis that the biologist would use for this test. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\n2 χ for the results of the F2 generation. Complete Table 2.2 calculate the value of (iii) and\n2 χ is: The equation for the calculation of\nO = observed result\nE)2 (O –\nE = expected result 2 χ ∑ =\nE\n∑ = sum of\nTable 2.2\n2 (O – E)\noffspring phenotype O E\nE\npurple, axial flowers 1756\npurple, terminal flowers 653\nwhite, axial flowers 702\nwhite, terminal flowers 234\n2 χ =\n[3]\n2 χ Table 2.3 shows some critical values of at different probability levels.\nTable 2.3\nprobability (p)\ndegrees of\nfreedom\n0.10 0.05 0.01 0.001\n1 2.71 3.84 6.64 10.83\n2 4.61 5.99 9.21 13.82\n3 6.25 7.82 11.34 16.27\n4 7.78 9.49 13.28 18.46\n2 χ test. ....................................................... [1] State the critical value at p < 0.05 for this (iv)\n2) (χ to: Use your calculated value of chi-squared (v)\n• explain whether the null hypothesis should be accepted or rejected\n....................................................................................................................................\n..................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-51-q02-9c9fbb509d.pdf"
    },
    {
      "id": "9700-2021-mj-52-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 18,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_52.pdf?download=true",
      "source_pages": [
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-52-q01-p01-3d6ccddfbd.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q01-p02-7a17ce76c0.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q01-p03-dc8b119f21.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q01-p04-66713108d3.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q01-p05-645a7be8f3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "The enzyme catalase catalyses the reaction shown in Fig. 1.1. 1\ncatalase\nwater + oxygen hydrogen peroxide\nFig. 1.1\ncopper(II) sulfate can act as an inhibitor, reducing the activity of catalase. A student had read that\ncopper(II) sulfate as an inhibitor on the activity of The student wanted to investigate the effect of\ncatalase to test the hypothesis:\ncopper(II) sulfate, the lower the volume of The higher the concentration of the inhibitor\noxygen produced in the presence of catalase.\nState the independent variable dependent variable in this investigation. (a) and\nindependent variable ................................................................................................................\n...................................................................................................................................................\ndependent variable ...................................................................................................................\n...................................................................................................................................................\n[2]\n–3 copper(II) solution of sulfate. The student was provided with a 0.04 mol dm (b) (i)\nDescribe how the student could use serial dilution to make a suitable range of\nconcentrations to test their hypothesis.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe student used the apparatus shown in Fig. 1.2 to collect the data needed to test the (ii)\ncopper(II) sulfate on the production of oxygen in the effect of different concentrations of\npresence of catalase.\ngraduated tube\ncm3 5 of 10%\nto collect oxygen\nhydrogen\nperoxide\nmixture of\ncatalase and\ncopper(II) sulfate\nwater\nFig. 1.2\nDescribe a method the student could use to collect the data needed to test the effect of\ncopper(II) sulfate on the production of oxygen in the presence different concentrations of\nof catalase.\nThe description of your method should be set out in a logical way and be detailed enough\nfor another person to follow.\n–3 You should repeat the details fro",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-52-q01-6910a42486.pdf"
    },
    {
      "id": "9700-2021-mj-52-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p01-01e547c979.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p02-d1e9821c7c.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p03-4962a42f30.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p04-239319fe90.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p05-f5a6120cbe.png",
        "9700-experimental-skills/assets/9700-2021-mj-52-q02-p06-9218a3c2d6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Inheritance of petal colour and pollen shape in sweet pea plants are controlled by two genes. 2 (a)\n• Gene P/p controls petal colour. Allele for purple petals is dominant to allele for red P p\npetals.\n• Gene G/g controls the shape of pollen grains. Allele for long pollen grains is dominant G\nto allele for round pollen grains. g\nFig. 2.1 shows the structures inside a flower from a sweet pea plant.\npetals\nstigma – female\nanther – male\novary containing\nfemale gametes\nFig. 2.1\nSweet pea plants are normally self-fertilising. Pollen grains containing the male gametes\nare produced in the anthers and released into the flower. The pollen grains fall on the stigma\nin the flower, leading to fertilisation of the female gametes.\nA plant breeder two sweet pea plants, and B. cross-fertilised A\nA homozygous dominant plant with purple petals and long pollen grains (plant A) was crossed\nwith a homozygous recessive plant with red petals and round pollen grains (plant B).\nThe breeder transferred pollen grains from the flowers on plant to the flowers on plant B. A\nSuggest way in which the breeder could prevent the self-fertilisation of plant B. (i) one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSuggest way in which the breeder could transfer pollen grains from an anther to a (ii) one\nstigma.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe plant breeder then: (b)\n• collected the seeds from plant B\n• grew the seeds in a glasshouse.\nThe environmental conditions were standardised.\nState variables that should have been standardised in the glasshouse. three\n1 ................................................................................................................................................\n...................................................................................................................................................\n2 ................................................................................................................................................\n...................................................................................................................................................\n3 ................................................................................................................................................\n...................................................................................................................................................\n[3]\nAll the F1 plants grown from the seeds collected from plant had purple petals and lon",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-52-q02-ed3d8abd45.pdf"
    },
    {
      "id": "9700-2021-mj-53-q01",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 21,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p01-e56bac3f18.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p02-f8662bbcea.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p03-b05766b576.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p04-6f6d80f20b.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p05-a168059ea6.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q01-p06-30402e891d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "When bananas ripen, they turn brown in colour. This is due to the formation of a brown product 1\ncalled melanin.\nThe enzyme catechol oxidase acts on its substrate, catechol, leading to the formation of a brown\nproduct, melanin, as shown in Fig. 1.1.\ncatechol oxidase\ncatechol melanin\n(pale pink substrate) (brown product)\nFig. 1.1\nAs melanin is produced, the colour of the reaction mixture changes to brown. The intensity of\nthe brown colour produced is proportional to the concentration of melanin. The more active the\nenzyme, the more intense the brown colour.\nA colorimeter is used to measure the absorbance of the reaction mixture. Absorbance is a measure\nof the light absorbed by a coloured solution. With the reaction shown in Fig. 1.1, the more intense\nthe brown colour, the higher the absorbance.\nCatechol oxidase can be extracted from bananas.\nSome students were asked to extract catechol oxidase by grinding a slice of banana with (a)\nsand and water, using a mortar and pestle. The resulting mixture was then filtered through a\npiece of cloth.\nSuggest a reason for:\ngrinding with sand\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\nfiltering through cloth.\n..........................................................................................................................................................\n..........................................................................................................................................................\n.................................................................................................................................................... [2]\nThe students investigated the effect of the substrate concentration on the enzyme-catalysed (b)\nreaction shown in Fig. 1.1.\nThe students were provided with:\n• the extracted catechol oxidase enzyme solution prepared in step (a), which was kept\ncold until needed\n• a stock solution of 1.0% catechol solution made up in a buffered solution of pH 7.0\n• a buffer solution of pH 7.0.\nThe procedure used by the students is outlined in steps 1 to 3.\n1. The extracted catechol oxidase enzyme solution was mixed with the 1.0% catechol\nsolution.\n2. After 2 minutes a colorimeter was used to measure the absorbance of this mixture.\n3. Step 1 and step 2 were repeated using different concentrations of catechol solution.\nSuggest why the students used a colorimeter to measure the absorbance rather than (i)\njudging the intensity of the colour by eye.\n..................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-53-q01-2f438a97e1.pdf"
    },
    {
      "id": "9700-2021-mj-53-q02",
      "subject": "9700",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-may-june/9700_s21_qp_53.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-May-June/qp/9700_s21_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-mj-53-q02-p01-7eb6667e0f.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q02-p02-09dc5c1c7c.png",
        "9700-experimental-skills/assets/9700-2021-mj-53-q02-p03-741185d323.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Inheritance of wing shape and eye colour in the fruit fly, Drosophila melanogaster, is controlled by 2\ntwo genes.\n• Gene N/n controls wing shape. Allele for wrinkled wings is dominant to allele for normal N n\nwings.\n• Gene E/e controls eye colour. Allele for rosy eyes is dominant to allele for red eyes. E e\nA biologist predicted that, if the genes are on chromosomes, the ratio of the phenotypes different\nof the F2 generation would be 9:3:3:1.\nThe biologist carried out a breeding experiment.\n• Homozygous dominant fruit flies with wrinkled wings and rosy eyes were crossed with\nhomozygous recessive fruit flies with normal wings and red eyes.\n• All the F1 fruit flies had wrinkled wings and rosy eyes.\n• The F1 fruit flies were crossed with each other.\nTable 2.1 shows the results for the F2 generation.\nTable 2.1\nF2 phenotype frequency\nwrinkled wings 44\nrosy eyes\nwrinkled wings 2\nred eyes\nnormal wings 2\nrosy eyes\nnormal wings 16\nred eyes\ntotal 64\n2 (χ The chi-squared test test) was used to analyse the data in Table 2.1. (a)\n2 (χ test) was used. State reason why the chi-squared test (i) one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState the null hypothesis that the biologist would use for this test. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\n2 χ Complete Table 2.2 calculate the value of for the results of the F2 generation. (iii) and\n2 χ is: The equation for the calculation of\nO = observed result\nE)2 (O –\n2 χ = !\nE E = expected result\n= sum of !\nTable 2.2\n2 (O – E )\noffspring phenotype O E\nE\nwrinkled wings\n44\nrosy eyes\nwrinkled wings\n2\nred eyes\nnormal wings\n2\nrosy eyes\nnormal wings\n16\nred eyes\n2 χ =\n[3]\n2 χ at different probability levels. Table 2.3 shows some critical values of\nTable 2.3\ndegrees of probability (p)\nfreedom\n0.10 0.05 0.01 0.001\n1 2.71 3.84 6.64 10.83\n2 4.61 5.99 9.21 13.82\n3 6.25 7.82 11.34 16.27\n4 7.78 9.49 13.28 18.46\nState why the student used the critical value of in this investigation. (iv) 7.82\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\n2 χ Use your calculated",
      "pdf_path": "9700-experimental-skills/print/9700-2021-mj-53-q02-dee44a1219.pdf"
    },
    {
      "id": "9700-2021-on-41-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3,
        4
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-on-41-q01-p01-110efde62f.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-41-q01-p02-27f94c8307.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-41-q01-p03-c6f721a298.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a diagram of a kidney nephron and some of its blood vessels. 1 (a)\nB\nF\nC\nA\nG\nD\nE\nFig. 1.1\nWith reference to Fig. 1.1, complete Table 1.1 using the letters – G. A\nEach letter may be used once, more than once or not at all.\nTable 1.1\ndescription letter\nefferent blood vessel\n..................\npart of nephron containing cells that respond to ADH\n..................\npart of nephron where podocyte cells are located\n..................\npart of nephron containing cells that are located in the medulla\n..................\n[4]\nDescribe explain how the cells of the proximal convoluted tubule are adapted to carry out (b) and\nselective reabsorption.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nFig. 1.2 shows the concentrations of ADH in the blood at different percentage changes in (c)\nwater potential of the blood.\n16\n14\n12\n10\nconcentration\nof ADH in the blood 8\narbitrary units /\n6\n4\n2\n0\n– – 20 10 0 +10\npercentage change in\nwater potential of the blood\nFig. 1.2\nDescribe the trend shown in Fig. 1.2. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..............................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-on-41-q01-57628e725e.pdf"
    },
    {
      "id": "9700-2021-on-41-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-on-41-q02-p01-a40bce2150.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-on-41-q02-p02-044b4c667a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Cotton, Gossypium hirsutum, and false flax, Camelina sativa, are crop plants that are grown 2 (a)\nin different parts of the world.\nRubisco activase is an enzyme in the stroma of chloroplasts that is needed to maintain the\nactivity of a second enzyme, rubisco.\nScientists measured the activity of rubisco activase in cotton and in false flax at a range of\ntemperatures.\nFig. 2.1 shows the results.\n0.12\nKey\nfalse flax\n0.10\ncotton\n0.08\nrubisco activase\nactivity\n0.06\narbitrary units /\n0.04\n0.02\n0.00\n35 40 45 20 25 30\n/ °C temperature\nFig. 2.1\nWith reference to Fig. 2.1, compare the results obtained for cotton and false flax. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nSuggest reasons for the differences shown in Fig. 2.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-on-41-q02-7439a4bf53.pdf"
    },
    {
      "id": "9700-2021-on-41-q03",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-41-q03-p01-c2a3587c14.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-41-q03-p02-6240e7aa35.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-41-q03-p03-cbeba6f970.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Ecological surveys are conducted before conservation decisions are made. For example, surveys 3\ncan be carried out before deciding whether to reintroduce a species to its former habitat.\nOutline how an ecological survey can measure the biodiversity of a terrestrial habitat. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nTable 3.1 lists some mammal species that became extinct in Great Britain or were introduced (b)\ninto Great Britain in the last 12 400 years.\nThe reasons for each extinction or introduction and the time of each event are shown.\nTable 3.1\nnumber of years\nspecies event reason\nbefore present time\narctic fox 12 400 extinction climate change\nsheep 5 400 introduction farming\nhouse mouse 3 500 introduction accidental\nlynx 1 500 extinction hunting\nfallow deer 900 introduction food\nSouth American coypu 86 introduction fur\nSouth American coypu 33 extinction conservation culling\nWith reference to Table 3.1, state explain the factors that have had a negative impact on and\nbiodiversity in Great Britain.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-on-41-q03-a84924e8eb.pdf"
    },
    {
      "id": "9700-2021-on-41-q04",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-on-41-q04-p01-a81cb81736.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-on-41-q04-p02-d52a3089f4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "In 1984, the geneticist Alec Jeffreys invented a DNA testing technique, known as DNA 4 (a)\nprofiling, that produces a DNA banding pattern on a gel. The DNA banding pattern (profile) is\nunique to each individual.\nDNA profiling can be used in police forensic work to catch criminals.\nSince 1987, police in many countries have collected and stored DNA from crime scenes to\ncreate DNA profiles, which they try to match with the DNA profiles of criminal suspects.\nDNA at a crime scene may be obtained from hairs and traces of blood, semen and saliva. (i)\nExplain why PCR may be needed before DNA from a crime scene can be profiled.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nExplain why electrophoresis produces a DNA banding pattern on a gel. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nGEDmatch is described as ‘an open data personal genomics website’. It can be used by (b)\npeople who want to upload their DNA data to trace their ancestors and other relatives.\nIn 2018, police in the USA solved a large number of serious crimes. Some of these crimes\nhad been unsolved for over thirty years. The police used GEDmatch to profile DNA taken\nfrom crime scenes and to look for matching DNA profiles. In many cases the police found\npartial matches to the relatives of criminals. This allowed the criminals to be identified and\nthen charged on the basis of a complete DNA profile match.\nSuggest why the police strate",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-on-41-q04-70df7526f4.pdf"
    },
    {
      "id": "9700-2021-on-41-q05",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-41-q05-p01-9eb75fab30.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-41-q05-p02-d290d547f6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Gibberellin is a plant growth regulator involved in barley seed germination. Production of 5 (a)\ngibberellin is stimulated by the uptake of water.\nState the location of gibberellin synthesis in a barley seed during germination.\n............................................................................................................................................. [1]\nBarley seeds germinate when placed on blotting (absorbent) paper soaked in water. (b)\nThe germination of barley seeds placed on blotting paper soaked in solutions of different\nwater potential was investigated.\nThe success of germination was measured as a germination index for:\n• barley seeds placed on blotting paper soaked in water\n• barley seeds placed on blotting paper soaked in 5 solutions of different water\npotential.\nThe results are shown in Fig. 5.1.\nThe higher the germination index value, the more successful the germination of the barley\nseeds.\n50\n40\n30\ngermination\nindex\n20\n10\n0\n– – – – – 0.0 0.5 1.0 1.5 2.0 2.5\nwater potential / MPa\nFig. 5.1\nWith reference to Fig. 5.1, describe the relationship between the germination index of (i)\nbarley seeds and water potential.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSuggest explanations for the relationship shown in Fig. 5.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nDuring barley seed germination, gibberellin stimulates the synthesis of enzymes. (c)\nState the name of one of these enzymes the precise location of its synthesis. and\nenzyme .....................................................................................................................................\nlocation ................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-41-q05-a077cdb54a.pdf"
    },
    {
      "id": "9700-2021-on-41-q06",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-41-q06-p01-33a3a6b847.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-41-q06-p02-8a84417086.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-41-q06-p03-e0b5b383b8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Fig. 6.1 outlines the first three stages of respiration in aerobic conditions. 6\nglucose\nATP × 2\ntriose phosphate × 2\nreduced NAD × 2\nATP × 4\npyruvate × 2\nmolecule × 2 Y\nreduced NAD × 2\nacetyl coenzyme A × 2\nKrebs cycle × 2\nATP × 2\nmolecule × 4 Y\nreduced NAD × 6\nreduced FAD × 2\nFig. 6.1\nName molecule in Fig. 6.1. (a) Y\n............................................................................................................................................. [1]\nExplain how Fig. 6.1 shows that glycolysis involves oxidation. (b)\n...................................................................................................................................................\n............................................................................................................................................. [1]\nAt one time it was thought that the oxidative phosphorylation of: (c)\n• one molecule of reduced NAD results in the synthesis of 3 ATP molecules\n• one molecule of reduced FAD results in the synthesis of 2 ATP molecules.\nUsing Fig. 6.1, a theoretical value for the net number of ATP molecules that are synthesised\nfor each molecule of glucose can be calculated.\nModern research has shown that the actual net number of ATP molecules synthesised for\neach glucose molecule respired is much lower than this theoretical value.\nUsing Fig. 6.1, calculate the theoretical value for the net number of ATP molecules that (i)\nare synthesised for each molecule of glucose respired in phosphorylation reactions. all\nShow your working.\nanswer = ......................................................... [2]\nSuggest reasons why the actual net number of ATP molecules synthesised is less (ii) two\nthan the theoretical number.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nOutline the roles of NAD and FAD in aerobic respiration. (d)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-on-41-q06-7550f49ccc.pdf"
    },
    {
      "id": "9700-2021-on-41-q07",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-on-41-q07-p01-9499c7c2d8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The function of the enzyme iduronate 2-sulfatase is to break down certain complex molecules. 7 (a)\nHunter syndrome is a rare, inherited condition in humans, caused by iduronate 2-sulfatase\nnot functioning properly or by iduronate 2-sulfatase not being produced. This will cause the\ncomplex molecules to build up, leading to progressive damage to many organs.\nExplain how a gene mutation can lead to iduronate 2-sulfatase not functioning properly or not\nbeing produced.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nHunter syndrome is caused by a recessive allele located on the chromosome. (b) X\nUsing appropriate symbols, draw a genetic diagram to show the probability of a normal male\nand a heterozygous normal female having a child with Hunter syndrome.\nkey\nparental normal male × normal female\nphenotype\nparental\ngenotype\ngametes\noffspring\ngenotypes\noffspring\nphenotypes\nprobability ..................................... [5]\n[Total: 9]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-on-41-q07-f411d4dcbe.pdf"
    },
    {
      "id": "9700-2021-on-41-q08",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-41-q08-p01-6d2c9fed32.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-41-q08-p02-b42208b5bc.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-41-q08-p03-774b3682cf.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "California salamanders are terrestrial amphibians. 8\nAn original population of California salamanders occupied an area of forest in northern California.\nApproximately 10 million years ago, two separate populations, and B, started to migrate from A\nthe original population.\n• Population travelled along the coast of California, to the west of the Great Central A\nValley.\n• Population travelled east of the Great Central Valley. B\nThe two populations now live close to each other in southern California.\nFig. 8.1 outlines the migratory routes of populations and B. A\noriginal\npopulation\nGreat Central\nValley\nmigration\nroute\nmigration\nroute\npopulation B\nnow\npopulation A\nnow\nFig. 8.1\nSalamanders from population rarely interbreed with salamanders of population B. If they do A\ninterbreed, the offspring are infertile.\nSuggest explain the sequence of events that have resulted in these two populations becoming and\nreproductively isolated from each other.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n.................................................................................................................................................... [5]\n[Total: 5]\nSection B\nAnswer question. one",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-on-41-q08-4bd1f96b42.pdf"
    },
    {
      "id": "9700-2021-on-41-q09",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-on-41-q09-p01-7a10a5e806.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Describe explain the mechanism by which guard cells open stomata. [7] 9 (a) and\nExplain how the anatomy physiology of the leaves of maize or sorghum are able to (b) and\nmaximise carbon dioxide fixation at high temperatures. [8]\n[Total: 15]",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-on-41-q09-27e01f8724.pdf"
    },
    {
      "id": "9700-2021-on-41-q10",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-41-q10-p01-47603010cc.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-41-q10-p02-e874c07125.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-41-q10-p03-dfc93b82ba.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-41-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Describe explain the transmission of an action potential in a myelinated neurone. [9] 10 (a) and\nExplain what is meant by homeostasis in a mammal explain why it is important to (b) and\nmaintain body temperature, blood glucose concentration and the water potential of blood. [6]\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n......................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-41-q10-b58eb5f6f7.pdf"
    },
    {
      "id": "9700-2021-on-42-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3,
        4
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-on-42-q01-p01-726c97825c.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-42-q01-p02-db4bc4f43f.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-42-q01-p03-3df6f68e71.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The Bowman’s capsule of a nephron is involved in ultrafiltration. 1 (a)\nFig. 1.1 is a diagram of part of a Bowman’s capsule and glomerulus.\nB C D E F\nA\nFig. 1.1\nWith reference to Fig. 1.1, complete Table 1.1 using the letters – F. A\nEach letter may be used once, more than once or not at all.\nTable 1.1\nfeature letter\nglomerular filtrate\n...................\nbasement membrane\n...................\npodocyte cell\n...................\ncapillary endothelial cell\n...................\n[4]\nDescribe explain how the structures in the Bowman’s capsule and its associated blood (b) and\nsupply are adapted to allow ultrafiltration to take place.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nThe glomerular filtration rate (GFR) is the rate at which blood plasma is filtered in the (c)\nBowman’s capsule.\nFig. 1.2 shows the relationship between GFR and mean renal arterial blood pressure.\n200\n180\n160\n140\n120\nGFR\n100\ncm3 min–1 /\n80\n60\n40\n20\n0\n20 25 0 5 10 15\n/ kPa mean renal arterial blood pressure\nFig. 1.2\nComment on the relationship between GFR and mean renal arterial blood pressure. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n.............................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-on-42-q01-c81556a2e0.pdf"
    },
    {
      "id": "9700-2021-on-42-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-on-42-q02-p01-21b38fe983.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-on-42-q02-p02-994564027f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Scientists are researching new ways to reduce the global atmospheric carbon dioxide (CO ) 2\n2\nconcentration may lead to concentration. There are concerns that an increasing atmospheric CO\n2\neffects that decrease biodiversity.\nGive example of a human activity, other than deforestation, that contributes greatly to the (a) one\nconcentration. increase in global atmospheric CO\n2\n............................................................................................................................................. [1]\nAlgae are aquatic photosynthetic protoctists. Some researchers genetically modified the unicellular\nalga, Chlorella vulgaris, to try to increase the rate of the light independent stage of photosynthesis.\nC. vulgaris was modified to increase the expression of the gene coding for aldolase. Aldolase is an\nenzyme that causes an increase in the concentration of rubisco.\nTwo cultures of C. vulgaris, one that was not genetically modified (unmodified) and one genetically\nmodified, were grown under controlled conditions for 14 days. Samples were taken from the\ncultures at regular intervals during the 14 days to obtain measurements of dry mass.\nThe results are shown in Fig. 2.1.\nKey\nunmodified\ngenetically modified\n2.0\n1.6\n1.2\ndry mass\ndm–3 g /\n0.8\n0.4\n0.0\n10 12 14 0 2 4 6 8\n/ days time\nFig. 2.1\nWith reference to Fig. 2.1, describe the differences between the results for the two cultures. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nExplain how the Calvin cycle was affected by the genetic modification of C. vulgaris. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-on-42-q02-20512154fb.pdf"
    },
    {
      "id": "9700-2021-on-42-q03",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q03-p01-a516738d43.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q03-p02-186beaa41c.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q03-p03-9513c4d08b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q03",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "During the germination of barley seeds, amylase is produced. 3\nDescribe the sequence of events that lead to the production of amylase during germination of (a)\nbarley seeds.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nMalting is a process involved in the production of a drink called beer. During malting, barley (b)\nseed germination is controlled so that the sugars produced during germination can be used in\nthe production of beer.\nFig. 3.1 shows two features of a germinating barley seed during the first five days of malting:\n• activity of the amylase enzyme\n• the percentage of starch reserves remaining in the barley seed.\nKey\nstarch reserves\namylase activity\n100 10\n95 8\n90 6\namylase activity percentage of starch\narbitrary units / reserves remaining\n85 4\n80 2\n75 0\n0 1 2 3 4 5\ntime from start of malting / days\nFig. 3.1\nState the precise location of the starch reserves in the barley seed. (i)\n..................................................................................................................................... [1]\nWith reference to Fig. 3.1, describe explain the effect of malting on amylase activity (ii) and\nand the percentage of starch reserves remaining in the germinated barley seed.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-42-q03-4706f851f7.pdf"
    },
    {
      "id": "9700-2021-on-42-q04",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-on-42-q04-p01-98f85b616f.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-on-42-q04-p02-6290224246.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-on-42-q04-p03-3f700b948d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genome-wide association studies find links between single nucleotide polymorphisms (SNPs) 4 (a)\nand phenotypic features such as human diseases. SNPs are points on the DNA that vary in\nthe population because of DNA base substitutions.\nA genome-wide association study investigates the effect of genetic variation on a disease.\nA large number of people with the disease and a large number of healthy control individuals\nprovide DNA. Microarray chips are used to identify the genotype of each individual at many\nSNPs.\nThe Wellcome Trust Case Control Consortium (WTCCC) study was an important genome-\nwide association study.\n• The study used a microarray chip that identified each person’s genotype at 500 000\ndifferent SNPs.\n• The study looked for links between SNPs and 7 different diseases.\n• For each disease, 2000 people with the disease were tested.\n• Their results were compared with the results of 3000 healthy control individuals.\nOutline how microarrays are used in the analysis of genomes. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nExplain why bioinformatics was important to the WTCCC study. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n....................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-on-42-q04-0dc4963642.pdf"
    },
    {
      "id": "9700-2021-on-42-q05",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-42-q05-p01-023b7b65dc.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-42-q05-p02-22dd5541c9.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-42-q05-p03-8da8aedc08.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q05",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Biodiversity can be considered at three different levels. 5\nFig. 5.1, Fig. 5.2 and Fig. 5.3 are images that show these three different levels of biodiversity.\nFig. 5.1\nFig. 5.2\nFig. 5.3\nDescribe the level of biodiversity most clearly visible in each image. (a)\nFig. 5.1 ......................................................................................................................................\nFig. 5.2 ......................................................................................................................................\nFig. 5.3 ......................................................................................................................................\n[3]\nPrairie strips are restored habitat areas planted with native grasses and wildflowers. Prairie (b)\nstrips are usually located between or around the edges of fields of maize or soy bean.\nStudents in North America wanted to investigate biodiversity along a prairie strip.\nOutline how the students could measure the biodiversity of plants and insects along a (i)\nprairie strip.\nplants .................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\ninsects ...............................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n[4]\nMathematical methods and statistical tests help to summarise and compare biodiversity (ii)\ndata.\nName the mathematical method statistical test that the students can use: or\nto estimate the total biodiversity of a prairie strip\n...........................................................................................................................................\nto test whether plant biodiversity is associated with insect biodiversity.\n...........................................................................................................................................\n[2]\nPrairie strips have environmental and economic effects. (c)\nScientists have made long-term measurements of the environmental effects of prairie strips.\nThe scientists found that when prairie strips formed 10% of a crop-field:",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-on-42-q05-9737628c62.pdf"
    },
    {
      "id": "9700-2021-on-42-q06",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-42-q06-p01-8a13b25e45.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-42-q06-p02-39b696d6d9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "ATP is needed for many metabolic processes in living organisms. 6 (a)\nDescribe the properties of ATP that make it suitable for its role as the universal energy (i)\ncurrency.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nSuggest why ATP is needed for protein synthesis. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nFig. 6.1 is a diagram of mitochondrion. (b)\nA B C\ndiameter\nμm 0.72\nE D\nμm length 5.25\nFig. 6.1\nComplete Table 6.1 using the letters to from Fig. 6.1. (i) A E\nEach letter may be used once, more than once, or not at all.\nTable 6.1\nstatement letter\nthe site of the Krebs Cycle\n................\n+ ions a phospholipid bilayer impermeable to H\n................\nthe site of translation\n................\n[2]\nAssume that the mitochondrion in Fig. 6.1 is a cylinder. (ii)\nCalculate the surface area of this mitochondrion.\n2 2πrh 2πr + Use the formula: Surface area of cylinder =\nShow your working.\n2 µm [2] surface area = .................................................\nThe inner membrane of the mitochondrion has a much larger surface area than the outer (iii)\nmembrane because of the presence of cristae.\nDifferent cell types vary in the number of cristae present per mitochondrion. Cardiac\nmuscle cells have mitochondria with a very large number of cristae.\nSug",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-on-42-q06-6fc1dbdfdb.pdf"
    },
    {
      "id": "9700-2021-on-42-q07",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-on-42-q07-p01-934e995d8e.png",
        "9700-topic-16-inheritance/assets/9700-2021-on-42-q07-p02-c0beb426ec.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The Labrador is a variety of domestic dog. Labradors have fur that can be brown, black or 7 (a)\nyellow.\nIn Labradors, TYRP1 is one gene that codes for fur colour. This gene has two alleles, B\nand b.\n• The dominant allele, B, codes for the enzyme tyrosinase that functions in the pathway to\nproduce melanin, leading to black fur.\n• The production of melanin in Labradors is very similar to the production of melanin in\nhumans.\n• The recessive allele, b, codes for an enzyme that results in the production of a brown\nform of melanin, leading to brown fur.\nOutline how melanin may be produced in Labradors to produce black fur.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nAnother gene, MC1R, interacts with TYRP1. (b)\nMC1R has two alleles, and e. E\n• The dominant allele, E, allows the alleles of TYRP1 to be expressed.\n• The recessive allele, e, prevents the alleles of TYRP1 from being expressed.\n• When no form of melanin is produced the Labrador will have yellow fur.\nConstruct a genetic diagram to show the ratio of possible offspring from a cross between (i)\na black male Labrador, heterozygous for both genes, and a yellow female Labrador,\nheterozygous for TYRP1.\nparental black × yellow\nphenotype\nparental\ngenotype\ngametes\noffspring\ngenotypes\noffspring\nphenotypes\nratio ...................................................................................................................................\n[6]\nState the term used to describe a protein that is involved in the control of gene expression (ii)\nin eukaryotes.\n..................................................................................................................................... [1]\n[Total: 9]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-on-42-q07-0dc4146423.pdf"
    },
    {
      "id": "9700-2021-on-42-q08",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-42-q08-p01-f62d190195.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-42-q08-p02-8b36f4af9a.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-42-q08-p03-5fb0be917c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Twenty million years ago, an ocean covered the area where the country of Panama is now located. 8\nThere was a gap between the continents of North America and South America through which the\nwaters of the Atlantic and Pacific Oceans flowed freely.\nThe porkfish, Anisotremus sp, lived in this area between North America and South America.\nFig. 8.1 shows a porkfish.\nFig. 8.1\nAbout 3 million years ago, volcanic activity and sedimentation formed a narrow strip of land,\nPanama, joining North America and South America.\nFig. 8.2 shows the area 20 million years ago and now.\n20 million years ago now\nNorth America North America\nAtlantic Ocean Atlantic Ocean\nPacific Ocean Pacific Ocean\nSouth America South America\nPanama\nFig. 8.2\nTwenty million years ago, porkfish in the Atlantic and Pacific Oceans were able to breed\nsuccessfully and produce fertile offspring.\nExplain why Atlantic porkfish and Pacific porkfish are now able to breed successfully to not\nproduce fertile offspring.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n.................................................................................................................................................... [5]\n[Total: 5]\nSection B\nAnswer question. one",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-on-42-q08-439c3750ba.pdf"
    },
    {
      "id": "9700-2021-on-42-q09",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        24
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-on-42-q09-p01-69ab13ba98.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Describe the roles of ADH the collecting ducts in osmoregulation. [9] 9 (a) and\nDescribe the structure of a motor neurone. [6] (b)\n[Total: 15]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-on-42-q09-6ba5b70d62.pdf"
    },
    {
      "id": "9700-2021-on-42-q10",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_42.pdf?download=true",
      "source_pages": [
        24,
        25,
        26
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q10-p01-54dcd30e80.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q10-p02-e874c07125.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-42-q10-p03-1ddc4eb48d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-42-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Explain how dip sticks function to test for glucose in a sample of urine. [7] 10 (a)\nExplain the control of gibberellin synthesis outline how gibberellin stimulates stem (b) and\nelongation. [8]\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n.......................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-42-q10-90eb07eb89.pdf"
    },
    {
      "id": "9700-2021-on-43-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3,
        4
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2021-on-43-q01-p01-110efde62f.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-43-q01-p02-27f94c8307.png",
        "9700-topic-14-homeostasis/assets/9700-2021-on-43-q01-p03-c6f721a298.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a diagram of a kidney nephron and some of its blood vessels. 1 (a)\nB\nF\nC\nA\nG\nD\nE\nFig. 1.1\nWith reference to Fig. 1.1, complete Table 1.1 using the letters – G. A\nEach letter may be used once, more than once or not at all.\nTable 1.1\ndescription letter\nefferent blood vessel\n..................\npart of nephron containing cells that respond to ADH\n..................\npart of nephron where podocyte cells are located\n..................\npart of nephron containing cells that are located in the medulla\n..................\n[4]\nDescribe explain how the cells of the proximal convoluted tubule are adapted to carry out (b) and\nselective reabsorption.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nFig. 1.2 shows the concentrations of ADH in the blood at different percentage changes in (c)\nwater potential of the blood.\n16\n14\n12\n10\nconcentration\nof ADH in the blood 8\narbitrary units /\n6\n4\n2\n0\n– – 20 10 0 +10\npercentage change in\nwater potential of the blood\nFig. 1.2\nDescribe the trend shown in Fig. 1.2. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..............................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2021-on-43-q01-df3d4d7847.pdf"
    },
    {
      "id": "9700-2021-on-43-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-on-43-q02-p01-a40bce2150.png",
        "9700-topic-13-photosynthesis/assets/9700-2021-on-43-q02-p02-044b4c667a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Cotton, Gossypium hirsutum, and false flax, Camelina sativa, are crop plants that are grown 2 (a)\nin different parts of the world.\nRubisco activase is an enzyme in the stroma of chloroplasts that is needed to maintain the\nactivity of a second enzyme, rubisco.\nScientists measured the activity of rubisco activase in cotton and in false flax at a range of\ntemperatures.\nFig. 2.1 shows the results.\n0.12\nKey\nfalse flax\n0.10\ncotton\n0.08\nrubisco activase\nactivity\n0.06\narbitrary units /\n0.04\n0.02\n0.00\n35 40 45 20 25 30\n/ °C temperature\nFig. 2.1\nWith reference to Fig. 2.1, compare the results obtained for cotton and false flax. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nSuggest reasons for the differences shown in Fig. 2.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-on-43-q02-94cbc4fe05.pdf"
    },
    {
      "id": "9700-2021-on-43-q03",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-43-q03-p01-c2a3587c14.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-43-q03-p02-6240e7aa35.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2021-on-43-q03-p03-cbeba6f970.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Ecological surveys are conducted before conservation decisions are made. For example, surveys 3\ncan be carried out before deciding whether to reintroduce a species to its former habitat.\nOutline how an ecological survey can measure the biodiversity of a terrestrial habitat. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nTable 3.1 lists some mammal species that became extinct in Great Britain or were introduced (b)\ninto Great Britain in the last 12 400 years.\nThe reasons for each extinction or introduction and the time of each event are shown.\nTable 3.1\nnumber of years\nspecies event reason\nbefore present time\narctic fox 12 400 extinction climate change\nsheep 5 400 introduction farming\nhouse mouse 3 500 introduction accidental\nlynx 1 500 extinction hunting\nfallow deer 900 introduction food\nSouth American coypu 86 introduction fur\nSouth American coypu 33 extinction conservation culling\nWith reference to Table 3.1, state explain the factors that have had a negative impact on and\nbiodiversity in Great Britain.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2021-on-43-q03-04f070647e.pdf"
    },
    {
      "id": "9700-2021-on-43-q04",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2021-on-43-q04-p01-a81cb81736.png",
        "9700-topic-19-genetic-technology/assets/9700-2021-on-43-q04-p02-d52a3089f4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "In 1984, the geneticist Alec Jeffreys invented a DNA testing technique, known as DNA 4 (a)\nprofiling, that produces a DNA banding pattern on a gel. The DNA banding pattern (profile) is\nunique to each individual.\nDNA profiling can be used in police forensic work to catch criminals.\nSince 1987, police in many countries have collected and stored DNA from crime scenes to\ncreate DNA profiles, which they try to match with the DNA profiles of criminal suspects.\nDNA at a crime scene may be obtained from hairs and traces of blood, semen and saliva. (i)\nExplain why PCR may be needed before DNA from a crime scene can be profiled.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nExplain why electrophoresis produces a DNA banding pattern on a gel. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nGEDmatch is described as ‘an open data personal genomics website’. It can be used by (b)\npeople who want to upload their DNA data to trace their ancestors and other relatives.\nIn 2018, police in the USA solved a large number of serious crimes. Some of these crimes\nhad been unsolved for over thirty years. The police used GEDmatch to profile DNA taken\nfrom crime scenes and to look for matching DNA profiles. In many cases the police found\npartial matches to the relatives of criminals. This allowed the criminals to be identified and\nthen charged on the basis of a complete DNA profile match.\nSuggest why the police strate",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2021-on-43-q04-2e0d17861b.pdf"
    },
    {
      "id": "9700-2021-on-43-q05",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-43-q05-p01-9eb75fab30.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-43-q05-p02-d290d547f6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Gibberellin is a plant growth regulator involved in barley seed germination. Production of 5 (a)\ngibberellin is stimulated by the uptake of water.\nState the location of gibberellin synthesis in a barley seed during germination.\n............................................................................................................................................. [1]\nBarley seeds germinate when placed on blotting (absorbent) paper soaked in water. (b)\nThe germination of barley seeds placed on blotting paper soaked in solutions of different\nwater potential was investigated.\nThe success of germination was measured as a germination index for:\n• barley seeds placed on blotting paper soaked in water\n• barley seeds placed on blotting paper soaked in 5 solutions of different water\npotential.\nThe results are shown in Fig. 5.1.\nThe higher the germination index value, the more successful the germination of the barley\nseeds.\n50\n40\n30\ngermination\nindex\n20\n10\n0\n– – – – – 0.0 0.5 1.0 1.5 2.0 2.5\nwater potential / MPa\nFig. 5.1\nWith reference to Fig. 5.1, describe the relationship between the germination index of (i)\nbarley seeds and water potential.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSuggest explanations for the relationship shown in Fig. 5.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nDuring barley seed germination, gibberellin stimulates the synthesis of enzymes. (c)\nState the name of one of these enzymes the precise location of its synthesis. and\nenzyme .....................................................................................................................................\nlocation ................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-43-q05-9b4ffb4e54.pdf"
    },
    {
      "id": "9700-2021-on-43-q06",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-43-q06-p01-33a3a6b847.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-43-q06-p02-8a84417086.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2021-on-43-q06-p03-e0b5b383b8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Fig. 6.1 outlines the first three stages of respiration in aerobic conditions. 6\nglucose\nATP × 2\ntriose phosphate × 2\nreduced NAD × 2\nATP × 4\npyruvate × 2\nmolecule × 2 Y\nreduced NAD × 2\nacetyl coenzyme A × 2\nKrebs cycle × 2\nATP × 2\nmolecule × 4 Y\nreduced NAD × 6\nreduced FAD × 2\nFig. 6.1\nName molecule in Fig. 6.1. (a) Y\n............................................................................................................................................. [1]\nExplain how Fig. 6.1 shows that glycolysis involves oxidation. (b)\n...................................................................................................................................................\n............................................................................................................................................. [1]\nAt one time it was thought that the oxidative phosphorylation of: (c)\n• one molecule of reduced NAD results in the synthesis of 3 ATP molecules\n• one molecule of reduced FAD results in the synthesis of 2 ATP molecules.\nUsing Fig. 6.1, a theoretical value for the net number of ATP molecules that are synthesised\nfor each molecule of glucose can be calculated.\nModern research has shown that the actual net number of ATP molecules synthesised for\neach glucose molecule respired is much lower than this theoretical value.\nUsing Fig. 6.1, calculate the theoretical value for the net number of ATP molecules that (i)\nare synthesised for each molecule of glucose respired in phosphorylation reactions. all\nShow your working.\nanswer = ......................................................... [2]\nSuggest reasons why the actual net number of ATP molecules synthesised is less (ii) two\nthan the theoretical number.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nOutline the roles of NAD and FAD in aerobic respiration. (d)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2021-on-43-q06-5672f59256.pdf"
    },
    {
      "id": "9700-2021-on-43-q07",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2021-on-43-q07-p01-9499c7c2d8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The function of the enzyme iduronate 2-sulfatase is to break down certain complex molecules. 7 (a)\nHunter syndrome is a rare, inherited condition in humans, caused by iduronate 2-sulfatase\nnot functioning properly or by iduronate 2-sulfatase not being produced. This will cause the\ncomplex molecules to build up, leading to progressive damage to many organs.\nExplain how a gene mutation can lead to iduronate 2-sulfatase not functioning properly or not\nbeing produced.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nHunter syndrome is caused by a recessive allele located on the chromosome. (b) X\nUsing appropriate symbols, draw a genetic diagram to show the probability of a normal male\nand a heterozygous normal female having a child with Hunter syndrome.\nkey\nparental normal male × normal female\nphenotype\nparental\ngenotype\ngametes\noffspring\ngenotypes\noffspring\nphenotypes\nprobability ..................................... [5]\n[Total: 9]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2021-on-43-q07-8c0b2923b4.pdf"
    },
    {
      "id": "9700-2021-on-43-q08",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-43-q08-p01-6d2c9fed32.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-43-q08-p02-b42208b5bc.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2021-on-43-q08-p03-774b3682cf.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "California salamanders are terrestrial amphibians. 8\nAn original population of California salamanders occupied an area of forest in northern California.\nApproximately 10 million years ago, two separate populations, and B, started to migrate from A\nthe original population.\n• Population travelled along the coast of California, to the west of the Great Central A\nValley.\n• Population travelled east of the Great Central Valley. B\nThe two populations now live close to each other in southern California.\nFig. 8.1 outlines the migratory routes of populations and B. A\noriginal\npopulation\nGreat Central\nValley\nmigration\nroute\nmigration\nroute\npopulation B\nnow\npopulation A\nnow\nFig. 8.1\nSalamanders from population rarely interbreed with salamanders of population B. If they do A\ninterbreed, the offspring are infertile.\nSuggest explain the sequence of events that have resulted in these two populations becoming and\nreproductively isolated from each other.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n.................................................................................................................................................... [5]\n[Total: 5]\nSection B\nAnswer question. one",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2021-on-43-q08-cd610a87db.pdf"
    },
    {
      "id": "9700-2021-on-43-q09",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2021-on-43-q09-p01-7a10a5e806.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Describe explain the mechanism by which guard cells open stomata. [7] 9 (a) and\nExplain how the anatomy physiology of the leaves of maize or sorghum are able to (b) and\nmaximise carbon dioxide fixation at high temperatures. [8]\n[Total: 15]",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2021-on-43-q09-2b4d224787.pdf"
    },
    {
      "id": "9700-2021-on-43-q10",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_43.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-43-q10-p01-47603010cc.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-43-q10-p02-e874c07125.png",
        "9700-topic-15-control-and-coordination/assets/9700-2021-on-43-q10-p03-dfc93b82ba.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-43-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Describe explain the transmission of an action potential in a myelinated neurone. [9] 10 (a) and\nExplain what is meant by homeostasis in a mammal explain why it is important to (b) and\nmaintain body temperature, blood glucose concentration and the water potential of blood. [6]\n[Total: 15]\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n..................................................................................................................................................................\n......................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2021-on-43-q10-2ecd82c7d8.pdf"
    },
    {
      "id": "9700-2021-on-51-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p01-6ae1f0e792.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p02-b9bda965fb.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p03-19bf41e222.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p04-f549320bc4.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p05-529d7efb31.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q01-p06-b7db59866c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Baker’s yeast, Saccharomyces cerevisiae, has two metabolic pathways for the production of ATP, 1\naerobic respiration and fermentation. Oxygen is used in fermentation. not\nThese pathways are summarised in Fig. 1.1.\naerobic respiration fermentation\nglucose glucose\nglycolysis glycolysis ATP ATP\npyruvate pyruvate\ncarbon dioxide\ncarbon dioxide\nacetyl coenzyme A ethanal\nKrebs cycle\noxidative ATP\nphosphorylation\ncarbon dioxide ethanol\nand water\nFig. 1.1\nSome students were researching details of the respiration of yeast. In their research, the students\nfound information about the Crabtree effect. This effect occurs when yeast is kept in high\nconcentrations of glucose. The yeast gains sufficient ATP from substrate-linked phosphorylation\nduring glycolysis. Yeast does not carry out the Krebs cycle or oxidative phosphorylation and\ntherefore does not use oxygen.\nThe students wanted to investigate the effect of temperature on the activity of yeast and decided\nto use yeast solutions in which the yeast obtained all of its ATP by fermentation.\ncm3 • The students made a suspension of yeast from 1 g of dried yeast and 25 of water.\n• The suspension was left for 2 hours at 20 °C.\ncm3 • After 2 hours, 25 of glucose solution was added to the yeast suspension and the mixture\nwas stirred.\n• Some of the mixture was added immediately to the apparatus shown in Fig. 1.2.\nThe students measured the volume of carbon dioxide produced by yeast fermentation.\nstop-watch gas syringe\nthermometer\nyeast suspension\nand glucose solution\nwater\nFig. 1.2\nState the independent variable the dependent variable in this investigation. (a) (i) and\nindependent variable .........................................................................................................\n...........................................................................................................................................\ndependent variable ............................................................................................................\n...........................................................................................................................................\n[2]\nExplain why the students stirred the mixture of yeast and glucose before adding it to the (ii)\ntube in the apparatus in Fig. 1.2.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-51-q01-89b952ca00.pdf"
    },
    {
      "id": "9700-2021-on-51-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_51.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-51-q02-p01-7128de2fbf.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q02-p02-e225119b5c.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q02-p03-972c64d226.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q02-p04-2120d9e785.png",
        "9700-experimental-skills/assets/9700-2021-on-51-q02-p05-a8692330e0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Species diversity is often used as a simple way to compare the biodiversity of different regions of 2\nthe world.\nA student read that species diversity decreases with increasing distance from the equator. The\nlatitude at the equator is 0°. Latitudes increase north and south of the equator.\nThe student selected data from a database which holds records of the annual bird count in North\nAmerica.\nThis annual count is carried out each year during one day in the winter. Volunteers count all the\nspecies of birds that they observe in each location. The area of each location is the same size.\nThe student found data from 17 sample locations within the Delmarva Peninsula on the east coast\nof the USA for 2005.\nFig. 2.1 shows the 17 locations within the Delmarva Peninsula.\nN\n39\nlatitude / ° north\n38\n37\n0 50 100\nkm\nFig. 2.1\nThe student used the data to see if there was any relationship between the latitudes of the sample\nlocations and the number of species recorded at each location.\nThe data are shown in Fig. 2.2.\n180\n160\n140\n120\n100\nnumber of\nbird species\n80\n60\n40\n20\n0\n37.0 37.5 38.0 38.5 39.0 39.5 40.0\nlatitude / ° north\nFig. 2.2\nDescribe the trend shown by the scatter graph in Fig. 2.2. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe student analysed the data using Spearman’s rank correlation test. (b)\nState reasons why this statistical test is a suitable way to analyse the data about (i) three\nspecies diversity in Fig. 2.2.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nState the null hypothesis for this investigation. (ii)\n................................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-51-q02-a7c3d2e4bc.pdf"
    },
    {
      "id": "9700-2021-on-52-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 17,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-52-q01-p01-bfd35d394f.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q01-p02-216d43a7c2.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q01-p03-5dc58b74bb.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q01-p04-a3e0d04808.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q01-p05-37fad8ec9c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Some students researched how yeast, Saccharomyces cerevisiae, is used in the food industry. 1\nThe students found that in different manufacturing processes yeast is provided with different types\nof sugar. These sugars are either monosaccharides or disaccharides.\nThe students decided to compare the effect of supplying yeast with glucose or with\nmaltose.\nGlucose is a monosaccharide. Maltose is a disaccharide composed of two glucose molecules\njoined by a glycosidic bond.\ndm3 • The students made a standard yeast suspension using 100 g of dried yeast and 1 of\ndeionised water.\n• The suspension was left at 20 °C for two hours. During this time the yeast became active.\n• After two hours the yeast culture was divided into two samples.\n• Sample was provided with glucose and sample was provided with maltose. A B\n• The apparatus shown in Fig. 1.1 was set up for each sample separately.\ntube filled\nlarge test-tube with yeast\nculture and\nsugar\ninverted small\ntest-tube\nwater\nFig. 1.1\n• As the yeast respired, gas collected at the top of the small test-tubes, as shown in Fig. 1.2.\ngas collected\nyeast culture\nand sugar\nwater and\nyeast culture\nand sugar\nFig. 1.2\nThe students used the apparatus to test the hypothesis:\nYeast has a higher rate of respiration when supplied with glucose than when supplied with\nmaltose.\nState the independent variable the dependent variable in this investigation. (a) (i) and\nindependent variable .........................................................................................................\n...........................................................................................................................................\ndependent variable ............................................................................................................\n...........................................................................................................................................\n[2]\nThe students stirred the yeast suspension before dividing it into the two samples. (ii)\nExplain why it is important to stir the suspension.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe a method using the apparatus set up as shown in Fig. 1.1 that the students (iii)\ncould use to test their hypothesis.\nYour method should be set out in a logical order and be detailed enough for another\nperson to follow.\nYou should include details of how to make the yeast suspension or how to set up the not\napparatus.\n...........................................................................................................................................\n................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-52-q01-9371b55bc0.pdf"
    },
    {
      "id": "9700-2021-on-52-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-52-q02-p01-def7a90c76.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q02-p02-e66ada875c.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q02-p03-34ed41b0a3.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q02-p04-c2cae8f0bf.png",
        "9700-experimental-skills/assets/9700-2021-on-52-q02-p05-40f9ab2934.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Fig. 2.1 shows a gray tree frog, Dryophytes versicolor. 2\nFig. 2.1\nMales of this species make a very distinctive mating call.\nSome male gray tree frogs were kept in the laboratory at different temperatures. Their mating calls\nwere recorded. Each call consists of a number of pulses. The calls are repeated at intervals.\nFig. 2.2 shows four calls recorded at 20 °C. Each vertical line represents a single pulse.\none call call interval\n2\nsound\nfrequency / 1\nkHz\n0\n0 2 4 6 8 10 12 14 16 18\ntime / s\nFig. 2.2\nA scientist wanted to find out if there was a relationship between body temperature and the interval\nbetween calls.\nThe scientist put each frog in a temperature-controlled environment. The scientist took the body\ntemperature of each frog before recording the mating calls.\nState variable, temperature, that should be standardised while making the (a) one other than\nrecordings of the male gray tree frogs.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nRecordings were made of 50 frogs with body temperatures between 9.0 °C and 33.5 °C. (b)\nThe scientist measured the length of time of the intervals between the calls made by each\nfrog. The results are shown in the scatter graph in Fig. 2.3.\n8\n7\n6\n5\ninterval\nbetween 4\n/ s calls\n3\n2\n1\n0\n5 10 15 20 25 30 35 0\nbody temperature / °C\nFig. 2.3\nDescribe the trend shown by the scatter graph in Fig. 2.3.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe scientist analysed the data in Fig. 2.3 by using Pearson’s linear correlation test. (c)\nState reasons why this statistical test is a suitable way to analyse the data. (i) three\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-52-q02-c670c91cdd.pdf"
    },
    {
      "id": "9700-2021-on-53-q01",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p01-6ae1f0e792.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p02-b9bda965fb.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p03-19bf41e222.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p04-f549320bc4.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p05-529d7efb31.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q01-p06-b7db59866c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Baker’s yeast, Saccharomyces cerevisiae, has two metabolic pathways for the production of ATP, 1\naerobic respiration and fermentation. Oxygen is used in fermentation. not\nThese pathways are summarised in Fig. 1.1.\naerobic respiration fermentation\nglucose glucose\nglycolysis glycolysis ATP ATP\npyruvate pyruvate\ncarbon dioxide\ncarbon dioxide\nacetyl coenzyme A ethanal\nKrebs cycle\noxidative ATP\nphosphorylation\ncarbon dioxide ethanol\nand water\nFig. 1.1\nSome students were researching details of the respiration of yeast. In their research, the students\nfound information about the Crabtree effect. This effect occurs when yeast is kept in high\nconcentrations of glucose. The yeast gains sufficient ATP from substrate-linked phosphorylation\nduring glycolysis. Yeast does not carry out the Krebs cycle or oxidative phosphorylation and\ntherefore does not use oxygen.\nThe students wanted to investigate the effect of temperature on the activity of yeast and decided\nto use yeast solutions in which the yeast obtained all of its ATP by fermentation.\ncm3 • The students made a suspension of yeast from 1 g of dried yeast and 25 of water.\n• The suspension was left for 2 hours at 20 °C.\ncm3 • After 2 hours, 25 of glucose solution was added to the yeast suspension and the mixture\nwas stirred.\n• Some of the mixture was added immediately to the apparatus shown in Fig. 1.2.\nThe students measured the volume of carbon dioxide produced by yeast fermentation.\nstop-watch gas syringe\nthermometer\nyeast suspension\nand glucose solution\nwater\nFig. 1.2\nState the independent variable the dependent variable in this investigation. (a) (i) and\nindependent variable .........................................................................................................\n...........................................................................................................................................\ndependent variable ............................................................................................................\n...........................................................................................................................................\n[2]\nExplain why the students stirred the mixture of yeast and glucose before adding it to the (ii)\ntube in the apparatus in Fig. 1.2.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-53-q01-18461da4cf.pdf"
    },
    {
      "id": "9700-2021-on-53-q02",
      "subject": "9700",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2021-oct-nov/9700_w21_qp_53.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2021-Oct-Nov/qp/9700_w21_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2021-on-53-q02-p01-7128de2fbf.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q02-p02-e225119b5c.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q02-p03-972c64d226.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q02-p04-2120d9e785.png",
        "9700-experimental-skills/assets/9700-2021-on-53-q02-p05-a8692330e0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2021-on-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Species diversity is often used as a simple way to compare the biodiversity of different regions of 2\nthe world.\nA student read that species diversity decreases with increasing distance from the equator. The\nlatitude at the equator is 0°. Latitudes increase north and south of the equator.\nThe student selected data from a database which holds records of the annual bird count in North\nAmerica.\nThis annual count is carried out each year during one day in the winter. Volunteers count all the\nspecies of birds that they observe in each location. The area of each location is the same size.\nThe student found data from 17 sample locations within the Delmarva Peninsula on the east coast\nof the USA for 2005.\nFig. 2.1 shows the 17 locations within the Delmarva Peninsula.\nN\n39\nlatitude / ° north\n38\n37\n0 50 100\nkm\nFig. 2.1\nThe student used the data to see if there was any relationship between the latitudes of the sample\nlocations and the number of species recorded at each location.\nThe data are shown in Fig. 2.2.\n180\n160\n140\n120\n100\nnumber of\nbird species\n80\n60\n40\n20\n0\n37.0 37.5 38.0 38.5 39.0 39.5 40.0\nlatitude / ° north\nFig. 2.2\nDescribe the trend shown by the scatter graph in Fig. 2.2. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe student analysed the data using Spearman’s rank correlation test. (b)\nState reasons why this statistical test is a suitable way to analyse the data about (i) three\nspecies diversity in Fig. 2.2.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nState the null hypothesis for this investigation. (ii)\n................................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2021-on-53-q02-085ef11ea6.pdf"
    },
    {
      "id": "9700-2022-m-42-q01",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-m-42-q01-p01-597d8214c8.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-m-42-q01-p02-d8890e0c07.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q01",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The European hedgehog, Erinaceus europaeus, is a small omnivorous mammal. Its body is 1\ncovered in spines, which are usually brown. A rare variant, which lacks the brown pigment, has\nblonde (pale yellow) spines. This characteristic is coded for by a recessive allele.\nFig. 1.1 shows a European hedgehog with blonde spines.\nFig. 1.1\nAlderney is a small island between the UK and France. Hedgehogs were found on (a) not\nAlderney until the 1960s, when three pairs of hedgehogs were introduced to the island. The\nhedgehogs started to breed and some of the offspring had blonde spines.\nBy 2017, the population of hedgehogs, including individuals with brown spines and individuals\nwith blonde spines, had increased to approximately 600.\nSuggest reasons why the population of hedgehogs increased to such a large number. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nA survey taken in 2017 showed that 60% of the hedgehogs on Alderney had blonde spines. (ii)\nSuggest explain reasons why the proportion of hedgehogs with blonde spines was and\nso high.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nThe hedgehogs on Alderney are an example of an invasive alien species. (b)\nExplain why it is sometimes important to control invasive alien species.\n...................................................................................................................................................\n...........................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-m-42-q01-b557bf3904.pdf"
    },
    {
      "id": "9700-2022-m-42-q02",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-m-42-q02-p01-4e3423c26b.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-m-42-q02-p02-9f49297ee1.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-m-42-q02-p03-9620a70354.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-m-42-q02-p04-4b61b8c0b2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q02",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Giant axonal neuropathy (GAN) is a rare autosomal recessive disease, which affects neurones. 2\nMitochondria can move freely within the axons of neurones. (a)\nIn GAN, the axons of neurones become enlarged and blocked by the accumulation of specific\nproteins. This prevents the free movement of mitochondria and affects the transmission of\naction potentials along the axon membrane.\nSuggest explain why preventing the free movement of mitochondria within the (i) and\ncytoplasm of the axon affects the transmission of action potentials along the axon\nmembrane.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nA nerve conduction velocity (NCV) test can be used to measure the speed of transmission (ii)\nof nerve impulses along neurones in different parts of the body.\nNCV tests were carried out on three people with GAN. The speed of transmission of\nnerve impulses was measured in neurones in two nerves:\n• the median nerve, one of the main nerves in the arm\n• the peroneal nerve, one of the main nerves in the leg.\nTable 2.1 shows the NCV test results for the three people with GAN and the expected\nrange for a person without GAN.\nTable 2.1\n–1 speed of transmission of nerve impulse / m s\nlocation of\nexpected neurones\nperson 1 person 2 person 3\nrange\nmedian nerve 47 39 43 50–65\nperoneal nerve 44 22 28 40–45\nWith reference to Table 2.1, describe the effect of GAN on the speed of transmission of\nnerve impulses.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nWalking requires nervous control to coordinate movements. One of the first signs of GAN (iii)\nis having problems with walking.\nExplain how the speed of transmission of nerve impulses in people with GAN can affect\nwalking.\n...................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-m-42-q02-3bf798c1e0.pdf"
    },
    {
      "id": "9700-2022-m-42-q03",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-m-42-q03-p01-908dd4c801.png",
        "9700-topic-14-homeostasis/assets/9700-2022-m-42-q03-p02-5145dc4673.png",
        "9700-topic-14-homeostasis/assets/9700-2022-m-42-q03-p03-15f80e4bb0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q03",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The red blood cells of people with sickle cell anaemia have reduced oxygen-carrying capacity. 3\nExplain the relationship between the gene mutation that causes sickle cell anaemia and the (a)\nreduced oxygen-carrying capacity of red blood cells.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nPeople with sickle cell anaemia may need a blood transfusion. (b)\nOne risk associated with a blood transfusion is a condition known as transfusion-associated\ncirculatory overload (TACO). TACO is caused by a large increase in blood volume over a\nshort period of time. This increase in blood volume can be harmful.\nPredict the effect of an increase in blood volume on ADH secretion state (i) and one\nconsequence for kidney function of this change in ADH secretion.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nOne feature of TACO is an increase in fluid entering the alveoli, which makes it difficult (ii)\nto breathe. T",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-m-42-q03-6b70a8972e.pdf"
    },
    {
      "id": "9700-2022-m-42-q04",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-m-42-q04-p01-b6cd4a49d0.png",
        "9700-topic-16-inheritance/assets/9700-2022-m-42-q04-p02-1fb53e6056.png",
        "9700-topic-16-inheritance/assets/9700-2022-m-42-q04-p03-11788debfb.png",
        "9700-topic-16-inheritance/assets/9700-2022-m-42-q04-p04-fd22bac13a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q04",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Meiosis is an important process that contributes to genetic variation in a population. 4 (a)\nI of meiosis. Fig. 4.1 shows a pair of homologous chromosomes during prophase\nFig. 4.1\nOn Fig. 4.1, use the letter with a label line to identify where crossing over occurs. [1] (i) C\nDescribe how crossing over produces genetic variation in a population. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nWing pattern in the butterfly species Heliconius melpomene is controlled by genes on (b)\nautosomal chromosomes.\nThe gene for banding pattern on the upper wing has two alleles:\n• a dominant allele coding for a full band\n• a recessive allele coding for a broken band.\nThe gene for ray pattern on the lower wing has two alleles:\n• a dominant allele coding for rays\n• a recessive allele coding for no rays.\nScientists crossed a butterfly that was homozygous dominant for both genes with a butterfly\nthat was homozygous recessive for both genes. The scientists wanted to check whether the\nphenotypic ratio for offspring in the F2 generation agreed with the expected phenotypic ratio\nof 9 : 3 : 3 : 1.\nThe results of this genetic cross are shown in Fig. 4.2.\nfull broken\nband band\nparental\n×\ngeneration\nrays no rays\nF1\ngeneration\nF2\ngeneration\nnumber of\noffspring of each 99 21 8 8\nphenotype\nFig. 4.2\nExplain the term F1 generation. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-m-42-q04-bd78a6cd83.pdf"
    },
    {
      "id": "9700-2022-m-42-q05",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-m-42-q05-p01-0dd9a47d26.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-m-42-q05-p02-69ed3fed89.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "VIII Factor can be made as a recombinant human protein. 5\nVIII. Name the disease that is treated using recombinant human factor (a) (i)\n..................................................................................................................................... [1]\nVIII was available, this disease was treated with Before recombinant human factor (ii)\nVIII from donated blood. factor\nVIII, VIII instead of factor Give advantages of using recombinant human factor two\nfrom donated blood, to treat this disease.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nVIII can be synthesised from messenger RNA purified The gene that codes for human factor (b)\nfrom human liver cells.\nName the enzyme that uses messenger RNA as a template to produce complementary (i)\nDNA.\n..................................................................................................................................... [1]\nOutline sources, other than messenger RNA, from which genes can be obtained for (ii) two\ngenetic engineering.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nVIII The gene that codes for human factor can be transferred into mammalian cells in tissue (c)\nculture.\nExplain why a promoter also needs to be transferred into the mammalian cells so that human\nVIII can be synthesised. factor\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-m-42-q05-4a906534a4.pdf"
    },
    {
      "id": "9700-2022-m-42-q06",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-m-42-q06-p01-692ebe1da7.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-m-42-q06-p02-e725bfe602.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-m-42-q06-p03-45efcb077b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q06",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "The banana plant, Musa acuminata, is a tall herbaceous plant with very large leaves. 6 (a)\nFig. 6.1 shows a banana plant.\nFig. 6.1\nAn investigation was carried out to measure the net carbon dioxide uptake by a banana plant\nat different light intensities.\nFig. 6.2 shows the results of the investigation.\n8\n6\n4\nmean net\ncarbon dioxide\n2\nuptake\nm–2 s–1 μmol /\n– 0\n– 2\n– 4\n2000 4000 6000 0 8000 10 000 12 000\nlight intensity / lux\nFig. 6.2\nWith reference to Fig. 6.2, describe explain the results at a light intensity of 1000 lux. (i) and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nWith reference to Fig. 6.2, describe explain what can be concluded from the graph (ii) and\nat light intensities of between 2000 lux and 7000 lux.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-m-42-q06-0ab4f2d82d.pdf"
    },
    {
      "id": "9700-2022-m-42-q07",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-m-42-q07-p01-dcd1cecf20.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-m-42-q07-p02-4d15321131.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q07",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "In aerobic respiration, ATP is synthesised by substrate-linked reactions and by oxidative 7 (a)\nphosphorylation.\nOutline the process of oxidative phosphorylation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [7]\nIn an experiment on respiration, two different populations of yeast cells were used: and B. (b) A\n• Yeast cells in population had no mitochondria in their cells. A\n• Yeast cells in population did have mitochondria in their cells. B\nBoth populations were provided with glucose in solution and the concentration of ATP was\nmeasured every minute for seven minutes.\nFig. 7.1 shows the results of the experiment.\npopulation B 120\n100\n80\nATP\nconcentration 60\ndm–3 μmol /\n40\npop",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-m-42-q07-68906fe2ab.pdf"
    },
    {
      "id": "9700-2022-m-42-q08",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-m-42-q08-p01-5d00a11b62.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-m-42-q08-p02-7918f58ad6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q08",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Natural selection and artificial selection (selective breeding) both have important roles in 8\nevolutionary change.\nSeahorses are fish with an unusual appearance and many specialised features. (a)\nFig. 8.1 shows a seahorse.\nFig. 8.1\nTwo species of seahorse, Hippocampus erectus and H. zosterae, are found in the coastal\nwaters of the Gulf of Mexico and the Caribbean Sea.\n• It is thought that these two species of seahorse have a common ancestor.\n• The ranges of the two species overlap in many areas.\n• Male and female seahorses stay together for the whole of the breeding season.\n• H. erectus is much larger than H. zosterae.\n• Few seahorses occur that are intermediate in size between the two species.\nUse this information to name explain the type of speciation that may have occurred in the and\nevolution of these two species.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nFor many years, farmers have used selective breeding to improve the milk yield of dairy cattle (b)\nand to improve the yield of crops.\nExplain why improving milk yields in cattle by selective breeding can be more challenging\nthan improving yields of crops by selective breeding.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.....................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-m-42-q08-53fcd33a92.pdf"
    },
    {
      "id": "9700-2022-m-42-q09",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        26
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-m-42-q09-p01-34a4923ec8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Describe the biological species concept. 9 (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nSuggest when the morphological species concept is more useful than the biological species (b)\nconcept.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nCompare the characteristics of members of the domain Archaea with the characteristics of (c)\nmembers of the domain Bacteria.\nInclude similarities and differences.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...........................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-m-42-q09-ac2580fff7.pdf"
    },
    {
      "id": "9700-2022-m-42-q10",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_42.pdf?download=true",
      "source_pages": [
        27
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-m-42-q10-p01-1c36e61afd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-42-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 10.1 shows the structure of two guard cells. 10 (a)\nA\nB\nC\nFig. 10.1\nName the structures labelled A, and in Fig. 10.1. B C\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n[3]\nIn times of water stress, abscisic acid is released. This results in the closure of stomata to (b)\nreduce water loss by transpiration.\nDescribe the role of abscisic acid in stomatal closure.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 7]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-m-42-q10-f8cd7419d6.pdf"
    },
    {
      "id": "9700-2022-m-52-q01",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-m-52-q01-p01-6e3ca6e2fa.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q01-p02-b427a608de.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q01-p03-0c69fb09e5.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q01-p04-88eceb13c4.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q01-p05-c5db44f652.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Some students wanted to investigate the effect of light intensity on the rate of photosynthesis in 1\nthe Brazilian waterweed, Egeria densa. This plant is found in freshwater ponds.\nThe students used the apparatus shown in Fig. 1.1 in their investigation. A stand and clamp were\nused to hold the apparatus in a vertical position.\nplunger of syringe\ngas collected\nbubbles of gas given\noff by Egeria densa syringe barrel\npond water Egeria densa\ncontaining sodium\nhydrogencarbonate\nrubber tubing\ncm\nmeniscus\n0.0\ncapillary tubing\n2.0\n4.0\n6.0\n8.0\nFig. 1.1\nThe students carried out these steps to set up the apparatus shown in Fig. 1.1.\n• Some sodium hydrogencarbonate was added to a sample of pond water as a source of\ncarbon dioxide.\n• The syringe barrel was filled with the pond water containing sodium hydrogencarbonate.\n• A fresh piece of E. densa was cut under water.\n• The cut stem was quickly placed in the pond water in the syringe barrel.\n• The plunger was replaced, as shown in Fig. 1.1, and adjusted until the meniscus in the\ncapillary tubing was near the top of the scale.\n• A bench lamp was used as a light source.\nThe students noticed that bubbles of gas were given off from the E. densa. As the gas collected in\nthe syringe barrel, the meniscus moved down the capillary tube. The plunger of the syringe did not\nmove.\nThe students used the apparatus to investigate the effect of different light intensities on the rate of\nphotosynthesis in E. densa.\nSuggest a suitable control for this investigation. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nDescribe a method, using the apparatus shown in Fig. 1.1, that the students could use to (b) (i)\ninvestigate the effect of light intensity on the rate of photosynthesis of E. densa.\nYour method should be set out in a logical order and be detailed enough to allow another\nperson to follow it.\nThe steps carried out to set up the apparatus shown in Fig. 1.1 should be included. not\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.........................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-m-52-q01-776eab7348.pdf"
    },
    {
      "id": "9700-2022-m-52-q02",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_52.pdf?download=true",
      "source_pages": [
        7,
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p01-c61a90dcfd.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p02-fe0b43321d.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p03-e90b90fbe1.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p04-0d8d0c7064.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p05-8045564db6.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q02-p06-509661e924.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "An invasive alien species is a species that has been introduced into an ecosystem where it is not 2\nnormally found and causes harm to this ecosystem. Invasive alien species can change habitats,\nreduce biodiversity and cause the extinction of native species. Saltcedar trees, Tamarix spp., are\ninvasive alien species introduced to North America from Europe and Asia in the early 1800s.\nA biologist carried out an investigation in a woodland ecosystem next to the Virgin River in Arizona,\nNorth America. In the woodland, there were areas that contained only saltcedar trees and areas\nthat contained a mixture of native tree species and saltcedar trees.\nThe biologist wanted to test the hypothesis that:\nThe diversity of rodents (mice and rats) is lower in areas with only saltcedar trees than in\nareas with a mixture of native tree species and saltcedar trees.\nIdentify the variable in this investigation. (a) independent\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe biologist decided to trap rodents in an area with only saltcedar trees and to trap rodents (b)\nin an area with a mixture of native tree species and saltcedar trees.\nThe biologist:\n• placed 25 small mammal traps at random sites in each area of woodland\n• baited each trap with 10 g of food for rodents\n• checked each trap after 24 hours\n• identified any rodents caught in the trap\n• marked the rodents with an ear tag and released them back into the woodland\n• carried out this trapping process on four days in April.\nThe number of individuals of each rodent species trapped during the four trapping different\nsessions was recorded.\nIdentify variables, the small mammal trap and ear tag, that the biologist (i) three other than\nstandardised in this investigation.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe a method that the biologist could have used to select the trap sites randomly. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n........................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-m-52-q02-4086824261.pdf"
    },
    {
      "id": "9700-2022-m-52-q03",
      "subject": "9700",
      "year": 2022,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 4,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-feb-march/9700_m22_qp_52.pdf?download=true",
      "source_pages": [
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-March/qp/9700_m22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-m-52-q03-p01-5851ae718b.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q03-p02-062e271a56.png",
        "9700-experimental-skills/assets/9700-2022-m-52-q03-p03-3261a5f894.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-m-52-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Mussels, Mytilus edulis, are molluscs that live in sea water on the shorelines of coastal regions 3\naround the world. Mussels are a popular seafood and so are widely cultivated. After harvesting\nthe mussels, the mussel farmer must replace them with young mussels. This is often done by\ncollecting young mussels from wild marine ecosystems.\nFig. 3.1 shows mature mussels after harvesting.\nFig. 3.1\nThe slipper limpet, Crepidula fornicata, is an invasive alien species introduced into Europe from\nNorth America. Slipper limpets compete with the mussels and reduce the yield of mussels.\nSometimes slipper limpets are accidentally collected with young mussels. Mussel farmers want to\nprevent the introduction of slipper limpets when re-stocking their mussel farms.\nFig. 3.2 shows slipper limpets.\nFig. 3.2\nScientists investigated the best way to kill slipper limpets without harming the mussels. The\nscientists put 30 slipper limpets (length > 30 mm) into each of four trays. Each tray of slipper\nlimpets was exposed to a different test condition for a period of three days at a temperature of\n12 °C to 13 °C. The four test conditions are shown in Table 3.1.\nTable 3.1\ntest condition details of test condition\n1 air exposure leave organisms exposed to the air\n2 brine rinse at the start of day 1, rinse organisms with a saturated salt (sodium chloride)\nsolution for 5 minutes and then leave exposed to the air\n3 repeat brine rinse at the start of day 1, day 2 and day 3, rinse organisms with a saturated salt\n(sodium chloride) solution for 5 minutes and then leave exposed to the air\n4 brine soak at the start of day 1, soak organisms in a saturated salt (sodium chloride)\nsolution for 1 hour and then leave exposed to the air\nThis procedure was repeated with one tray of 30 slipper limpets (length < 30 mm) and one tray of\n30 mussels (length > 30 mm) for each of the four test conditions.\nAfter three days, the scientists counted the number of dead slipper limpets and dead mussels in\neach tray.\nThe scientists repeated the whole investigation at a temperature of 4 °C to 5 °C, using fresh\nsamples of slipper limpets and mussels.\nThe results of these tests are shown in Fig. 3.3A and Fig. 3.3B.\nA mussel farmer concluded from the results of the investigation shown in Fig. 3.3 that:\nYoung mussels used to re-stock the mussel farm should be given a brine soak treatment to\nkill any slipper limpets that are present.\nEvaluate this conclusion.\nYou should use the data in Fig. 3.3A and Fig. 3.3B to support your answer.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n......................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-m-52-q03-7eec3eabb5.pdf"
    },
    {
      "id": "9700-2022-mj-41-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-41-q01-p01-a66a29c817.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-41-q01-p02-08774b8c3c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "The golden mantella, Mantella aurantiaca, is a small terrestrial frog found in Madagascar. 1 (a)\nFig. 1.1 shows a golden mantella.\nFig. 1.1\nName the domain and kingdom to which the golden mantella belongs. (i)\ndomain ..............................................................................................................................\nkingdom .............................................................................................................................\n[2]\nThe skin of the golden mantella is brightly coloured and contains a toxin. (ii)\nSuggest a benefit to the frog of being brightly coloured.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe toxin in the skin of the golden mantella affects the action of the sarcomeres in muscle (b)\nCa2+ATPase, fibres (muscle cells) of mammalian striated muscle. The toxin inhibits a protein,\nfound in the membrane of the sarcoplasmic reticulum.\nCa2+ATPase pumps calcium ions from the cytoplasm into the sarcoplasmic reticulum when\nthe fibre is no longer stimulated.\nSuggest the consequences to the sarcomere of the action of the golden mantella toxin.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDescribe the role of calcium ions in a cholinergic synapse. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n......................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-41-q01-76a5c0977e.pdf"
    },
    {
      "id": "9700-2022-mj-41-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        4,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-41-q02-p01-bb29b7e8d4.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-41-q02-p02-24ccb988ba.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-41-q02-p03-26ab84f4a9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "In eukaryotes, the cells of plants and photosynthetic protoctists contain chloroplasts. 2 (a)\nDescribe the relationship between the function of a chloroplast and its structure.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nOutline the main features of cyclic photophosphorylation. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-mj-41-q02-237607fdda.pdf"
    },
    {
      "id": "9700-2022-mj-41-q03",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-mj-41-q03-p01-7d4f2ea13a.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-mj-41-q03-p02-70597dbd39.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Respiration is a process that results in the synthesis of ATP. The ATP can be used within the cell 3\nfor energy-requiring reactions and processes.\nThere are four stages in aerobic respiration: glycolysis, the link reaction, the Krebs cycle and\noxidative phosphorylation.\nThe ATP synthesised in respiration can be used to make larger and more complex biological (a)\nmolecules from smaller molecules.\nName the type of reaction that occurs when larger more complex biological molecules are\nmade from smaller molecules.\n............................................................................................................................................. [1]\nThe first part of glycolysis uses ATP. (b)\nExplain why ATP is needed in the first part of glycolysis.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nState the precise locations of substrate-linked phosphorylation reactions in aerobic respiration. (c)\n...................................................................................................................................................\n............................................................................................................................................. [1]\nExplain what happens to pyruvate in the link reaction in aerobic respiration. (d)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nChemiosmosis is a process that occurs in mitochondria during aerobic respiration and in (e)\nchloroplasts during photosynthesis.\nDescribe the differences between the process of chemiosmosis in mitochondria and the\nprocess of chemiosmosis in chloroplasts.\n............................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-mj-41-q03-f565fcf34f.pdf"
    },
    {
      "id": "9700-2022-mj-41-q04",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-41-q04-p01-b20f13ae0f.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-41-q04-p02-4cb33cdab4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "In 1973 a technique for genetic engineering was used for the first time. Recombinant DNA was 4\nmade using a plasmid and this was successfully transferred into an organism.\nIn 2012 a new technique for genetic engineering, called gene editing, was developed.\nTable 4.1 lists some statements about the two genetic engineering techniques. (a)\nComplete Table 4.1 to compare the original genetic engineering technique using a plasmid\n3 vector with the newer technique of gene editing. For each row, place a tick ( ) in the correct\ncolumn if the statement applies and leave a blank if the statement does not apply.\nTable 4.1\nstatement genetic engineering gene editing\nusing a plasmid\nIt can produce a transgenic\norganism.\nIt can modify the\ncharacteristics of an organism.\nIt can delete unwanted DNA.\nIt uses an enzyme that cuts\nDNA.\nIt can use RNA to precisely\nlocate the target gene.\n[5]\nOrange trees, Citrus sinensis, produce fruits that are an important food crop. The functional (b)\nleaf area of orange trees may be reduced by the growth of citrus canker bacteria. These\nbacteria cause citrus canker disease.\nScientists used gene editing to develop two types of orange tree with different mutations\n(changes to the DNA). The mutant orange tree leaves showed resistance to citrus canker\ndisease.\nFig. 4.1 shows the area of leaf with citrus canker disease in wild type (not gene edited) and\ngene edited orange tree leaves after they have been exposed to citrus canker bacteria.\n3.5\n3.0\n2.5\n2.0 area of leaf\nwith canker\nmm2 disease / 1.5\n1.0\n0.5\n0.0\nA B C\ntype of orange tree\nFig. 4.1\nIdentify the letter that represents the wild type orange trees on Fig. 4.1. (i)\n..................................................................................................................................... [1]\nExplain the social benefits of this example of gene editing. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\n[Total: 9]",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-mj-41-q04-df5c1a3df7.pdf"
    },
    {
      "id": "9700-2022-mj-41-q05",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-41-q05-p01-126d2e6c9e.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-41-q05-p02-f16783f2cf.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-41-q05-p03-09837a9214.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Two subspecies of reindeer, Rangifer tarandus, live in North America. Members of the different 5\nsubspecies belong to the same species but have some morphological differences and are found\nin different geographical locations.\nFig. 5.1 shows a reindeer.\nFig. 5.1\nTable 5.1 compares the features of the two North American reindeer subspecies.\nTable 5.1\nwoodland subspecies, barren ground subspecies,\nfeature\nR. tarandus caribou R. tarandus groenlandicus\nsouthern woodland northern tundra\nhabitat\n(warmer) (colder)\ntype of food tree leaves, grass lichens, moss\nsummer and winter feeding\nyes no\ngrounds overlap\ncarry out long migrations no yes\nbody size large small\ncolour of fur dark light\nDuring the last ice age an ice sheet separated southern and northern populations of (a)\nR. tarandus in North America.\nExplain how this ice sheet affected the evolution of R. tarandus to result in the two different\nsubspecies.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nAssess the relative importance of natural selection and genetic drift in producing: (b)\nthe different colours of fur of the two subspecies of reindeer (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..........................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-mj-41-q05-eed7670bba.pdf"
    },
    {
      "id": "9700-2022-mj-41-q06",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-mj-41-q06-p01-a066f71e95.png",
        "9700-topic-14-homeostasis/assets/9700-2022-mj-41-q06-p02-47ee630559.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "In mammals, the blood glucose concentration must be maintained within narrow limits so that 6 (a)\nthe body cells can function efficiently.\nName the mechanism by which the blood glucose concentration is maintained within\nnarrow limits.\n............................................................................................................................................. [1]\n(α) Glucagon is released by the alpha cells of the pancreas when the blood glucose (b)\nconcentration decreases below the set point.\nFig. 6.1 outlines the response of liver cells to glucagon.\nglucagon\ncell enzyme\nsurface receptor A\nmembrane\nATP G-protein\nsecond\nmessenger\nB\ncytoplasm\nactivated\nkinase\nenzymes\nenzyme\ncascade\nactivated\nglycogen\nphosphorylase\nFig. 6.1\nState how glucagon reaches the liver cells. (i)\n..................................................................................................................................... [1]\nWith reference to Fig. 6.1, name enzyme and second messenger B. (ii) A\n........................................................................................................................................ A\n........................................................................................................................................ B\n[2]\nState the role of the enzyme cascade. (iii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState the function of the final enzyme in the pathway, glycogen phosphorylase. (iv)\n..................................................................................................................................... [1]\nA biosensor is used to measure blood glucose concentration to check that it is within the (c)\nnormal range.\nDescribe how a glucose biosensor works.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.......................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-mj-41-q06-8791fc9cde.pdf"
    },
    {
      "id": "9700-2022-mj-41-q07",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-mj-41-q07-p01-6b74c3ac33.png",
        "9700-topic-16-inheritance/assets/9700-2022-mj-41-q07-p02-bfbf845509.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The fruit fly, Drosophila melanogaster, usually has red eyes. A gene for eye colour has four 7 (a)\nalleles: red, apricot, honey and white.\nDefine the terms gene and allele.\ngene ..........................................................................................................................................\n...................................................................................................................................................\nallele .........................................................................................................................................\n...................................................................................................................................................\n[2]\nR, • The allele for red eyes, is dominant to the other three alleles. (b) C\nA, H. • The allele for apricot eyes, is dominant to the allele for honey eyes, C C\nW, • The allele for white eyes, is recessive to the other three alleles. C\nConstruct a genetic diagram to show the genotypes, phenotypes and ratio of the offspring\nR H, A W. and a fruit fly with apricot eyes, from a cross between a fruit fly with red eyes, C C C C\nparents phenotypes red eyes apricot eyes\nR H A W parents genotypes C C C C\ngametes\noffspring genotypes\noffspring phenotypes\nratio\n[3]\nDescribe how you would carry out a test cross use it to determine the genotype of a red- (c) and\neyed fruit fly.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-mj-41-q07-693ee72df6.pdf"
    },
    {
      "id": "9700-2022-mj-41-q08",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-41-q08-p01-006356ccc8.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-41-q08-p02-4880c38603.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Over the past 200 years, many species of animals and plants have become extinct. 8 (a)\nFig. 8.1 shows the changes between the years 1800 and 2000 in:\n• the number of species becoming extinct\n• the size of the world human population.\n40 000 8000\n7000\n30 000 6000\n5000\nnumber of world human\n20 000 4000 known extinct population\n(×106) species\n3000\n000 10 2000\n1000\n0 0\n1800 1850 1900 1950 2000\nyear\nKey\nnumber of species\nhuman population\nFig. 8.1\nIt has been suggested that there is a correlation between the number of species (i)\nbecoming extinct and the size of the world human population.\nSuggest reasons for this possible correlation.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nCalculate the rate of species extinction per year between 1950 and 2000. (ii)\nShow your working.\nanswer = ......................................................... [2]\nExtinction of animal and plant species reduces biodiversity. (b)\nExplain why it is important to maintain biodiversity.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-mj-41-q08-6eaae99741.pdf"
    },
    {
      "id": "9700-2022-mj-41-q09",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-41-q09-p01-e6e7ed712d.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-41-q09-p02-3312d4bb89.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Stomata are involved in both transpiration and photosynthesis in plants. 9 (a)\nFig. 9.1 is a diagram of an open stoma, its guard cells and surrounding epidermal cells.\nA\nD\nB\nC\nFig. 9.1\nComplete Table 9.1 by choosing the correct letter from Fig. 9.1 to match the feature stated in\nTable 9.1.\nEach letter may be used once, more than once or not at all.\nTable 9.1\nletter feature\nlocation of Calvin cycle\n..................\nmade mainly of cellulose\n..................\n[2]\nThe opening and closing of stomata are due to changes in environmental conditions. (b)\n(H+) (K+) and potassium ions are involved in the opening of stomata. Hydrogen ions\nDescribe how hydrogen ions and potassium ions are involved in the opening of a stoma.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 6]",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-mj-41-q09-4439aaa213.pdf"
    },
    {
      "id": "9700-2022-mj-41-q10",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-41-q10-p01-41688034e1.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-41-q10-p02-6b792c35c0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-41-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Motor neurones and sensory neurones have different roles in nervous coordination. 10 (a)\nContrast the structure function of sensory neurones and motor neurones. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nMammals have many types of neurones, which vary in axon diameter and myelination. (b)\nTable 10.1 shows the axon diameter and mean impulse transmission speed of four different\ntypes of mammalian neurone.\nTable 10.1\nmean impulse neurone type myelination mean axon diameter\nµm transmission /\n–1 speed / m s\n3 α motor 13.0–20.0 80–120\n3 β motor 6.0–12.0 35–75\n3 δ motor 1.0–5.0 5–35\n7 motor C 0.2–1.5 0.5–2.0\nUsing the data shown in Table 10.1, comment on the relationship between:\n• myelination and mean impulse transmission speed\n• axon diameter and mean impulse transmission speed.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-41-q10-bcf659d0ff.pdf"
    },
    {
      "id": "9700-2022-mj-42-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-mj-42-q01-p01-53ab84e4ee.png",
        "9700-topic-16-inheritance/assets/9700-2022-mj-42-q01-p02-0b29c1e1cc.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q01",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The jaguar, Panthera onca, is a large cat that lives mainly in South America. The majority of 1\njaguars have light brown fur with black spots, as shown in Fig. 1.1. Some jaguars have completely\nblack fur, as shown in Fig. 1.2.\nFig. 1.1 Fig. 1.2\nThe pigments involved in fur colour are produced as a result of biochemical pathways that (a)\ntake place in cells called melanocytes. These pathways are similar to those that occur in\nhuman melanocytes.\nThe melanocortin 1 receptor (MC1R) is located on the cell surface membrane of melanocytes\nand is coded for by the MC1R gene.\nFig. 1.3 outlines the processes that occur in jaguar melanocytes.\nphaeomelanin\n(light brown)\nMC1R\nnot activated\ntyrosine substance X\nMC1R\nactivated\neumelanin MC1R\n(black)\ncell surface\nmembrane\nFig. 1.3\nName the substance represented by X. (i)\n..................................................................................................................................... [1]\nWhen MC1R is activated a second messenger is produced in the cell. (ii)\nGive an example of a second messenger.\n..................................................................................................................................... [1]\nSubstance is also produced in humans, but a mutation of the TYR gene can result in (iii) X\nsubstance not being produced. X\nDescribe the phenotype of a person with this mutation.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe MC1R gene has two alleles and is located on an autosome. (b)\n• When two jaguars with light brown fur mate all the offspring have light brown fur.\n• When two jaguars with black fur mate either all the offspring will have black fur or some\noffspring will have black fur and some will have light brown fur.\nUsing symbols, construct a genetic diagram to show how two jaguars with black fur can\nproduce some offspring with black fur and some offspring with light brown fur.\n[4]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-mj-42-q01-6b47f42f6d.pdf"
    },
    {
      "id": "9700-2022-mj-42-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-42-q02-p01-51a609e6c1.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-42-q02-p02-e9b22bd73d.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-42-q02-p03-730669c331.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-42-q02-p04-a2fbf9b26f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is affected by many environmental factors. 2 (a)\nExplain why light intensity can be a limiting factor in photosynthesis. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\n) can also be a limiting factor. It has an effect The concentration of carbon dioxide (CO (ii)\n2\non the Calvin cycle in the light-independent stage of photosynthesis.\nRibulose bisphosphate (RuBP), triose phosphate (TP) and glycerate 3-phosphate (GP)\nare three important molecules in the Calvin cycle.\nis Fig. 2.1 shows how the concentration of GP changes when the concentration of CO\n2\nreduced from 0.04% (atmospheric) to 0.008%.\nComplete Fig. 2.1 by sketching the lines for RuBP and TP when the concentration of\nis reduced from 0.04% to 0.008%. CO\n2\n0.04% 0.008%\nreduced CO concentration atmospheric CO\n2 2\nGP\nTP\nrelative\nconcentration\nof Calvin\nRuBP\ncycle molecules\nGP\ntime\n[2] Fig. 2.1\nIn very dry conditions, CO concentration can become the main limiting factor of (iii)\n2\nphotosynthesis in plants.\nconcentration to become the main limiting Explain how very dry conditions cause CO\n2\nfactor of photosynthesis in plants.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nA factor that can limit the rate of photosynthesis is the rate of regeneration of RuBP. (b)\nSedoheptulose-1,7-bisphosphatase (SBPase) is an enzyme in the Calvin cycle that controls\nthe rate of regeneration of RuBP. SBPase is coded for by the gene SBPase.\nIn an experiment, wheat plants were genetically modified to make more SBPase by introducing\nthe SBPase gene from another grass species, Brachypodium distachyon. The resulting GM\nwheat plants were named Sox4.\n• Wild type plants (not GM) and Sox4 plants were grown in a greenhouse.\n• Lig",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-mj-42-q02-738bd479f3.pdf"
    },
    {
      "id": "9700-2022-mj-42-q03",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        8
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-mj-42-q03-p01-676c4bbb56.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Aerobic respiration occurs when oxygen is available. A much greater energy yield is obtained from 3\naerobic respiration than from respiration in anaerobic conditions.\nExplain why the link reaction only occurs when oxygen is available. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nOutline the stage of aerobic respiration that occurs in the cytoplasm of eukaryotic cells. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-mj-42-q03-bb92aaf064.pdf"
    },
    {
      "id": "9700-2022-mj-42-q04",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-42-q04-p01-8c8f0c9509.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-42-q04-p02-99d1d90d1a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "In 1973, a technique for genetic engineering was used for the first time. Recombinant DNA was 4\nmade using a plasmid and this was successfully transferred into an organism.\nIn 2012, a new technique for genetic engineering, called gene editing, was developed.\nTable 4.1 lists some statements about the two genetic engineering techniques. (a)\nComplete Table 4.1 to compare the original genetic engineering technique using a plasmid\n3 vector with the newer technique of gene editing. For each row, place a tick ( ) in the correct\ncolumn if the statement applies and leave a blank if the statement does not apply.\nTable 4.1\nstatement genetic engineering gene editing\nusing a plasmid\nIt may cause the organism to\nproduce a different protein.\nIt may cause a single base pair\nin a gene to be changed.\nThe success of the technique\ncan be evaluated using marker\ngenes.\nIt may use the CRISPR system.\nIt uses DNA ligase.\n[5]\nCassava plants, Manihot esculenta, produce roots that have a high starch content. These (b)\nroots are an important food source in tropical regions. The growth of cassava plants is\nreduced by competition from weeds.\nScientists used gene editing to develop two types of cassava plant with different mutations\n(changes to the DNA). The gene edited cassava plants showed resistance to the herbicide\nglyphosate. In susceptible plants, glyphosate prevents synthesis of three amino acids from a\nprecursor molecule called shikimate.\nFig. 4.1 shows the concentration of shikimate in the wild type (not gene edited) and the two\ntypes of gene edited cassava plant after they were exposed to three different concentrations\nof glyphosate.\n30\nKey:\ndm–3 μmol / glyphosate\n0\n50\n20\n500 shikimate\n/ concentration\narbitrary units\n10\n0\nA B C\ntype of cassava plant\nFig. 4.1\nIdentify the letter in Fig. 4.1 that represents the wild type cassava plant. (i)\n..................................................................................................................................... [1]\nExplain the social benefit of this example of gene editing. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.............................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-mj-42-q04-9848d5e1fd.pdf"
    },
    {
      "id": "9700-2022-mj-42-q05",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-42-q05-p01-f2ade81e88.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-42-q05-p02-661c1ae954.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-42-q05-p03-1d26d5f9bb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Bison are a type of large wild cattle. Ancestors of modern bison appeared in Asia 2.5 million years 5\nago. This ancestral bison species increased its range into Europe and North America.\nWhile the ancestral species is now extinct, its descendants include Bison bonasus, the European\nbison, and Bison bison, the American bison.\nFig. 5.1 shows an American bison.\nFig. 5.1\nAfter the end of the last ice age, populations of the ancestral bison were separated by sea (a)\nand by forests that were not suitable as habitats. The separation resulted in the evolution of\nthe European bison and the American bison.\nExplain how this separation resulted in the evolution of the two bison species.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nTable 5.1 compares features of European bison and American bison. (b)\nTable 5.1\nfeature European bison, American bison,\nB. bonasus B. bison\ntype of food grass and higher vegetation grass and low vegetation\nsuch as leaves of bushes and\ntrees\nheight / m 2.1 2.0\nmaximum mass / kg 1000 1270\nAssess the relative importance of natural selection and genetic drift in producing the different\nheights and masses of the two species of bison.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-mj-42-q05-bd0637a203.pdf"
    },
    {
      "id": "9700-2022-mj-42-q06",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-42-q06-p01-ce02a6ed76.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-42-q06-p02-f569f43fed.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-42-q06-p03-0d81b9642c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q06",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Gibberellin is a plant hormone that has an important role in seed germination. 6 (a)\nFig. 6.1 is a diagram of a section through a barley seed (grain).\naleurone\nlayer\nendosperm\nembryo\nFig. 6.1\nOn Fig. 6.1:\n• Draw arrow, labelled G, to show the movement of gibberellin during germination. one\n• Draw arrow, labelled A, to show the movement of amylase during germination. one\n[2]\nOutline the role of amylase in seed germination. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nThe germination of three groups of seeds of the plant Penstemon digitalis was investigated. (c)\nThe seeds were soaked for 24 hours in distilled water or in a solution of gibberellin. They\nwere then sown on filter paper in dishes and kept moist for 10 days.\nFig. 6.2 shows the results for each group.\n100\ngroup C\ndm–3 90 1000 mg\ngibberellin solution\n80\n70\n60\ngroup B\npercentage\ndm–3 50 500 mg\ngermination\ngibberellin solution\n40\n30\ngroup A\n20 distilled water\n10\n0\n0 2 4 1 3 5 6 7 8 9 10\ntime after soaking / days\nFig. 6.2\nDescribe the results shown in Fig. 6.2.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.......................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-42-q06-fea746999a.pdf"
    },
    {
      "id": "9700-2022-mj-42-q07",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-mj-42-q07-p01-decb60feca.png",
        "9700-topic-16-inheritance/assets/9700-2022-mj-42-q07-p02-4795ce39e8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The bacterium, Escherichia coli, can use glucose or disaccharides, such as lactose, in its 7\nβ-galactosidase metabolism. Lactose needs to be hydrolysed by the enzyme to form glucose and\ngalactose, which can then be used by E. coli.\nβ-galactosidase The production of is controlled by a length of DNA called the lac operon.\nFig. 7.1 shows the lac operon when lactose is absent. (a)\nRNA polymerase\nrepressor\nP O lacZ lacY lacA\nFig. 7.1\nOn Fig. 7.2, draw the positions of RNA polymerase and the repressor molecule when lactose\nis present.\nP O lacZ lacY lacA\n[2] Fig. 7.2\nThe protein coded for by lacY is not involved in the control of gene expression. (b)\nName the type of gene represented by lacY. (i)\n..................................................................................................................................... [1]\nName the protein product coded for by lacY state the precise role of this protein. (ii) and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nIn an investigation into the growth of E. coli, a sample of the bacterium was grown in a (c)\nmedium that contained limited concentrations of glucose and lactose. The population size of\nE. coli was measured at regular intervals.\nFig. 7.3 shows the population growth curve obtained for this investigation.\nE. coli\npopulation\nsize\ntime\nFig. 7.3\nDescribe suggest explanations for the population growth curve shown in Fig. 7.3. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-mj-42-q07-d4e50c8b12.pdf"
    },
    {
      "id": "9700-2022-mj-42-q08",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-42-q08-p01-19a011c43c.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-42-q08-p02-96f92dc089.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "The role of sensory receptor cells in mammals is to detect stimuli and generate action potentials in 8\nsensory neurones.\nHuman taste buds on the tongue contain chemoreceptor cells. Different chemoreceptor cells\nrespond to different chemical stimuli.\nFig. 8.1 is a diagram of chemoreceptor cells in a taste bud.\nY\ntight junction\nsupport cell\nchemoreceptor cell A\nsensory\nneurone B\nFig. 8.1\nName the structures in the region Y. (a)\n............................................................................................................................................. [1]\nSuggest a reason for the tight junctions between the chemoreceptor cells. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\n(Na+) Chemoreceptor cell responds to sodium ions in salt. (c) A\nNa+ can result in an action potential in sensory Describe how the contact of cell with A\nneurone B.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-42-q08-f63a7aa7dd.pdf"
    },
    {
      "id": "9700-2022-mj-42-q09",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-mj-42-q09-p01-4aa2510c81.png",
        "9700-topic-14-homeostasis/assets/9700-2022-mj-42-q09-p02-5546292724.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Ultrafiltration in the kidney takes place between the glomerulus and the Bowman’s capsule. 9 (a)\nThe afferent blood vessel carrying blood to the glomerulus has a wider lumen than the efferent\nblood vessel.\nExplain why the lumen of the afferent blood vessel needs to be wider than the lumen of the\nefferent blood vessel.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nFig. 9.1 is a diagram of part of the glomerulus and Bowman’s capsule. (b)\nB A\nblood capillary lumen of Bowman’s\ncapsule\nFig. 9.1\nName and B. (i) A\n........................................................................................................................................ A\n........................................................................................................................................ B\n[2]\nDescribe the roles of and in the formation of the glomerular filtrate. (ii) A B\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\n[Total: 7]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-mj-42-q09-6f4858d81d.pdf"
    },
    {
      "id": "9700-2022-mj-42-q10",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_42.pdf?download=true",
      "source_pages": [
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-42-q10-p01-f349fcd4b4.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-42-q10-p02-f5a50118f0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-42-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The African penguin, Spheniscus demersus, lives along the coast of South Africa. 10\nFig. 10.1 shows two African penguins.\nFig. 10.1\nOne way of estimating the size of a population of African penguins is to use the mark-release- (a)\nrecapture method.\nSuggest assumptions that must be made for the mark-release-recapture method to be three\nvalid.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nThe International Union for the Conservation of Nature (IUCN) Red List of Threatened (b)\nSpecies has categorised the African penguin as endangered and could become extinct.\nSuggest reasons why the African penguin has become endangered and could become\nextinct.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nFig. 10.2 shows the estimated numbers of African penguins in the years of 1800, 1900 and (c)\n2000.\n4 000 000\n3 000 000\nestimated\n2 000 000\npopulation\nsize\n1 000 000\n0\n1800 1900 2000\nyear\nFig. 10.2\nCalculate the mean yearly decrease in population size of the African penguin between 1900\nand 2000.\n–1 [1] answer = .................................................. yr\nThe African penguin is a member of the kingdom Animalia. (d)\nOutline the characteristic features of the kingdom Animalia.\n...................................................................................................................................................\n......................................................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-mj-42-q10-55e52c8f0c.pdf"
    },
    {
      "id": "9700-2022-mj-43-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-mj-43-q01-p01-7803dc30c2.png",
        "9700-topic-14-homeostasis/assets/9700-2022-mj-43-q01-p02-7aefcca483.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The water potential of mammalian blood needs to be maintained within narrow limits so that 1 (a)\ncells function efficiently. This process is called osmoregulation.\nThe relative medullary thickness (RMT) indicates the proportion of a kidney that is composed\nof medullary tissue.\nthickness of medulla\n× RMT = 10\nkidney size\nTable 1.1 shows the relationship between the RMT and the concentration of urine produced\nby four mammals from different habitats.\nTable 1.1\nmammal habitat RMT urine\nconcentration / arbitrary\nunits\nbeaver rivers and lakes 1.4 0.90\nwarthog savannah 2.8 2.35\nhuman variable 3.2 2.50\nkangaroo rat desert 8.6 10.50\nName the parts of the nephron that are located in the medulla. (i)\n...........................................................................................................................................\n..................................................................................................................................... [2]\nName a hormone involved in osmoregulation. (ii)\n..................................................................................................................................... [1]\nDescribe the relationship between the RMT and the concentration of urine produced (iii) and\nexplain the differences between the data for the beaver and the kangaroo rat.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nThe warthog, Phacochoerus africanus, is a member of the pig family. The warthog lives in dry (b)\nsavannah areas of sub-Saharan Africa.\nFig. 1.1 shows a warthog.\nFig. 1.1\nA warthog and a human have similar values of RMT and concentration of urine. A human\ncan survive only a few days without drinking water, whereas a warthog can live for several\nmonths without drinking water.\nSuggest how a warthog is able to survive several months without drinking water.\n............................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-mj-43-q01-18b70ad765.pdf"
    },
    {
      "id": "9700-2022-mj-43-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-43-q02-p01-76ef22ea0d.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-43-q02-p02-b4fd9ec82a.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-mj-43-q02-p03-1b737a3ab5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is an energy transfer process that results in the production of carbohydrate. It has 2\ntwo stages: the light-dependent stage and the light-independent stage.\nCyclic photophosphorylation and non-cyclic photophosphorylation are essential pathways in\nphotosynthesis that occur in the light-dependent stage.\nDescribe the similarities and differences between cyclic photophosphorylation and (a) (i)\nnon-cyclic photophosphorylation.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nExplain why herbicides that prevent cyclic photophosphorylation and non-cyclic (ii)\nphotophosphorylation stop carbohydrate being produced in the chloroplast.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe rate of regeneration of RuBP in the Calvin cycle is known to limit the rate of photosynthesis. (b)\nSedoheptulose-1,7-bisphosphatase (SBPase) is an enzyme in the Calvin cycle that controls\nthe rate of regeneration of RuBP. SBPase is coded for by the gene SBPase.\nIn an experiment, wheat plants were genetically modified to make more SBPase by introducing\nthe SBPase gene from another grass species, Brachypodium distachyon. The resulting GM\n",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-mj-43-q02-fc04e691e2.pdf"
    },
    {
      "id": "9700-2022-mj-43-q03",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        8
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-mj-43-q03-p01-a89a38f5c2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "ATP is synthesised from ADP and P in a phosphorylation reaction. 3 (a)\ni\nState the different ways in which this phosphorylation reaction occurs in aerobic respiration. two\n...................................................................................................................................................\n............................................................................................................................................. [2]\nCoenzymes are important in all four stages of aerobic respiration. (b)\nDescribe explain the role of the coenzymes NAD and FAD in aerobic respiration. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [6]\nThe enzyme pyruvate dehydrogenase catalyses the link reaction. Pyruvate dehydrogenase is (c)\ninhibited when the ratio of acetyl coenzyme A to coenzyme A increases.\nSuggest the importance of this inhibition to the functioning of the cell.\n...................................................................................................................................................\n...................................................................................................................................................\n........................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-mj-43-q03-3138a7b109.pdf"
    },
    {
      "id": "9700-2022-mj-43-q04",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-43-q04-p01-3ad4a6957b.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-mj-43-q04-p02-16006a4eec.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "In 1973, a technique for genetic engineering was used for the first time. Recombinant DNA was 4\nmade using a plasmid and this was successfully transferred into an organism.\nIn 2012, a new technique for genetic engineering, called gene editing, was developed.\nTable 4.1 lists some statements about the two genetic engineering techniques. (a)\nComplete Table 4.1 to compare the original genetic engineering technique using a plasmid\nvector with the newer technique of gene editing.\n3 For each row, place a tick ( ) in the correct column if the statement applies and leave a blank\nif the statement does not apply.\nTable 4.1\nstatement genetic engineering gene editing\nusing a plasmid\nIt can add a new phenotypic\ncharacteristic to an organism.\nIt can change an A–T base pair\nto C–G.\nIt can inactivate a desired\nselected gene in an organism.\nIt may change DNA in a way\nthat cannot be distinguished\nfrom a natural mutation.\nIt requires a DNA donor and a\nrecipient.\n[5]\nCamelina sativa is a fast-growing plant with oil-rich seeds. (b)\nC. sativa grows in dry and poor soils and so it may be important as a food crop in the future.\nThe oil from its seeds has a high content of polyunsaturated fatty acids. This shortens the\ntime that the oil can be stored for, which is a disadvantage.\nScientists used gene editing to develop two types of C. sativa with different genetic changes.\nThe gene edited C. sativa seeds produced oil with longer storage times.\nFig. 4.1 shows the percentage composition of fatty acids in the oil extracted from seeds of\ngene edited and wild type (not gene edited) C. sativa.\ntype type type A Key B C\n60\n40\npercentage composition\nof fatty acids 20\n0\n18:3 20:1 22:1 16:0 18:0 18:1 18:2\nfatty acids shown as number\nof carbons:number of\nC=C double bonds\nFig. 4.1\nIdentify the letter that represents the oil of the wild type C. sativa on Fig. 4.1. (i)\n..................................................................................................................................... [1]\nWith reference to Fig. 4.1, discuss the social benefits of this example of gene editing. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-mj-43-q04-251f20bdae.pdf"
    },
    {
      "id": "9700-2022-mj-43-q05",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        11,
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-43-q05-p01-d7edc13b2f.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-43-q05-p02-0bc5325674.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-43-q05-p03-55cc0381cd.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-mj-43-q05-p04-626404ce8a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The puma, Puma concolor, lives in North and South America. 5\nFig. 5.1 shows a puma.\nFig 5.2 shows the distribution of the puma species.\nFig. 5.1\nA\nFlorida\nTexas\npuma distribution\nB\nFig. 5.2\nMembers of different subspecies belong to the same species but have some morphological (a)\ndifferences and are found in different geographical locations.\nIn the past the puma has been divided into 32 subspecies. The subspecies of puma varied in\nbody size, coat colour and behaviour to adapt each population to its environment.\nExplain how the different subspecies of puma evolved.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIn 2016, genetic analysis concluded that there are only two genetically distinct subspecies of\npuma, one in North and Central America and one in South America.\nOutline how practical techniques could be used to conduct a genetic analysis of the (b)\npuma species.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-mj-43-q05-1270be9852.pdf"
    },
    {
      "id": "9700-2022-mj-43-q06",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-43-q06-p01-97ccc072e6.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-43-q06-p02-ec94bb66e8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q06",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "The role of sensory receptor cells in mammals is to detect stimuli and generate action potentials in 6\nsensory neurones.\nHuman taste buds on the tongue contain chemoreceptor cells. Different chemoreceptor cells\nrespond to different chemical stimuli.\nFig. 6.1 is a diagram of chemoreceptor cells in a taste bud.\nY\ntight junction\nsupport cell\nchemoreceptor cell A\nsensory neurone B\nFig. 6.1\nName the structures in the region describe their function in a chemoreceptor cell. (a) Y and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nWhen a sugar molecule binds to a receptor protein on the cell surface membrane of cell A, (b)\ncalcium ions are released into the cytoplasm of the cell by the endoplasmic reticulum.\nExplain how the release of calcium ions will lead to an action potential being generated in\nsensory neurone B.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n........",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-43-q06-4d679ab02c.pdf"
    },
    {
      "id": "9700-2022-mj-43-q07",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-mj-43-q07-p01-d01d09278f.png",
        "9700-topic-16-inheritance/assets/9700-2022-mj-43-q07-p02-53e076aff2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Epistasis occurs when a gene at one locus can affect the expression of a gene at 7 (a)\nanother locus.\nDefine the terms gene and locus.\ngene ..........................................................................................................................................\n...................................................................................................................................................\nlocus .........................................................................................................................................\n...................................................................................................................................................\n[2]\nFur colour in mice, Mus musculus, is determined by a number of genes. One example is the (b)\nresult of epistatic interaction between two genes, and B. A\n• Allele codes for the production of pigment in the fur. A\n• Allele does not code for the production of pigment and results in white fur (albino). a\n• Allele codes for the production of brown fur. B\n• Allele codes for the production of black fur. b\nConstruct a genetic diagram to show the results, including the ratio, of a cross between two\nmice heterozygous for both genes.\nparent genotypes AaBb x AaBb\nparent phenotypes\ngametes\nratio ...................................................................................................................................... [6]\nWhite fur is due to a mutation of the TYR gene. This is called albinism. (c)\nExplain how a mutation of the TYR gene can result in albinism.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 11]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-mj-43-q07-5af9c07ce7.pdf"
    },
    {
      "id": "9700-2022-mj-43-q08",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-43-q08-p01-08975d4187.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-mj-43-q08-p02-6b80d568f1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The Malayan tapir, Tapirus indicus, lives in the rainforest of South East Asia. 8 (a)\nFig. 8.1 shows a Malayan tapir and her calf.\nFig. 8.1\nOn the International Union for the Conservation of Nature (IUCN) Red List of Threatened\nSpecies, the Malayan tapir is categorised as endangered and could become extinct. One\nproblem is the illegal trade in the Malayan tapir.\nApart from illegal trading, suggest explain reasons why the Malayan tapir has become and\nendangered and could become extinct.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe trade in Malayan tapirs is regulated by the Convention on International Trade in (b)\nEndangered Species (CITES).\nSuggest ways by which CITES attempts to regulate the trade in wild fauna and flora.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..........................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-mj-43-q08-bd9dda3ffa.pdf"
    },
    {
      "id": "9700-2022-mj-43-q09",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        21
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-mj-43-q09-p01-e9f0838dc0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 9.1 shows a cross-section through a myelinated neurone. 9 (a)\nA\nB\nFig. 9.1\nIdentify structures and B. A\n............................................................................................................................................... A\n............................................................................................................................................... B\n[2]\nWith reference to voltage-gated sodium ion channels, explain the difference in speed of (b)\ntransmission of an action potential along a myelinated neurone and a non-myelinated neurone.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 6]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-mj-43-q09-a1e90c0cf4.pdf"
    },
    {
      "id": "9700-2022-mj-43-q10",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_43.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-mj-43-q10-p01-71f8978728.png",
        "9700-topic-14-homeostasis/assets/9700-2022-mj-43-q10-p02-e9aafe7779.png",
        "9700-topic-14-homeostasis/assets/9700-2022-mj-43-q10-p03-fd3717b171.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-43-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Insulin has an important role in the maintenance of blood glucose concentration. 10 (a)\nAn investigation measured how blood glucose concentration and blood insulin concentration\nchanged after a glucose-rich meal had been eaten.\nThe results are shown in Fig. 10.1.\n7.5 300\n7.0 280\n6.5 260\n6.0 240\n5.5 220\n5.0 200\n4.5 180\nblood glucose blood insulin 4.0 160\nconcentration concentration\ndm–3 dm–3 mmol / / pmol\n3.5 140\n3.0 120\n2.5 100\n2.0 80\n1.5 60\n1.0 40\n0.5 20\n0.0 0\n07.00 08.00 09.00 10.00 11.00 12.00\ntime of day\nKey\nglucose-rich meal eaten\nblood glucose\nconcentration\nblood insulin\nconcentration\nFig. 10.1\nDescribe explain how the results shown in Fig. 10.1 indicate a relationship between (i) and\nblood glucose concentration and blood insulin concentration after the consumption of a\nglucose-rich meal.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nSuggest explain how the results shown in Fig. 10.1 would change if the meal was (ii) and\nmostly starch rather than glucose.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\n(α) Glucagon is synthesised by cells in the pancreas known as alpha cells. Glucagon binds (b)\nto G-protein-coupled receptors in the cell surface membrane of liver cells. This results in the\nactivation of G-proteins.\nOutline the sequence of events occurri",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-mj-43-q10-42219b690b.pdf"
    },
    {
      "id": "9700-2022-mj-51-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-51-q01-p01-9cba5ef262.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q01-p02-5dc25aed40.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q01-p03-90292ccbeb.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q01-p04-5a04351c09.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q01-p05-ed110a50b8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "A group of students investigated two habitats: dense woodland and open grassland. 1\n• Trees are the dominant plants in woodland. The trees form a canopy under which flowering\nherbs, small shrubs and grasses grow.\n• Grasses are the dominant plants in grassland. The grasses and small flowering plants form a\ncontinuous ground cover with only a few small shrubs present.\nFig. 1.1 shows photographs of the two areas.\nFig. 1.1\nThe students decided to compare the adaptations for reducing water loss in the plants growing in\neach area.\nThe students suggested the following hypothesis:\nThe density of stomata will be higher in plants found in the woodland habitat than in the\ngrassland habitat.\nState the independent dependent variables in this investigation. (a) (i) and\nindependent variable .........................................................................................................\ndependent variable ...................................................................................................... [2]\nThe students collected leaves from a number of different plant species in each area. To study\nthe density of stomata on leaves, impressions of the epidermis can be made using clear\nvarnish.\n• One surface of each leaf is painted with a thin layer of varnish.\n• The varnish is left to dry.\n• Clear sticky tape is applied to the leaf over the varnished area, to make a leaf impression.\n• The tape and varnish are removed and stuck to a microscope slide.\n• The leaf impression on the slide is viewed using a microscope.\nFig. 1.2 shows the microscope image of one leaf impression prepared in this way.\nstoma\nFig. 1.2\nState the measurements that the students need to make to determine the stomatal (ii)\ndensity in Fig. 1.2.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nCalculate the stomatal density in Fig. 1.2. (iii)\nSpace for working.\n–2 [2] stomatal density = ............................................... mm\nDescribe how the students could gather data to compare the stomatal density of plants (iv)\ngrowing in the woodland and the grassland habitat.\nYour method should be set out in a logical order and be detailed enough to allow another\nperson to follow it.\nYou should repeat details of the method for making the leaf impressions. not\n...........................................................................................................................................\n...........................................................................................................................................\n..........................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-51-q01-52282728d3.pdf"
    },
    {
      "id": "9700-2022-mj-51-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_51.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-51-q02-p01-153b295ac1.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q02-p02-dc18866519.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q02-p03-ec5f046d37.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q02-p04-8c9dae1a9c.png",
        "9700-experimental-skills/assets/9700-2022-mj-51-q02-p05-3719475005.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Human immunodeficiency virus (HIV) is an example of a virus that spreads from other animals to 2\nhumans where it causes disease.\nFig. 2.1 shows a chimpanzee, Pan troglodytes troglodytes. Chimpanzees can carry the simian\nimmunodeficiency virus (SIV) which is similar to HIV.\nIt is thought that chimpanzees who carry antibodies for SIV do not become ill if infected with HIV.\nThis has been investigated by scientists developing potential vaccines for HIV.\nFig. 2.1\nTests were carried out to see if antibodies against SIV present in chimpanzees bind to HIV\nantigens.\nTest strips were prepared which contained several different HIV antigens.\nWhen samples are applied to the test strip a line will appear in the control region. If the sample\ncontains antibodies to the HIV antigens present on the strip, additional lines will also appear.\nSamples of chimpanzee faeces were collected from a number of sites in Gabon in central Africa.\nThe faecal samples were prepared and then applied to the test strips.\nSuggest why the investigators collected faecal samples from chimpanzees rather than (a) (i)\nplasma samples.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSuggest a method for preparing the faecal samples before applying them to the test (ii)\nstrips.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe scientists collected 608 faecal samples from 224 individual chimpanzees. (iii)\nSuggest a laboratory method that could be used to identify whether two faecal samples\nbelong to the same individual or not.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSamples from humans and chimpanzees were applied to the ",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-51-q02-07bc7374b3.pdf"
    },
    {
      "id": "9700-2022-mj-52-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 18,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p01-44064fc319.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p02-a73e6e8b62.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p03-a49bf3b613.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p04-80e87dc73e.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p05-7ae3f6ca43.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q01-p06-892378255c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Fig. 1.1 shows the root vegetable beetroot, Beta vulgaris. 1 A\nFig. 1.1 shows a single cell from the beetroot tissue. The vacuole contains the red pigment, B\nbetalain.\nThe pigment molecules are too large to diffuse out of the vacuole.\nvacuole containing betalain\ncell membrane\ntonoplast\ncytoplasm\nred tissue\nnucleus\ncell wall\nA B\nFig. 1.1\nCa(ClO) , the beetroot When pieces of beetroot are soaked in a solution of calcium hypochlorite,\n2\ntissue changes from red to white. This is caused by the calcium hypochlorite diffusing into the\nbeetroot cell vacuoles and decolourising the betalain pigment.\nFig. 1.2 shows a cube of beetroot before soaking in the solution of calcium hypochlorite and the\nhalved cube of beetroot after soaking in the solution of calcium hypochlorite.\nbeetroot cube before soaking halved beetroot cube after soaking\nwidth of red zone\nred zone\ndecolourised zone\nwidth of cube width of cube\nFig. 1.2\nIn a class experiment:\n• Cubes of beetroot of different dimensions were placed in a beaker containing a 10% solution\nof calcium hypochlorite.\n• The cubes were left for 30 minutes.\n• The cubes were then cut in half, and the width of the tissue remaining red (the red zone) was\nmeasured.\nA group of students investigated the effect of changing the surface area:volume ratio on the\ndiffusion of calcium hypochlorite into beetroot cubes. The students used cubes of different sizes\nand measured the width of the red zone after soaking the beetroot in calcium hypochlorite.\nThe students suggested the hypothesis:\nThe width of red tissue remaining in the beetroot after soaking in calcium hypochlorite is\ninversely proportional to the surface area:volume ratio of the beetroot cubes.\nIdentify the variable in this investigation. (a) (i) dependent\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSketch a curve on Fig. 1.3 to show how you expect the width of red tissue to change as (ii)\nthe surface area:volume ratio changes if the hypothesis is correct.\nLabel the axes.\nFig. 1.3\n[2]\nDescribe a method the students could use to collect the data needed to test their (b) (i)\nhypothesis.\nThe students were provided with beetroot tissue, a 10% calcium hypochlorite solution\nand standard laboratory equipment.\nYour method should be set out in a logical order and be detailed enough to let another\nperson follow it.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-52-q01-6129c2d2d4.pdf"
    },
    {
      "id": "9700-2022-mj-52-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p01-aafe143477.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p02-2f1bd608f0.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p03-258bbf4272.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p04-2d3b245de7.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p05-e4abafed11.png",
        "9700-experimental-skills/assets/9700-2022-mj-52-q02-p06-e41b325373.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "The European badger, Meles meles, pictured in Fig. 2.1, is a mammal found in England. 2\nBadgers can contract bovine tuberculosis (bTB) caused by the pathogen, Mycobacterium bovis.\n• Researchers estimated that 5% of the badger population in England are infected with bTB.\n• bTB can be spread between badger populations and other populations of wild animals and\nfarmed animals.\n• Herds of dairy cows produce milk for human consumption. If a dairy herd becomes infected\nwith bTB, the pathogen can be transmitted to human populations in milk.\n• A variety of measures have been introduced in an attempt to reduce the bTB rate in badgers\nand so prevent infection of dairy cows.\n• These measures include vaccination and the removal of badger populations (culling).\nFig. 2.1\nSuggest how researchers estimated the percentage of the badger population that was (a)\ninfected with bTB.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nIt is possible that wild badgers infect dairy cows with bTB. (b)\nA group of scientists investigated the correlation between bTB infections in badgers and bTB\ninfections in dairy cows.\n• 30 study areas in England were identified.\nkm2. • Each of the areas was approximately 100\n• 10 of the 30 study areas were randomly selected and given the codes A–J.\n• The number of infected cows the number of infected badgers were counted in each and\narea.\n• The number of infected badgers was determined by testing for the presence of\nMycobacterium bovis in tissue samples taken from the badgers.\n• The number of infected cows was determined by testing for a response (inflammation) to\nthe introduction of antigenic material into the skin.\nData was collected as shown in Table 2.1.\nTable 2.1\ndata set areas whe",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-52-q02-64c5b0ef39.pdf"
    },
    {
      "id": "9700-2022-mj-53-q01",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 19,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p01-0e1f64dd83.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p02-eaa14464b3.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p03-3f8bf1c9c1.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p04-d4bb6a88ae.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p05-911c207878.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q01-p06-42d89927b4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "There are a variety of methods used to estimate the water potential of plant tissues. 1\nA student decided to estimate the water potential of potato tuber tissue using the method (a)\ndescribed.\n• A number of test-tubes were set up, each containing the same volume of\nsucrose solution.\n• Each test-tube contained a different concentration of sucrose solution.\n• Pieces of potato tissue were placed in each test-tube.\n• After a period of time, the potato tissue was removed from the test-tubes.\nThe student then measured the change in density of the sucrose solutions left in each\ntest-tube. The more water present in a sucrose solution, the less dense it is.\n• Sucrose solutions of the same concentrations as those of the test-tubes original\nwere prepared.\n• Each of these sucrose solutions was coloured with methylene blue.\n• One drop of the coloured sucrose solution of a known concentration was released into\nthe middle of the test-tube whose sucrose concentration was originally the same as that\nof the coloured sucrose drop.\nThe three possible outcomes that the student would see are shown in Fig. 1.1.\n1 The drop will rise if the coloured sucrose drop is less dense than the sucrose solution in\nthe test-tube.\n2 The drop will remain where it is released if the coloured sucrose drop is similar in density\nto the sucrose solution in the test-tube.\n3 The drop will sink if the coloured sucrose drop is more dense than the sucrose solution in\nthe test-tube.\ndropper\ntest-tube\nsucrose solution\nafter potato tuber\ntissue has been\nsoaked in it point of\nrelease\ncoloured sucrose\ndrop\n1. coloured 2. coloured 3. coloured\nsucrose drop less sucrose drop sucrose drop\ndense than same density as more dense than\nsucrose solution sucrose solution sucrose solution\nFig. 1.1\nIdentify the independent variable in the experiment. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nPredict the result if the water potential of the potato tissue was (less negative) (ii) higher\nthan the surrounding sucrose solution.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nExplain the prediction that you gave in (a)(ii). (iii)\n...........................................................................................................................................\n...........................................................................................................................................\n..............",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-53-q01-1d36ec9121.pdf"
    },
    {
      "id": "9700-2022-mj-53-q02",
      "subject": "9700",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-may-june/9700_s22_qp_53.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-May-June/qp/9700_s22_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-mj-53-q02-p01-52b5c156c3.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q02-p02-05c586d2da.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q02-p03-b98888cd2e.png",
        "9700-experimental-skills/assets/9700-2022-mj-53-q02-p04-a300bcf2fd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Respiration in blowfly larvae, Calliphora erythrocephala, can be studied by determining the 2\nrespiratory quotient (RQ).\nA student used the apparatus set up as shown in Fig. 2.1 to measure oxygen uptake by blowfly\nlarvae during respiration.\nFive replicates were carried out.\nclip to close tube\nlarge test-tube\nsubstance X\ncotton wool\nreservoir\nblowfly larvae\nmeniscus\ncm / mm scale\nmanometer containing\ncoloured fluid\nFig. 2.1\nState the role of substance in Fig. 2.1. (a) X\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nPredict which direction the meniscus in the manometer will move during the experiment. (b)\nExplain your answer.\ndirection ....................................................................................................................................\nexplanation ...............................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nTo calculate the RQ, the student determined the volume of carbon dioxide produced by the (c)\nblowfly larvae.\nDescribe how the apparatus in Fig. 2.1 was modified so that the volume of carbon dioxide\nproduced was determined.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nState variables that should be standardised each time the student carried out (d) two\nthe experiments.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-mj-53-q02-dccce7a03a.pdf"
    },
    {
      "id": "9700-2022-on-41-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-41-q01-p01-89193abaf7.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-41-q01-p02-d35189fe4f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q01",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The wild Bactrian camel, Camelus ferus, lives only in the desert regions of Mongolia and 1\nnorthern China.\nFig. 1.1 shows a wild Bactrian camel.\nFig. 1.1\nThe wild Bactrian camel is at risk of extinction in the wild and is categorised as critically (a)\nendangered by the International Union for Conservation of Nature (IUCN). There are only\n950 wild Bactrian camels left in their natural habitat.\nSuggest reasons why the wild Bactrian camel has become critically endangered.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nOutline the role of the IUCN. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nSome zoos use assisted reproduction techniques, such as embryo transfer, in their captive (c)\nbreeding programmes for endangered species. Embryo transfer has resulted in domesticated\ndromedary camels giving birth to wild Bactrian camel calves.\nDescribe the procedure of embryo transfer in a mammal such as a camel.\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-on-41-q01-91e5c76832.pdf"
    },
    {
      "id": "9700-2022-on-41-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-on-41-q02-p01-fc7904cc7a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is an important energy transfer process. 2\nName chloroplast pigment. (a) one\n...................................................................................................................................................\n............................................................................................................................................. [1]\nExplain the relationship between the spectrum of the main chloroplast pigments (b) absorption\nin a species of plant and the spectrum for photosynthesis for that species. action\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nExplain why temperature can be a limiting factor of photosynthesis. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nOutline how the light-independent stage of photosynthesis leads to the production of (d)\ncarbohydrates such as starch in plant leaves.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-on-41-q02-ab11180cef.pdf"
    },
    {
      "id": "9700-2022-on-41-q03",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-41-q03-p01-7d97edc9f9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-41-q03-p02-1be8c609c9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-41-q03-p03-71cceeaf81.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The results of investigations carried out on mitochondria show how the structure of a 3 (a)\nmitochondrion is related to its role in aerobic respiration.\n• Intact mitochondria (not damaged) were removed from cells.\n• A technique was used to remove the outer mitochondrial membrane, leaving the\ninner membrane intact.\n• The inner mitochondrial membrane was separated from the contents of the matrix\nso that both could be analysed.\nThe removal of the outer membranes of mitochondria involves placing the organelles (i)\nin pure water. This results in the rupture (bursting) of the outer membrane. The inner\nmitochondrial membrane does not rupture and remains intact.\nSuggest explain why the inner membrane of a mitochondrion remains intact when and\nthe organelle is placed in pure water.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nName molecules, other than coenzymes, that are found in the mitochondrial matrix (ii) three\nexplain their role in aerobic respiration. and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nThe inner membrane contains a very high proportion of the molecule cardiolipin. (iii)\nCardiolipin makes the membrane impermeable to some ions.\nSuggest why the inner membrane contains a very high proportion of cardiolipin.\n...........................................................................................................................................\n.......................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-on-41-q03-a70ade46f3.pdf"
    },
    {
      "id": "9700-2022-on-41-q04",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-on-41-q04-p01-62befd151f.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-on-41-q04-p02-cd6072e58c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Adenosine deaminase (ADA) deficiency is an immune system disorder caused by a recessive 4\nautosomal mutation.\nSevere combined immunodeficiency caused by a lack of ADA is called ADA-SCID.\nGenetic engineering is used to make a recombinant human protein to treat people with (a)\nADA-SCID.\nOutline the principles of genetic engineering.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIn 2016 gene therapy to cure ADA-SCID was approved in Europe. The gene therapy involves (b)\nthree main steps.\n• Blood (haematopoietic) stem cells are taken from the bone marrow of the person\nwith ADA-SCID.\n• The functional gene and its promoter are inserted into the blood stem cells.\n• A single infusion (injection) of the gene-corrected cells is given to the patient.\nExplain why a single infusion of gene-corrected stem cells is enough to cure the disease. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nExplain why a promoter has to be transferred as well as the desired gene. (ii)\n................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-on-41-q04-f327f112b6.pdf"
    },
    {
      "id": "9700-2022-on-41-q05",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q05-p01-8eaa2afb99.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q05-p02-6e1b819223.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "A large number of alpine plant species grow in the mountains of New Zealand’s South Island. 5\nAlpine plants are defined as plants that live above the treeline, which is the height above which\ntrees cannot grow. The current treeline for most of South Island is 1200 m.\nThe alpine areas nearest to South Island are 1500–2000 km away across the sea. Many of (a)\nSouth Island’s alpine species live nowhere else in the world.\nExplain how a large number of alpine plant species developed on South Island.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nFig. 5.1 shows two aspects of the history of South Island over the last 3.9 million years. (b)\n• The dashed line shows how the mean height of mountains in the Clyde region of\nSouth Island increased over time. The mountains in this range have a mean height\nof 2400 m at the present time.\n• The solid line models the height of the treeline over time based on geological climate\ndata. The treeline was higher when the climate was warmer, and the treeline was\nlower when the climate was colder, during ice ages.\n2400\n1800\n1200\n/ m height\n600\n0\n0 1.0 3.9 2.0 3.0\npresent time\ntime before present / million years\nFig. 5.1\nWith reference to Fig. 5.1, identify with reasons the time period when South Island’s (i)\nalpine plant species developed.\n.........................................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-on-41-q05-0e3b5795b8.pdf"
    },
    {
      "id": "9700-2022-on-41-q06",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q06-p01-ce5ba47e63.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q06-p02-b966b45123.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q06-p03-bcaa905a7c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-41-q06-p04-0751b6a8c2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q06",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Selective breeding has been used to improve the characteristics of crop plants, including maize 6\nand rice.\nOutline how vigorous, uniform varieties of maize were developed by selective breeding. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIR8 is a variety of rice plant that was developed in the 1960s. (b)\nIR8 was developed by breeding together two different varieties of rice:\n• PETA, which produces a high yield\n• DGWG, which is a dwarf variety.\nFig. 6.1 is a diagram showing the three varieties of rice.\nPETA edible\ngrains\nIR8\nDGWG\nwater line\nFig. 6.1\nSuggest why IR8 is an improvement on the PETA variety of rice. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nIn 2009, scientists produced a new variety of rice, IR64-Sub1, by breeding together: (ii)\n• FR13A, a variety which has a low yield but has an allele for flood tolerance\n• IR64, a variety which produces a high yield.\nThe scientists sequenced the DNA of these three rice varieties.\nSuggest the benefit of sequencing the DNA of IR64-Sub1.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nAfter a few generations of breeding, the scientists crossed IR64-Sub1 with IR64. (iii)\nExplain why the scientists crossed IR64-Sub1 with IR64.\n....................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-on-41-q06-0c86ea8927.pdf"
    },
    {
      "id": "9700-2022-on-41-q07",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-on-41-q07-p01-882c30c9cd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "In some varieties of domestic cats the gene for fur colour is located on the chromosome. 7 (a) X\nThis gene has two alleles. One allele codes for black fur and the other allele codes for ginger\nfur. The two alleles are codominant so a heterozygous cat will have fur with patches of black\nand ginger colours. A cat with fur of two colours is known as a tortoiseshell.\nUsing appropriate symbols, construct a genetic diagram to show the results of a cross\nbetween a female tortoiseshell cat and a male ginger cat.\nsymbols\nparent phenotypes tortoiseshell female ginger male\nparent genotypes\ngametes\noffspring genotypes\noffspring phenotypes\n[5]\nIn humans the TYR gene is involved in the production of a dark pigment, melanin, in (b)\nsome cells.\nDescribe how the expression of the TYR gene leads to the production of melanin.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-on-41-q07-6ec9a28fc2.pdf"
    },
    {
      "id": "9700-2022-on-41-q08",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        17,
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-41-q08-p01-a1d616feb0.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-41-q08-p02-e397aa2b31.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-41-q08-p03-7eef7d3d02.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q08",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 8.1 is a diagram of a kidney nephron. 8 (a)\nBowman’s\ncapsule\nA\nB\nC medulla\ncollecting\nduct\nrenal pelvis\nFig. 8.1\nWith reference to Fig. 8.1, name A, and C. B\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n[3]\nAntidiuretic hormone (ADH) is involved in the maintenance of the water potential of the blood. (b)\nFig. 8.2 shows the relationship between blood ADH concentration, urine concentration and\nthe flow rate of urine.\nThe flow rate of urine is the rate of production of urine by the kidneys.\n1500 15\n1400 14\n1300 13\n1200 12\n1100 11\n1000 10\n900 9\nurine urine\n8 800\nconcentration flow rate\ndm–3 cm3 min–1 / mmol /\n7 700\n6 600\n5 500\n4 400\n3 300\n2 200\n100 1\n0 0\n10 20 30 40 50 60 70 80 90 100 0\n/ arbitrary units blood ADH concentration\nKey\nurine concentration\nurine flow rate\nFig. 8.2\nDescribe the relationships shown in Fig. 8.2.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDescribe explain the action of ADH on the cells of the collecting duct when the water (c) and\npotential of the blood decreases.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-41-q08-7f0155a505.pdf"
    },
    {
      "id": "9700-2022-on-41-q09",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-41-q09-p01-8205344882.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-41-q09-p02-0078ce1377.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-41-q09-p03-5884858a28.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "A neuromuscular junction allows the transmission of an action potential from a motor neurone 9 (a)\nto a striated muscle fibre, causing it to contract.\nFig. 9.1 is a graph of an action potential in a motor neurone.\n+40\n+20\n0\n–20\nmembrane\npotential –40\nmV /\n–60\n–80\n–100\n0 1 2 3 4 5\n/ ms time\nFig. 9.1\nFig. 9.2 is a graph of an action potential in a striated muscle fibre.\n+40\n+20\n0\n–20\nmembrane\npotential –40\nmV /\n–60\n–80\n–100\n0 1 2 3 4 5 6 7 8 9 10\n/ ms time\nFig. 9.2\nWith reference to Fig. 9.1 and Fig. 9.2, describe the differences between the action potential\nin a motor neurone and the action potential in a striated muscle fibre.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThere are three phases in the contraction of a striated muscle: latent phase, contraction (b)\nphase and relaxation phase.\nThe tension in a muscle represents the degree of contraction of its fibres.\nFig. 9.3 is a graph of the tension in a striated muscle during the three phases of contraction.\nlatent contraction relaxation\nphase phase phase\ntension in\nmuscle\n0 10 20 30 40 50 60 70 80 90 100\ntime / ms\nstimulation\nFig. 9.3\nWith reference to Fig. 9.3, explain what is happening in the striated muscle fibre during (i)\nthe latent phase.\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-on-41-q09-892ef5bb99.pdf"
    },
    {
      "id": "9700-2022-on-41-q10",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_41.pdf?download=true",
      "source_pages": [
        23
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-41-q10-p01-15e8152d99.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-41-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The passage below outlines homeostasis. 10\nComplete the passage by using the most appropriate scientific terms.\nHomeostasis, in mammals, is the process of keeping the ........................................... environment\nof the body in optimum conditions so that cells can function efficiently. Blood water potential, core\ntemperature and blood glucose concentration are all factors that need to be kept at optimum\nvalues or set-point.\nWhen a condition deviates from its set-point, a corrective mechanism is triggered. An increase\nin blood glucose concentration triggers processes to decrease it and vice versa. This corrective\nmechanism is called ........................................... .\nThe pancreas is involved in the control of blood glucose concentration. Glucose binds\nto ........................................... on the cell surface membrane of pancreatic cells. These\nare ........................................... cells, which secrete hormones such as insulin and glucagon. The\ntwo hormones have opposite effects on the blood glucose concentration. For example the action\nof one hormone stimulates the uptake of glucose by cells for respiration and the action of the other\nhormone stimulates the breakdown of ........................................... to glucose in the liver.\n[Total: 5]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-41-q10-7ba1ee91ae.pdf"
    },
    {
      "id": "9700-2022-on-42-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-42-q01-p01-206b1aab95.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-42-q01-p02-ebe554c6a1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q01",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The Asiatic lion, Panthera leo persica, is found only in the Gir Forest National Park in the Gujarat 1\nregion of western India.\nFig. 1.1 shows a female Asiatic lion.\nFig. 1.1\nThe Asiatic lion is at risk of extinction in the wild and is categorised as endangered by the (a)\nInternational Union for Conservation of Nature (IUCN).\nOutline the role of the IUCN.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nThe Maldhari community is a tribe of herdsmen that lives in the Gir Forest. The Maldhari have (b)\nco-existed with the Asiatic lions for thousands of years. The Maldhari place old and weak\ncattle at the edges of their cattle enclosures.\nSuggest why the Maldhari place old and weak cattle at the edges of their cattle enclosures.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nSome zoos use assisted reproduction techniques, such as IVF, in their captive breeding (c)\nprogrammes for endangered species.\nDescribe how IVF can be used with an endangered species such as the Asiatic lion.\n...................................................................................................................................................\n...................................................................................................................................................\n......................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-on-42-q01-3470fa2a15.pdf"
    },
    {
      "id": "9700-2022-on-42-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-42-q02-p01-21e9a77897.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-42-q02-p02-ddf9041ad9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-42-q02-p03-2c52070b51.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q02",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Yeast cells can respire in anaerobic conditions. 2\nOutline how yeast carries out respiration in anaerobic conditions. (a) (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nExplain why respiration in anaerobic conditions is an advantage to yeast. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nBioethanol is a type of biofuel produced from maize starch on an industrial scale. In the high\ntemperature method, heating to 120 °C is used to break apart starch molecules. The enzymes\nα-amylase and glucoamylase are added to the resulting starch suspension once it has cooled\ndown. These enzymes hydrolyse the starch to glucose. Yeast cells are then added and maintained\nin anaerobic conditions to produce ethanol.\nThe high temperature method is expensive to carry out, so a new method has been developed\nwhich heats starch to a lower temperature of 80 °C.\nIn the lower temperature method, enzymes catalysing the hydrolysis of starch do need to be not\nadded to the starch suspension. A genetically modified (GM) strain of the same yeast species is\nα-amylase used. The GM strain of yeast has genes that allow the cell to produce and glucoa",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-on-42-q02-e3b98c4b32.pdf"
    },
    {
      "id": "9700-2022-on-42-q03",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-on-42-q03-p01-a12d1c8e4f.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-on-42-q03-p02-331f96ccf9.png",
        "9700-topic-13-photosynthesis/assets/9700-2022-on-42-q03-p03-c8ef0b9582.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q03",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is an energy transfer process that takes place in chloroplasts of plant cells. 3 (a)\nFig. 3.1 shows a diagram of a chloroplast. (i)\nA\nB\nC\nD not to scale\nFig. 3.1\nUsing the letters A, B, or D, identify the structure that: C\n• contains genes that code for some of the enzymes used in photosynthesis\n................................................................\n• is the site of synthesis of some of the enzymes used in photosynthesis.\n................................................................\n[2]\nExplain how grana are adapted for their specific role in photosynthesis. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nAn experiment was carried out to investigate the effect of changing light conditions on the (b)\npH of the chloroplast stroma. Scientists followed pH changes in chloroplast stroma using\nfluorescent chemicals that can be used as pH indicators.\n• Chloroplasts were isolated from cells.\n• A suspension of chloroplasts was prepared and kept in the dark for 180 seconds.\n• The chloroplasts were exposed to a period of light of fixed intensity for 240 seconds,\nthen returned to dark conditions.\n• The pH of chloroplast stroma was continuously measured and recorded.\nFig. 3.2 shows the results of this experiment.\ndark light dark\n7.7\n7.6\n7.5 pH of\nchloroplast\nstroma 7.4\n7.3\n7.2\n0 120 240 360 480 600 720\ntime / s\nFig. 3.2\nDescribe the results shown in Fig. 3.2. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n.........................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-on-42-q03-66faf64e52.pdf"
    },
    {
      "id": "9700-2022-on-42-q04",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-on-42-q04-p01-3978077f7b.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-on-42-q04-p02-14fbe0fa28.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-on-42-q04-p03-4a5023302c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Haemophilia is a blood clotting disorder in humans caused by a mutant allele on the X chromosome. 4\nTable 4.1 compares two forms of haemophilia: haemophilia A and haemophilia B.\nTable 4.1\nhaemophilia A haemophilia B\ngene F8 F9\nclotting factor protein factor VIII factor IX\nproportion of males born with\n1 in 5000 1 in 30 000\nhaemophilia\nlength of functional gene (exons\n7.0 1.6\nonly) / kilobase pairs\nGenetic engineering is used to make recombinant human proteins to treat people with (a)\nhaemophilia A and haemophilia B.\nOutline the principles of genetic engineering.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nScientists are working towards a goal of treating haemophilia by gene therapy. They plan to (b)\nuse a common, harmless virus to introduce the functional gene. The virus has a genome that\nis 4.7 kilobase pairs long.\nWith reference to Table 4.1 and the introduction to (a), assess: (i)\n• which form of haemophilia, A or B, scientists should try to treat first\n• whether they should attempt to treat haemophilia with gene therapy at all.\nExplain your reasoning.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-on-42-q04-e4d151ef15.pdf"
    },
    {
      "id": "9700-2022-on-42-q05",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-42-q05-p01-5c5cdd423f.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-42-q05-p02-a0dec9dbe3.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-42-q05-p03-6aef493d3b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Myosotis is a genus of small flowering plants. Many different Myosotis species grow on the islands 5\nof New Zealand, which are an important site of Myosotis evolution. Lowland Myosotis species\ngrow at low altitude while alpine Myosotis species grow at high altitude at the tops of mountains.\nScientists wanted to obtain molecular data to determine the evolutionary relationships of (a)\nNew Zealand’s Myosotis species. They extracted DNA from individuals of Myosotis species\ncollected from three different islands in New Zealand.\nTo carry out a polymerase chain reaction before DNA sequencing, the DNA samples were\nmixed with primers, deoxynucleotides and Taq polymerase and put through 35 cycles of\ntreatment. Each treatment cycle involved one minute at 95 °C, followed by one minute at\n50 °C and then four minutes at 72 °C.\nDescribe what happened to the DNA at each temperature.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nFig. 5.1 shows the three largest New Zealand islands. (b)\nNorth Island\nSouth Island\n0 200\nkm\nStewart Island\nFig. 5.1\nNorth Island has mostly lowland habitat. South Island and Stewart Island have mountains\nwith alpine habitats that are above the tree line.\nDNA sequence data for three Myosotis species were compared. The results are described in\nthe bullet points.\n• In the alpine species M. pygmaea, individuals on South Island showed genetic differences\nfrom individuals of M. pygmaea on Stewart Island.\n• In the alpine species M. pulvinaris, individuals from different mountains on South Island\nshowed genetic differences.\n• In the lowland species M. pottsiana, individuals from different areas of North Island\nshowed overall genetic similarity.\nDiscuss reasons for the results in the three species.\n...................................................................................................................................................\n...................................................................................................................................................\n........................................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-on-42-q05-2658f180fb.pdf"
    },
    {
      "id": "9700-2022-on-42-q06",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-on-42-q06-p01-7a175d74df.png",
        "9700-topic-16-inheritance/assets/9700-2022-on-42-q06-p02-2586bc3020.png",
        "9700-topic-16-inheritance/assets/9700-2022-on-42-q06-p03-9026709263.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q06",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The lac operon of prokaryotes contains a group of structural genes that are under the same 6 (a)\ncontrol and are transcribed together.\nAnother operon found in prokaryotes is the trp operon.\nFig. 6.1 summarises the structure and control of the trp operon.\ninactive repressor\nRNA polymerase binds to promoter\ntrpR promoter operator trpE trpD trpC trpB trpA\nattenuator\n(regulates the structural genes\nextent of are transcribed\ntranscription)\ntryptophan\nactive repressor\ntrpR promoter operator trpE trpD trpC trpB trpA\nattenuator\nno transcription of\nstructural genes\nFig. 6.1\nDescribe the differences in structure control between the lac operon and (i) and\nthe trp operon.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nSuggest why structural genes in operons are transcribed together. (ii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\ntrpA is an example of a structural gene and trpR is an example of a regulatory gene. (iii)\nDescribe the differences between the functions of structural genes and regulatory genes.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n............................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-on-42-q06-0752e8ae6b.pdf"
    },
    {
      "id": "9700-2022-on-42-q07",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-on-42-q07-p01-aa0cfe0fdd.png",
        "9700-topic-16-inheritance/assets/9700-2022-on-42-q07-p02-f5a8742b9f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The determination of sex in domestic turkeys is different from that in humans. The sex 7\nchromosomes in turkeys are named and W. Male turkeys are and female turkeys are ZW. Z ZZ\nThe gene for feather colour is located only on the chromosome. Z\n• The dominant allele codes for bronze feathers.\n• The recessive allele codes for brown feathers.\nDefine the terms dominant and recessive. (a)\ndominant ...................................................................................................................................\n...................................................................................................................................................\nrecessive ..................................................................................................................................\n...................................................................................................................................................\n[2]\nUsing suitable symbols, construct a genetic diagram to show the results of a cross between a (b)\nheterozygous bronze male turkey and a brown female turkey.\nsymbols\nparent phenotypes bronze male brown female\nparent genotypes\ngametes\noffspring genotypes\noffspring phenotypes\n[5]\nExplain how you would carry out a test cross to determine the genotype of a (c)\nbronze male turkey.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 10]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-on-42-q07-9891a03f2a.pdf"
    },
    {
      "id": "9700-2022-on-42-q08",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-42-q08-p01-2d3a169164.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-42-q08-p02-ce1ecdda18.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q08",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Selective reabsorption takes place in the proximal convoluted tubule of a kidney nephron. 8 (a)\nFig. 8.1 is a diagram of two cells of the proximal convoluted tubule and part of the adjacent\nblood capillary.\nblood\nlumen of proximal plasma\nconvoluted tubule\nproximal convoluted\ntubule cell\nFig. 8.1\nThe cells of the proximal convoluted tubule are adapted to carry out selective reabsorption.\nOn Fig. 8.1, use label lines and letters to indicate:\n• C, where cotransporter proteins are located\n• P, where sodium-potassium pumps are located. [2]\nExplain how the cells of the proximal convoluted tubule are adapted to carry out (b)\nselective reabsorption.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nThe plasma, the glomerular filtrate and urine are composed of various substances. (c)\nTable 8.1 shows the percentage composition of plasma, the glomerular filtrate and urine.\nTable 8.1\nsubstance percentage composition\nplasma glomerular filtrate urine\nwater 90.00 90.00 94.00\nglucose 0.10 0.10 0.00\namino acids 0.05 0.05 0.00\nplasma proteins 8.00 0.00 0.00\nurea 0.03 0.03 2.00\nother substances 1.82 9.82 4.00\nAs a result of the action of the cells of the proximal convoluted tubule, glucose and amino acids\nhave been reabsorbed by the blood, whereas only some of the urea has been reabsorbed.\nCalculate how many times the percentage composition of urea has increased in the urine\ncompared with the glomerular filtrate.\nShow your working and give your answer to the nearest whole number.\nanswer ......................................................... [2]\n[Total: 9]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-42-q08-e7fefe5d7d.pdf"
    },
    {
      "id": "9700-2022-on-42-q09",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-42-q09-p01-37060c61a2.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-42-q09-p02-19d23ec6e8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Voltage-gated channels are involved in the generation of an action potential. 9 (a)\nFig. 9.1 is a diagram of the voltage-gated channels of sodium ions and potassium ions in the\nmembrane of an axon. The channels are shown in three different states, 1, and 3. 2\ntissue fluid\nstate\n1\naxon cytoplasm\nKey state\n2 Na+ voltage-gated channel\nK+ voltage-gated channel\nNa+\nK+\nstate\n3\nFig. 9.1\nFig. 9.2 is a diagram of different phases of an action potential in an axon. The phases are\nlabelled A, B, C, D, E, and G. F\n+40\n+20\nD\nC\n0\n–20\nmembrane\npotential –40 E\nB\nmV /\n–60\n–80\nG\nA\nF\n4 5 0 1 2 3\n/ ms time\nFig. 9.2\nComplete Table 9.1 to match each of the listed phases of the action potential with the\nappropriate state of the voltage-gated channels: 1, or 3. 2\nTable 9.1\nphase of action potential state of voltage-gated channels\nA\n............................\nC\n............................\nE\n............................\nF\n............................\nG\n............................\n[3]\nMany neurones are surrounded by myelin sheaths. (b)\nDescribe explain the role of the myelin sheath in the transmission of an action potential. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 6]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-on-42-q09-1e78bf5e7f.pdf"
    },
    {
      "id": "9700-2022-on-42-q10",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_42.pdf?download=true",
      "source_pages": [
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-42-q10-p01-16e309ee7f.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-42-q10-p02-d87441ef68.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-42-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Glucagon has a role in the maintenance of blood glucose concentration. 10 (a)\nFig. 10.1 shows the relationship between blood glucose concentration and blood glucagon\nconcentration, measured over 90 minutes in a healthy person. The person did not have any\nfood in the three hours before the measurements were taken.\nKey\nblood glucose concentration\nblood glucagon concentration\n5 50\n4 40\nblood blood\n3 30\nglucose glucagon\nconcentration concentration\n2 20 dm–3 dm–3 / mmol pmol /\n1 10\n0 0\n0 15 30 45 60 75 90\n/ min time\nFig. 10.1\nExplain the relationship shown in Fig. 10.1.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThe passage below outlines cell signalling. (b)\nComplete the passage by using the most appropriate scientific terms.\nCells need to interact with their environment and other cells around them. This is called\ncell signalling. Cells in mammals are involved in a complex system of communication with\neach other. For example, liver cells detect signals using receptors in their cell surface\nmembranes. Glucagon binds to the receptors because its shape is ........................................ .\nThis binding activates a ........................................ , leading to stimulation of the enzyme\n........................................ . [3]\n[Total: 7]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-42-q10-bda573f924.pdf"
    },
    {
      "id": "9700-2022-on-43-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-43-q01-p01-89193abaf7.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2022-on-43-q01-p02-d35189fe4f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q01",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The wild Bactrian camel, Camelus ferus, lives only in the desert regions of Mongolia and 1\nnorthern China.\nFig. 1.1 shows a wild Bactrian camel.\nFig. 1.1\nThe wild Bactrian camel is at risk of extinction in the wild and is categorised as critically (a)\nendangered by the International Union for Conservation of Nature (IUCN). There are only\n950 wild Bactrian camels left in their natural habitat.\nSuggest reasons why the wild Bactrian camel has become critically endangered.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nOutline the role of the IUCN. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nSome zoos use assisted reproduction techniques, such as embryo transfer, in their captive (c)\nbreeding programmes for endangered species. Embryo transfer has resulted in domesticated\ndromedary camels giving birth to wild Bactrian camel calves.\nDescribe the procedure of embryo transfer in a mammal such as a camel.\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2022-on-43-q01-f0f5518d23.pdf"
    },
    {
      "id": "9700-2022-on-43-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2022-on-43-q02-p01-fc7904cc7a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q02",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is an important energy transfer process. 2\nName chloroplast pigment. (a) one\n...................................................................................................................................................\n............................................................................................................................................. [1]\nExplain the relationship between the spectrum of the main chloroplast pigments (b) absorption\nin a species of plant and the spectrum for photosynthesis for that species. action\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nExplain why temperature can be a limiting factor of photosynthesis. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nOutline how the light-independent stage of photosynthesis leads to the production of (d)\ncarbohydrates such as starch in plant leaves.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2022-on-43-q02-2dc35ac985.pdf"
    },
    {
      "id": "9700-2022-on-43-q03",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-43-q03-p01-7d97edc9f9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-43-q03-p02-1be8c609c9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2022-on-43-q03-p03-71cceeaf81.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The results of investigations carried out on mitochondria show how the structure of a 3 (a)\nmitochondrion is related to its role in aerobic respiration.\n• Intact mitochondria (not damaged) were removed from cells.\n• A technique was used to remove the outer mitochondrial membrane, leaving the\ninner membrane intact.\n• The inner mitochondrial membrane was separated from the contents of the matrix\nso that both could be analysed.\nThe removal of the outer membranes of mitochondria involves placing the organelles (i)\nin pure water. This results in the rupture (bursting) of the outer membrane. The inner\nmitochondrial membrane does not rupture and remains intact.\nSuggest explain why the inner membrane of a mitochondrion remains intact when and\nthe organelle is placed in pure water.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nName molecules, other than coenzymes, that are found in the mitochondrial matrix (ii) three\nexplain their role in aerobic respiration. and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nThe inner membrane contains a very high proportion of the molecule cardiolipin. (iii)\nCardiolipin makes the membrane impermeable to some ions.\nSuggest why the inner membrane contains a very high proportion of cardiolipin.\n...........................................................................................................................................\n.......................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2022-on-43-q03-9e01f782ea.pdf"
    },
    {
      "id": "9700-2022-on-43-q04",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2022-on-43-q04-p01-62befd151f.png",
        "9700-topic-19-genetic-technology/assets/9700-2022-on-43-q04-p02-cd6072e58c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Adenosine deaminase (ADA) deficiency is an immune system disorder caused by a recessive 4\nautosomal mutation.\nSevere combined immunodeficiency caused by a lack of ADA is called ADA-SCID.\nGenetic engineering is used to make a recombinant human protein to treat people with (a)\nADA-SCID.\nOutline the principles of genetic engineering.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIn 2016 gene therapy to cure ADA-SCID was approved in Europe. The gene therapy involves (b)\nthree main steps.\n• Blood (haematopoietic) stem cells are taken from the bone marrow of the person\nwith ADA-SCID.\n• The functional gene and its promoter are inserted into the blood stem cells.\n• A single infusion (injection) of the gene-corrected cells is given to the patient.\nExplain why a single infusion of gene-corrected stem cells is enough to cure the disease. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nExplain why a promoter has to be transferred as well as the desired gene. (ii)\n................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2022-on-43-q04-fe1fb8ab47.pdf"
    },
    {
      "id": "9700-2022-on-43-q05",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q05-p01-8eaa2afb99.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q05-p02-6e1b819223.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "A large number of alpine plant species grow in the mountains of New Zealand’s South Island. 5\nAlpine plants are defined as plants that live above the treeline, which is the height above which\ntrees cannot grow. The current treeline for most of South Island is 1200 m.\nThe alpine areas nearest to South Island are 1500–2000 km away across the sea. Many of (a)\nSouth Island’s alpine species live nowhere else in the world.\nExplain how a large number of alpine plant species developed on South Island.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nFig. 5.1 shows two aspects of the history of South Island over the last 3.9 million years. (b)\n• The dashed line shows how the mean height of mountains in the Clyde region of\nSouth Island increased over time. The mountains in this range have a mean height\nof 2400 m at the present time.\n• The solid line models the height of the treeline over time based on geological climate\ndata. The treeline was higher when the climate was warmer, and the treeline was\nlower when the climate was colder, during ice ages.\n2400\n1800\n1200\n/ m height\n600\n0\n0 1.0 3.9 2.0 3.0\npresent time\ntime before present / million years\nFig. 5.1\nWith reference to Fig. 5.1, identify with reasons the time period when South Island’s (i)\nalpine plant species developed.\n.........................................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-on-43-q05-310c1d2049.pdf"
    },
    {
      "id": "9700-2022-on-43-q06",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q06-p01-ce5ba47e63.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q06-p02-b966b45123.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q06-p03-bcaa905a7c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2022-on-43-q06-p04-0751b6a8c2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q06",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Selective breeding has been used to improve the characteristics of crop plants, including maize 6\nand rice.\nOutline how vigorous, uniform varieties of maize were developed by selective breeding. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIR8 is a variety of rice plant that was developed in the 1960s. (b)\nIR8 was developed by breeding together two different varieties of rice:\n• PETA, which produces a high yield\n• DGWG, which is a dwarf variety.\nFig. 6.1 is a diagram showing the three varieties of rice.\nPETA edible\ngrains\nIR8\nDGWG\nwater line\nFig. 6.1\nSuggest why IR8 is an improvement on the PETA variety of rice. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nIn 2009, scientists produced a new variety of rice, IR64-Sub1, by breeding together: (ii)\n• FR13A, a variety which has a low yield but has an allele for flood tolerance\n• IR64, a variety which produces a high yield.\nThe scientists sequenced the DNA of these three rice varieties.\nSuggest the benefit of sequencing the DNA of IR64-Sub1.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nAfter a few generations of breeding, the scientists crossed IR64-Sub1 with IR64. (iii)\nExplain why the scientists crossed IR64-Sub1 with IR64.\n....................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2022-on-43-q06-3aa3c2f3d6.pdf"
    },
    {
      "id": "9700-2022-on-43-q07",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2022-on-43-q07-p01-882c30c9cd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "In some varieties of domestic cats the gene for fur colour is located on the chromosome. 7 (a) X\nThis gene has two alleles. One allele codes for black fur and the other allele codes for ginger\nfur. The two alleles are codominant so a heterozygous cat will have fur with patches of black\nand ginger colours. A cat with fur of two colours is known as a tortoiseshell.\nUsing appropriate symbols, construct a genetic diagram to show the results of a cross\nbetween a female tortoiseshell cat and a male ginger cat.\nsymbols\nparent phenotypes tortoiseshell female ginger male\nparent genotypes\ngametes\noffspring genotypes\noffspring phenotypes\n[5]\nIn humans the TYR gene is involved in the production of a dark pigment, melanin, in (b)\nsome cells.\nDescribe how the expression of the TYR gene leads to the production of melanin.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2022-on-43-q07-a3468d305c.pdf"
    },
    {
      "id": "9700-2022-on-43-q08",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        17,
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-43-q08-p01-a1d616feb0.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-43-q08-p02-e397aa2b31.png",
        "9700-topic-14-homeostasis/assets/9700-2022-on-43-q08-p03-7eef7d3d02.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q08",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 8.1 is a diagram of a kidney nephron. 8 (a)\nBowman’s\ncapsule\nA\nB\nC medulla\ncollecting\nduct\nrenal pelvis\nFig. 8.1\nWith reference to Fig. 8.1, name A, and C. B\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n[3]\nAntidiuretic hormone (ADH) is involved in the maintenance of the water potential of the blood. (b)\nFig. 8.2 shows the relationship between blood ADH concentration, urine concentration and\nthe flow rate of urine.\nThe flow rate of urine is the rate of production of urine by the kidneys.\n1500 15\n1400 14\n1300 13\n1200 12\n1100 11\n1000 10\n900 9\nurine urine\n8 800\nconcentration flow rate\ndm–3 cm3 min–1 / mmol /\n7 700\n6 600\n5 500\n4 400\n3 300\n2 200\n100 1\n0 0\n10 20 30 40 50 60 70 80 90 100 0\n/ arbitrary units blood ADH concentration\nKey\nurine concentration\nurine flow rate\nFig. 8.2\nDescribe the relationships shown in Fig. 8.2.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDescribe explain the action of ADH on the cells of the collecting duct when the water (c) and\npotential of the blood decreases.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-43-q08-ebd1ee04a3.pdf"
    },
    {
      "id": "9700-2022-on-43-q09",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-43-q09-p01-8205344882.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-43-q09-p02-0078ce1377.png",
        "9700-topic-15-control-and-coordination/assets/9700-2022-on-43-q09-p03-5884858a28.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "A neuromuscular junction allows the transmission of an action potential from a motor neurone 9 (a)\nto a striated muscle fibre, causing it to contract.\nFig. 9.1 is a graph of an action potential in a motor neurone.\n+40\n+20\n0\n–20\nmembrane\npotential –40\nmV /\n–60\n–80\n–100\n0 1 2 3 4 5\n/ ms time\nFig. 9.1\nFig. 9.2 is a graph of an action potential in a striated muscle fibre.\n+40\n+20\n0\n–20\nmembrane\npotential –40\nmV /\n–60\n–80\n–100\n0 1 2 3 4 5 6 7 8 9 10\n/ ms time\nFig. 9.2\nWith reference to Fig. 9.1 and Fig. 9.2, describe the differences between the action potential\nin a motor neurone and the action potential in a striated muscle fibre.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThere are three phases in the contraction of a striated muscle: latent phase, contraction (b)\nphase and relaxation phase.\nThe tension in a muscle represents the degree of contraction of its fibres.\nFig. 9.3 is a graph of the tension in a striated muscle during the three phases of contraction.\nlatent contraction relaxation\nphase phase phase\ntension in\nmuscle\n0 10 20 30 40 50 60 70 80 90 100\ntime / ms\nstimulation\nFig. 9.3\nWith reference to Fig. 9.3, explain what is happening in the striated muscle fibre during (i)\nthe latent phase.\n...........................................................................................................................................\n...........................................................................................................................................\n.......................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2022-on-43-q09-c4172e4d57.pdf"
    },
    {
      "id": "9700-2022-on-43-q10",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_43.pdf?download=true",
      "source_pages": [
        23
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2022-on-43-q10-p01-15e8152d99.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-43-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The passage below outlines homeostasis. 10\nComplete the passage by using the most appropriate scientific terms.\nHomeostasis, in mammals, is the process of keeping the ........................................... environment\nof the body in optimum conditions so that cells can function efficiently. Blood water potential, core\ntemperature and blood glucose concentration are all factors that need to be kept at optimum\nvalues or set-point.\nWhen a condition deviates from its set-point, a corrective mechanism is triggered. An increase\nin blood glucose concentration triggers processes to decrease it and vice versa. This corrective\nmechanism is called ........................................... .\nThe pancreas is involved in the control of blood glucose concentration. Glucose binds\nto ........................................... on the cell surface membrane of pancreatic cells. These\nare ........................................... cells, which secrete hormones such as insulin and glucagon. The\ntwo hormones have opposite effects on the blood glucose concentration. For example the action\nof one hormone stimulates the uptake of glucose by cells for respiration and the action of the other\nhormone stimulates the breakdown of ........................................... to glucose in the liver.\n[Total: 5]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2022-on-43-q10-2d0ee6cfb8.pdf"
    },
    {
      "id": "9700-2022-on-51-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p01-63a75ec7de.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p02-bf42377f7d.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p03-c7294e6b46.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p04-88cd2837fe.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p05-1b4df645ea.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p06-514eb46dd7.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p07-67eae30457.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q01-p08-8ca8239549.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Antibiotic resistance in bacteria is a global problem that has caused scientists to research 1\ninto antibacterial substances other than antibiotics. Honey has properties that make it a good\nantibacterial substance. For example, honey contains hydrogen peroxide, which is known to kill\nbacteria.\nThe most effective honey tested so far for antibacterial activity is Manuka honey. It contains less\nhydrogen peroxide than many other types of honey, but it does contain an antibacterial compound,\nmethylglyoxal (MGO), which is not found in other types of honey.\nA student decided to investigate the effect of two antibacterial substances on the bacterium\nBacillus subtilis, which respires aerobically:\n• MGO in Manuka honey\n• an antibiotic solution used in cell cultures to prevent contamination.\nThe student wanted to find the lowest concentration of each antibacterial substance that would kill\nor inhibit the growth of B. subtilis.\nThe student used a broth culture for the investigation. To make a broth culture, a small (a)\nquantity of B. subtilis is added to a clear nutrient solution. A fresh (newly made) broth culture\nof B. subtilis is also clear.\nFig. 1.1 is a diagram of a fresh broth culture of B. subtilis.\ncotton wool bung\nfresh broth culture\nFig. 1.1\nA sterile cotton wool bung was used in the top of the flask containing the fresh broth (i)\nculture of B. subtilis to protect the culture from contamination.\nExplain why it is better to use a sterile cotton wool bung in a flask containing broth culture\nof B. subtilis, rather than using a sterile rubber bung.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nBefore comparing the two antibacterial substances, the student carried out a (ii)\ntrial experiment.\nThe student transferred a sample of fresh broth culture to a culture tube and incubated\nthe tube at 25 °C for 24 hours.\nFig. 1.2 summarises the results of the trial experiment.\ncotton wool bung\nincubate at 25°C\ncm3 15 culture tube turbid (cloudy)\ncontaining broth culture broth culture\nFig. 1.2\nThe student decided that turbidity of the broth culture is a measure of bacterial population\ngrowth (bacterial growth).\nExplain how this concept can be used in an investigation to measure the extent of\nbacterial growth.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student decided to test the antibiotic solution before testing the Manuka honey. (b)\nIn addition to normal laborator",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-51-q01-db61951a8c.pdf"
    },
    {
      "id": "9700-2022-on-51-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_51.pdf?download=true",
      "source_pages": [
        11,
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-51-q02-p01-e4729cf4ee.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q02-p02-e78a2a267a.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q02-p03-27240082a5.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q02-p04-859e7cb875.png",
        "9700-experimental-skills/assets/9700-2022-on-51-q02-p05-9fabd3c732.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Holstein cattle have been selectively bred for their high milk production. 2\nThe seed of the cotton plant, Gossypium spp. is a good source of fibre, protein and carbohydrate.\nTo improve milk production, whole cottonseed or cottonseed meal can be added to livestock feed.\nCottonseed meal is cottonseed that has been processed by grinding.\nFig. 2.1A shows whole cotton seed and Fig. 2.1B shows cottonseed meal.\nA B\nFig. 2.1\nFree gossypol is a toxin found in cotton seeds. Some of this free gossypol binds to protein to form\nbound gossypol. This occurs during cottonseed meal production and in the rumen (forestomach)\nof the cow during microorganism fermentation.\nFree gossypol is easily absorbed. Bound gossypol is absorbed less easily and is not toxic.\nAn investigation was carried out on the effect of different dry-matter diets on lactating\n(milk-producing) Holstein cows:\n• 30 healthy cows were fed on the same cottonseed-free diet for 14 days.\n• The cows were then divided equally into five groups, to E, and fed on one of five A\nexperimental diets for 42 days.\n• The quantity of milk produced each day was recorded.\n• The cows had no visible signs of illness during the 42 days.\nFor each diet, Table 2.1 shows:\n• the cottonseed content and total gossypol (free and bound) content\n• the mean daily dry matter taken in during feeding (intake)\n• the mean daily lactation performance (milk yield).\nTable 2.1\nSE = standard error\ncottonseed content /\ntotal dry matter milk\npercentage of dry matter\ndescription of dry gossypol intake yield /\ngroup\n–1 –1 matter diet in diet / kg d kg d\nwhole cottonseed\n–1 SE = 1.4 SE = 1.3 mg kg\ncottonseed meal\ncottonseed replaced by\n0 0 0 24.8 27.6 A\nsoybean meal\nwhole cottonseed 15.0 0 1039.7 23.6 29.7 B\ncottonseed meal 0 7.0 900.1 23.2 27.9 C\nmixed 1 (whole\ncottonseed and 7.5 3.5 959.7 22.6 28.7 D\ncottonseed meal)\nmixed 2 (whole\ncottonseed and 15.0 7.0 1922.0 24.0 32.6 E\ncottonseed meal)\nThe differences in the dry matter intake for groups to were statistically significant. A E not\nExplain why the cows were all fed on the same cottonseed-free diet for 14 days. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nUsing the information provided and the results shown in Table 2.1, a student made some (b)\nstatements about the effect of feeding different diets to lactating Holstein cows.\nThe student stated that if the investigation is repeated, the groups of cows will produce (i)\nmilk in this order:\n(lowest quantity). (highest quantity) B D C A E\nWith reference to Table 2.1, explain why the statement made by the student is not\nsupported by the data.\n..........",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-51-q02-950099c1d7.pdf"
    },
    {
      "id": "9700-2022-on-52-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-52-q01-p01-94ad9c0b99.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q01-p02-fb6d20c61e.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q01-p03-a972d5f3f2.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q01-p04-4fad441a4a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Egg albumen (egg white) is a good source of protein. Some eggs contain up to 1\ndm–3) protein. 10.0% w/v (100 g\nA biuret assay can be used to determine the concentration of protein in egg albumen.\n• copper(II) Biuret reagent is a mixture of sulfate solution and sodium hydroxide solution.\n• In a biuret assay, biuret reagent is added to a solution.\n(Cu2+) • copper(II) If protein is present, ions in the biuret reagent bind with the nitrogen atoms\nthat form peptide bonds in the protein.\n• This binding causes the reagent to change from a clear, pale blue to a purple colour.\n• Different concentrations of protein result in different intensities of purple colour.\nA colorimeter can be used to measure differences in the intensity of the purple colour in\nsamples tested.\nFig. 1.1 shows a colorimeter that measures the absorbance of light when it passes through\ndifferent coloured solutions.\ncm3 3\ncuvette\nFig. 1.1\nOutline how a colorimeter is set up so that a test cuvette can be placed in the slot to obtain a (a)\ncorrect measurement of intensity of a coloured solution.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nA student carried out a biuret assay to determine the concentration of protein in egg albumen (b)\nfrom eggs produced by a chicken that had been fed on a new type of chicken feed. The eggs\nproduced by the chicken before eating the new feed contained 6.0% w/v of protein in the egg\nalbumen.\nEgg albumen is diluted to obtain a concentration that is suitable for testing in the biuret assay.\nFor example, egg albumen with 6.0% w/v protein is diluted by a factor of 10 to obtain a\n0.6% w/v solution for a biuret assay.\nTo carry out the biuret assay on egg albumen of unknown protein concentration, a calibration\ncurve needs to be produced. This involves using standard solutions of known concentrations\nof protein.\nThe range of concentrations that can be measured using a colorimeter in a biuret assay (i)\nis 0.1 – 1.0% w/v.\nA stock solution of 1.0% w/v protein can be diluted using distilled water to",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-52-q01-dfabfd855b.pdf"
    },
    {
      "id": "9700-2022-on-52-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_52.pdf?download=true",
      "source_pages": [
        7,
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-52-q02-p01-ceff68c5ea.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q02-p02-cbfb238b7d.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q02-p03-1094617171.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q02-p04-1aa1b15604.png",
        "9700-experimental-skills/assets/9700-2022-on-52-q02-p05-174d3b53f3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "A main feature of Huntington’s disease (HD) is a loss of neurones from a region of the brain known 2\nas the striatum. A study of brain tissue taken from people with HD showed higher than normal\n(increased) urea concentrations in the anterior striatum and cerebellum regions of the brain.\nUrea is a toxic waste product of amino acid and protein metabolism. It is produced from ammonia,\nwhich is also toxic, in an enzyme-controlled cycle known as the urea cycle.\nScientists carried out a study using transgenic (genetically modified) sheep, known as OVT73,\nto investigate the increased urea concentrations. OVT73 have the mutant allele of the Huntingtin\n(HTT) gene. At the time of the study, OVT73 were 5 years old and had no detectable loss of\nstriatal neurones or disease symptoms.\nSamples of tissue were taken from the same region of the anterior striatum in:\n• six OVT73 (three females and three males)\n• a control group of non-transgenic sheep.\nDescribe features of an appropriate control group for this investigation. (a) two\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nA whole-genome RNA-sequencing experiment was carried out on the tissue samples using (b)\na commercially available diagnostic testing kit. This experiment produced gene expression\nprofiles for comparison.\nFrom the hundreds of genes that showed differential gene expression, 24 genes were\ndescribed as genes of interest because they were involved in urea metabolism or in urea\ntransport out of cells. None of these were genes coding for enzymes in the urea cycle.\nThe 24 genes of interest were identified from RNA sequencing using mRNA transcripts.\nExplain why it was more appropriate to use RNA sequencing rather than DNA sequencing for\nthe identification of these genes of interest.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.........................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-52-q02-5142a019c9.pdf"
    },
    {
      "id": "9700-2022-on-53-q01",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p01-63a75ec7de.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p02-bf42377f7d.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p03-c7294e6b46.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p04-88cd2837fe.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p05-1b4df645ea.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p06-514eb46dd7.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p07-67eae30457.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q01-p08-8ca8239549.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Antibiotic resistance in bacteria is a global problem that has caused scientists to research 1\ninto antibacterial substances other than antibiotics. Honey has properties that make it a good\nantibacterial substance. For example, honey contains hydrogen peroxide, which is known to kill\nbacteria.\nThe most effective honey tested so far for antibacterial activity is Manuka honey. It contains less\nhydrogen peroxide than many other types of honey, but it does contain an antibacterial compound,\nmethylglyoxal (MGO), which is not found in other types of honey.\nA student decided to investigate the effect of two antibacterial substances on the bacterium\nBacillus subtilis, which respires aerobically:\n• MGO in Manuka honey\n• an antibiotic solution used in cell cultures to prevent contamination.\nThe student wanted to find the lowest concentration of each antibacterial substance that would kill\nor inhibit the growth of B. subtilis.\nThe student used a broth culture for the investigation. To make a broth culture, a small (a)\nquantity of B. subtilis is added to a clear nutrient solution. A fresh (newly made) broth culture\nof B. subtilis is also clear.\nFig. 1.1 is a diagram of a fresh broth culture of B. subtilis.\ncotton wool bung\nfresh broth culture\nFig. 1.1\nA sterile cotton wool bung was used in the top of the flask containing the fresh broth (i)\nculture of B. subtilis to protect the culture from contamination.\nExplain why it is better to use a sterile cotton wool bung in a flask containing broth culture\nof B. subtilis, rather than using a sterile rubber bung.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nBefore comparing the two antibacterial substances, the student carried out a (ii)\ntrial experiment.\nThe student transferred a sample of fresh broth culture to a culture tube and incubated\nthe tube at 25 °C for 24 hours.\nFig. 1.2 summarises the results of the trial experiment.\ncotton wool bung\nincubate at 25°C\ncm3 15 culture tube turbid (cloudy)\ncontaining broth culture broth culture\nFig. 1.2\nThe student decided that turbidity of the broth culture is a measure of bacterial population\ngrowth (bacterial growth).\nExplain how this concept can be used in an investigation to measure the extent of\nbacterial growth.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student decided to test the antibiotic solution before testing the Manuka honey. (b)\nIn addition to normal laborator",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-53-q01-9e431c2506.pdf"
    },
    {
      "id": "9700-2022-on-53-q02",
      "subject": "9700",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2022-oct-nov/9700_w22_qp_53.pdf?download=true",
      "source_pages": [
        11,
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2022-Oct-Nov/qp/9700_w22_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2022-on-53-q02-p01-e4729cf4ee.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q02-p02-e78a2a267a.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q02-p03-27240082a5.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q02-p04-859e7cb875.png",
        "9700-experimental-skills/assets/9700-2022-on-53-q02-p05-9fabd3c732.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2022-on-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Holstein cattle have been selectively bred for their high milk production. 2\nThe seed of the cotton plant, Gossypium spp. is a good source of fibre, protein and carbohydrate.\nTo improve milk production, whole cottonseed or cottonseed meal can be added to livestock feed.\nCottonseed meal is cottonseed that has been processed by grinding.\nFig. 2.1A shows whole cotton seed and Fig. 2.1B shows cottonseed meal.\nA B\nFig. 2.1\nFree gossypol is a toxin found in cotton seeds. Some of this free gossypol binds to protein to form\nbound gossypol. This occurs during cottonseed meal production and in the rumen (forestomach)\nof the cow during microorganism fermentation.\nFree gossypol is easily absorbed. Bound gossypol is absorbed less easily and is not toxic.\nAn investigation was carried out on the effect of different dry-matter diets on lactating\n(milk-producing) Holstein cows:\n• 30 healthy cows were fed on the same cottonseed-free diet for 14 days.\n• The cows were then divided equally into five groups, to E, and fed on one of five A\nexperimental diets for 42 days.\n• The quantity of milk produced each day was recorded.\n• The cows had no visible signs of illness during the 42 days.\nFor each diet, Table 2.1 shows:\n• the cottonseed content and total gossypol (free and bound) content\n• the mean daily dry matter taken in during feeding (intake)\n• the mean daily lactation performance (milk yield).\nTable 2.1\nSE = standard error\ncottonseed content /\ntotal dry matter milk\npercentage of dry matter\ndescription of dry gossypol intake yield /\ngroup\n–1 –1 matter diet in diet / kg d kg d\nwhole cottonseed\n–1 SE = 1.4 SE = 1.3 mg kg\ncottonseed meal\ncottonseed replaced by\n0 0 0 24.8 27.6 A\nsoybean meal\nwhole cottonseed 15.0 0 1039.7 23.6 29.7 B\ncottonseed meal 0 7.0 900.1 23.2 27.9 C\nmixed 1 (whole\ncottonseed and 7.5 3.5 959.7 22.6 28.7 D\ncottonseed meal)\nmixed 2 (whole\ncottonseed and 15.0 7.0 1922.0 24.0 32.6 E\ncottonseed meal)\nThe differences in the dry matter intake for groups to were statistically significant. A E not\nExplain why the cows were all fed on the same cottonseed-free diet for 14 days. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nUsing the information provided and the results shown in Table 2.1, a student made some (b)\nstatements about the effect of feeding different diets to lactating Holstein cows.\nThe student stated that if the investigation is repeated, the groups of cows will produce (i)\nmilk in this order:\n(lowest quantity). (highest quantity) B D C A E\nWith reference to Table 2.1, explain why the statement made by the student is not\nsupported by the data.\n..........",
      "pdf_path": "9700-experimental-skills/print/9700-2022-on-53-q02-b5ef7447f8.pdf"
    },
    {
      "id": "9700-2023-m-42-q01",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-m-42-q01-p01-11b2ec9ad4.png",
        "9700-topic-14-homeostasis/assets/9700-2023-m-42-q01-p02-2da2614162.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a drawing of a longitudinal section (LS) of a human kidney. 1 (a)\nA\nB\nC\nD\nFig. 1.1\nUse the letters A, B, and in Fig. 1.1 to complete Table 1.1. C D\nEach letter may be used once, more than once or not at all.\nFor each description, list the letters that are correct. all\nTable 1.1\nregion of\ndescription\nkidney\nlocation of loops of Henle\n........................\nlocation of Bowman’s capsules\n........................\nlocation of glomeruli\n........................\ncontains urine at final concentration\n........................\n[4]\nThe volume and water potential of the urine produced by the kidney vary according to the (b)\nwater potential of the blood. This is a result of osmoregulation.\nDescribe the role of aquaporins in osmoregulation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDescribe the role of the brain in osmoregulation when the water potential of the blood (c)\nabove the set point. increases\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 10]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-m-42-q01-dc94ef8868.pdf"
    },
    {
      "id": "9700-2023-m-42-q02",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q02-p01-32c5828807.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q02-p02-a8ee041dc4.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q02-p03-01713d017d.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q02-p04-6e46fa554a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q02",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "(IFN-α) Interferon-alpha can be produced as a recombinant human protein to treat some types of 2\nIFN-α. cancer. The gene IFNA2 codes for\nIFN-α One method of producing recombinant uses genetically engineered Escherichia coli\nbacteria that contain recombinant plasmids. Each recombinant plasmid contains:\n• the gene IFNA2\nlacI) • three regulatory sequences of the lac operon (promoter, operator and\nAMPR. • a gene for antibiotic resistance,\nAMPR lacI gene and gene contains its own promoter. As a result, Each of the sequences for the\nthese genes are always expressed in E. coli bacteria that contain this recombinant plasmid.\nlacI gene and Fig. 2.1 is a diagram of the recombinant plasmid. The promoter regions of the\nAMPR gene are shown. not\npromoter op\ne r a\nt o\nr\nI F\nN\nA\n2\nla\nc\nI R\nP\nM\nA\nFig. 2.1\nThe start of transcription of the gene IFNA2 by E. coli with the recombinant plasmid shown in (a)\nIFN-α. Fig. 2.1 needs to be controlled to obtain an optimum yield of\nScientists investigated the effect of two inducers of transcription on the production of\nIFN-α: recombinant\n• lactose, which is converted to allolactose in E. coli\n• IPTG, which is a synthetic molecule with a very similar structure to allolactose. IPTG\nbe broken down by E. coli. cannot\nThe scientists grew three cultures of E. coli containing the recombinant plasmid in the same\ngrowth medium. The growth medium contained glucose, amino acids, essential vitamins and\nminerals. The growth medium did contain lactose. not\nAfter four hours, either lactose or IPTG at the same concentration was added to two of the\ncultures of E. coli. As a control, the third culture of E. coli was grown without adding lactose or\nIPTG.\nIFN-α The concentration of recombinant in the cultures was measured at different times over\na period of 28 hours. The results are shown in Fig. 2.2.\n300\nkey\n200\nculture to which IPTG added\nconcentration\nIFN-α of\nculture to which lactose added\ndm–3 μg /\n100 control culture\n0\n30 0 10 20\n/ hours time\nFig. 2.2\nThe regulatory sequences of the lac operon contained in the recombinant plasmid are (i)\ninvolved in the control of transcription of the gene IFNA2.\nlacI in the control of transcription of the IFNA2 gene between Explain the role of the gene\nand hours. 0 hours 4\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nWith reference to Fig. 2.2, describe the changes",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-m-42-q02-9a1001d122.pdf"
    },
    {
      "id": "9700-2023-m-42-q03",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q03-p01-f7eb9c52e6.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q03-p02-4466709b45.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Salmon can be genetically modified (GM) to produce increased quantities of growth hormone, 3\nwhich is a protein. GM salmon modified in this way have a faster growth rate and reach their\nmaximum body mass at a younger age than non-GM salmon.\nWithin any population of salmon there is variation in body mass. This is an example of (a)\ncontinuous variation.\nExplain what is meant by continuous variation how it can be caused. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nScientists investigated whether injection of very young non-GM salmon with recombinant (b)\ngrowth hormone could cause an increase in the growth rate of the salmon.\nThe scientists used two groups of non-GM salmon:\n• a control group of salmon that were not injected with recombinant growth hormone\n• µg an experimental group of salmon that were injected with 1.0 of recombinant\ngrowth hormone at the start of the experiment and once a week for the next six\nweeks.\nThe mean body mass of the salmon in the two groups at the start of the experiment was the\nsame (5.3 g).\nAfter six weeks, the body mass of every salmon was measured again. The results are\nsummarised in Table 3.1.\nTable 3.1\nno injection with injected with\nrecombinant recombinant\ngrowth hormone growth hormone\n(n) number of non-GM salmon 28 27\nrange 6.5–8.6 7.2–12.7\nbody mass\n(xr 7.7 9.4 ) mean\n/ g\n(s) standard deviation 0.4 1.1\nA student decided that a t-test should be performed on the results shown in Table 3.1.\nCalculate the value of t for the results shown in Table 3.1 using the formula for the t-test: (i)\nxr -xr\n2 1 = t\n2 2 s s\n1 2 +\nfn n p\n1 2\nGive your answer to decimal places. two\nShow your working.\nt = ...............................................................\n[3]\nThe critical value at p = 0.05 for these data is 2.01. (ii)\nThe student used the results in Table 3.1 and the t-test to conclude that the injections\nof recombinant growth hormone cause an increase in the growth rate of the non-GM\nsalmon.\nComment on the extent to which the conclusion made by t",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-m-42-q03-9c95aa492b.pdf"
    },
    {
      "id": "9700-2023-m-42-q04",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q04-p01-c41b4c2329.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q04-p02-a899e08d74.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-m-42-q04-p03-7ac32d98d3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Array comparative genome hybridisation (aCGH) is a technique involving the use of a microarray 4\nto analyse a genome or sections of a genome.\nOutline the steps required to prepare the genome of an individual so that the genome is ready (a)\nfor analysis using a microarray chip.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDiGeorge syndrome is a dominant inherited disease in humans. DiGeorge syndrome is (b)\ncaused by deletion of a large number of nucleotides from chromosome 22.\nThe number of nucleotides deleted varies between individuals in a range from 800 000 to\n3 100 000. The largest deletions can cause the removal of up to 46 protein-coding genes from\nthe chromosome.\nFig. 4.1 shows the results of aCGH using a microarray specific for the section of\nchromosome 22 within which the DiGeorge syndrome deletion occurs. The microarray\nanalysed DNA from two individuals:\n• one with DiGeorge syndrome\n• one who did not have DiGeorge syndrome (control DNA for comparison).\nIn the aCGH results shown in Fig. 4.1:\n• Each small circle represents the results from a single probe on the microarray.\n• The x-axis shows the position of each probe on chromosome 22. The position is\nshown as distance along the chromosome in millions of nucleotides.\n• A result close to 100% fluorescence on the y-axis means that the DNA from the\nindividual with DiGeorge syndrome fluoresces at the same intensity as the control\nDNA for that probe.\n• A result close to 50% fluorescence on the y-axis means that the DNA from the\nindividual with DiGeorge syndrome fluoresces half as much as the control DNA for\nthat probe.\n150\nfluorescence of\n100\nDNA from an\nindividual with\nDiGeorge syndrome\nas a percentage\nof the fluorescence 50\nof control DNA\n0\n17.0 18.0 16.0 19.0 20.0 21.0\nposition of probe on chromosome 22\n/ millions of nucleotides\nFig. 4.1\nWith reference to Fig. 4.1, estimate the number of nucleotides deleted from the affected (i)\nchromosome 22 in the individual with DiGeorge syndrome.\nGive your answer to the nearest 100 00",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-m-42-q04-2c752b2a6d.pdf"
    },
    {
      "id": "9700-2023-m-42-q05",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-m-42-q05-p01-ad10d183ac.png",
        "9700-topic-16-inheritance/assets/9700-2023-m-42-q05-p02-56c95569bd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Meiosis is described as a reduction division because the number of chromosomes in the daughter 5\ncells is reduced by half.\nTable 5.1 describes some of the events that take place during four of the different stages of (a)\nmeiosis in an animal cell.\nTable 5.1\nstage of meiosis spindle fibres diagram\nattach to centromeres and\narrange homologous pairs of\nI metaphase\nchromosomes at the equator of\nthe cell\nI anaphase\nre-form spindle in daughter cells\nII disassemble telophase\nComplete Table 5.1 by:\nI • outlining the behaviour of the spindle fibres during anaphase\n• identifying the stage of meiosis in which spindle fibres re-form the spindle in\ndaughter cells\nII. • drawing a diagram to show telophase\nII. You do need to add labels to your diagram showing telophase not\n[4]\nExplain the need for a reduction division during meiosis. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-m-42-q05-c27955813a.pdf"
    },
    {
      "id": "9700-2023-m-42-q06",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-m-42-q06-p01-b7edf6ac83.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2023-m-42-q06-p02-37149a922a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Fig. 6.1 is a diagram of a section through a mitochondrion. 6 (a)\nB A\nD\nC\nFig. 6.1\nThe four arrows, A, B, and D, show the movement of molecules and ions. C\nUse the letters to identify the arrows (one or more) that show: all\nactive transport of protons (i)\n.......................................................... [1]\ndiffusion of carbon dioxide. (ii)\n.......................................................... [1]\nOutline the role of the mitochondrial matrix in respiration. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nExplain how a lack of oxygen affects oxidative phosphorylation. (c)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 9]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-m-42-q06-3fff0e9825.pdf"
    },
    {
      "id": "9700-2023-m-42-q07",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-m-42-q07-p01-df6c501da8.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-m-42-q07-p02-886cc1fd30.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 7.1 is a diagram representing a synapse between a chemoreceptor cell from a human 7 (a)\ntaste bud and a dendrite of a sensory neurone.\nmicrovilli\nchemoreceptor cell\nsynapse\ndendrite of\nsensory neurone\nFig. 7.1\nIn an experiment, different concentrations of sodium chloride solution were applied to the\nmicrovilli of the chemoreceptor cell. The membrane potential of the chemoreceptor cell and\nthe membrane potential of the dendrite of the sensory neurone were recorded for each\nconcentration.\nThe resting potential of this chemoreceptor cell is –50 mV and the resting potential of the\ndendrite of this sensory neurone is –70 mV.\nThe results are shown in Table 7.1.\nTable 7.1\nconcentration of membrane potential / mV\nsodium chloride\nchemoreceptor cell dendrite of sensory neurone –3 solution / g dm\n0.1 –50 –70\n1.0 +30 +40\n10.0 +30 +40\nExplain the results shown in Table 7.1.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...............................................................................................................................................[4]\nDescribe the differences in structure function between sensory neurones and motor (b) and\nneurones.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-m-42-q07-427967ab66.pdf"
    },
    {
      "id": "9700-2023-m-42-q08",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-m-42-q08-p01-3f2a26c553.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-m-42-q08-p02-75fbe75bc8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q08",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Describe the functions of the internal membranes of the chloroplast in photosynthesis. 8 (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [7]\nRubisco activase (RA) is an enzyme that has an effect on the activity of rubisco. (b)\nAn investigation was carried out on the effect of RA on the activity of rubisco.\n• Solutions of rubisco and RuBP were added to two tubes, and B. A\n• RA was added to tube A.\n• Both tubes were incubated at 25 °C for 6 minutes.\n• The activity of rubisco was measured every 30 seconds.\nAll conditions were kept the same, except for the addition of RA to tube A.\nThe results are shown in Fig. 8.1.\n12\nA\n10\n8\nrubisco\nactivity\n6\narbitrary /\nunits\n4\n2\nB\n0\n6 0 1 2 3 4 5\n/ min time\nFig. 8.1\nDescribe the results shown in Fig. 8.1 suggest an explanation for the effect of RA on the and\nactivity of rubisco.\n......",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-m-42-q08-ad453594b1.pdf"
    },
    {
      "id": "9700-2023-m-42-q09",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-m-42-q09-p01-806517a7b2.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-m-42-q09-p02-6346263916.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 9.1 is a diagram of a relaxed sarcomere in striated muscle. 9 (a)\nZ-line M-line Z-line\nI-band I-band A-band\nFig. 9.1\nOn Fig. 9.1, use label lines and letters to label: (i)\n• an actin filament with the letter P\n• a myosin filament with the letter R.\n[2]\nI-band State what happens to the A-band and the when the sarcomere contracts. (ii)\nA-band ...............................................................................................................................\nI-band ................................................................................................................................\n[2]\nThe plant Strychnos toxifera produces the toxin curare, which can cause muscle paralysis in (b)\nmammals.\nThe toxin acts by binding to receptors on the cell surface membranes (sarcolemma) of muscle\ncells at neuromuscular junctions.\nSuggest how binding of curare to receptors may cause muscle paralysis. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nSuggest why the action of curare may lead to the death of a mammal. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\n[Total: 10]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-m-42-q09-3a62d6b004.pdf"
    },
    {
      "id": "9700-2023-m-42-q10",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_42.pdf?download=true",
      "source_pages": [
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-m-42-q10-p01-ad81d2e4c4.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-m-42-q10-p02-3fdba3dccb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-42-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The passage in Fig. 10.1 is about biodiversity. 10 (a)\nComplete the passage by using the most appropriate scientific terms.\nBiodiversity within an area can be assessed at different levels, including the species\ndiversity, genetic diversity and ecological diversity.\nSpecies diversity can be assessed by determining the number of different species and\nthe relative .............................................. of different species in a given area. From this\ninformation, species diversity can be estimated using .............................................. index\nof diversity.\nOrganisms of the same species can show much genetic diversity even though they share the\nsame .............................................. . This is because they can have different combinations of\n.............................................. .\nThe greater the genetic diversity, the greater the ability of a species to\n.............................................. to a changing environment.\nEcological diversity is a measure of the number and range of different ecosystems and\n.............................................. within a given area.\nFig. 10.1\n[6]\nThe International Union for Conservation of Nature (IUCN) Red List of Threatened Species is (b)\nupdated regularly.\nTable 10.1 shows the numbers of endangered animal species counted every three years\nbetween 2007 and 2019.\nTable 10.1\nyear number of endangered species\n2007 7 851\n2010 9 618\n2013 11 212\n2016 12 630\n2019 14 234\nCalculate the rate of increase in the number of endangered species between 2007 and (i)\n2019.\nShow your working.\nGive your answer to the nearest whole number.\nrate of increase = ................................................. per year\n[2]\nMore species of fish were listed as endangered in 2019 than species of mammals. (ii)\nSuggest reasons why more fish species than mammal species are endangered.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..........",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-m-42-q10-ed60c799a9.pdf"
    },
    {
      "id": "9700-2023-m-52-q01",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p01-bd9a2e97b9.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p02-165f14b107.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p03-71e7c7ce4e.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p04-c3f095fbf8.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p05-8deebcfdbb.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p06-0111316a48.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p07-43306006a5.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p08-355d8853bc.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q01-p09-f45a63dd0a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Potatoes are underground organs made by the potato plant, Solanum tuberosum. 1\nFig. 1.1 shows a potato plant.\nsoil\npotato\nFig. 1.1\nPotatoes contain high quantities of starch and are a popular food. After harvesting, potatoes are\nstored. During storage, the starch content of potatoes gradually decreases due to the breakdown\nof starch molecules into glucose.\nA glucose assay using a Benedict’s test and a colorimeter can be used to determine the\nconcentration of glucose in potatoes.\nThe glucose assay uses these steps:\n• The outer skin is removed from a potato.\n• The potato is cut into small pieces that are then ground into a pulp using a mortar and\npestle.\n• The potato pulp is filtered to remove most of the solids and obtain potato juice.\n• Excess Benedict’s solution is added to the potato juice.\n• The mixture is heated in a water-bath at 90 °C for five minutes. During this five-minute\n(Cu2+) in the Benedict’s solution react with glucose in the potato period, copper ions\njuice to form an insoluble precipitate.\n• The mixture is filtered to remove the precipitate. The filtrate, which is blue, contains\ncopper ions that did not react with the glucose in the potato juice while the mixture was\nbeing heated.\n• A sample of the blue filtrate is transferred to a colorimeter tube (cuvette).\n• A colorimeter is used to measure the absorbance of the blue filtrate.\n• A calibration curve is used to determine the glucose concentration of the potato juice\nfrom the measurement of absorbance.\nFig. 1.2 shows one type of colorimeter.\nhole to hold\ncolorimeter tube\ndigital readout of\nabsorbance\n0.000\ncolorimeter\ntube\ncm3 3 sample\nto be tested\n(blue filtrate)\nFig. 1.2\nA clean colorimeter tube containing the sample to be tested is placed in the colorimeter. Light is\npassed through the sample and the absorbance of light by the sample is measured.\nOutline how colorimeters should be prepared carrying out measurements on (a) (i) before\nsamples so that correct absorbance readings in the glucose assay are obtained.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nTo determine the concentration of glucose in potato juice using the glucose assay, a calibration\ncurve needs to be produced. This involves using the glucose assay to obtain absorbance\nreadings for standard glucose solutions of known concentrations.\nA student was given a 2.0% stock solution of glucose from which to prepare a range of (ii)\nstandard glucose solutions of known concentrations.\nState the glucose concentrations the student could use to produce a calibration curve\n3 of each solution should be prepared by dilution describe how proportional and 20 cm\nof ",
      "pdf_path": "9700-experimental-skills/print/9700-2023-m-52-q01-2afabdefa6.pdf"
    },
    {
      "id": "9700-2023-m-52-q02",
      "subject": "9700",
      "year": 2023,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-march/9700_m23_qp_52.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-March/qp/9700_m23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p01-a434314bf9.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p02-29502ee298.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p03-00148d86a3.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p04-84f0d93893.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p05-5b70e02f7c.png",
        "9700-experimental-skills/assets/9700-2023-m-52-q02-p06-0e2f4277f6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Rice, Oryza sativa, is an important food crop. Rice plants are wind pollinated. Pollen containing the 2\nmale gamete is transferred by the wind to female reproductive organs of rice plants. Fertilisation\nand grain formation then occur.\nWeedy rice is a wild form of rice that grows in fields of cultivated rice. Weedy rice competes with\ncultivated rice for resources. Weedy rice plants are taller than cultivated rice plants and produce a\nlow yield of rice grains.\nSeveral genetically modified (GM) varieties of cultivated rice have been developed.\nOne concern about the use of GM rice is gene flow from GM rice plants to weedy rice. Gene flow\noccurs when the wind carries pollen from GM rice plants to weedy rice plants.\nSome scientists investigated gene flow from herbicide-resistant GM rice plants to weedy rice\nplants. The GM rice plants had a gene for herbicide resistance. Weedy rice plants do not have the\ngene for herbicide resistance.\nThe scientists wanted to test the hypothesis that:\nGene flow from GM rice to weedy rice decreases as the distance between the GM rice\ncrop and weedy rice increases.\nThe scientists planted GM rice and weedy rice in a field, as shown in Fig. 2.1.\nFig. 2.1 also shows that the wind normally blows in a north–west (NW) direction.\nN\nNW NE\nW E\ndistance from\nSW SE GM rice plants\nS\n1 m\nm 2\nm 5\nm 10\nnormal wind\nkey\ndirection\nGM rice plants\nweedy rice plants\nbare soil\nFig. 2.1\nAfter the plants had been pollinated and the rice grains had developed, the grains were collected\nfrom the rice plants only. weedy\n• Grains were collected from all weedy rice plants growing 1 m from the GM rice plants in\nthe directions N, NE, E, SE, S, SW, W and NW.\n• Grains were also collected in the same directions from the GM rice plants at distances of\n2 m, 5 m and 10 m.\n• Approximately 1000 grains from each collection point were planted and germinated in\ncontrolled conditions in a glasshouse.\n• The plants were grown for three weeks and then tested to determine whether they had\nthe gene for herbicide resistance.\nFig. 2.2 shows young rice plants three weeks after germination.\nFig. 2.2\nIdentify the independent variables in this investigation. (a) (i) two\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSome of the young weedy rice plants had the gene for herbicide resistance. This gene (ii)\nhad been carried to the parents of these plants in pollen from the GM rice plants.\nOutline a method the scientists could use in the glasshouse to determine how many of\nthe young weedy rice plants had the gene for herbicide resistance.\nYour method should require the extraction of nucleic acids. not\n...........................................................................................................................................\n..............",
      "pdf_path": "9700-experimental-skills/print/9700-2023-m-52-q02-7f145fe739.pdf"
    },
    {
      "id": "9700-2023-mj-41-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-41-q01-p01-2cb8188051.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-41-q01-p02-32b735fcf0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q01",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 shows a transmission electron micrograph of part of a chloroplast. 1\nA\nB\nC\nD\n×80 000\nFig. 1.1\nTable 1.1 describes some functions that occur in different parts of a chloroplast. (a)\nComplete Table 1.1 by identifying the letter on Fig. 1.1 that is a location matching the\ndescription. Each letter may be used once, more than once, or not at all.\nTable 1.1\ndescription letter\naccumulates (builds up) a high\nconcentration of protons\nmakes triose phosphate\nmakes some chloroplast proteins\npumps protons\n[4]\nMembranes of the type labelled in Fig. 1.1 were made into a liquid extract. Chromatography (b) C\nwas then used to separate and identify the coloured components (pigments) in this extract.\nThe resulting chromatogram showed that these membranes contain a yellow pigment, an\norange pigment, a green-brown pigment and two different green pigments.\nDescribe how you would carry out chromatography to separate and identify the coloured (i)\npigments in the liquid extract of C.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nExplain why membrane has many different coloured pigments to function efficiently. (ii) C\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..............................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-mj-41-q01-a061f293ac.pdf"
    },
    {
      "id": "9700-2023-mj-41-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-41-q02-p01-d1cecce151.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-41-q02-p02-c56b63609e.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-41-q02-p03-8f13edf055.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-41-q02-p04-5133a25a9f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q02",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The natural on Hawadax Island in Alaska was disrupted in the 1780s when brown 2 ecosystem\nrats, Rattus norvegicus, swam to the island from a sinking ship and then rapidly increased their\npopulation size.\nThe rats occupied a new on the island as predators. The rats ate the eggs and chicks niche\nof birds such as the black oystercatcher, Haematopus bachmani, and the glaucous-winged gull,\nLarus glaucescens. These birds make nests, lay eggs and rear their chicks on the beaches of the\nisland.\nDefine the terms ecosystem and niche. (a)\necosystem (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nniche (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nConservation ecologists carried out a project to try to restore the natural ecosystem of Hawadax\nIsland. In 2008, they removed all rats from the island.\nBefore removing the rats, the ecologists measured the abundance of birds, invertebrates and\nseaweeds on eight of the island’s beaches. Seaweeds are large algae that grow attached to rocks\non the beach. The ecologists repeated these measurements in 2013 and in 2019, so that they\ncould calculate the percentage change in abundance from 2008.\nTo measure the abundance of invertebrates and seaweeds, the ecologists used this method: (b)\n• They laid 30 m tapes from high-tide mark to low-tide mark on the beach.\n• They placed quadrats at 5 metre intervals next to the tapes.\n• They took a photograph of each quadrat.\n• They analysed the photographs to calculate the percentage cover of seaweeds and the\npercentage cover of invertebrates such as mussels and sea snails.\nState the name of the sampling technique used. (i)\n..................................................................................................................................... [1]\nBiodiversity can be assessed at a number of different levels. (ii)\nIdentify the levels of biodiversity:\n• that were assessed by this sampling technique\n• that were assessed by this sampling technique. not\n...........................................................................................................................................\n...........................................................................................................................................\n.............",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-mj-41-q02-b87e13d266.pdf"
    },
    {
      "id": "9700-2023-mj-41-q03",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-41-q03-p01-d8ac944ef6.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-41-q03-p02-31c7c18e68.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q03",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Oryza is a genus of grass plants that includes the rice, Oryza sativa, a food crop. 3\nFarmers flood fields of rice because this encourages faster growth and higher yields. (a)\nAn adaptation of rice plants that allows them to grow in water is the development of (i)\naerenchyma.\nState the function of aerenchyma.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState way in which the roots of rice are adapted to being submerged in water. (ii) one other\n...........................................................................................................................................\n..................................................................................................................................... [1]\nAnother adaptation in some varieties of rice is the fast growth of stems. (iii)\nDescribe how selective breeding could produce varieties of rice with fast-growing stems.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nAuxin is a plant growth hormone that affects the growth of rice stems. (iv)\nExplain how auxin affects the growth of rice stems.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-mj-41-q03-4d765e284a.pdf"
    },
    {
      "id": "9700-2023-mj-41-q04",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-41-q04-p01-93c76dad35.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-41-q04-p02-84ed0d232c.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-41-q04-p03-15e054df1e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Recombinant human proteins can be used to treat disease. 4 (a)\nDefine the term recombinant DNA. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nFrom the 1920s until the 1970s, insulin obtained from the bodies of animals was used to (ii)\ntreat diabetes. From the 1970s, recombinant human insulin was used instead.\nExplain the advantages of using recombinant human insulin to treat people with diabetes.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nInsulin is composed of two polypeptide chains, the A chain and the B chain, that are linked by (b)\ndisulfide bonds.\nVariations in amino acid sequence occur:\n• in the insulin molecules of different animals\n• in new versions of human insulin that have been engineered to control blood glucose\nconcentration more effectively than normal recombinant human insulin. These new\nversions of human insulin are called analogues.\nTable 4.1 shows the amino acid positions where variation occurs in different animal and\nhuman analogue insulin molecules. The dashes indicate a missing amino acid.\nTable 4.1\ntype of insulin\namino\nacid\nhuman cow pig cat short-acting long-acting\nposition\nanalogue analogue\nA8 threonine alanine threonine alanine threonine threonine\nA10 isoleucine valine isoleucine valine isoleucine isoleucine\nA18 asparagine – asparagine histidine asparagine asparagine\nA21 asparagine asparagine asparagine asparagine asparagine glycine\nB3 asparagine asparagine asparagine asparagine asparagine –\nB28 proline proline proline proline lysine proline\nB29 lysine lysine lysine lysine proline lysine\nB30 threonine alanine alanine alanine threonine threonine\nB31 – – – – – arginine\nB32 – – – – – arginine\nCats with diabetes can be successfully treated with insulin injections. Cat insulin is (i) not\navailable, but vets can choose from the other types of insulin shown in Table 4.1.\nIdentify the type of insulin that is most suitable for treating cats.\n..................................................................................................................................... [1]\nSuggest ways in which analogue insulin molecules can be produced by genetic (ii)",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-mj-41-q04-563a390128.pdf"
    },
    {
      "id": "9700-2023-mj-41-q05",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-mj-41-q05-p01-210856119a.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-41-q05-p02-948bb8b402.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-41-q05-p03-32cc9401f2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q05",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The kidney is an important organ of homeostasis. One role of the kidney is osmoregulation. 5\nFig. 5.1 is a photomicrograph of part of a kidney nephron. (a)\nB\nA\n×900\nFig. 5.1\nIdentify the structures labelled and in Fig. 5.1. (i) A B\n........................................................................................................................................ A\n........................................................................................................................................ B\n[2]\nDescribe how blood is filtered by the part of the kidney nephron shown in Fig. 5.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nThe cell surface membranes of kidney cells have receptors for many molecules, including (b)\nglucagon and antidiuretic hormone (ADH).\nGlucagon binds to G-protein-coupled receptors on kidney cells. (i)\nThe binding of glucagon to kidney cells activates a cell signalling pathway that is similar\nto the cell signalling pathway activated when glucagon binds to liver cells.\nFig. 5.2 is an outline of the cell signalling pathway activated when glucagon binds to\nkidney cells.\nglucagon\nC\ncell surface membrane activates\nof kidney cell\nstimulates\nD\nATP\nG-protein\nactivates\nprotein kinase A\nFig. 5.2\nName the molecules labelled and in Fig. 5.2. C D\n........................................................................................................................................ C\n........................................................................................................................................ D\n[2]\nSyndrome of inappropriate antidiuresis (SIAD) is a condition that affects osmoregulation (ii)\nin the kidney.\nFig. 5.3 shows how sodium ion concentration in the blood affects the ADH concentration\nin the blood in:\n• people with normal homeostasis\n• people with one type of SIAD, known as type C SIAD.\n12\ntype C Key\n(grey region) = normal range\n8\nof ADH conce",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-mj-41-q05-3120f26753.pdf"
    },
    {
      "id": "9700-2023-mj-41-q06",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-mj-41-q06-p01-c425b42944.png",
        "9700-topic-16-inheritance/assets/9700-2023-mj-41-q06-p02-11ae46b19e.png",
        "9700-topic-16-inheritance/assets/9700-2023-mj-41-q06-p03-0498ce7ec1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q06",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Domestic rabbits vary in the length and colour of their fur. 6\nFig. 6.1 shows a domestic rabbit with short fur and a fur colour pattern called Himalayan.\nFig. 6.1\nThe two genes that determine the length and colour of the fur of this rabbit occur at the locus A / a\nh locus. These two gene loci are on separate chromosomes. and the / b B / b\n• The allele results in short fur. A\n• The allele results in long fur. a\n• is dominant to a. A\n• The allele results in black fur all over the body. B\nh • The allele results in black fur on the nose, ears, paws and tail of the rabbit, and white fur b\non the rest of the body (Himalayan pattern).\n• The allele results in white fur all over the body (albino). b\nh h • is dominant to and is dominant to b. B b b\nList the possible genotypes of the rabbit shown in Fig. 6.1. (a) (i) four\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nOne phenotype of rabbit always breeds true. This means that when it is mated to a rabbit (ii)\nthat looks the same as itself, all the offspring look the same as the parents.\nDescribe the of the rabbit that breeds true. phenotype\n..................................................................................................................................... [1]\nA rabbit with long, black fur all over the body that was homozygous at both loci was crossed (b)\nwith a rabbit with short, white fur that was homozygous at both loci. The F1 offspring had\nshort, black fur. These F1 rabbits were mated together to become the parents of the F2\ngeneration.\nDraw a genetic diagram to predict the F2 offspring genotypes and the ratio of F2 offspring\nphenotypes.\nF1 genotypes:\ngametes:\nF2 offspring genotypes:\nratio of F2 offspring phenotypes:\n[5]\nA rabbit breeder performed multiple crosses of the type described in (b). This gave enough (c)\ndata to test the prediction that the genes for fur length and fur colour show independent\nassortment.\nExplain why the two genes assort independently. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.............................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-mj-41-q06-548935f00e.pdf"
    },
    {
      "id": "9700-2023-mj-41-q07",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        19,
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-41-q07-p01-73ca9e11aa.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-41-q07-p02-bbd1137d9b.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-41-q07-p03-e6c1a55092.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Amino acids are the monomers that are used to produce proteins in organisms. Amino acids also 7\nhave other, non-protein, roles in the body.\nFig. 7.1 shows the structures of five amino acids with varying numbers and arrangements of (a)\ncarbon atoms.\nH\nH H C H\nH\nH C H S\nH N O H H H\nH\nH C C H C O H H C H\nC C H\nH\nH H C H H C H H C H H C H\nH O O H O O H H C C H\nH N C C C N C N C O N C C C N C\nH O H O H O H O H H H H\nH H H H H O H\nasparagine methionine proline serine leucine\nFig. 7.1\nIn the liver, one of these amino acids can be converted to pyruvate and one of these amino\nacids can be converted to oxaloacetate.\nSuggest which of the amino acids shown in Fig. 7.1 would be most directly converted to:\n• pyruvate\n• oxaloacetate.\namino acid converted to pyruvate ...............................................................\namino acid converted to oxaloacetate ...............................................................\n[2]\nThe amino acid glycine can act as a neurotransmitter. (b)\nA glycinergic synapse is shown in Fig. 7.2.\nmitochondria\nglycine\npresynaptic\nneurone\nCa2+\npostsynaptic\n– Cl\n– Cl neurone\nFig. 7.2\nThe glycinergic synapse and a cholinergic synapse use different neurotransmitters and (i)\ndifferent postsynaptic receptors.\nDescribe differences between the glycinergic synapse shown in Fig. 7.2 and a cholinergic\nsynapse.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe binding of glycine to receptors, as shown in Fig. 7.2, makes an action potential less (ii)\nlikely to occur in the postsynaptic neurone.\nSuggest why an action potential is less likely to occur after the binding of glycine to\nreceptors.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-mj-41-q07-11a3134a32.pdf"
    },
    {
      "id": "9700-2023-mj-41-q08",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 4,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-mj-41-q08-p01-7df19290fb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q08",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "I Fig. 8.1 shows a cell from the testis of a locust at the late prophase stage of meiosis. 8\nFig. 8.1\nI Explain how the behaviour of the chromosomes in prophase of meiosis results in the appearance\nshown in Fig. 8.1.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n[Total: 4]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-mj-41-q08-ee18d8097d.pdf"
    },
    {
      "id": "9700-2023-mj-41-q09",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        23
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-mj-41-q09-p01-65c1e4a953.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Many factors affect the rate of cellular respiration. 9\nPyruvate carboxylase (PC) deficiency is a rare disease. The rate of respiration in the cells of (a)\na person with PC deficiency is much lower than normal.\nThe passage describes the occurrence of PC deficiency in human populations.\nComplete the passage by using the most appropriate scientific terms.\nPC deficiency affects one in 250 000 people. A person either has the disease or does not,\nso there are no intermediate forms of the disease. PC deficiency is caused by a genetic\nmutation in one gene. Whether a person develops the disease is not influenced by\n......................................... factors. The distribution of PC deficiency in human populations\nshows ......................................... variation. [2]\nCoenzymes are important in aerobic respiration. (b)\nOutline the roles of named coenzymes in aerobic respiration.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\n[Total: 7]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-mj-41-q09-8ed8b20350.pdf"
    },
    {
      "id": "9700-2023-mj-41-q10",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_41.pdf?download=true",
      "source_pages": [
        24
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-mj-41-q10-p01-0858b38739.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-41-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Sweet peas are garden plants that vary in height. 10\n• Tall sweet peas grow to 200 cm in height.\n• Dwarf sweet peas grow to 30 cm in height.\n• Tall sweet peas contain a dominant allele. Le\n• le Dwarf sweet peas are homozygous for the recessive allele.\nlele Explain how the genotype results in the dwarf phenotype in sweet peas, with reference to the\nlele effect of on:\n• enzyme synthesis\n• hormone production\n• the expression of genes affecting plant growth.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n[Total: 6]\nThe boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply\nofficial endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country, territory, or area or any of its\nauthorities, or of the delimitation of its front",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-mj-41-q10-f96707b864.pdf"
    },
    {
      "id": "9700-2023-mj-42-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-mj-42-q01-p01-c10e464198.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2023-mj-42-q01-p02-435ca05b31.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Glycolysis is a biochemical pathway that occurs in the cytoplasm of cells. 1 (a)\nIn glycolysis, a molecule of glucose is metabolised to two molecules of pyruvate. The process\nis outlined in Fig. 1.1.\nglucose\n2ATP\n2ADP\nfructose 1, 6-bisphosphate\ntriose phosphate triose phosphate\n2ADP 2ADP NAD NAD\nreduced NAD 2ATP reduced NAD 2ATP\npyruvate pyruvate\nFig. 1.1\nExplain why glucose is phosphorylated at the beginning of glycolysis. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSuggest use of the reduced NAD that is produced in glycolysis. (ii) one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nName the type of phosphorylation reaction by which ATP is made during glycolysis. (iii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nPyruvate can enter the mitochondrion by active transport. (b)\nDescribe the main conditions that are required for pyruvate to enter the mitochondrion by\nactive transport.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nPyruvate is involved in the link reaction in the matrix of the mitochondrion. (c)\nDescribe the link reaction.\n...................................................................................................................................................\n................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-mj-42-q01-7033635e8e.pdf"
    },
    {
      "id": "9700-2023-mj-42-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-mj-42-q02-p01-8f8300124f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "In sexual reproduction, meiosis occurs to produce haploid gametes from a diploid cell. 2 (a)\nExplain the meaning of the terms haploid diploid. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nI in animal cells. Describe the behaviour of during the main stages of meiosis (b) chromosomes\nI ................................................................................................................................ prophase\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nI .............................................................................................................................. metaphase\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nI ................................................................................................................................ anaphase\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nI ................................................................................................................................ telophase\n...................................................................................................................................................\n................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-mj-42-q02-084e137725.pdf"
    },
    {
      "id": "9700-2023-mj-42-q03",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-42-q03-p01-1cca23534b.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-42-q03-p02-d2adf6a82b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q03",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The sea blush, Plectritis congesta, is a flowering plant that grows on the west coast of North 3 (a)\nAmerica.\nIndividual sea blush plants produce fruit that is either winged or wingless. Investigations have\nshown that this characteristic is controlled by a single gene with two alleles:\n• a dominant winged fruit allele\n• a recessive wingless fruit allele.\nFig. 3.1 shows the difference in structure between winged fruit and wingless fruit phenotypes.\n2 mm\nwinged fruit wingless fruit\nFig. 3.1\nA large sample of sea blush fruits was collected and their fruit-wing characteristic was (i)\nrecorded.\nName the type of variation that is shown for the fruit-wing characteristic of the sea blush.\n..................................................................................................................................... [1]\nDifferent sea blush plants can have fruit with different colours. (ii)\nSuggest ways in which a new fruit colour could naturally occur in a sea blush two\npopulation.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nEarly taxonomists classified sea blush plants with winged fruits as a different species to (iii)\nsea blush plants with wingless fruits.\nSince this time, evidence from observations and experiments has confirmed that the\nplants belong to the same species.\nSuggest examples of the evidence obtained that helped to confirm that these sea three\nblush plants belong to the same species.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nNatural selection occurs in populations, such as in populations of sea blush plants. (b)\nE",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-mj-42-q03-142b8c3e75.pdf"
    },
    {
      "id": "9700-2023-mj-42-q04",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-42-q04-p01-a19bd6bf7f.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-42-q04-p02-7eea8f5511.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genetic technology uses many different enzymes and techniques. 4\nRestriction endonucleases are used in genetic modification. These enzymes occur naturally (a)\nin prokaryotic cells. More than 3500 different restriction endonucleases have been identified\nand it is thought there are many more to discover.\nName domains that are a source of restriction endonucleases. two\n...................................................................................................................................................\n............................................................................................................................................. [1]\nOriginally, the method used to obtain a restriction endonuclease was to: (b)\n• grow large numbers of the specific prokaryotic cells that are the source of the enzyme\n• break open the cells and extract and purify the restriction endonuclease.\nThis original method produced only a small quantity of restriction endonuclease and was not\neconomical.\nThe newer method for large-scale production is to:\n• obtain the gene coding for a specific restriction endonuclease\n• introduce the gene into Escherichia coli, with a promoter that allows the gene to be\nexpressed continuously.\nThe newer method increases the quantity of specific restriction endonuclease produced.\nSuggest explain the steps needed to carry out the newer method for large-scale and\nproduction of a specific restriction endonuclease.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nDescribe the advantages of databases for the study and use of res",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-mj-42-q04-6a2c20b4fe.pdf"
    },
    {
      "id": "9700-2023-mj-42-q05",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-42-q05-p01-a48952d3f8.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-42-q05-p02-1b371e0667.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Warfarin is a poison used to kill rats. Some rats are resistant to warfarin and can survive the 5 (a)\neffects of the poison. Warfarin resistance is due to a mutant dominant allele at a single gene\nlocus.\nResearchers investigated the population of brown rats on a large island where there are no\npredators of rats. On this island, warfarin is used to try to control the rat population.\nThe researchers found that 40% of the population of rats was resistant to warfarin.\nUse Hardy–Weinberg equation 1 and equation 2 to calculate the percentage of the (i)\npopulation of rats on the island that is heterozygous for warfarin resistance.\np + q = 1 equation 1\n2 q2 + 2pq + = 1 p equation 2\nShow your working.\nanswer ..................................................... % [3]\nIn natural populations, it is usual to find that only some of the conditions of the (ii)\nHardy–Weinberg principle are met.\nSuggest explain which of the conditions of the Hardy–Weinberg principle are most and\nlikely to be met for the island population of brown rats.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nDominant advantageous alleles and recessive advantageous alleles both naturally occur in (b)\npopulations.\nExplain why, when a new advantageous allele occurs, its frequency increases dominant\nmore quickly in the population than when a new advantageous allele occurs. recessive\n...................................................................................................................................................\n...............................................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-mj-42-q05-59fdc6f5e2.pdf"
    },
    {
      "id": "9700-2023-mj-42-q06",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-mj-42-q06-p01-b7d19616c0.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-42-q06-p02-cf4528094e.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-42-q06-p03-ad59eb3960.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 6.1 is a photomicrograph of a section through a Bowman’s capsule and a glomerulus. 6 (a)\nFig. 6.1\nOn Fig. 6.1, use label lines and letters to label:\n– the location of the glomerular filtrate F\n– the location of the blood plasma. [2] P\nThe glomerular filtration rate (GFR) is the rate at which fluid filters from the blood in the (b)\nglomerulus into the Bowman’s capsule.\nFig. 6.2 shows the mean GFR values for healthy males and females of different age groups.\nKey:\nmales\nfemales\n140\n120\n100\n80\nmean GFR\ncm3 min–1 /\n60\n40\n20\n0\n20–29 30–39 40–49 50–59 60–69 70–79 80–89\nage groups / years\nFig. 6.2\nUse Fig. 6.2 to calculate the yearly rate of decrease in mean GFR from the 20–29 age (i)\ngroup to the 80–89 age group for females.\nShow your working.\nGive your answer to decimal place and include units. one\nanswer ......................................................... [3]\nThe age of a person and whether they are male or female can affect GFR. (ii)\nSuggest explain other factors that can affect GFR. and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nThe glomerular filtrate passes from the Bowman’s capsule to the proximal convoluted tubule, (c)\nwhere selective reabsorption takes place.\nDescribe the process of selective reabsorption.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-mj-42-q06-4a55c89b8b.pdf"
    },
    {
      "id": "9700-2023-mj-42-q07",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-42-q07-p01-454ed1547f.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-42-q07-p02-a8c577b4c0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "A striated muscle cell (muscle fibre) will contract when stimulated by a motor neurone at a 7 (a)\nneuromuscular junction.\nAir temperature can affect the temperature of striated muscle cells.\nInvestigations have shown that the efficiency of contraction of striated muscle cells decreases\nwhen the air temperature decreases.\nSuggest reasons why a reduction in temperature can decrease the efficiency of contraction of\nstriated muscle cells.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThe greater blue-ringed octopus, Hapalochlaena lunulata, produces tetrodotoxin (TTX). TTX (b)\nis a neurotoxin. If a mammal is bitten by this octopus, the effect of TTX can cause the death\nof the mammal.\nFig. 7.1 shows a greater blue-ringed octopus.\nFig. 7.1\nTTX binds to voltage-gated sodium ion channels in the axon of a neurone and changes the\ntertiary structure of the channel protein.\nSuggest how TTX may affect the functioning of a motor neurone.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.......................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-mj-42-q07-38eeb57015.pdf"
    },
    {
      "id": "9700-2023-mj-42-q08",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-42-q08-p01-c70eb12c0d.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-42-q08-p02-b635b17d94.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q08",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "In plants such as rice, Oryza sativa, rubisco has a low rate of activity, which in turn affects the 8 (a)\nrate of photosynthesis. The cereal crop sorghum, Sorghum bicolor, has a high rate of activity\nof rubisco.\nA genetically modified (GM) variety of rice was produced. Parts of the quaternary structure of\nrubisco in rice were altered to be the same as the rubisco in sorghum.\nThe rate of activity of rubisco in non-GM rice and GM rice was measured at different\n). concentrations of atmospheric carbon dioxide (CO\n2\nThe results are shown in Fig. 8.1.\n6\nGGMM rriiccee\n4\nrate of activity\nof rubisco\n/ arbitrary units\n2\nnnoonn--GGMM rriiccee\n0\n0 200 400 600\nconcentration atmospheric CO\n2\ndm–3 / mg\nFig. 8.1\nCompare the two curves shown in Fig. 8.1 explain why the curve for non-GM rice (i) and\nlevels off.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nSuggest which part of the rubisco molecule was altered to produce the GM variety of (ii)\nrice.\n...........................................................................................................................................\n..................................................................................................................................... [1]\nFig. 8.2 shows an outline of the Calvin cycle. (b)\nCO\n2\nRuBP\nGP ADP + Pi\nATP\nATP\nADP + Pi\nreduced NADP\nNADP\nTP\nother\nmolecules\nFig. 8.2\nName the process that involves the enzyme rubisco. (i)\n..................................................................................................................................... [1]\nState the type of reaction that occurs when GP is converted to TP. (ii)\n..................................................................................................................................... [1]\nRuBP is regenerated from molecules of TP. (iii)\nState how many molecules of RuBP are produced from 10 molecules of TP.\n..............................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-mj-42-q08-001b0ca82a.pdf"
    },
    {
      "id": "9700-2023-mj-42-q09",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-mj-42-q09-p01-5e31cdbfba.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-42-q09-p02-45e79f02e3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "The passage outlines the endocrine system. 9 (a)\nComplete the passage by using the most appropriate scientific terms.\nThe endocrine system consists of tissues and ................................................................. ,\nwhich secrete hormones such as insulin, glucagon and antidiuretic hormone (ADH).\nGlucagon only affects target cells that have complementary receptors, which are located\non the ................................................................. . Glucagon binds to these receptors\nand this leads to the production of a ................................................................. , which\ntransfers the signal throughout the cytoplasm. The target cells for insulin are in the liver,\n................................................................. and adipose tissue. The target cells for ADH are\nthose of the distal convoluted tubule and the ................................................................. of the\nkidney nephron. These hormones are involved in ................................................................. ,\nwhich is the maintenance of a constant internal environment. [6]\nThe blood glucose concentration of a person was measured at regular intervals after the (b)\ningestion of a meal rich in glucose.\nFig. 9.1 shows the results of this investigation.\nA B\n11\n10\n9\n8\n7\nblood glucose\n6\nconcentration\n5\ndm–3 / mmol\n4 set point\n3\n2\n1\n0\n0 1 2 3 4 5\ntime / h\nglucose-rich\nmeal ingested\nFig. 9.1\nExplain how the blood glucose concentration is reduced during phase of the curve. (i) A\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nSuggest why the blood glucose concentration increases again during phase B. (ii)\n...........................................................................................................................................\n................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-mj-42-q09-6aa0da508f.pdf"
    },
    {
      "id": "9700-2023-mj-42-q10",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-42-q10-p01-d3b95436a6.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-42-q10-p02-bc323631ce.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-42-q10-p03-44ece3335d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-42-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Many zoos around the world are involved in captive breeding programmes for endangered 10 (a)\nspecies.\nOutline the role of zoos in the conservation of endangered species, captive other than\nbreeding programmes.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nChester Zoo, in the United Kingdom, controls the European captive breeding programme for (b)\nthe eastern black rhinoceros, Diceros bicornis michaeli.\nFig. 10.1 shows an eastern black rhinoceros.\nFig. 10.1\nIn the captive breeding programme for the eastern black rhinoceros, 15.8% of females breed\neach year while in the wild 23.7% of females breed each year.\nSuggest reasons why fewer females breed in captivity than in the wild.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...............................................................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-mj-42-q10-ec8faac1a0.pdf"
    },
    {
      "id": "9700-2023-mj-43-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-mj-43-q01-p01-66411d05e9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2023-mj-43-q01-p02-6007ffe30f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The Krebs cycle was named after the biochemist Sir Hans Krebs, who worked out the 1 (a)\nsequence in 1937.\nFig. 1.1 is an outline of the Krebs cycle.\nacetyl coenzyme A (2C)\noxaloacetate (4C)\nP\nS\n(6C)\nR\nQ Q\nU\nATP\nR ADP\nT\n5C\nR\ncompound\nQ\nS\nFig. 1.1\nIdentify substances P–U.\n............................................................................................................................................... P\n............................................................................................................................................... Q\n............................................................................................................................................... R\n............................................................................................................................................... S\n............................................................................................................................................... T\n............................................................................................................................................... U\n[3]\nDescribe how ATP is produced from ADP in the Krebs cycle. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nDescribe explain the features of ATP that make it suitable as the universal energy (c) and\ncurrency.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-mj-43-q01-4392a4671b.pdf"
    },
    {
      "id": "9700-2023-mj-43-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-mj-43-q02-p01-2e6d002622.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Humans use antibiotics to treat bacterial infections. The increased use of antibiotics has led to an 2\nincrease in the number of strains of bacteria that are resistant to antibiotics.\nThe evolution of antibiotic resistance in bacteria has resulted from natural selection.\nOutline how bacteria become resistant to antibiotics.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n[Total: 7]",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-mj-43-q02-594b374d37.pdf"
    },
    {
      "id": "9700-2023-mj-43-q03",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-43-q03-p01-dce451b783.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-43-q03-p02-69ee46f1ec.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Gene expression in a cell is controlled. When a gene is expressed (switched on), the gene is 3\ntranscribed. When a gene is expressed (switched off), the gene is transcribed. not not\nEnvironmental changes can cause some genes to be switched on or switched off.\nAn example of control of gene expression in prokaryotes is regulation in the lac operon. (a)\nThe lac operon is a length of DNA that is made up of different parts.\nlacI Fig. 3.1 shows a simple diagram representing the (regulatory) gene and the lac operon.\nlacI\nFig. 3.1\nOutline the main features of the lac operon. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nlacI Explain the role of the gene in the regulation of the lac operon. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nThe lac operon codes for inducible enzymes. Repressible operons code for repressible (b)\nenzymes.\nSuggest expl",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-mj-43-q03-9377f24a5a.pdf"
    },
    {
      "id": "9700-2023-mj-43-q04",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-43-q04-p01-4eb6e91367.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-mj-43-q04-p02-a5b14127a6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "One cause of the genetic disease severe combined immunodeficiency (SCID) is a mutation in the 4\nADA gene. This mutation results in a deficiency of the enzyme adenosine deaminase (ADA).\nAlthough ADA is found throughout the body, it is especially active in lymphocytes. The absence of\nfunctional ADA causes the build-up of toxic metabolites that kill lymphocytes and damage organs.\nBabies are often diagnosed with SCID by six months old. Treatment can greatly improve the life\nexpectancy of children with SCID.\nSome treatment options are available.\n• Enzyme replacement therapy with recombinant human ADA made by genetically modified\n(GM) Escherichia coli. Weekly intra-muscular injections are given.\n• Bone marrow transplant if a well-matched donor, such as a close relative, can be found.\n• Gene therapy.\nSuggest explain why it may be more appropriate to use enzyme replacement therapy to (a) and\ntreat SCID instead of a bone marrow transplant.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nOutline the procedure used for gene therapy treatment of a person with SCID. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-mj-43-q04-f084de6525.pdf"
    },
    {
      "id": "9700-2023-mj-43-q05",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-mj-43-q05-p01-90987bae01.png",
        "9700-topic-16-inheritance/assets/9700-2023-mj-43-q05-p02-06cb85704c.png",
        "9700-topic-16-inheritance/assets/9700-2023-mj-43-q05-p03-33b3b133a2.png",
        "9700-topic-16-inheritance/assets/9700-2023-mj-43-q05-p04-d1c55022cd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "A freshwater fish species, Oryzias latipes, has individuals with four body colour patterns, as shown 5\nin Table 5.1.\nTable 5.1\nphenotype body colour pattern\nred\nwhite\nred with black spots\nwhite with black spots\nTwo unlinked genes determine the body colour patterns shown in Table 5.1.\nOne gene controls whether the body colour is red or white:\n• dominant allele = red R\n• recessive allele = white. r\nThe other gene controls whether black spots are present or present: not\n• dominant allele = with black spots B\n• recessive allele = without black spots. b\nA fish that is homozygous recessive at both loci is white.\nGenetic crosses were carried out to investigate the inheritance of the four different body colour\npatterns.\nMales that were red with black spots, and homozygous at both loci, were crossed with females\nthat were white. The F1 offspring were all red with black spots.\nThese F1 offspring were then crossed to produce the F2 generation.\nTable 5.2 shows the observed numbers obtained of each of the four different phenotypes for (a)\nthe F2 generation.\nTable 5.2\n2 (O –E)\n2 phenotype observed expected O–E (O–E)\nE\nred with black\n279 281.25\nspots\nwhite with\n95 93.75 1.25 1.5625 0.017\nblack spots\nred 96 93.75 2.25 5.0625 0.054\nwhite 30 31.25\n|2 = ......................\nTable 5.2 compares the observed numbers with the numbers that would be expected in the\nF2 generation for a normal dihybrid ratio.\n|2 for the F2 generation by completing Table 5.2. Calculate\n|2 is: The formula for\n–E)2 (O Σ\n|2 =\n[3] E\nThe critical value at p = 0.05 and 3 degrees of freedom is 7.815. (b)\nComment on whether the null hypothesis should be accepted or rejected.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nFurther analysis of the results from the F2 generation in Table 5.2 showed that there were no\nwhite males or white males with black spots.\nIn O. latipes, females have two chromosomes and males have an and a chromosome. X X Y\nIt was deduced that, in O. latipes:\n• the gene that controls body colour is located on the chromosome the chromosome X and Y\n• the gene that controls whether black spots are present or is located on an autosome. not\nr R To produce the F2 generation, red males with black spots, Bb, were crossed with red (c) X Y\nR r Bb. females with black spots, X X\nComplete the Punnett square in Fig",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-mj-43-q05-a5d4d1c295.pdf"
    },
    {
      "id": "9700-2023-mj-43-q06",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-mj-43-q06-p01-5918d4f689.png",
        "9700-topic-14-homeostasis/assets/9700-2023-mj-43-q06-p02-57c8b45e47.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 6.1 shows part of the Bowman’s capsule of a kidney nephron. 6 (a)\nA\nlumen of\ncapillary\nB\nC\nlumen of\nBowman’s capsule\nnot to scale\nFig. 6.1\nName structures A, and C. B\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n[3]\nFluid is forced into the Bowman’s capsule by ultrafiltration to form the glomerular filtrate. (b)\nDescribe the role of structure in ultrafiltration. B\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe main component of the glomerular filtrate formed in the Bowman’s capsule is water. Most (c)\nof this water is reabsorbed as the filtrate passes along the nephron.\ndm3 • 180 of water forms part of the glomerular filtrate each day.\ndm3 • 1.4 of water leaves the body in urine each day.\nCalculate the percentage of water in the glomerular filtrate that is reabsorbed as the\nglomerular filtrate passes along the nephron.\nShow your working and give your answer to decimal place. one\nanswer ...................................................... % [2]\nDescribe how the action of antidiuretic hormone (ADH) on the cells of the collecting duct (d)\nleads to the reabsorption of water in the glomerular filtrate.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-mj-43-q06-85e373f6b5.pdf"
    },
    {
      "id": "9700-2023-mj-43-q07",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-43-q07-p01-cbb850bd3a.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-43-q07-p02-d1b0ec6dd6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Sarcomeres are the functioning units of the myofibrils of the muscle fibres (muscle cells) of 7 (a)\nstriated muscle.\nThe arrival of an action potential at the sarcoplasmic reticulum of a striated muscle fibre can\nlead to the shortening of a sarcomere. This shortening occurs by the sliding filament model.\nOutline the sequence of events occurring after stimulation of the sarcoplasmic reticulum that\nresults in the shortening of a sarcomere.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nScientists compared the diameter of samples of striated muscle fibres taken from young mice (b)\nand adult mice.\nThe results are shown in Fig. 7.1.\n40\nKey\nyoung mice\n30\nadult mice\nnumber of\n20\nmuscle fibres\n10\n0\n10 30 50 70 90\nμm / muscle fibre diameter\nFig. 7.1\nWith reference to Fig. 7.1:\n• describe differences between the muscle fibres of young mice and adult mice two\n• suggest how these differences may affect the sliding filament model.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-mj-43-q07-807e34563d.pdf"
    },
    {
      "id": "9700-2023-mj-43-q08",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-43-q08-p01-9a9eb49cde.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-mj-43-q08-p02-511b4b4f4c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q08",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Fig. 8.1 is a diagram of a chloroplast. 8 (a)\nK\nL\nM\nN\nFig. 8.1\nUsing the letters K–N, complete Table 8.1.\nEach letter may be used once, more than once, or not at all.\nTable 8.1\nletter\nhigh concentration of protons\n......................\nlocation of photosynthetic pigments\n......................\nsite of light-independent stage\n......................\nsite of light-dependent stage\n......................\n[4]\nChlorophyll a is the main photosynthetic pigment in plants. (b)\nDescribe the role of other photosynthetic pigments found in plant chloroplasts.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...............................................................................................................................................[4]\nThe absorption of different wavelengths of light by chloroplast pigments can be represented (c)\nby an absorption spectrum.\nFig. 8.2 is an absorption spectrum of an extract of chloroplast pigments and an absorption\nspectrum of whole chloroplasts.\n10\nwhole chloroplasts\n9\n8\n7\n6\nabsorbance\n5\narbitrary units /\n4\n3\npigment\n2\nextract\n1\n0\n400 600 650 700 450 500 550\n/ nm wavelength\nFig. 8.2\nWith reference to Fig. 8.2, describe the differences between the two spectra suggest and\nexplanations for the differences.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-mj-43-q08-f6c6baa4f5.pdf"
    },
    {
      "id": "9700-2023-mj-43-q09",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-43-q09-p01-11043d709c.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-mj-43-q09-p02-d5d56c1deb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Dopamine is a neurotransmitter released in some synapses in the brain. The release and 9 (a)\naction of dopamine is similar to that of acetylcholine.\nFig. 9.1 is a diagram of a brain synapse where dopamine is the neurotransmitter.\ndopamine\nvesicle\npresynaptic\nmembrane\npostsynaptic\nmembrane\nFig. 9.1\nDescribe how the release of dopamine from the presynaptic neurone can lead to an (i)\naction potential in the postsynaptic neurone.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nIn brain cells, the amino acid tyrosine is changed into DOPA, which then is converted to (ii)\ndopamine.\nName another compound in the body produced from DOPA.\n..................................................................................................................................... [1]\nIn some brain synapses, the neurotransmitter gamma-aminobutyric acid (GABA) is released. (b)\nThis results in an influx of chloride ions into the postsynaptic neurone.\nSuggest explain whether an action potential would be generated in the postsynaptic and\nneurone if GABA is released into a brain synapse.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n........................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-mj-43-q09-f8e231b6f3.pdf"
    },
    {
      "id": "9700-2023-mj-43-q10",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_43.pdf?download=true",
      "source_pages": [
        23,
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-43-q10-p01-86aeb794a4.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-43-q10-p02-d06dc161d8.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-mj-43-q10-p03-e05d2b1dd5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-43-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The moose, Alces alces, is a large member of the deer family. It lives in temperate forests in 10 (a)\nNorth America and northern Europe, where snow is present for large parts of the year.\nFig. 10.1 shows an adult male moose feeding in a lake.\nFig. 10.1\nThe moose feeds on a plant in the lake called watermilfoil, Myriophyllum aquaticum.\nThe moose and the watermilfoil belong to the domain Eukarya, which includes the kingdoms\nAnimalia and Plantae.\nDescribe the main differences between the kingdom Animalia and the kingdom Plantae.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nMeasurements of the surface temperature of land and oceans can be taken from locations (b)\naround the world. The mean global surface temperature for land and ocean combined can be\ncalculated for a fixed time period.\nScientists calculated:\n• the mean global temperature for the twentieth century\n• the mean global temperature for each decade (ten years) from 1880 to 2020.\nThe mean temperature for each decade was compared to the mean for the twentieth century.\nFor each decade, the difference in temperature was calculated.\nThe calculated differences are shown in Fig. 10.2.\n+0.9\n+0.8\n+0.7\n+0.6\n+0.5\n+0.4\ncalculated +0.3\ntemperature\ndifference +0.2\n/ °C\n+0.1\n0.0\nmean for the\ntwentieth –0.1\ncentury\n–0.2\n–0.3\n–0.4\n1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010 2020\nyear\nFig. 10.2\nCalculate the rate of increase in temperature per between 1980 and 2020. (i) decade\nShow your working.\nWrite your answer to decimal places. two\nanswer ................................. °C per decade [2]\nMoose populations have decreased in North America since 1980. (ii)\nSuggest explain reasons for the decrease in moose populations. a",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-mj-43-q10-f0da531977.pdf"
    },
    {
      "id": "9700-2023-mj-51-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p01-3240677978.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p02-a23e87517b.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p03-cef0f03935.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p04-7fcef297a2.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p05-da4694d210.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p06-9084c3aa07.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p07-291659827f.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q01-p08-8cb9e8db43.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Catalase is an enzyme that is found in many plant cells and animal cells. Catalase breaks down 1\nhydrogen peroxide, which is a waste product from cell metabolism.\nFig. 1.1 shows the breakdown of hydrogen peroxide by catalase.\ncatalase\nO 2 H O + O 2 H\n2 2 2 2\nhydrogen peroxide water oxygen\nFig. 1.1\nA student decided to investigate how catalase activity varies along the length of a carrot root.\nFig. 1.2 shows a photograph of some carrots.\nleaves\nshoot\ncarrot root\nFig. 1.2\nThe student extracted catalase from different parts of the carrot root tissue.\n• Carrot root tissue was mixed with a small volume of pH7 buffer and some fine sand.\n• The mixture was ground with a pestle and mortar.\n• The mixture was placed into a centrifuge tube and spun in a variable speed centrifuge for a\nfew minutes.\n• This caused the mixture to be separated into a solid pellet at the bottom of the centrifuge tube\nand a liquid carrot extract at the top, as shown in Fig. 1.3.\n• The liquid carrot extract was poured into a clean test-tube. This liquid contained catalase.\nFig. 1.3 shows a centrifuge and centrifuge tube with its contents, after spinning in the centrifuge\nfor a few minutes.\nlid of centrifuge tube\nplastic centrifuge tube\ncm3) (2.0\nliquid carrot extract\nsolid pellet\nFig. 1.3\nState variable that should be standardised in the procedure to obtain the liquid carrot (a) one\nextract, the carrot tissue, fine sand and apparatus used. other than\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe student carried out these steps to investigate the activity of catalase in the different samples\nof carrot root tissue.\n• A disc of filter paper of 5 mm diameter was dipped into the liquid carrot extract using forceps.\n• The filter paper disc was placed at the bottom of a beaker of 1% hydrogen peroxide solution\nusing forceps.\n• The time taken for the filter paper disc to rise to the surface of the hydrogen peroxide solution\nwas measured.\nFig. 1.4 shows the apparatus used to measure the time taken for the filter paper disc to rise.\n0:12\n00\ncm3 beaker 10\nstop-clock\n1% hydrogen peroxide solution\n5 mm filter paper disc\nFig. 1.4\nState the relationship between the time taken for the filter paper disc to rise to the surface of (b)\nthe hydrogen peroxide solution and the catalase activity of the liquid carrot extract.\n...................................................................................................................................................\n...................................................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-51-q01-7123bc1077.pdf"
    },
    {
      "id": "9700-2023-mj-51-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_51.pdf?download=true",
      "source_pages": [
        10,
        11,
        12,
        13,
        14,
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p01-b6691f7cff.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p02-9ac83662db.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p03-5b3e2f6e8f.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p04-8602e2f180.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p05-9384ae9f3a.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p06-3d616fce21.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p07-c979733606.png",
        "9700-experimental-skills/assets/9700-2023-mj-51-q02-p08-1d475fd145.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Sea trout, Salmo trutta, are found in rivers and coastal seawaters in many regions of the world. 2\nSea trout lay their eggs in rivers and the young fish remain in this freshwater environment. After\nliving for 1–8 years in freshwater rivers, sea trout migrate to the sea. Adult sea trout feed in coastal\nseawater regions, usually no more than 80 km from their home river.\nFish farms that produce Atlantic salmon, Salmo salar, may be located in the coastal seawater\nregions where wild sea trout are also found.\nFig. 2.1 shows one cage at a fish farm.\nnet to stop birds\nAIR\nSEAWATER\nfish cage\nfarmed Atlantic salmon\nwild sea trout\nnot to scale\nFig. 2.1\nThe increase of fish farming is thought to have caused more infections by salmon lice,\nLepeophtheirus salmonis, in wild fish such as sea trout. When sea trout enter coastal seawater\nregions, they can become infected with salmon lice.\nSalmon lice larvae live in seawater and can move freely between fish cages and open seawater.\nSalmon lice attach to the skin of sea trout and feed on their skin and blood. The resulting tissue\ndamage can sometimes cause the death of sea trout.\nFig. 2.2 shows a sea trout infected with salmon lice.\nbody of sea trout\nsalmon lice\nFig. 2.2\nSome scientists found out from a published source that 13 or more attached salmon lice (a)\nusually cause the death of a sea trout kept in laboratory conditions.\nThe scientists decided that each sea trout with 13 or more salmon lice is at risk of wild\nphysiological stress and potential death.\nOutline a laboratory method that could be used to determine the number of salmon lice\nneeded to cause the death of a sea trout.\nSuggest a suitable control that could be used to confirm that it is the salmon lice that are\ncausing the death of the sea trout.\nmethod that could be used .......................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nsuitable control .........................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n........................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-51-q02-fd58a9af34.pdf"
    },
    {
      "id": "9700-2023-mj-52-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p01-e36148394a.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p02-84bd37c561.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p03-a3439194d0.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p04-32592c8283.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p05-b5514a9ded.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q01-p06-c9cbafa78e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Polyphenol oxidase is an enzyme involved in the browning of fruit. 1\nFig. 1.1 shows browning of the skin of bananas.\nbrown areas\nnon-brown areas\nFig. 1.1\nA student extracted polyphenol oxidase enzyme from ripened bananas. The student used this\nenzyme extract to investigate the factors affecting the activity of the enzyme.\nThe student initially found that the enzyme extract was too concentrated, so the student made a\n1:10 dilution of the enzyme extract.\nDescribe how the student made a 1:10 dilution of the enzyme extract. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe student studied the effect of pH on the activity of polyphenol oxidase enzyme. (b)\nDopamine hydrochloride solution was used as a substrate for the enzyme. The product of this\nreaction is red in colour.\nA colorimeter can be used to follow the progress of this enzyme-catalysed reaction.\nThe student set up:\n• 7 colorimeter tubes containing a solution of the substrate, dopamine hydrochloride\n• 6 colorimeter tubes containing diluted polyphenol oxidase extract, each in a different\nbuffered solution: pH 3.0, pH 4.0, pH 5.0, pH 6.0, pH 7.0, pH 8.0\n• 1 colorimeter tube containing water in pH 7.0 buffer.\nThe colorimeter tubes were equilibrated at a set temperature.\nA colorimeter was set up with a filter transmitting light at 470 nm.\nAt time 0 seconds, the substrate solution and the polyphenol oxidase solution at pH 3.0\nwere mixed in one colorimeter tube. The colorimeter tube was then immediately put into the\ncolorimeter.\nThe absorbance was measured at 15 second intervals over the first three minutes of the\nreaction.\nThe procedure was repeated for the remaining buffered solutions and the water at pH 7.0.\nIdentify the variable in this experiment. (i) independent\n...........................................................................................................................................\n..................................................................................................................................... [1]\nIdentify variable that the student has standardised. (ii) one\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student then plotted a graph of the results, showing the increase in absorbance. (c)\nAbsorban",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-52-q01-0a4e2ddd19.pdf"
    },
    {
      "id": "9700-2023-mj-52-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-52-q02-p01-1501cf93b2.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q02-p02-a5910991b5.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q02-p03-2e1052d291.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q02-p04-a5cd85409a.png",
        "9700-experimental-skills/assets/9700-2023-mj-52-q02-p05-ea748c130c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Air pollution due to motor vehicles is an environmental issue that needs constant monitoring. 2\nLichens consist of protoctists and fungi living in close association. The protoctists and fungi each\nbenefit from this association. Lichens are found in a large range of habitats, including on the\nsurface of tree trunks (tree bark), as shown in Fig. 2.1.\ntree bark\nwith lichen\ntree bark\nwithout lichen\nFig. 2.1\nLichens are sensitive to air pollution. The abundance of lichens decreases with increasing levels\nof air pollution and so lichens can be used to monitor levels of air pollution.\nSome scientists investigated the relationship between distance from a road and levels of air\npollution. The scientists investigated this relationship by measuring the abundance of lichens on\nthe tree trunks in a woodland beside a busy road.\nThe scientists used the grid shown in Fig. 2.2 to determine the abundance of lichens on the tree\ntrunks.\ntransparent plastic\nsheet with printed grid\n15 cm\ncircles for sampling\n15 cm\nFig. 2.2\nFor each tree sampled, the scientists:\n• placed the grid against the tree trunk\n• counted the number of circles that contained lichen\n• used the number of circles that contained lichen to calculate the percentage cover of lichen\non the tree\n• measured the distance from the tree to the road.\nState variables that the scientists should standardise in this investigation. (a) (i) two\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState risk the safety precaution that the scientists should take when measuring (ii) one and\nthe abundance of the lichens on the tree trunks.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe scientists recorded the presence of lichens by shading in t",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-52-q02-3eea535193.pdf"
    },
    {
      "id": "9700-2023-mj-53-q01",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p01-d10035c13d.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p02-ef19fcb5c7.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p03-dd93072a1c.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p04-eb7cea98ae.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p05-ca236fa2b3.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p06-174e0ddb55.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q01-p07-3c1b895b39.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Hydrogencarbonate indicator is a water-soluble solution that can act as a source of carbon dioxide 1\nfor aquatic photosynthetic organisms. The solution changes colour depending on the concentration\nof carbon dioxide in the solution. These colours are related to different pH values, as shown in\nTable 1.1.\nTable 1.1\ncolour of hydrogencarbonate pH concentration of carbon dioxide\nindicator solution in the solution\nyellow 7.6\nincreasing carbon dioxide\nyellow-orange 7.8\nconcentration\norange 8.0\norange-red 8.2\nred 8.4 atmospheric concentration\nred-magenta 8.6\ndecreasing carbon dioxide\nmagenta 8.8\nconcentration\nmagenta-purple 9.0\npurple 9.2\nChlorella vulgaris is a protoctist that is single-celled, aquatic and photosynthetic. It can be\nimmobilised in alginate beads.\nAlginate beads with immobilised C. vulgaris can be used to measure the rate of photosynthesis.\nA student noticed that a colour change occurred, from red to magenta, when the alginate (a)\nbeads with immobilised C. vulgaris were left in a container of hydrogencarbonate indicator\nsolution and exposed to light.\nExplain why this colour change occurred.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe student used the alginate beads with immobilised C. vulgaris in hydrogencarbonate (b)\nindicator solution to investigate the rate of photosynthesis in different light intensities.\nFig. 1.1 shows some of the apparatus and reagents the student used.\nwater\nsmall beaker\nlarge test-tube\nhydrogencarbonate\nindicator solution\nalginate beads containing\nC. vulgaris\nFig. 1.1\nIdentify the variable in this investigation. (i) independent\n..................................................................................................................................... [1]\nThe student was provided with a supply of alginate beads containing immobilised (ii)\nC. vulgaris.\nDescribe a method the student could use to collect data to determine the effect of light\nintensity on the rate of photosynthesis of C. vulgaris using hydrogencarbonate indicator\nand the experimental set-up in Fig. 1.1.\nYour method should be set out in a logical order and be detailed enough to let another\nperson follow.\n...........................................................................................................................................\n................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-53-q01-13c5aee3fc.pdf"
    },
    {
      "id": "9700-2023-mj-53-q02",
      "subject": "9700",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-may-june/9700_s23_qp_53.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2023-May-June/qp/9700_s23_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-mj-53-q02-p01-e1ef35b7d3.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q02-p02-7f2af15a6b.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q02-p03-7c445801c2.png",
        "9700-experimental-skills/assets/9700-2023-mj-53-q02-p04-859e3a6a20.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Lichens consist of protoctists and fungi living in close association. Both types of organism benefit 2\nfrom this association. Lichens are found in a large range of habitats, including on the trunks of\ntrees.\nFig. 2.1 shows an example of a lichen, the salted shield lichen, Parmelia saxatilis, on the trunk of\na tree.\nsalted shield lichen\ntree trunk with no lichen\nFig. 2.1\nThe red deer, Cervus elaphus, is found in many parts of the world including north-west Europe.\nRed deer are found in a large range of habitats, including mixed woodlands. Mixed woodlands\ncontain many different tree species.\nFig. 2.2 shows a male red deer.\nFig. 2.2\nRed deer have increased in numbers in the mountainous regions of Scotland in north-west Europe\nsince the 1960s.\nScientists think that grazing of plants by red deer has affected the abundance of certain lichens\nfound growing on the trunks of trees in mixed woodlands.\nTo investigate this, the scientists excluded red deer from certain mixed woodland areas\n(exclosures) in 1995, allowing the exclosures to remain ungrazed.\nIn 2013, the scientists compared the abundance of the salted shield lichen in the exclosures and\nin the grazed areas.\nThe scientists used the grid shown in Fig. 2.3 to determine the abundance of the salted shield\nlichen on the tree trunks.\ntransparent plastic\nsheet with printed grid\n15 cm\ncircles for sampling\n15 cm\nFig. 2.3\nFor each tree sampled:\n• The grid was placed against the tree trunk at three different heights: base, middle and upper.\n• The scientists counted the number of circles that contained the salted shield lichen.\n• The number of circles that contained the salted shield lichen was used to calculate the\npercentage cover of salted shield lichen on the tree.\nThis study was carried out over the course of the three months (June, July and August) in the\nsummer of 2013 in the exclosures and the grazed areas.\nSuggest variables that the scientists should have standardised in this investigation. (a) (i) two\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState risk the safety precaution that the scientists should take when measuring (ii) one and\nthe abundance of salted shield lichen on the tree trunks.\n...........................................................................................................................................\n...........................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-mj-53-q02-52810c60af.pdf"
    },
    {
      "id": "9700-2023-on-41-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-on-41-q01-p01-e62d650da2.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-on-41-q01-p02-f81d1beb18.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q01",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Chloroplasts carry out photosynthesis. 1\nFig. 1.1 shows some structural features of a chloroplast and some processes that occur within it.\nlight\nCO\n2\nstroma\nO H\n2\ncomponents\ncycle C\nmake up A\ngranum\nproduct B\nproduct chloroplast envelope D\nFig. 1.1\nIdentify the structures labelled in Fig. 1.1. (a) (i) A\n.......................................................... [1] A\nExplain how the structure appearance of the granum, and the components labelled (ii) and\nA, relate to their function.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nIdentify the metabolic pathway labelled cycle in Fig. 1.1. (b) (i) C\n.......................................................... [1] C\nExplain why pathway is described as a cycle. (ii) C\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nIdentify the products of photosynthesis labelled and in Fig. 1.1. (c) (i) B D\n.......................................................... B\n.......................................................... [2] D\nSuggest explain the importance of glucose and the product labelled in Fig. 1.1 to (ii) and B\necosystems.\n.............................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-on-41-q01-f6a90bf664.pdf"
    },
    {
      "id": "9700-2023-on-41-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-41-q02-p01-aa7279b8b3.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-41-q02-p02-b071a44e09.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-41-q02-p03-3691fc0416.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-41-q02-p04-966102730d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q02",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Biodiversity can be assessed at three different levels. One of these is the genetic variation within 2\neach species.\nOutline other levels at which biodiversity can be measured. (a) two\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nTo calculate the genetic variation that exists within a species, scientists:\n• obtain DNA sequences from many individuals of one species\n• count the number of nucleotides that differ when the sequences of two individuals are\ncompared\n• repeat this with different pairs of individuals.\nThis allows scientists to calculate the number of differences at every nucleotide position mean\nalong the sequence (mean number of nucleotide differences per site).\nExplain why scientists use databases and computers to calculate the mean number of (b)\nnucleotide differences per site.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nTable 2.1 shows the mean number of nucleotide differences per site of some species. (c)\nTable 2.1\nspecies mean number of nucleotide\ndifferences per site\nDrosophila melanogaster, fruit fly 0.0087\nAnopheles gambiae, mosquito vector of malaria 0.0301\nPlasmodium falciparum, malarial pathogen 0.0015\nZea mays, wild maize 0.0139\nState the name of the species that shows the most genetic variation. (i) genus\n..................................................................................................................................... [1]\nState how many kingdoms of organisms are represented in Table 2.1. (ii)\n..................................................................................................................................... [1]\nGenetic variation is considered important in the conservation of species. Low genetic variation (d)\nis assumed to decrease the chance of the long-term survival of a s",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-on-41-q02-10ba9b90d8.pdf"
    },
    {
      "id": "9700-2023-on-41-q03",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-on-41-q03-p01-754696c42f.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-41-q03-p02-0b715c6901.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-41-q03-p03-458c3f63e2.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-41-q03-p04-42b7b27e9c.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-41-q03-p05-55b7f0a3f5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q03",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "There are more than 600 plant species in the genus Ipomoea. Many species are grown for their 3\nattractive flowers, and some species are used as crop plants.\nFig. 3.1 shows Ipomoea purpurea, the common morning glory. (a)\nFig. 3.1\nThe gene that determines flower colour in I. purpurea has two alleles:\n• a dominant allele that results in purple flowers\n• a recessive allele that results in red flowers.\nA student recorded the flower colour of all the I. purpurea plants in a field.\nThe student recorded 660 plants with purple flowers and 440 plants with red flowers.\nAssuming the Hardy-Weinberg principle applies to this population, calculate the number of\nplants in the field that are heterozygous.\nUse the equations:\np + q = 1\n2 q2 + 2pq + = 1 p\nShow your working and give your answer to the nearest whole number.\nnumber of heterozygous plants ......................................................... [3]\nThe Japanese morning glory, I. nil, has over 20 different flower colour phenotypes, including (b)\nshades of blue, purple, red and pink.\nThe flower colour of I. nil is controlled by at least four genes. The flower colour can change\ngradually after the flowers open each morning and can change with fluctuations in the carbon\ndioxide concentration of the surrounding air.\nA student concluded that the flower colour phenotype in I. nil shows continuous variation.\nSuggest reasons why the student made this conclusion. two\n1 ................................................................................................................................................\n...................................................................................................................................................\n2 ................................................................................................................................................\n............................................................................................................................................. [2]\n) Scientists investigated the response of stomata to changing carbon dioxide (CO (c)\n2\nconcentrations in the beach morning glory, I. pes-caprae.\nThe scientists placed I. pes-caprae plants in chambers. They measured the width of\nopen stomata (stomatal apertures) after the plants had been exposed to different CO\n2\nconcentrations for 40 minutes. Light intensity and temperature were kept constant.\nconcentration and the mean width of stomatal apertures is The relationship between CO\n2\nshown in Fig. 3.2.\n2.0\n1.5\nmean width of\n1.0 stomatal aperture\nμm /\n0.5\n0.0\n300 400 500 600 700 800 900 1000\nmol–1 μmol CO concentration /\n2\nFig. 3.2\nmol–1. μmol In 2016, a study measured the atmospheric CO concentration as 400 (i)\n2\nIn the future, climate change may reduce water availability and increase atmospheric\nconcentrations in some habitats. CO\n2\nSuggest how the stomatal response shown in Fig. 3.2 would allow I. pes-caprae to\nsurvive the effects of climate change.\n.........................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-on-41-q03-cac9c5579a.pdf"
    },
    {
      "id": "9700-2023-on-41-q04",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-41-q04-p01-ad273ad2a5.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-on-41-q04-p02-4b2a3d0f68.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "The potato plant, Solanum tuberosum, is an important food crop. Crop yield is reduced if the 4\nleaves of the plant are eaten by the larvae (immature stages) of the Colorado beetle, Leptinotarsa\ndecemlineata.\nCrop scientists used recombinant DNA technology to create two genetically modified (GM)\nvarieties of potato plant. These plants produce proteins that are poisonous to insects.\n• GM potato variety contains two new genes, SN and Bt. A\n• GM potato variety contains two new genes, SN and OCII. B\nThe new varieties were tested by having a constant number of Colorado beetle larvae introduced\nto the plants at time 0 hours. The number of larvae that were alive after 24, 48 and 72 hours was\nrecorded. The percentage of the larvae that had died in each time interval was calculated. This\nwas repeated for potato plants that had not been genetically modified (non-GM).\nTable 4.1 shows the percentage of Colorado beetle larvae that had died on the GM potato plant\nvarieties and on non-GM potato plants.\nTable 4.1\npercentage of Colorado beetle larvae that had died\ntype of potato plant\n24 h 48 h 72 h\nGM potato variety 50 93 100 A\nGM potato variety 37 70 93 B\nnon-GM potato 0 0 0\nSuggest what is meant by recombinant DNA technology. (a) (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSuggest why the scientists created two different types of GM potato plant. (ii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState why the scientists also performed the test on non-GM potato plants. (iii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDiscuss how the results in Table 4.1 provide information that could help to solve the global (b)\ndemand for food.\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-41-q04-40cbf6fab0.pdf"
    },
    {
      "id": "9700-2023-on-41-q05",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-on-41-q05-p01-e7007638b5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q05",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The respiratory quotient (RQ) values for different respiratory substrates can be calculated. 5 (a)\nTable 5.1 shows:\n• the formulae of four respiratory substrates\n• the number of oxygen molecules (O ) needed to completely respire each substrate to\n2\ncarbon dioxide and water.\nTable 5.1\nrespiratory substrate formula number of oxygen molecules\nneeded for respiration\nH O C 4.5 beta-hydroxybutyric acid\n4 8 3\nH O C 6.0 glucose\n6 12 6\nH O C 3.0 malic acid\n4 6 5\nH O C 25.5 oleic acid\n18 34 2\nUsing Table 5.1, name the respiratory substrate with the highest RQ and the respiratory\nsubstrate with the lowest RQ.\nrespiratory substrate with the RQ ............................................................... highest\nrespiratory substrate with the RQ ............................................................... lowest\n[2]\nIn anaerobic conditions, the production of ATP in mammals and yeast involves glycolysis and (b)\nfermentation.\nDescribe the similarities differences between fermentation in mammals and in yeast. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\n[Total: 7]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-on-41-q05-ea3d01c45c.pdf"
    },
    {
      "id": "9700-2023-on-41-q06",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-on-41-q06-p01-066924cc44.png",
        "9700-topic-16-inheritance/assets/9700-2023-on-41-q06-p02-1d2db248e3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q06",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The tiger barb, Puntigrus tetrazona, is a South American fish that is popular worldwide as an 6\naquarium fish. Fig. 6.1 shows the appearance (phenotype) of a normal (wild-type) tiger barb.\nFig. 6.1\n• Tiger barbs that show a wild-type phenotype are gold with black stripes.\n• Tiger barbs that show an albino phenotype are gold with white stripes.\n• In 2012, a fish breeder discovered a tiger barb with a new, transparent, phenotype. This fish\nhad a transparent body and black stripes.\nThe fish breeder crossed the tiger barb showing the new transparent phenotype with a tiger barb\nshowing the albino phenotype.\nAll the F1 offspring were wild-type. These F1 offspring were crossed with each other.\nTable 6.1 shows the phenotypes obtained in the F2 generation and the number of fish showing\neach phenotype.\nTable 6.1\nF2 phenotype number of fish\nwild-type (gold with black stripes) 173\nalbino (gold with white stripes) 57\ntransparent with black stripes 58\ntransparent with white stripes 19\nThe fish breeder concluded that the new transparent phenotype had occurred because (a) (i)\nof a mutation.\nExplain how the results in Table 6.1 support this conclusion.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState the approximate whole-number ratio shown by the results in Table 6.1. (ii)\n..................................................................................................................................... [1]\nExplain what the results in Table 6.1 show about the genes and alleles that determine (iii)\nthe wild-type, albino and the two different transparent phenotypes in tiger barbs.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-on-41-q06-aaa3481d69.pdf"
    },
    {
      "id": "9700-2023-on-41-q07",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-41-q07-p01-ec5f750442.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-41-q07-p02-cec4abe8ff.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "When an impulse arrives at a neuromuscular junction, it stimulates a muscle fibre of striated 7\nmuscle to contract.\nOutline the in structure between a neuromuscular junction and a cholinergic (a) (i) similarities\nsynapse.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\n(P). The hydrolysis of ATP during muscle contraction releases inorganic phosphate (ii)\ni\n(Ca2+) can combine with P in the sarcoplasmic reticulum to form insoluble Calcium ions\ni\ncalcium phosphate. This may result in fewer power strokes occurring in sarcomeres.\nSuggest why calcium phosphate formation in the sarcoplasmic reticulum may result in\nfewer power strokes occurring in sarcomeres.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nAdrenaline is a hormone that can affect muscle contraction. Adrenaline binds to (b)\nG-protein-coupled receptors on T-tubule membranes.\nThe cell signalli",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-on-41-q07-f81ed86752.pdf"
    },
    {
      "id": "9700-2023-on-41-q08",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-41-q08-p01-30d0a929c9.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-on-41-q08-p02-924d69c965.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Environmental conditions such as light intensity affect plant physiology. 8\nUse the letter to identify a point on the sketch graph in Fig. 8.1 where light intensity is acting (a) X\nas a limiting factor on the rate of photosynthesis. [1]\nrate of\nphotosynthesis\narbitrary units /\nlight intensity / arbitrary units\nFig. 8.1\nAn experiment investigated how light intensity affected gene expression in kale, Brassica (b)\noleracea sabellica.\n• Two groups of kale plants were grown, with one group in high light intensity and one\ngroup in low light intensity. All other conditions were standardised.\n• After the same period of time, messenger RNA (mRNA) was extracted from each group\nof kale plants.\n• The mRNA was used to produce cDNA.\n• The cDNA was hybridised with probes for 89 621 kale genes on a microarray.\n• The microarrays showed which genes were switched on (expressed) in each set of\nconditions.\nThe results showed that expression of 18% of the genes was affected by light intensity.\n• 14% of the genes were switched on only in high light intensity.\n• 4% were switched on only in low light intensity.\nState the name of the enzyme that produces cDNA from an mRNA template. (i)\n..................................................................................................................................... [1]\nState the name of the type of proteins that control gene expression in plants. (ii)\n..................................................................................................................................... [1]\nSuggest why genes were switched on only in plants growing in light intensity (iii) more high\ncompared to fewer genes that were switched on only in plants growing in low light\nintensity.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\n[Total: 6]",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-41-q08-9ff770a128.pdf"
    },
    {
      "id": "9700-2023-on-41-q09",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        24,
        25,
        26
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-41-q09-p01-3a42fbc20c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-41-q09-p02-a3c3ce2b82.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-41-q09-p03-ea3b69e5d5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q09",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Green lacewings are a family of insects with more than 1300 species. 9\nThe common green lacewing, Chrysoperla carnea, is shown in Fig. 9.1.\nFig. 9.1\nGreen lacewings have sense organs, known as tympanal organs, that detect sound. (a)\nThe tympanal organ of green lacewings has evolved to detect the high frequency sounds that\nbats make when they are hunting. Bats eat green lacewings.\nWhen a green lacewing senses the presence of a bat, it moves away or closes its wings in\nflight to escape.\nOutline how the tympanal organ of green lacewings could have evolved by natural (i)\nselection.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nWhen high frequency sound is detected, the receptor cells in the tympanal organ (ii)\nstimulate the transmission of impulses in sensory neurones.\nDescribe the sequence of events that results in an action potential in a sensory neurone.\nThe first event in the sequence has been given for you.\nCalcium ions enter the cytoplasm of the receptor cell .................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..............................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-on-41-q09-ec67a873cd.pdf"
    },
    {
      "id": "9700-2023-on-41-q10",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_41.pdf?download=true",
      "source_pages": [
        27
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-41-q10-p01-fed266eddd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-41-q10",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Medicine is defined as the diagnosis, treatment and prevention of disease. 10\nOutline the different ways in which genetic technology can be applied to medicine, with reference\nto named diseases.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n...............",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-41-q10-9093dd4822.pdf"
    },
    {
      "id": "9700-2023-on-42-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-on-42-q01-p01-e510e0a621.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2023-on-42-q01-p02-3513e53211.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Some organisms carry out respiration in anaerobic conditions when oxygen is not available or 1 (a)\nwhen there is a low concentration of oxygen. In yeast and some plants, this is called ethanol\nfermentation. In mammals, it is called lactate fermentation.\nFig. 1.1 outlines ethanol fermentation and lactate fermentation.\nglucose\npyruvate\nethanol lactate\nfermentation fermentation\nA\nB\nC\nD\nB\nC\nethanol lactate\nFig. 1.1\nIdentify substances A–D.\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n............................................................................................................................................... D\n[3]\nExplain how processes such as ethanol fermentation and lactate fermentation allow cells to (b)\ncontinue to function in the absence of oxygen.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nThe cereal crop rice, Oryza sativa, grows in fields that are flooded with water. The roots of the (c)\nrice plants are submerged in water that contains very little oxygen.\nDescribe explain how rice plants are adapted to grow in flooded fields. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.........................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-on-42-q01-c9298802bf.pdf"
    },
    {
      "id": "9700-2023-on-42-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-on-42-q02-p01-dc3ccf1bcb.png",
        "9700-topic-16-inheritance/assets/9700-2023-on-42-q02-p02-17580de8ea.png",
        "9700-topic-16-inheritance/assets/9700-2023-on-42-q02-p03-96e09e3c4c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Yeasts are unicellular organisms from the kingdom Fungi. Saccharomyces cerevisiae is one 2\nspecies of yeast that can carry out either asexual reproduction by mitosis or sexual reproduction\nby meiosis.\nBudding in S. cerevisiae is a process where a small daughter cell forms as a bud on the parent\ncell. The bud contains a copy of the parent cell nucleus and it eventually separates from the\nparent cell to form a new cell.\nS. cerevisiae can exist in two forms: haploid cells or diploid cells.\n• α. Haploid cells can be one of two different mating types: and a\n• Haploid cells can only mate with other haploid cells of the opposite mating type.\nFig. 2.1 shows the life cycle of S. cerevisiae with its asexual and sexual reproductive stages.\nα mating type a a Key: a\nα\na budding budding\nα α mating type\nα\na\n1 1\nα a\nα\na\nmating\nα a\n2 formation of\nzygote\nα a\naα budding\naα\n3\naα\naα\nlack of\nnutrients\naα\nan ascus -\na sac\n4 containing\nfour spores\n5 5\nα a\nα a\ngermination when conditions\nhave improved\nFig. 2.1\nWith reference to Fig. 2.1, state the numbers of the stages 1–5 that: (a)\ninvolve mitosis ....................\ninvolve meiosis ....................\nproduces new genetic variation ....................\nshows only haploid cells ....................\nshows only diploid cells ....................\n[5]\nWhen there is a lack of nutrients, cells made in stage will carry out stage to make spores, (b) 3 4\nwhich germinate only when conditions improve.\nSuggest explain how the type of reproduction that makes spores during stage is and 4\nadvantageous for S. cerevisiae in a changing environment.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nHaploid and diploid cells of S. cerevisiae can carry out reproduction. (c) asexual\nSuggest why a new harmful recessive mutation may have a damaging effect on: not\n• an asexually reproducing population of haploid cells of S. cerevisi",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-on-42-q02-3170604b7c.pdf"
    },
    {
      "id": "9700-2023-on-42-q03",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        7
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-42-q03-p01-47d1893fa5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q03",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Research suggests that selective breeding began about 10 000 years ago. Selective breeding has 3\nbeen important in satisfying the global demand for food.\nThe introduction of disease resistance to varieties of wheat is an example of selective (a)\nbreeding.\nOutline the steps involved in the introduction of disease resistance to varieties of wheat.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nSuggest possible disadvantages of selective breeding. (b) two\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-on-42-q03-849a097e43.pdf"
    },
    {
      "id": "9700-2023-on-42-q04",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-42-q04-p01-72508412bf.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-on-42-q04-p02-99fc1e1bf5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "A marker gene can be used when a gene of interest is introduced into a plant by genetic 4 (a)\nengineering.\nDescribe explain how the use of a marker gene coding for a fluorescent product can and\nshow that the introduced gene of interest is being expressed in plants.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nGive advantages of genetically engineering herbicide resistance in crop plants such as (b) two\nsoybean.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nDNA microarray analysis is a technique used in genetic technology that involves fluorescence. (c)\nA DNA microarray has single-stranded probes attached to its surface. These probes hybridise\nto the fluorescently tagged single-stranded DNA that is added.\nExplain why D",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-42-q04-38a90d5b6e.pdf"
    },
    {
      "id": "9700-2023-on-42-q05",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-on-42-q05-p01-d0cb8fb47f.png",
        "9700-topic-16-inheritance/assets/9700-2023-on-42-q05-p02-4ca6250a7f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The fruit fly, Drosophila melanogaster, has autosomal genes for body colour and wing shape. 5\nGene is involved in the production of body colour: B/b\n• = dominant allele for brown body colour B\n• = recessive allele for black body colour. b\nGene is involved in wing shape: D/d\n• = dominant allele for straight wing D\n• = recessive allele for curved wing. d\nA dihybrid test cross was carried out between flies heterozygous for body colour and for wing\nshape and flies homozygous recessive for body colour and for wing shape.\nTable 5.1 shows the number of offspring of each phenotype obtained in the test cross. (a)\nTable 5.1\nobserved expected\nphenotype\nnumber number\nbrown body colour, straight wings 2843\nbrown body colour, curved wings 855\nblack body colour, straight wings 842\nblack body colour, curved wings 2768\nUse Table 5.1 to calculate the expected number of each phenotype if the two genes are on\nautosomes. Write your answers in the table. [1] different\n(χ2) test was carried out to compare the observed results with the results that A chi-squared (b)\nwould be expected from a dihybrid cross involving genes on different autosomes.\nχ2 = 2097.836. The value of\nχ2 distribution. Table 5.2 shows the critical values for the\nTable 5.2\np value\ndegrees of\nfreedom\n0.05 0.01 0.001\n1 3.841 6.635 10.828\n2 5.991 9.210 13.816\n3 7.815 11.345 16.266\n4 9.488 13.277 18.467\nχ2 Explain how the value of and Table 5.2 can be used to assess the significance of the\ndifference between the observed results and the expected numbers in Table 5.1.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n......................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-on-42-q05-13410c9aa9.pdf"
    },
    {
      "id": "9700-2023-on-42-q06",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-on-42-q06-p01-19a7034cc2.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-42-q06-p02-8375d76da1.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-42-q06-p03-d95f0321ab.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "When blood glucose concentration decreases, glucagon is released by the pancreas 6 (a) (i)\ninto the blood and is transported to the cells.\nFig. 6.1 outlines the effect of glucagon on liver cells.\ntissue fluid\nglucagon\nP Q\ncell surface\nmembrane\nactivated ATP\nR S\ncytoplasm\nenzyme\ncascade\ncellular\nresponse\nFig. 6.1\nIdentify P, Q, and shown in Fig. 6.1. R S\n........................................................................................................................................ P\n....................................................................................................................................... Q\n........................................................................................................................................ R\n........................................................................................................................................ S\n[4]\nTable 6.1 shows four processes carried out by liver cells that are affected by a decrease (ii)\nin blood glucose concentration. Some of these are cellular responses from cell signalling\nby glucagon.\nComplete Table 6.1 by stating whether the rate of each process or increases decreases\nwhen blood glucose concentration decreases.\nTable 6.1\neffect on rate of process when blood glucose\nprocess\nconcentration decreases\nglycogenolysis\n(glycogen breakdown) .........................................................................................\nglycogenesis\n(glycogen formation) .........................................................................................\nglycolysis\n.........................................................................................\nfatty acid synthesis\n.........................................................................................\n[3]\nType 2 diabetes is a condition affecting the homeostatic control of blood glucose concentration. (b)\nIn type 2 diabetes, target cells have a much lower response to insulin and some glucose is\nexcreted in the urine.\nTest strips can be used to measure the concentration of glucose in a sample of urine, but\nmany people with type 2 diabetes measure blood glucose concentration using a biosensor.\nExplain the principles of operation of a test strip for glucose suggest advantages for a and\nperson with type 2 diabetes of using a biosensor instead of a test strip.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-on-42-q06-faf4e3982d.pdf"
    },
    {
      "id": "9700-2023-on-42-q07",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-42-q07-p01-f0ddf2bc0f.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-42-q07-p02-0d345658c1.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-42-q07-p03-badbd22006.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "There are many mitochondria located in the presynaptic knobs of neuromuscular junctions 7 (a)\nand in the sarcomeres of muscle fibres of striated muscle.\nExplain the need for many mitochondria in the presynaptic knobs and in the sarcomeres.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nFig. 7.1 is a diagram of synapses between a postsynaptic neurone and two presynaptic (b)\nneurones, and Y. X\nY X\npostsynaptic\nneurone\nFig. 7.1\n• Neurone releases the neurotransmitter acetylcholine. X\n• Neurone releases the neurotransmitter glutamate. Y\n• Both neurotransmitters bind to channel proteins in the membrane of the postsynaptic\nneurone.\n• Acetylcholine binding results in an influx (entry) of sodium ions.\n• Glutamate binding results in an influx of chloride ions.\nA student made three statements:\n1. When only neurone is stimulated, an action potential occur in the postsynaptic X will\nneurone.\n2. When only neurone is stimulated, an action potential occur in the postsynaptic Y will\nneurone.\n3. W hen neurone and neurone are stimulated at the same time, an action potential X Y will\noccur in the postsynaptic neurone. not\nExplain whether or not you agree with these statements.\n...................................................................................................................................................\n...................................................................................................................................................\n................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-on-42-q07-fcee21b42e.pdf"
    },
    {
      "id": "9700-2023-on-42-q08",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-on-42-q08-p01-2db23ffee6.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-on-42-q08-p02-fa4d6a96e1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q08",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Non-cyclic photophosphorylation occurs in the light-dependent stage of photosynthesis. 8 (a)\nOutline the main features of non-cyclic photophosphorylation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nSome Calvin cycle intermediates are used to produce other molecules that are part of the (b) not\nCalvin cycle.\nName of these intermediates. two\nFor each intermediate, state example of the type of molecule that is produced. one\nintermediate ..............................................\nmolecule produced ..............................................\nintermediate ..............................................\nmolecule produced ..............................................\n[2]\nFig. 8.1 shows how the rate of uptake of carbon dioxide by a plant varies with temperature, at (c)\nan optimum intensity of light on a clear day.\n4\nrate of\nuptake of\ncarbon dioxide\n3\narbitrary units (au) /\n2\n1\n0\n0 10 20 30 40\n°C temperature /\nFig. 8.1\nCalculate the mean increase in rate of uptake of carbon dioxide as the temperature (i)\n°C °C. increases from 5 to 20\nShow your working.\nGive your answer to decimal places. two\n–1 °C [2] .............................................. au\n°C. Suggest why the rate of carbon dioxide uptake levels off and then decreases after 20 (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n.........................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-on-42-q08-99435b8992.pdf"
    },
    {
      "id": "9700-2023-on-42-q09",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-on-42-q09-p01-291093e0a0.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-42-q09-p02-5c7191497b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Antidiuretic hormone (ADH) is involved in the maintenance of a constant blood water potential. 9 (a)\ndm3 of water and the concentration of ADH in their blood was measured A student drank 0.5\nevery 30 minutes for 3 hours.\nFig. 9.1 shows the results of this investigation.\n10\nblood ADH\nconcentration 8\narbitrary /\nunits\n6\n4\n2\n0\n0 1 2 3\n/ hours time\nwater drunk\nFig. 9.1\nExplain the results shown in Fig. 9.1.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nOutline the differences between the endocrine system and the nervous system. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..................................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-on-42-q09-fde7cca666.pdf"
    },
    {
      "id": "9700-2023-on-42-q10",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-42-q10-p01-043c51a850.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-42-q10-p02-5d518f71c3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-42-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "The passage outlines one method of estimating the size of an animal population. 10 (a)\nComplete the passage by using the most appropriate scientific terms.\nThe size of a population of animals can be estimated using the mark-release-recapture\nmethod. A sample of animals is captured using a humane (harmless) trap and is counted.\nThe animals are marked using a method that does not harm them. This can be done using\na small tag or ................................................. . The marked animals are released into the\narea where they were captured. Time is allowed for the marked animals to mix into the\npopulation. This period of time must be short so that emigration, immigration, migration or\n........................................ do not occur.\nA second sample of animals is then captured and the number of marked and unmarked\nanimals is counted. The population size can be estimated using the ........................................\nIndex. For reliability, the method should be ........................................ .\n[4]\nHimalayan balsam, Impatiens glandulifera, is an annual plant native to Pakistan, India and (b)\nNepal. It was first introduced into the United Kingdom (UK) in 1839 because of its attractive\nflowers. Individual plants can produce thousands of mature seeds. These can be dispersed\nfor long distances when the capsules that contain the seeds burst.\nFig. 10.1 shows Himalayan balsam growing in a woodland.\nFig. 10.1\nHimalayan balsam has now become established in the UK. It is listed as an invasive alien\nspecies and attempts are being made to eradicate (remove) the plant.\nNot all alien plant species are considered to be invasive.\nSuggest why Himalayan balsam has been listed as an invasive alien species in the UK.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-on-42-q10-59af5909db.pdf"
    },
    {
      "id": "9700-2023-on-43-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2023-on-43-q01-p01-e62d650da2.png",
        "9700-topic-13-photosynthesis/assets/9700-2023-on-43-q01-p02-f81d1beb18.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q01",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Chloroplasts carry out photosynthesis. 1\nFig. 1.1 shows some structural features of a chloroplast and some processes that occur within it.\nlight\nCO\n2\nstroma\nO H\n2\ncomponents\ncycle C\nmake up A\ngranum\nproduct B\nproduct chloroplast envelope D\nFig. 1.1\nIdentify the structures labelled in Fig. 1.1. (a) (i) A\n.......................................................... [1] A\nExplain how the structure appearance of the granum, and the components labelled (ii) and\nA, relate to their function.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nIdentify the metabolic pathway labelled cycle in Fig. 1.1. (b) (i) C\n.......................................................... [1] C\nExplain why pathway is described as a cycle. (ii) C\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nIdentify the products of photosynthesis labelled and in Fig. 1.1. (c) (i) B D\n.......................................................... B\n.......................................................... [2] D\nSuggest explain the importance of glucose and the product labelled in Fig. 1.1 to (ii) and B\necosystems.\n.............................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2023-on-43-q01-f1431d5406.pdf"
    },
    {
      "id": "9700-2023-on-43-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-43-q02-p01-aa7279b8b3.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-43-q02-p02-b071a44e09.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-43-q02-p03-3691fc0416.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2023-on-43-q02-p04-966102730d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q02",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Biodiversity can be assessed at three different levels. One of these is the genetic variation within 2\neach species.\nOutline other levels at which biodiversity can be measured. (a) two\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nTo calculate the genetic variation that exists within a species, scientists:\n• obtain DNA sequences from many individuals of one species\n• count the number of nucleotides that differ when the sequences of two individuals are\ncompared\n• repeat this with different pairs of individuals.\nThis allows scientists to calculate the number of differences at every nucleotide position mean\nalong the sequence (mean number of nucleotide differences per site).\nExplain why scientists use databases and computers to calculate the mean number of (b)\nnucleotide differences per site.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nTable 2.1 shows the mean number of nucleotide differences per site of some species. (c)\nTable 2.1\nspecies mean number of nucleotide\ndifferences per site\nDrosophila melanogaster, fruit fly 0.0087\nAnopheles gambiae, mosquito vector of malaria 0.0301\nPlasmodium falciparum, malarial pathogen 0.0015\nZea mays, wild maize 0.0139\nState the name of the species that shows the most genetic variation. (i) genus\n..................................................................................................................................... [1]\nState how many kingdoms of organisms are represented in Table 2.1. (ii)\n..................................................................................................................................... [1]\nGenetic variation is considered important in the conservation of species. Low genetic variation (d)\nis assumed to decrease the chance of the long-term survival of a s",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2023-on-43-q02-85af693002.pdf"
    },
    {
      "id": "9700-2023-on-43-q03",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2023-on-43-q03-p01-754696c42f.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-43-q03-p02-0b715c6901.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-43-q03-p03-458c3f63e2.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-43-q03-p04-42b7b27e9c.png",
        "9700-topic-14-homeostasis/assets/9700-2023-on-43-q03-p05-55b7f0a3f5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q03",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "There are more than 600 plant species in the genus Ipomoea. Many species are grown for their 3\nattractive flowers, and some species are used as crop plants.\nFig. 3.1 shows Ipomoea purpurea, the common morning glory. (a)\nFig. 3.1\nThe gene that determines flower colour in I. purpurea has two alleles:\n• a dominant allele that results in purple flowers\n• a recessive allele that results in red flowers.\nA student recorded the flower colour of all the I. purpurea plants in a field.\nThe student recorded 660 plants with purple flowers and 440 plants with red flowers.\nAssuming the Hardy-Weinberg principle applies to this population, calculate the number of\nplants in the field that are heterozygous.\nUse the equations:\np + q = 1\n2 q2 + 2pq + = 1 p\nShow your working and give your answer to the nearest whole number.\nnumber of heterozygous plants ......................................................... [3]\nThe Japanese morning glory, I. nil, has over 20 different flower colour phenotypes, including (b)\nshades of blue, purple, red and pink.\nThe flower colour of I. nil is controlled by at least four genes. The flower colour can change\ngradually after the flowers open each morning and can change with fluctuations in the carbon\ndioxide concentration of the surrounding air.\nA student concluded that the flower colour phenotype in I. nil shows continuous variation.\nSuggest reasons why the student made this conclusion. two\n1 ................................................................................................................................................\n...................................................................................................................................................\n2 ................................................................................................................................................\n............................................................................................................................................. [2]\n) Scientists investigated the response of stomata to changing carbon dioxide (CO (c)\n2\nconcentrations in the beach morning glory, I. pes-caprae.\nThe scientists placed I. pes-caprae plants in chambers. They measured the width of\nopen stomata (stomatal apertures) after the plants had been exposed to different CO\n2\nconcentrations for 40 minutes. Light intensity and temperature were kept constant.\nconcentration and the mean width of stomatal apertures is The relationship between CO\n2\nshown in Fig. 3.2.\n2.0\n1.5\nmean width of\n1.0 stomatal aperture\nμm /\n0.5\n0.0\n300 400 500 600 700 800 900 1000\nmol–1 μmol CO concentration /\n2\nFig. 3.2\nmol–1. μmol In 2016, a study measured the atmospheric CO concentration as 400 (i)\n2\nIn the future, climate change may reduce water availability and increase atmospheric\nconcentrations in some habitats. CO\n2\nSuggest how the stomatal response shown in Fig. 3.2 would allow I. pes-caprae to\nsurvive the effects of climate change.\n.........................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2023-on-43-q03-059fb15e65.pdf"
    },
    {
      "id": "9700-2023-on-43-q04",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-43-q04-p01-ad273ad2a5.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-on-43-q04-p02-4b2a3d0f68.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "The potato plant, Solanum tuberosum, is an important food crop. Crop yield is reduced if the 4\nleaves of the plant are eaten by the larvae (immature stages) of the Colorado beetle, Leptinotarsa\ndecemlineata.\nCrop scientists used recombinant DNA technology to create two genetically modified (GM)\nvarieties of potato plant. These plants produce proteins that are poisonous to insects.\n• GM potato variety contains two new genes, SN and Bt. A\n• GM potato variety contains two new genes, SN and OCII. B\nThe new varieties were tested by having a constant number of Colorado beetle larvae introduced\nto the plants at time 0 hours. The number of larvae that were alive after 24, 48 and 72 hours was\nrecorded. The percentage of the larvae that had died in each time interval was calculated. This\nwas repeated for potato plants that had not been genetically modified (non-GM).\nTable 4.1 shows the percentage of Colorado beetle larvae that had died on the GM potato plant\nvarieties and on non-GM potato plants.\nTable 4.1\npercentage of Colorado beetle larvae that had died\ntype of potato plant\n24 h 48 h 72 h\nGM potato variety 50 93 100 A\nGM potato variety 37 70 93 B\nnon-GM potato 0 0 0\nSuggest what is meant by recombinant DNA technology. (a) (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSuggest why the scientists created two different types of GM potato plant. (ii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState why the scientists also performed the test on non-GM potato plants. (iii)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDiscuss how the results in Table 4.1 provide information that could help to solve the global (b)\ndemand for food.\n...................................................................................................................................................\n...................................................................................................................................................\n.............................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-43-q04-9961c79e4d.pdf"
    },
    {
      "id": "9700-2023-on-43-q05",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2023-on-43-q05-p01-e7007638b5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q05",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The respiratory quotient (RQ) values for different respiratory substrates can be calculated. 5 (a)\nTable 5.1 shows:\n• the formulae of four respiratory substrates\n• the number of oxygen molecules (O ) needed to completely respire each substrate to\n2\ncarbon dioxide and water.\nTable 5.1\nrespiratory substrate formula number of oxygen molecules\nneeded for respiration\nH O C 4.5 beta-hydroxybutyric acid\n4 8 3\nH O C 6.0 glucose\n6 12 6\nH O C 3.0 malic acid\n4 6 5\nH O C 25.5 oleic acid\n18 34 2\nUsing Table 5.1, name the respiratory substrate with the highest RQ and the respiratory\nsubstrate with the lowest RQ.\nrespiratory substrate with the RQ ............................................................... highest\nrespiratory substrate with the RQ ............................................................... lowest\n[2]\nIn anaerobic conditions, the production of ATP in mammals and yeast involves glycolysis and (b)\nfermentation.\nDescribe the similarities differences between fermentation in mammals and in yeast. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\n[Total: 7]",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2023-on-43-q05-b28314fd60.pdf"
    },
    {
      "id": "9700-2023-on-43-q06",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2023-on-43-q06-p01-066924cc44.png",
        "9700-topic-16-inheritance/assets/9700-2023-on-43-q06-p02-1d2db248e3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q06",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The tiger barb, Puntigrus tetrazona, is a South American fish that is popular worldwide as an 6\naquarium fish. Fig. 6.1 shows the appearance (phenotype) of a normal (wild-type) tiger barb.\nFig. 6.1\n• Tiger barbs that show a wild-type phenotype are gold with black stripes.\n• Tiger barbs that show an albino phenotype are gold with white stripes.\n• In 2012, a fish breeder discovered a tiger barb with a new, transparent, phenotype. This fish\nhad a transparent body and black stripes.\nThe fish breeder crossed the tiger barb showing the new transparent phenotype with a tiger barb\nshowing the albino phenotype.\nAll the F1 offspring were wild-type. These F1 offspring were crossed with each other.\nTable 6.1 shows the phenotypes obtained in the F2 generation and the number of fish showing\neach phenotype.\nTable 6.1\nF2 phenotype number of fish\nwild-type (gold with black stripes) 173\nalbino (gold with white stripes) 57\ntransparent with black stripes 58\ntransparent with white stripes 19\nThe fish breeder concluded that the new transparent phenotype had occurred because (a) (i)\nof a mutation.\nExplain how the results in Table 6.1 support this conclusion.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nState the approximate whole-number ratio shown by the results in Table 6.1. (ii)\n..................................................................................................................................... [1]\nExplain what the results in Table 6.1 show about the genes and alleles that determine (iii)\nthe wild-type, albino and the two different transparent phenotypes in tiger barbs.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2023-on-43-q06-520c34469d.pdf"
    },
    {
      "id": "9700-2023-on-43-q07",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-43-q07-p01-ec5f750442.png",
        "9700-topic-15-control-and-coordination/assets/9700-2023-on-43-q07-p02-cec4abe8ff.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "When an impulse arrives at a neuromuscular junction, it stimulates a muscle fibre of striated 7\nmuscle to contract.\nOutline the in structure between a neuromuscular junction and a cholinergic (a) (i) similarities\nsynapse.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\n(P). The hydrolysis of ATP during muscle contraction releases inorganic phosphate (ii)\ni\n(Ca2+) can combine with P in the sarcoplasmic reticulum to form insoluble Calcium ions\ni\ncalcium phosphate. This may result in fewer power strokes occurring in sarcomeres.\nSuggest why calcium phosphate formation in the sarcoplasmic reticulum may result in\nfewer power strokes occurring in sarcomeres.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nAdrenaline is a hormone that can affect muscle contraction. Adrenaline binds to (b)\nG-protein-coupled receptors on T-tubule membranes.\nThe cell signalli",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2023-on-43-q07-e1ee189fa3.pdf"
    },
    {
      "id": "9700-2023-on-43-q08",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-43-q08-p01-30d0a929c9.png",
        "9700-topic-19-genetic-technology/assets/9700-2023-on-43-q08-p02-924d69c965.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Environmental conditions such as light intensity affect plant physiology. 8\nUse the letter to identify a point on the sketch graph in Fig. 8.1 where light intensity is acting (a) X\nas a limiting factor on the rate of photosynthesis. [1]\nrate of\nphotosynthesis\narbitrary units /\nlight intensity / arbitrary units\nFig. 8.1\nAn experiment investigated how light intensity affected gene expression in kale, Brassica (b)\noleracea sabellica.\n• Two groups of kale plants were grown, with one group in high light intensity and one\ngroup in low light intensity. All other conditions were standardised.\n• After the same period of time, messenger RNA (mRNA) was extracted from each group\nof kale plants.\n• The mRNA was used to produce cDNA.\n• The cDNA was hybridised with probes for 89 621 kale genes on a microarray.\n• The microarrays showed which genes were switched on (expressed) in each set of\nconditions.\nThe results showed that expression of 18% of the genes was affected by light intensity.\n• 14% of the genes were switched on only in high light intensity.\n• 4% were switched on only in low light intensity.\nState the name of the enzyme that produces cDNA from an mRNA template. (i)\n..................................................................................................................................... [1]\nState the name of the type of proteins that control gene expression in plants. (ii)\n..................................................................................................................................... [1]\nSuggest why genes were switched on only in plants growing in light intensity (iii) more high\ncompared to fewer genes that were switched on only in plants growing in low light\nintensity.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\n[Total: 6]",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-43-q08-9dc09d215a.pdf"
    },
    {
      "id": "9700-2023-on-43-q09",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        24,
        25,
        26
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-43-q09-p01-3a42fbc20c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-43-q09-p02-a3c3ce2b82.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2023-on-43-q09-p03-ea3b69e5d5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q09",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Green lacewings are a family of insects with more than 1300 species. 9\nThe common green lacewing, Chrysoperla carnea, is shown in Fig. 9.1.\nFig. 9.1\nGreen lacewings have sense organs, known as tympanal organs, that detect sound. (a)\nThe tympanal organ of green lacewings has evolved to detect the high frequency sounds that\nbats make when they are hunting. Bats eat green lacewings.\nWhen a green lacewing senses the presence of a bat, it moves away or closes its wings in\nflight to escape.\nOutline how the tympanal organ of green lacewings could have evolved by natural (i)\nselection.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nWhen high frequency sound is detected, the receptor cells in the tympanal organ (ii)\nstimulate the transmission of impulses in sensory neurones.\nDescribe the sequence of events that results in an action potential in a sensory neurone.\nThe first event in the sequence has been given for you.\nCalcium ions enter the cytoplasm of the receptor cell .................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..............................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2023-on-43-q09-bd1b7e1784.pdf"
    },
    {
      "id": "9700-2023-on-43-q10",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_43.pdf?download=true",
      "source_pages": [
        27
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2023-on-43-q10-p01-fed266eddd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-43-q10",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Medicine is defined as the diagnosis, treatment and prevention of disease. 10\nOutline the different ways in which genetic technology can be applied to medicine, with reference\nto named diseases.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n...............",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2023-on-43-q10-9429acb79e.pdf"
    },
    {
      "id": "9700-2023-on-51-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p01-91bf518020.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p02-249c8f1796.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p03-3613af51b9.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p04-d0c1529823.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p05-d9ea4cc40b.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p06-b2fa98630e.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q01-p07-bfa8e0676c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Fruits in the diet can be a source of vitamins, such as vitamin C and vitamin A, which are important 1\nin maintaining good health.\nVitamin A is a group of molecules that includes retinol and retinal.\nVitamin C is an organic acid known as ascorbic acid.\nRipening occurs in fruit when molecular changes cause an increase in sugar content, colour\nchanges and softening of the fruit, over time.\nAscorbic acid is found in plant cells. The fruits of many plants have high concentrations of ascorbic\nacid.\nA student planned to investigate how ascorbic acid concentration changes with ripeness in the\nfruit of the apple tree, Malus domestica.\nThe fruit of M. domestica (apples) are shown in Fig. 1.1.\nFig. 1.1\nThe student removed an apple that was 5 cm in diameter from a tree. The student measured (a)\nthe sugar content of the apple, which was low. This indicated that the apple was at the early\nstages of ripening (unripe).\nFor the investigation, the student decided to select apples from trees when each apple was\n5 cm in diameter.\nThe student assumed that the ripeness of each apple would increase each day after it had\nbeen picked from a tree.\nSuggest why the diameter of the apples selected may give accurate estimates of (i) not\nripeness.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe an improvement to the method for selecting the apples that would give more (ii)\naccurate estimates of ripeness.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student researched how ascorbic acid concentration changes with ripeness in the fruit of (b)\nother species.\nBased on this research, the student predicted that ascorbic acid concentration in apples will:\n• be lowest in unripe apples when they are first picked from the tree\n• increase over several days and peak in partially ripe apples\n• decrease slightly in later days as the apples become fully ripe.\nSketch a line graph below, including labelled axes, to show the predicted results.\n[2]\nThe student used the redox indicator DCPIP to investigate how ascorbic acid concentration (c)\nchanges as the apples ripen.\nThe student tested the ascorbic acid concentration of the apples at intervals from the time\nthey were taken from the tree until the apples were fully ripe.\nFor each test, the student prepa",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-51-q01-ff033852ae.pdf"
    },
    {
      "id": "9700-2023-on-51-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_51.pdf?download=true",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-51-q02-p01-546386e38f.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q02-p02-0200d73d6a.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q02-p03-9c7ff4b9b1.png",
        "9700-experimental-skills/assets/9700-2023-on-51-q02-p04-5127844f96.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Trypsin is a protease enzyme. A student compared the activity of trypsin from two different species: 2\nthe Atlantic salmon, Salmo salar, and the domestic pig, Sus scrofa domesticus.\nFig. 2.1 shows an Atlantic salmon and Fig. 2.2 shows domestic pigs.\nFig. 2.1\nFig. 2.2\nAtlantic salmon are fish and are ectothermic, which means that the internal body temperature\nfluctuates with the surrounding environmental temperature.\nThe water temperature in the habitat of Atlantic salmon can decrease to – 0.5 °C.\nDomestic pigs are mammals and maintain an internal body temperature of approximately 38 °C.\nThe student was provided with trypsin from Atlantic salmon, trypsin from domestic pigs and cubes\nof gelatine. Gelatine is made from the protein collagen.\nThe student measured the time taken for the trypsin from each animal species to break down the\ngelatine cubes at 20 °C.\nThe student:\ncm3 • used 10 of 5% trypsin solution from each animal species\n• placed each trypsin solution in separate test-tubes, maintained at 20 °C in a water-bath\n• placed a gelatine cube in each test-tube of trypsin solution\n• recorded the time taken to break down the gelatine cube in each test-tube.\nThe student repeated this 12 times for each type of trypsin.\nIdentify the independent variable the dependent variable in this experiment. (a) (i) and\nindependent variable .........................................................................................................\ndependent variable ............................................................................................................\n[2]\nThe student standardised the temperature of the water-bath, the volume of the trypsin (ii)\nsolution and the concentration of the trypsin solution.\nState other variable that the student should standardise in this experiment. one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe results of the experiment are shown in Table 2.1. (b)\nTable 2.1\nreplicate time taken to break down gelatine cube / seconds\nAtlantic salmon domestic pigs\n1 125 130\n2 121 127\n3 128 134\n4 119 128\n5 132 135\n6 125 130\n7 131 132\n8 127 128\n9 126 128\n10 131 137\n11 131 133\n12 129 131\nmean 127.1\n(s) standard deviation 4.1\nUse the data in Table 2.1 to calculate the mean time taken for trypsin from domestic pigs (i)\nto break down the gelatine cube.\nmean = ........................................... seconds [1]\nThe formula for calculating standard deviation is: (ii)\nKey to symbols ˗ )2 /(x - x\n= s\nx = observation - n 1\n˗ x = mean\nn = sample size (number of observations)\n˗ )2 - /(x x The student calculated as for the domestic pigs. 110.92\n˗ )2 - /(x x and the data in Table 2.1 to complete the calculation Use the calculated value of\nof the standard deviation for domestic pigs.\ns = ............................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-51-q02-82ac043922.pdf"
    },
    {
      "id": "9700-2023-on-52-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 16,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-52-q01-p01-c9c25b18e7.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q01-p02-bb08abd9b9.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q01-p03-01fab8e18d.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q01-p04-4a922c1d58.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q01-p05-f5296317eb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Rocky shore ecosystems are found in some coastal regions. These ecosystems are inhabited by 1\nmany different species of algae and animals. Twice each day the sea level rises to the high tide\nlevel and decreases to the low tide level in a regular pattern. The organisms in these ecosystems\nare exposed to very different conditions during the course of each day.\nAlgae are photosynthetic protoctists. Multicellular algae are found in many marine ecosystems.\nRock pools are formed in depressions in the shore as the water level lowers when the tide goes\nout.\nFig. 1.1 is a profile of a rocky shore ecosystem showing the positions of some rock pools. The\nmid-shore zone lies between the lower-shore zone and the upper-shore zone.\ncliff\nheight\nrock pool\nfilled with sea\nHWM\nwater\nmid-shore zone LWM\nsea\ndistance between the cliff and sea level at low tide\nHWM = high-water mark – the highest point on the shore that is covered by sea water at high tide\nLWM = low-water mark – the lowest point on the shore that is exposed to the air at low tide\nFig. 1.1\nFig. 1.2 shows a large rock pool on the mid-shore zone surrounded by smaller rock pools at low tide.\nsea\nsmall\nrock\npool\nlarge\nrock\npool\nFig. 1.2\nA student investigated the distribution and abundance of large, multicellular algae on a rocky\nshore in North Wales in the UK at low tide on one day in October 2011.\nThe student completed a preliminary survey of the shore at low tide and made some observations.\n1. There were several different species of algae growing in the mid-shore zone.\n2. Some algae were exposed to the air at low tide, but most grew in rock pools.\n3. The distribution and abundance of the algae differed on the shore between the low-water\nmark (LWM) and the high-water mark (HWM).\n4. Large rock pools appeared to have more species of algae than small rock pools.\n5. Large rock pools were deeper than small rock pools.\nThe student decided to investigate the relationship between the depth of rock pools and the\nspecies diversity of the algae in the rock pools in the mid-shore zone.\nThe student selected 32 rock pools at random. Approximately half the rock pools were small and\nhalf were large. The depth of each rock pool at its deepest point was measured and the number of\ndifferent species of algae in each rock pool was counted.\nState variables that have been standardised in the investigation. (a) two\n1 ................................................................................................................................................\n...................................................................................................................................................\n2 ................................................................................................................................................\n...................................................................................................................................................\n[2]\nThe results of the investigatio",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-52-q01-d0db0d79ce.pdf"
    },
    {
      "id": "9700-2023-on-52-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_52.pdf?download=true",
      "source_pages": [
        7,
        8,
        9,
        10,
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p01-c6958daba6.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p02-131ee07ce0.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p03-959a43731f.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p04-918e38809a.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p05-26cb2d0bb6.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p06-b4be9fe003.png",
        "9700-experimental-skills/assets/9700-2023-on-52-q02-p07-ac94083ea7.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Abscisic acid (ABA) is a plant hormone that influences the growth and development of plants. 2\nSalicylic acid (SA) is a plant hormone that interacts with other plant hormones, such as ABA, to\ninfluence growth and development in plants.\nScientists investigated the effects of ABA and SA on shoot growth in seedlings of rice, Oryza\nsativa.\nThe scientists germinated some rice grains and divided the young seedlings into four batches, A\nto D, each containing 9 seedlings of the same length. The seedlings were put into trays containing\na seedling growth medium. The shoots in each batch of seedlings were sprayed once with a\ndifferent solution:\nbatch – water A\n–3 μmol ABA batch – 2 dm B\ndm–3 dm–3 μmol ABA + 1.0 mmol SA batch – 2 C\ndm–3 SA batch – 1.0 mmol D\nThe seedlings were then put into a growth chamber with constant conditions of light, temperature\nand humidity. The length of the shoot on each seedling was measured one day after spraying and\nfive days after spraying.\nFig. 2.1 shows two seedlings taken at random from each batch five days after being sprayed.\n/ mm cm\n6\n5\n4\n3\n2\n1\n0\nwater ABA ABA + SA SA A B C D\nFig. 2.1\nThe mean results of the investigation are shown in Fig. 2.2.\n45\n40\n35\n30\nmean\n25\nshoot\nlength\n20\n/ mm\n15\n10\n5\n0\nwater ABA ABA + SA SA water ABA ABA + SA SA A B C D A B C D\nday 1 day 5\nFig. 2.2\nSeedlings do not always grow straight. (a)\nSuggest how the measurements of shoot length of the rice seedlings can be made as\naccurately as possible.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThere were nine seedlings in each batch. (b)\nThe standard deviation for the length of seedlings in batch (sprayed with SA only) at day 5 D\nis 5 mm.\nThe formula for calculating standard error (SE) is:\ns = sample standard deviation s\nSE =\nn = sample size (number of observations) n\nThe formula for calculating the 95% confidence interval (95% CI) is:\n˗ ˗ × x x ± (2 SE) = mean 95% CI =\nCalculate the 95% CI for batch (treated with SA only) at day 5. (i) D\n95% CI = .................................................... mm [1]\nThe values for 95% CI were used to add error bars for batches and at day 5. These (ii) C D\nerror bars did not overlap. State what can be concluded from this.\n...........................................................................................................................................\n...........................................................................................................................................\n...............................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-52-q02-c360fe55fb.pdf"
    },
    {
      "id": "9700-2023-on-53-q01",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p01-91bf518020.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p02-249c8f1796.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p03-3613af51b9.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p04-d0c1529823.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p05-d9ea4cc40b.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p06-b2fa98630e.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q01-p07-bfa8e0676c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Fruits in the diet can be a source of vitamins, such as vitamin C and vitamin A, which are important 1\nin maintaining good health.\nVitamin A is a group of molecules that includes retinol and retinal.\nVitamin C is an organic acid known as ascorbic acid.\nRipening occurs in fruit when molecular changes cause an increase in sugar content, colour\nchanges and softening of the fruit, over time.\nAscorbic acid is found in plant cells. The fruits of many plants have high concentrations of ascorbic\nacid.\nA student planned to investigate how ascorbic acid concentration changes with ripeness in the\nfruit of the apple tree, Malus domestica.\nThe fruit of M. domestica (apples) are shown in Fig. 1.1.\nFig. 1.1\nThe student removed an apple that was 5 cm in diameter from a tree. The student measured (a)\nthe sugar content of the apple, which was low. This indicated that the apple was at the early\nstages of ripening (unripe).\nFor the investigation, the student decided to select apples from trees when each apple was\n5 cm in diameter.\nThe student assumed that the ripeness of each apple would increase each day after it had\nbeen picked from a tree.\nSuggest why the diameter of the apples selected may give accurate estimates of (i) not\nripeness.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe an improvement to the method for selecting the apples that would give more (ii)\naccurate estimates of ripeness.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student researched how ascorbic acid concentration changes with ripeness in the fruit of (b)\nother species.\nBased on this research, the student predicted that ascorbic acid concentration in apples will:\n• be lowest in unripe apples when they are first picked from the tree\n• increase over several days and peak in partially ripe apples\n• decrease slightly in later days as the apples become fully ripe.\nSketch a line graph below, including labelled axes, to show the predicted results.\n[2]\nThe student used the redox indicator DCPIP to investigate how ascorbic acid concentration (c)\nchanges as the apples ripen.\nThe student tested the ascorbic acid concentration of the apples at intervals from the time\nthey were taken from the tree until the apples were fully ripe.\nFor each test, the student prepa",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-53-q01-90ca48d8d7.pdf"
    },
    {
      "id": "9700-2023-on-53-q02",
      "subject": "9700",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2023-oct-nov/9700_w23_qp_53.pdf?download=true",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2023-Oct-Nov/qp/9700_w23_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2023-on-53-q02-p01-546386e38f.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q02-p02-0200d73d6a.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q02-p03-9c7ff4b9b1.png",
        "9700-experimental-skills/assets/9700-2023-on-53-q02-p04-5127844f96.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2023-on-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Trypsin is a protease enzyme. A student compared the activity of trypsin from two different species: 2\nthe Atlantic salmon, Salmo salar, and the domestic pig, Sus scrofa domesticus.\nFig. 2.1 shows an Atlantic salmon and Fig. 2.2 shows domestic pigs.\nFig. 2.1\nFig. 2.2\nAtlantic salmon are fish and are ectothermic, which means that the internal body temperature\nfluctuates with the surrounding environmental temperature.\nThe water temperature in the habitat of Atlantic salmon can decrease to – 0.5 °C.\nDomestic pigs are mammals and maintain an internal body temperature of approximately 38 °C.\nThe student was provided with trypsin from Atlantic salmon, trypsin from domestic pigs and cubes\nof gelatine. Gelatine is made from the protein collagen.\nThe student measured the time taken for the trypsin from each animal species to break down the\ngelatine cubes at 20 °C.\nThe student:\ncm3 • used 10 of 5% trypsin solution from each animal species\n• placed each trypsin solution in separate test-tubes, maintained at 20 °C in a water-bath\n• placed a gelatine cube in each test-tube of trypsin solution\n• recorded the time taken to break down the gelatine cube in each test-tube.\nThe student repeated this 12 times for each type of trypsin.\nIdentify the independent variable the dependent variable in this experiment. (a) (i) and\nindependent variable .........................................................................................................\ndependent variable ............................................................................................................\n[2]\nThe student standardised the temperature of the water-bath, the volume of the trypsin (ii)\nsolution and the concentration of the trypsin solution.\nState other variable that the student should standardise in this experiment. one\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe results of the experiment are shown in Table 2.1. (b)\nTable 2.1\nreplicate time taken to break down gelatine cube / seconds\nAtlantic salmon domestic pigs\n1 125 130\n2 121 127\n3 128 134\n4 119 128\n5 132 135\n6 125 130\n7 131 132\n8 127 128\n9 126 128\n10 131 137\n11 131 133\n12 129 131\nmean 127.1\n(s) standard deviation 4.1\nUse the data in Table 2.1 to calculate the mean time taken for trypsin from domestic pigs (i)\nto break down the gelatine cube.\nmean = ........................................... seconds [1]\nThe formula for calculating standard deviation is: (ii)\nKey to symbols ˗ )2 /(x - x\n= s\nx = observation - n 1\n˗ x = mean\nn = sample size (number of observations)\n˗ )2 - /(x x The student calculated as for the domestic pigs. 110.92\n˗ )2 - /(x x and the data in Table 2.1 to complete the calculation Use the calculated value of\nof the standard deviation for domestic pigs.\ns = ............................",
      "pdf_path": "9700-experimental-skills/print/9700-2023-on-53-q02-4b22194bec.pdf"
    },
    {
      "id": "9700-2024-m-42-q01",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-m-42-q01-p01-4a45ad8b97.png",
        "9700-topic-14-homeostasis/assets/9700-2024-m-42-q01-p02-1e8a5022eb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 is a diagram of a nephron. 1 (a)\nFig. 1.1\nLabel Fig. 1.1 using:\n• one labelling line and the letter to identify a region that contains urine A\n• one labelling line and the letter to identify a region that contains podocytes B\n• one labelling line and the letter to identify a region of the nephron that is within the C\nmedulla of the kidney\n• one labelling line and the letter to identify the afferent arteriole. D\n[4]\nThe cells of the proximal convoluted tubule are adapted to carry out selective reabsorption. (b)\nDescribe explain how these cells are adapted to carry out selective reabsorption. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\n[Total: 8]",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-m-42-q01-5f6933eb30.pdf"
    },
    {
      "id": "9700-2024-m-42-q02",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q02-p01-4f4d70ae9d.png",
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q02-p02-eabb48b760.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The scientist Gregor Mendel investigated differences in the length of the stem in the pea plant, 2\nPisum sativum. In 1866, he published the results of his investigation into this trait (characteristic).\nFig. 2.1 shows a diagram of a pea plant.\ninternode\nFig. 2.1\nMendel observed that the pea plants he grew either had tall stems or dwarf (short) stems. In his\ninvestigation, Mendel carried out crosses using pea plants with these two phenotypes.\nFrom the results of these crosses, Mendel demonstrated that tall stems were dominant to (a)\ndwarf stems in pea plants.\nIt is now known that the stem length trait in pea plants is controlled by one gene that has two\nalleles:\n• a dominant allele, Le\n• a recessive allele, le.\nDescribe a cross that could be carried out how the results of the cross could be analysed and\nto determine the genotype of a pea plant with a tall stem.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThe scientists P W Brian and H G Hemming identified that the difference in the length of (b)\nthe stem in pea plants was associated with the presence or absence of gibberellin. They\npublished their findings in 1955.\nGibberellin leads to a response in plant cells by binding to specific receptor molecules. (i)\nState the term used to describe a molecule, such as gibberellin, that binds to specific\nreceptor molecules and leads to a response in cells.\n..................................................................................................................................... [1]\nSuggest the response of the cells in the internode region of the stem, as labelled in (ii)\nFig. 2.1, to the presence of gibberellin describe how this response affects the trait and\ninvestigated by Mendel.\n....",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-m-42-q02-3df3d2cc36.pdf"
    },
    {
      "id": "9700-2024-m-42-q03",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q03-p01-87c9133a8b.png",
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q03-p02-16c1eafc64.png",
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q03-p03-ca9ac884a7.png",
        "9700-topic-16-inheritance/assets/9700-2024-m-42-q03-p04-aa2a7c2975.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Cystic fibrosis is an autosomal recessive genetic disease. People with cystic fibrosis have a 3\nhomozygous recessive genotype.\nExplain the meaning of the terms homozygous and recessive. (a)\nhomozygous .............................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\nrecessive ..................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n[2]\nIn 2020: (b) (i)\n• there were 10 800 people with cystic fibrosis in the UK\n• the UK population was estimated to be 67 100 000 people.\nA proportion of people in the UK population are heterozygous for the gene that causes\ncystic fibrosis and do have symptoms of the disease. not\nUse the Hardy–Weinberg principle to calculate the number of people in the UK population\nwho are expected to be heterozygous for the gene that causes cystic fibrosis.\nThe two equations for the Hardy–Weinberg principle are provided.\nequation 1\np + q = 1\nequation 2\n2 q2 + 2pq + = 1 p\np = frequency of the dominant allele\nq = frequency of the recessive allele\n2 = frequency of the homozygous dominant genotype p\n2pq = frequency of the heterozygous genotype\nq2 = frequency of the homozygous recessive genotype\nThe first stage of the calculation has been completed for you.\n10 800\nq2 =\n67 100 000\nq =\np =\n2pq =\nnumber of people in the UK expected to be heterozygous for the gene = ..............................................\n[3]\nThe Hardy–Weinberg principle provides a useful estimate of the number of people in (ii)\nthe UK who are heterozygous for cystic fibrosis. However, the estimate is lower than\nthe actual number. This underestimation occurs because all the conditions of the not\nHardy–Weinberg principle apply.\nIn the UK in 2020, the mean life expectancy of:\n• people with cystic fibrosis was approximately 50 years\n• people was approximately 80 years. all\nExplain how this information accounts for the underestimation of the number of people in\nthe UK that are heterozygous for cystic fibrosis.\n...........................................................................................................................................\n...........................................................................................................................................\n.............................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-m-42-q03-c082106191.pdf"
    },
    {
      "id": "9700-2024-m-42-q04",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-m-42-q04-p01-3f77efa6fa.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-m-42-q04-p02-e42cf4bace.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-m-42-q04-p03-8ebc584d4a.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-m-42-q04-p04-1156eb7a5d.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-m-42-q04-p05-3dc1ca0583.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Holstein Friesian cattle are a breed of cattle used by dairy farmers in many countries of the world 4\nfor the high milk yield of their cows.\nFig. 4.1 shows Holstein Friesian cattle.\nFig. 4.1\nMilk yield in Holstein Friesian cattle is affected by heat stress. Heat stress occurs when homeostatic\nmechanisms are enough to keep the body temperature down to normal levels. not\nOne of the factors that contributes to heat stress is air temperature.\nFig. 4.2 shows:\n• the mean daily air temperature in Central Europe\n• the mean monthly milk yield per cow of Holstein Friesian cattle in Central Europe.\n30 850\n25 800\n20 750\nmean\nmean daily air\nmonthly milk\ntemperature 15 700\nyield per cow\n°C /\ncow–1 / kg\n10 650\n5 600\n0 550\nJan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec\nmonth\nkey\nmean daily air temperature\nmean monthly milk yield per cow\nstandard error (SE) bar\nFig. 4.2\nWith reference to Fig. 4.2, describe the trends in air temperature and milk yield from April to (a)\nAugust.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nMany dairy farmers in tropical regions use cattle breeds that are tolerant to heat stress (b)\n(heat-tolerant cattle). These heat-tolerant cattle:\n• can tolerate higher air temperatures than Holstein Friesian cattle before heat stress\noccurs\n• have milder symptoms of heat stress than Holstein Friesian cattle for the same high\nair temperatures.\nWhere heat stress does occur, heat-tolerant cattle produce a lower milk yield than not\nHolstein Friesian cattle under the same conditions.\nScientists compared DNA sequences of Holstein Friesian cattle and heat-tolerant cattle for a\nnumber of genes known to have an effect on body temperature.\nTwenty genes were found that had alleles associated with heat-tolerant cattle. only\nWith reference to the information provided, including the data in Fig. 4.2:\n• state the type (pattern) of phenotypic variation shown by milk yield in cattle\n• identify factors that cause phenotypic variation in milk yield in cattle.\nIn each case, give a reason for your choice.\ntype (pattern) of phenotypic variation and reason for choice ....................................................\n...................................................................................................................................................\n.........................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-m-42-q04-9b61fbce38.pdf"
    },
    {
      "id": "9700-2024-m-42-q05",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        17
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-m-42-q05-p01-4c2fdb490e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q05",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Complete the following paragraphs using the most appropriate word or words. 5\nThe theory of evolution describes a process that can lead to the formation of new species from\npre-existing species over .............................................. .\nDNA sequence data of different species can be compared to show evolutionary relationships. Two\nspecies that have a more recent common ancestor share more .............................................. in\nthe DNA nucleotide sequences of their genomes than two species that are more distantly related.\nMitochondrial DNA can also be used in the study of evolutionary relationships. Mitochondrial\nDNA is inherited only from the female gamete, and its nucleotide sequence is unaffected by\n.............................................. during the production of gametes.\nDNA sequence data can be stored in large biological .............................................. , allowing faster\ncomparison of the nucleotide sequences of genomes using computer software. DNA sequence\ndata can also be used to predict the .............................................. sequences of proteins produced\nby a species.\nA .............................................. can be used to detect many different mRNA molecules at the\nsame time in studies that compare gene expression between different species.\n[6]",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-m-42-q05-4bb6cbfb6a.pdf"
    },
    {
      "id": "9700-2024-m-42-q06",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19,
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-m-42-q06-p01-95eca39a39.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-m-42-q06-p02-d4b9471c33.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-m-42-q06-p03-07b5692360.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-m-42-q06-p04-ea9d064e2b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "A respirometer is a piece of apparatus that can be used to measure the rate of respiration of living 6\ntissue such as germinating peas.\nA simple respirometer is shown in Fig. 6.1.\nsyringe\nwire\ngauze\nruler\ngerminating\ncm\npeas 0 U-shaped\ntube 1\ntest-tube\n2\ncoloured\n3\nliquid\npotassium 4\nhydroxide\n5\nsolution\nFig. 6.1\nA student carried out an investigation to determine the effect of temperature on the rate of\nrespiration of germinating peas.\n• The student set up the respirometer as shown in Fig. 6.1 and placed the respirometer in\na water-bath at 10 °C.\n• After five minutes, the student used the syringe to adjust the position of the coloured\nliquid in the right-hand side of the U-shaped tube so that it lined up with 0 cm on the ruler.\nThe student immediately started a timer.\n• The germinating peas used up oxygen, causing the coloured liquid in the U-shaped tube\nto move.\n• The student measured the distance moved by the coloured liquid after 20 minutes.\n• The student repeated the experiment at temperatures of 20 °C, 30 °C, 40 °C and 50 °C.\nState the function of the potassium hydroxide solution used in the investigation. (a)\n...................................................................................................................................................\n............................................................................................................................................. [1]\nSuggest how the validity of the results could be assessed. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nExplain why the respirometer was left in the water-bath for five minutes before starting the (c)\nexperiment.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe rate of movement of the coloured liquid in the U-shaped tube, calculated from the results, (d)\nis shown in Table 6.1.\nTable 6.1\n–1 temperature / °C rate of movement / mm min\n10 0.40\n20 0.70\n30 1.30\n40 1.15\n50 0.60\nPl",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-m-42-q06-09ba15e5e3.pdf"
    },
    {
      "id": "9700-2024-m-42-q07",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-m-42-q07-p01-e6da5de6d4.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-m-42-q07-p02-dab37c6fd6.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-m-42-q07-p03-004dc37304.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "The light-dependent stage of photosynthesis occurs within chloroplasts. In this stage, 7 (a)\nelectrons are emitted from the chlorophyll a molecules and passed to electron acceptors.\nIf a redox indicator, such as DCPIP, is added to a suspension of illuminated chloroplasts,\nelectrons will be transferred to DCPIP, causing the colour of the DCPIP to change from blue\nto colourless.\nA student investigated the effect of the wavelength of light (colour of light) on the rate of\nphotosynthesis.\n• DCPIP was added to three colorimeter tubes, each containing a suspension of\nchloroplasts. The chloroplast suspensions were kept in the dark until required.\n• The colorimeter tubes were each exposed to light of a different colour: red, blue or\ngreen. The intensity of light was the same for all tubes, and each was exposed to\nlight for four minutes. All other conditions were kept the same.\n• The absorbance of each chloroplast suspension was measured at one-minute\nintervals using a colorimeter.\nThe results are shown in Fig. 7.1.\n1.6\n1.5\ngreen\nlight\n1.4\n1.3\n1.2\nabsorbance 1.1\n1.0\nblue\n0.9\nlight\n0.8\nred\n0.7\nlight\n0.6\n0 1 2 3 4\ntime / min\nFig. 7.1\nExplain why the chloroplast suspensions were kept in the dark until required. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nDescribe the results shown in Fig. 7.1. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nWith reference to the light-dependent stage of photosynthesis, explain the differences (iii)\nbetween the results shown in Fig. 7.1 for red light and for green light.\n...........................................................................................................................................\n..............................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-m-42-q07-91fe301591.pdf"
    },
    {
      "id": "9700-2024-m-42-q08",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        25,
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-m-42-q08-p01-9b081fde89.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-m-42-q08-p02-7ca9996f12.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-m-42-q08-p03-46cad5ff96.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "109 × Approximately 2 people in the world are currently infected with the bacterial disease 8 (a)\ntuberculosis (TB) caused by Mycobacterium tuberculosis. Early diagnosis is important so that\ntreatment can begin.\nAPOPO is a non-profit organisation that has trained African giant pouched rats, Cricetomys\ngambianus, to use their sense of smell to detect M. tuberculosis. They do this by sniffing a\nsample of thick mucus from the lungs of people who may have TB. The African giant pouched\nrats are able to detect the presence of M. tuberculosis with an accuracy of 87–93%.\nFig. 8.1 shows an African giant pouched rat.\nFig. 8.1\nThe type of receptor cell used by African giant pouched rats to detect M. tuberculosis is (i)\nthe same as that used in human taste buds.\nName this type of receptor cell.\n..................................................................................................................................... [1]\nSuggest why African giant pouched rats trained to detect M. tuberculosis may also be (ii)\nable to detect other species of Mycobacterium that cause TB.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nThe African giant pouched rat belongs to the kingdom Animalia in the domain Eukarya. (b)\nComplete Table 8.1 to show the full classification of the African giant pouched rat.\nTable 8.1\nkingdom Animalia\nChordata\n.............................\nclass Mammalia\nRodentia\n.............................\nfamily Nesomyidae\n............................. .............................\nspecies gambianus\n[2]\nDifferences between members of the domain Eukarya and members of the domain Bacteria (c)\ninclude the presence or absence of particular membrane-bound cell structures.\nOutline differences in the characteristic features of members of the domain Eukarya other\nand members of the domain Bacteria.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n........................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-m-42-q08-ec3002c28b.pdf"
    },
    {
      "id": "9700-2024-m-42-q09",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        28,
        29
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-m-42-q09-p01-18175ac541.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-m-42-q09-p02-9230b2206a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "An investigation was carried out to study the effect of different intensities of blue light on the 9 (a)\npercentage germination of barley seeds. Barley seeds were exposed to blue light for a period\nof seven days. All other variables were kept constant.\nThe results are shown in Table 9.1.\nTable 9.1\nlight intensity percentage\n/ arbitrary units (au) germination\n0 (dark) 98.0\n36 76.9\n48 45.0\n57 14.7\nThe effect of blue light on the concentration of abscisic acid (ABA) was also investigated.\nABA concentration was measured at intervals over seven days in barley seeds exposed to\nblue light at an intensity of 57 arbitrary units.\nThe results are shown in Table 9.2.\nTable 9.2\nconcentration of ABA\nday\n/ arbitrary units (au)\n0 100\n1 90\n3 350\n5 351\n7 381\nFor comparison, in the dark the concentration of ABA in barley seeds fell from 100 au at the\nstart (day 0) to 45 au on day 1 and did increase from day 1 to day 7. not\nABA is thought to affect gibberellin synthesis or activity.\nUsing the information in Table 9.1 and Table 9.2, describe the effect of blue light on the\ngermination of barley seeds suggest an explanation for this effect. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nAfter germination, auxin is important in the growth of barley plants. (b)\nDescribe explain the role of auxin in cell elongation. and\n...................................................................................................................................................\n...................................................................................................................................................\n....................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-m-42-q09-826ead6eab.pdf"
    },
    {
      "id": "9700-2024-m-42-q10",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_42.pdf?download=true",
      "source_pages": [
        30,
        31
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-m-42-q10-p01-cbbbe2af1e.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-m-42-q10-p02-f52a32d3b3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-42-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Striated muscle is composed of myofibrils. Myofibrils contain several structural proteins 10 (a)\nincluding troponin, tropomyosin, actin and myosin.\nOutline the roles of these four structural proteins in the contraction of a sarcomere.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nDrugs that cause muscle paralysis (paralytic drugs) are used during surgery to stop the patient (b)\nmoving. One commonly used paralytic drug is succinylcholine, which works by preventing\ncontraction of muscles. Succinylcholine is able to prevent muscles contracting because it has\na similar shape to acetylcholine.\nSuggest how succinylcholine is able to prevent muscles contracting.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDuchenne muscular ",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-m-42-q10-bee6d323af.pdf"
    },
    {
      "id": "9700-2024-m-52-q01",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 17,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p01-0ebda9a35f.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p02-8272a8e57f.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p03-7a57391ad4.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p04-b3596c3e85.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p05-c604c81331.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p06-db691d893f.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q01-p07-34f1a90888.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Onions are vegetables that are harvested from the plant Allium cepa. In many countries, small 1\nonions are used to make pickled onions. Pickled onions can be stored for a long time because the\ngrowth of microorganisms is prevented.\nPickled onions are made using this method.\n• Remove the dry outer layer from small onions.\n• Place the onions in a sodium chloride solution for several hours.\n• Remove the sodium chloride solution and place the onions in a solution of ethanoic acid\n(vinegar).\nFig. 1.1 shows a container of pickled onions and some fresh onions that have been pickled. not\ncontainer of\npickled onions\nfresh onion\ndry outer\nlayer of onion\nFig. 1.1\nSome students decided to investigate how placing small onions in different concentrations of (a)\nsodium chloride solution affects the mass of the onions.\nIdentify the variable in this investigation. (i) independent\n..................................................................................................................................... [1]\nThe students were given a 20.0% stock solution of sodium chloride. The students (ii)\ndecided to use the stock solution to make sodium chloride solutions with concentrations\nof 0.0%, 1.0%, 5.0%, 10.0%, 15.0% and 20.0%.\n3 of these solutions in separate beakers. The students made 140 cm\n3 of these solutions could be made by Complete Table 1.1 to show how 140 cm\ndilution of the 20.0% stock solution of sodium chloride. proportional\nTable 1.1\nvolume of\npercentage concentration volume of 20.0%\nof sodium chloride sodium chloride solution\n............................................\n3 /cm solution\n3 /cm\n0.0\n1.0\n5.0\n10.0\n15.0\n20.0\n[2]\nThe students carried out this procedure. (b)\n• The dry outer layer was removed from 30 small onions.\n• Five of the small onions were placed into each of the six beakers containing the\nsodium chloride solutions prepared by the students.\n• The onions were left in the sodium chloride solutions for hours. 2\n• The mean percentage change in mass of the five small onions in each beaker was\ncalculated.\nThe students then repeated the whole procedure using a new set of 30 small onions. This\ntime the onions were left in the sodium chloride solutions for hours. 48\nFig. 1.2 shows the results of the investigation.\n+20.0\nkey\nafter 2 hours\n+18.0\nafter 48 hours\n+16.0\n+14.0\n+12.0\n+10.0\n+8.0\nmean\npercentage\n+6.0\nchange in\nmass\n+4.0\n+2.0\n0.0\n–2.0\n–4.0\n–6.0\n–8.0\n0.0 5.0 10.0 15.0 20.0\npercentage concentration of\nsodium chloride solution\nFig. 1.2\nExplain why the students calculated the percentage change in mass of the onions. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nOne of the st",
      "pdf_path": "9700-experimental-skills/print/9700-2024-m-52-q01-4261653009.pdf"
    },
    {
      "id": "9700-2024-m-52-q02",
      "subject": "9700",
      "year": 2024,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-march/9700_m24_qp_52.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12,
        13,
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2024-March/qp/9700_m24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p01-87b24f3f87.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p02-1d322235de.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p03-6b0de0fc74.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p04-f52f66b13f.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p05-447dd558e2.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p06-d05da35fe3.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p07-8fc304c66d.png",
        "9700-experimental-skills/assets/9700-2024-m-52-q02-p08-5d6711ebc4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Drosophila melanogaster is a small fruit fly that is often used in research on genetics. 2\nWild fruit flies normally have dark red eyes due to the presence of a brown pigment, ommochrome,\nand a bright red pigment, drosopterin.\nFig. 2.1 shows an adult female fruit fly and an adult male fruit fly.\nfemale fruit fly male fruit fly\neye\n3mm\npointed rounded\nabdomen abdomen\nFig. 2.1\nA biologist carried out an investigation to determine the roles of two genes that are involved in the\ndetermination of eye colour in adult fruit flies. Both genes have two alleles.\nEarlier research by other scientists suggested that:\n• the synthesis of the brown pigment ommochrome depends on the gene B/b\n• the synthesis of the bright red pigment drosopterin depends on the gene R/r.\nThe biologist obtained the parent fruit flies shown in Table 2.1.\nTable 2.1\nparent fruit flies genotype phenotype\nfemale BBrr brown eyes\nmale bbRR bright red eyes\nThe biologist carried out the procedure shown in Fig. 2.2 to cross the parent fruit flies and obtain first\ngeneration offspring.\nSome agar jelly containing nutrients and water was\nspecimen\nplaced in the bottom of several specimen tubes.\ntube\nagar jelly\ncontaining\nOn day 1, four adult female fruit flies with brown eyes\nnutrients\nand four adult male fruit flies with bright red eyes were\nand water\nplaced in each specimen tube.\ncotton\nwool\nA piece of cotton wool was placed in the top of each\nspecimen tube.\nfruit\nThe specimen tubes were left in a warm room for\nflies\nseveral days. During this time, the parent fruit flies\nreproduced and laid fertilised eggs on the agar jelly.\nThe fertilised eggs hatched into larvae that fed on the\nagar jelly.\nOn day 7, the parent fruit flies were removed from each\nspecimen tube.\nBy day 12, the first generation of adult fruit flies had\ndeveloped from the larvae and were ready for crossing\nto produce the second generation.\nFig. 2.2\nSuggest a reason why the biologist removed the parent fruit flies from the specimen tubes on (a)\nday 7.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe biologist wanted to cross the first generation of fruit flies with one another to produce the\nsecond generation of fruit flies.\nThe first generation of adult fruit flies in the specimen tubes on day 12 were a mixture of females\nand males and had yet mated. not\nThe biologist crossed the first generation of fruit flies with one another by:\n• preparing fresh specimen tubes in which to produce the second generation of fruit flies\n• using a chemical to anaesthetise the first generation of adult fruit flies so that they were\ntemporarily unable to move\n•",
      "pdf_path": "9700-experimental-skills/print/9700-2024-m-52-q02-7e4c789e16.pdf"
    },
    {
      "id": "9700-2024-mj-41-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-41-q01-p01-2c640995f6.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-41-q01-p02-4e539f3e83.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Oxygen is needed for aerobic respiration in cells. 1\nIn eukaryotic cells, the mitochondrion is the organelle used for aerobic respiration.\nFig. 1.1 shows a transmission electron micrograph of a mitochondrion. (a)\nA\nB\nC\nD\nFig. 1.1\nTable 1.1 lists the four stages of aerobic respiration.\nWith reference to Fig. 1.1, complete Table 1.1 using the letters to show where each A– D\nstage occurs.\nEach letter may be used once, more than once or not at all.\nTable 1.1\nstage of respiration letter\nglycolysis\n………\nlink reaction\n………\nKrebs cycle\n………\noxidative phosphorylation\n………\n[3]\nIn anaerobic conditions, no ATP can be synthesised by oxidative phosphorylation because (b)\nthe process stops.\nExplain why ATP synthesis by oxidative phosphorylation stops in anaerobic conditions.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nUp to 60% of the ATP that is produced in cancer cells comes from lactate fermentation of (c)\nglucose, even though oxygen is present.\nScientists are developing cancer treatments to inhibit the enzyme that catalyses the last step\nof lactate fermentation.\nExplain how the inhibition of this enzyme reduces the production of ATP in cancer cells.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.....................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-mj-41-q01-43b6012924.pdf"
    },
    {
      "id": "9700-2024-mj-41-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-41-q02-p01-191a3ed640.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-41-q02-p02-3e309adaca.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q02",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Coordination in humans involves two main systems: the nervous system and the endocrine 2 (a)\nsystem. Paracrine cell signalling is a third way in which coordination occurs. In paracrine\nsignalling, one cell secretes a chemical that diffuses a short distance to act upon cells that\nare very near to the secreting cell.\nIdentify similarity and difference between paracrine cell signalling and cell (i) one one\nsignalling that occurs as part of the endocrine system.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nExplain why a neurotransmitter such as acetylcholine could be described as a paracrine (ii)\nsignalling molecule.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nHuman muscle cells show paracrine signalling. After muscle cells have been exposed to a (b)\nsubstance called palmitate they:\n• produce a signalling molecule\n• show an increase in the expression of genes involved in a stress response (stress genes).\nScientists carried out three experiments to investigate the signalling molecule produced by\nmuscle cells after they have been exposed to palmitate.\nExperiment A\n• Muscle cells were exposed to palmitate.\n• The palmitate‑exposed muscle cells were removed from the medium containing palmitate\nand then cultured in a nutrient medium for six days.\n• The palmitate‑exposed cells were removed from the nutrient medium and muscle new\ncells were placed in this used nutrient medium.\n• The expression of stress genes in the muscle cells was measured after 24 hours. new\n• A using cells that had been exposed to palmitate was also carried out. control not\nExperiment B\nExperiment was repeated but the used nutrient medium for the palmitate‑exposed cells A\nand the control was boiled and then cooled before adding the new muscle cells.\nExperiment C\nExperiment was repeated but the used nutrient medium from the palmitate‑exposed cells A\nand the control was treated to separate the lipid part. This lipid part was added to a new\nnutrient medium to culture the new muscle c",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-mj-41-q02-00d2cba90e.pdf"
    },
    {
      "id": "9700-2024-mj-41-q03",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-mj-41-q03-p01-1f1bbe3b24.png",
        "9700-topic-14-homeostasis/assets/9700-2024-mj-41-q03-p02-17718ec3b5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q03",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Stomata in leaves respond to changes in environmental conditions by opening and closing. This 3\nregulates carbon dioxide uptake and water loss.\nDescribe the mechanism by which stomata open in sunlight. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nStomatal conductance is a measure of the rate of water vapour loss from the intercellular air (b)\nspaces of leaves to the atmosphere through the stomata.\nA student investigated the effect of different concentrations of abscisic acid (ABA) on stomatal\nconductance in Helianthus annuus, the common sunflower.\nThe student treated the roots of 15 sunflower plants with different concentrations of ABA\nsolution. The plants were left in standardised conditions for 24 hours.\nAfter 24 hours, the student measured:\n• the concentration of ABA in the xylem sap of each plant\n• the stomatal conductance of the leaves of each plant.\nThe results are shown in Fig. 3.1.\n2.5\n2.0\nstomatal 1.5\nconductance\nm−2 s−1 / mol 1.0\n0.5\n0\n0 250 500 750 1000\nm-3 μmol concentration of ABA in the xylem sap /\nFig. 3.1\nWith reference to Fig. 3.1, describe the relationship between the concentration of ABA in (i)\nthe xylem sap and stomatal conductance.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n........................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-mj-41-q03-326b074917.pdf"
    },
    {
      "id": "9700-2024-mj-41-q04",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-41-q04-p01-3fb1845c35.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-41-q04-p02-054ef6ff1a.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-41-q04-p03-b0b4c843b5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "White‑clawed crayfish, Austropotamobius pallipes, live in rivers and lakes in Europe. 4\nIn the 1850s, the North American signal crayfish, Pacifastacus leniusculus, was introduced to\nEurope. The introduced species carried a pathogen that causes a disease known as crayfish\nplague. This disease kills A. pallipes.\nSince 1850, the population size of A. pallipes has reduced in many areas of Europe due to the\nspread of crayfish plague.\nNorth American P. leniusculus can carry the crayfish plague pathogen without showing (a)\nsymptoms because they have evolved resistance to it.\nExplain how P. leniusculus could have evolved resistance to the crayfish plague pathogen.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nIt is difficult to locate crayfish because they live underwater. (b)\nAfter an outbreak of crayfish plague in one country in Europe, researchers used an\nenvironmental DNA technique to find the locations in a river where populations of A. pallipes\nwere still surviving.\nWater samples were taken from locations along the river. Each water sample was filtered to\nobtain any cells or DNA that had been released by organisms into the water. The polymerase\nchain reaction (PCR) was carried out on this DNA using primers specific to A. pallipes DNA.\nOutline explain the steps that occur in PCR. (i) and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..........................................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-mj-41-q04-696455735e.pdf"
    },
    {
      "id": "9700-2024-mj-41-q05",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-41-q05-p01-77982f89da.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-41-q05-p02-01ae0d230b.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-41-q05-p03-b6e2973cb4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q05",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Photosynthesis is the energy transfer process that occurs in chloroplasts. 5\nFig. 5.1 shows some biochemical events that occur in a chloroplast during the light‑dependent\nI II (PSI PSII) stage of photosynthesis. Photosystems and and and some associated proteins of\nthe thylakoid membrane are shown.\nlight light ADP + Pi\nD ATP A NADP\nC\nE\nII PS PSI\nthylakoid membrane\nB\nH+ H+\nFig. 5.1\nand are areas of the chloroplast. Name areas and B. (a) (i) A B A\n........................................................................................................................................ A\n........................................................................................................................................ B\n[2]\nPSI The group of proteins labelled C, and the protein labelled are involved in a specific (ii) E\nbiochemical process during the light‑dependent stage of photosynthesis.\nName this specific biochemical process and the protein labelled E.\nprocess ..............................................................................................................................\n........................................................................................................................................ E\n[2]\nProduct is used during the Calvin cycle. (iii) D\nIdentify product describe its specific role in the Calvin cycle. D and\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nThe rate at which plants use photosynthesis to transfer light energy into chemical energy (b)\nthat is stored in plant biomass can be measured in megajoules per square metre per year\nm–2 yr–1). (MJ\nFig. 5.2 compares the rates of energy transfer for different ecosystems.\n40\n30\nrate of energy\ntransfer 20\nm−2 yr−1 / MJ\n10\n0\ntropical temperate snow tropical temperate desert\nforest forest forest grassland grassland\necosystem\nFig. 5.2\nLimiting factors affect the rate of photosynthesis. (i)\nDescribe suggest how the results for tropical forest, temperate forest and snow and\nforest in Fig. 5.2 show the effect of limiting factors on photosynthesis.\n...........................................................................................................................................\n...........................................................................................................................................\n...........",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-mj-41-q05-84274eaac4.pdf"
    },
    {
      "id": "9700-2024-mj-41-q06",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-41-q06-p01-2f32841407.png",
        "9700-topic-16-inheritance/assets/9700-2024-mj-41-q06-p02-fc46570b00.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q06",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Many different genes are involved in the production of pigments in mammals. One example is the 6\nTYR gene.\nIn humans the TYR gene is located on chromosome 11. (a)\nFig. 6.1 shows the homologous pair for chromosome 11.\nTYR gene TYR gene\nFig. 6.1\nHomologous chromosomes have the same genes located at the same loci. (i)\nState one feature shared by homologous chromosomes. other\n...........................................................................................................................................\n..................................................................................................................................... [1]\nAlbinism in humans can be caused by recessive mutations of the TYR gene. (ii)\nExplain why a person who is homozygous recessive for the TYR gene shows albinism.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nFur colour in rabbits involves a number of different genes. Some of these genes interact. (b)\nGene 1 has two alleles, and b. B\n• The dominant allele, B, results in black fur.\n• The recessive allele, b, results in brown fur.\nGene 2 has two alleles, and f. F\n• The dominant allele, F, codes for a protein that allows the expression of gene 1.\n• The recessive allele, f, codes for a protein that does not allow the expression of gene 1,\nresulting in white fur.\nThe two genes are on different pairs of autosomes.\nComplete the genetic diagram for a cross between two black rabbits that are heterozygous\nfor both genes and show the ratio of possible offspring phenotypes.\n× parental phenotypes black black\n× parental genotypes BbFf BbFf\nPunnett square\nratio of offspring phenotypes:\n............................................................................................................................................. [4]\n[Total: 8]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-41-q06-406ef6ce98.pdf"
    },
    {
      "id": "9700-2024-mj-41-q07",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-mj-41-q07-p01-85bffaa365.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-mj-41-q07-p02-61628b2f59.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q07",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "It has been hypothesised that the mutation rate of an animal species may affect how fast animals 7\nof that species age and how long they live (lifespan).\nTable 7.1 compares the mutation rate and lifespan of five species of mammal.\nTable 7.1\nmutation rate per million mean lifespan\nspecies\nbase pairs per year / years\ncattle 0.08 20\nferret 0.20 8\nhorse 0.05 30\nhuman 0.02 75\nhouse mouse 0.28 3.6\nDescribe the relationship shown in Table 7.1. (a) (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nUse Table 7.1 to explain how these species may differ in the rate at which they can adapt (ii)\nif environmental selection pressures change.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nCompare the timescales needed for selective breeding and genetic engineering to (b) (i)\nincrease the frequency of a rare allele in cattle. Give a reason for your answer.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nGenetic engineering could be used to improve farmed animals, such as dairy cattle. (ii)\nState features of dairy cattle that could be improved by ge",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-mj-41-q07-e71a8b9bb5.pdf"
    },
    {
      "id": "9700-2024-mj-41-q08",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        18,
        19,
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-41-q08-p01-def7253799.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-41-q08-p02-2798a4b2f5.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-41-q08-p03-1ad95f941f.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-41-q08-p04-7592beaefe.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Bears are several species of mammal in the family Ursidae. 8\nClimate change and loss of habitats due to human activity have reduced the size of some bear\npopulations in recent years.\nThe International Union for Conservation of Nature (IUCN) has a role in the conservation of (a)\nspecies, such as species of bear.\nOutline the role of the IUCN in the conservation of species.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nGrizzly bears, Ursus arctos horribilis, and polar bears, Ursus maritimus, are found in North (b)\nAmerica.\nFig. 8.1 shows a grizzly bear and a polar bear.\ngrizzly bear polar bear\nFig. 8.1\nGrizzly bears have:\n• a varied diet that includes grasses, roots, berries, nuts, rodents, insects and fish\n• a habitat of woodland or grassland\n• a maximum mass of 270 kg.\nPolar bears have:\n• a diet that consists of other animals, such as seals\n• a habitat of land and sea ice\n• a maximum mass of 450 kg.\nClimate change has caused populations of grizzly bears to spread further north and\npopulations of polar bears to spread further south. In some areas there is habitat overlap and,\non rare occasions, breeding between grizzly bears and polar bears has occurred.\nThe offspring of grizzly bears and polar bears are known as pizzly bears.\nPizzly bears can breed together in the wild.\nFig. 8.2 shows a pizzly bear.\nFig. 8.2\nWith reference to different species concepts, discuss whether grizzly bears and polar three\nbears should be classified as separate species.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-mj-41-q08-37a085b1b0.pdf"
    },
    {
      "id": "9700-2024-mj-41-q09",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        22
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-41-q09-p01-dc9f72f276.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q09",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "The harlequin ladybird, Harmonia axyridis, is an insect that shows discontinuous variation in colour 9\npattern and continuous variation in body size.\nWithin a species, the variation that is observed for a particular characteristic can be described (a)\nas discontinuous or continuous.\nDiscontinuous variation has a different genetic basis from continuous variation.\nState differences between discontinuous variation and continuous variation, two other than\nhaving a different genetic basis.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nFig. 9.1 shows three of the colour patterns seen in H. axyridis and the percentage of each in (b)\na population.\ncolour\npattern\nconspicua spectabilis succinea\npercentage 5 13 82\nFig. 9.1\nSuggest the genetic basis of the variation in colour pattern in H. axyridis.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nThe body size of adult H. axyridis varies from 5 mm to 8 mm. Adult ladybirds do grow or (c) not\nchange size. Adults develop from larvae that hatch from eggs.\nState environmental factors that may affect adult body size in H. axyridis. two\n...................................................................................................................................................\n............................................................................................................................................. [2]\n[Total: 6]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-41-q09-615133530e.pdf"
    },
    {
      "id": "9700-2024-mj-41-q10",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_41.pdf?download=true",
      "source_pages": [
        23
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-41-q10-p01-8fc733fb72.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-41-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Humans detect the sweet taste of sucrose sugar using chemoreceptor cells in taste buds on 10\nthe tongue. The red admiral butterfly, Vanessa atalanta, detects sucrose using chemoreceptor\ncells located on its antennae and on its tarsi (feet). The mode of action of the chemoreceptors in\nV. atalanta is similar to that in humans.\nFig. 10.1 shows the locations of the chemoreceptor cells on V. atalanta.\nantennae\ntarsi\nFig. 10.1\nDescribe how the presence of sucrose causes an action potential in the sensory neurone\nassociated with a chemoreceptor cell in V. atalanta.\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n..........................................................................................................................................................\n.................................................................................................................................................... [6]\n[Total: 6]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-mj-41-q10-76608137b3.pdf"
    },
    {
      "id": "9700-2024-mj-42-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-mj-42-q01-p01-1ca8461dc2.png",
        "9700-topic-14-homeostasis/assets/9700-2024-mj-42-q01-p02-85a15c5676.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:2)(cid:1)(cid:3)(cid:44)\n1 Osmoregulation is the control of the water potential of body fluids such as blood.\nSIHT\nNI (a) Osmoreceptors and antidiuretic hormone (ADH) have an important role in osmoregulation.\nETIRW\n(i) Name the location of osmoreceptors.\nTON\n..................................................................................................................................... [1]\nOD\n(ii) State the stimulus that is detected by osmoreceptors.\n..................................................................................................................................... [1]\nNIGRAM\n(iii) Name the structure that secretes ADH into the blood.\nSIHT\n..................................................................................................................................... [1]\nNI\nETIRW\n(b) ADH acts on the cells of the collecting duct, resulting in changes in the volume and\nconcentration of urine.\nTON\nOD Fig. 1.1 shows the relationship between the concentration of ADH in the blood and the rate of\nproduction of urine by the kidneys.\n20\nNIGRAM\nSIHT\n15\nNI\nETIRW\nrate of\nproduction\nTON\n10 of urine\nOD cm3 min–1 /\n5\nNIGRAM\nSIHT\n0\nNI\n0 1 2 3 4 5 6 7\nETIRW\nconcentration of ADH in the blood\n/ arbitrary units\nTON\nOD\nFig. 1.1\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) As the concentration of ADH in the blood increases, the rate of production of\nSIHT\nurine decreases.\nNI\nETIRW\nUsing Fig. 1.1, calculate the percentage decrease in the rate of production of urine\nbetween ADH concentrations of 1 and 3 arbitrary units.\nTON\nShow your working.\nOD\nGive your answer to one decimal place.\nNIGRAM\nSIHT\nNI\nanswer = ......................................................... [2]\nETIRW\n(ii) Explain the relationship shown in Fig. 1.1.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n..................................................................................................................................... [3]\n[Total: 8]\nN",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-mj-42-q01-1535328b46.pdf"
    },
    {
      "id": "9700-2024-mj-42-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-mj-42-q02-p01-a275c69a7e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:2)(cid:1)(cid:5)(cid:44)\n2 Variation exists in populations of a species and this may provide the opportunity for evolution\nSIHT\nto occur.\nNI\nETIRW\n(a) Phenotypic variation exists in many forms and has a number of possible causes.\nTON Describe the main factors that are the cause of phenotypic variation.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n............................................................................................................................................. [3]\nETIRW\n(b) (i) Outline the theory of evolution.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n..................................................................................................................................... [3]\nTON\n(ii) The theory of evolution is supported by DNA sequence data.\nOD\nExplain how DNA sequences are used to show evolutionary relationships\nbetween species.\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n........................................................................................................................................... OD\n................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-mj-42-q02-3a7fadb5f5.pdf"
    },
    {
      "id": "9700-2024-mj-42-q03",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-42-q03-p01-b69b110ecf.png",
        "9700-topic-16-inheritance/assets/9700-2024-mj-42-q03-p02-a80cadb073.png",
        "9700-topic-16-inheritance/assets/9700-2024-mj-42-q03-p03-f928cfe158.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:2)(cid:1)(cid:7)(cid:44)\n3 The tortoise beetle, Chelymorpha alternans, is an insect found in Panama that has seevral different\nSIHT\ncolour patterns. Fig. 3.1 shows a tortoise beetle.\nNI\nETIRW pronotum, protective outer cover\nTON\nOD elytra, protective forewings\nNIGRAM\nFig. 3.1\nSIHT\nResearchers have identified a gene, L, that controls colour pattern in the pronotum and elytra.\nLV, LT, LR Lr. and Gene L has four different alleles:\nNI\nETIRW\nTable 3.1 shows five different colour pattern phenotypes of tortoise beetles and their genotypes.\nTON\nTable 3.1\nOD\nphenotype genotypes\nblack\nLR LR NIGRAM\nrufipennis LR LV\nLR Lr\nred SIHT\nNI\nETIRW\nLT LT\ndarien f. militaris-a\nblack and\nLT LV\nTON\n(dfm-a)\nred striped\nLT Lr\nOD\nblack\nLT LR\ndarien f. militaris-b\nNIGRAM black and\n(dfm-b)\nred striped\nSIHT\nNI\nLV LV ETIRW\nred veraguensis LV Lr\nTON\nOD\nLr Lr\nmetallic and\nmetallic\nred striped\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) Explain why the inheritance of colour pattern in tortoise beetles can be described as involving\nSIHT\nmultiple alleles.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n............................................................................................................................................. [1]\n(b) A tortoise beetle with dfm‑b phenotype was crossed with another tortoise beetle with\ndfm‑b phenotype.\nNIGRAM\nConstruct a genetic diagram to show the results of this cross, including the ratio of\nSIHT offspring phenotypes.\nNI\nparental phenotypes: dfm‑b × dfm‑b\nETIRW\nparental genotypes:\nTON\nOD\ngametes:\nNIGRAM\nSIHT\nNI\nETIRW\nof fspring genotypes: TON\nOD\nof fspring phenotypes:\nNIGRAM\nratio of of fspring phenotypes:\nSIHT\nNI [4]\nETIRW\n(c) Colour pattern phenotype involves alleles that show codominance. There is also an order of\nTON\ndominance of alleles (dominance hierarchy).\nOD\nUse the information in Table 3.1 to:\n• identify the codominant alleles\n• list the dominance hierarchy with alleles from the most dominant to the least dominant. NIGRAM\ncodominant alleles ....................................................................................................................\nSIHT\ndominance hierarchy .................................................................................................................\nNI\n[2] ETIRW\nTON\nNIGRAM\n(d) Researchers carried out two crosses.\nSIHT\nNI Cross 1: female veraguensis tortoise beetles were crossed with male metallic tortoise beetles.\nETIRW\nThe results are shown in Table 3.2.\nTON\nTable 3.2\nOD\nnumber of observed offspring phenotypes\nratio of observed\ncolour pattern\nmale female male female NIGRAM\nphenotypes\nveraguensis veraguensis metallic metallic\n139 153 136 ",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-42-q03-77562777ef.pdf"
    },
    {
      "id": "9700-2024-mj-42-q04",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-42-q04-p01-780afdd777.png",
        "9700-topic-16-inheritance/assets/9700-2024-mj-42-q04-p02-5ed2abdf8a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q04",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:2)(cid:1)(cid:10)(cid:44)\n4 Genes have a role in determining the overall phenotype of an organism.\nSIHT\nNI (a) Using haemophilia as an example, ex plain the relationship between a gene, a protein and\nETIRW\na phenotype.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [5]\nTON\n(b) In the pea plant, Pisum sativum, the genotype lele results in plants with short stems\nOD\n(dwarf plants).\nExplain how the lele genotype and its gene product results in dwarf plants.\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n........................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-42-q04-af8f078263.pdf"
    },
    {
      "id": "9700-2024-mj-42-q05",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-42-q05-p01-f339e7e0fe.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-42-q05-p02-5e084c1bff.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-42-q05-p03-6a13f3d73a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:2)(cid:2)(cid:2)(cid:44)\n5 The gene BRCA1 is expressed in breast tissue and in several other tissues of the body . BRCA1\nSIHT\ncodes for a tumour suppressor protein. This protein is involved in either repairing damaged DNA,\nNI or in triggering the death of a cell if DNA cannot be repaired.\nETIRW\nSome mutations in BRCA1 are associated with an increased risk of developing breast cancer.\nTON\n(a) State one reason why a doctor would recommend that a person has genetic testing for a\nOD\nmutation in the gene BRCA1 and explain why this mutation would put the person at a greater\nrisk of breast cancer.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n............................................................................................................................................. [4] TON\nOD\n(b) DNA sequencing is one method used to test for mutations in BRCA1. Before DNA sequencing\noccurs, the DNA sample goes through the polymerase chain reaction (PCR).\nSuggest why PCR is used for testing for mutations in BRCA1.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n.....................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-mj-42-q05-d7a4e611d9.pdf"
    },
    {
      "id": "9700-2024-mj-42-q06",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-42-q06-p01-2b8dc4abd6.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-42-q06-p02-0892f47854.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:3)(cid:2)(cid:5)(cid:44)\n6 (a) Fig. 6.1 is a diagram of part of a mitochondrion.\nSIHT\ncytoplasm NI\nETIRW\nouter\nmitochondrial TON\nmembrane\nOD\nprotein\nintermembrane\nspace\nNIGRAM\ninner SIHT\nA B A B\nmitochondrial\nC C\nNI\nmembrane\nETIRW\nD D\nmitochondrial\nTON\nmatrix\nOD\nFig. 6.1\nOn Fig. 6.1, use arrows to show the route taken by oxygen from the cytoplasm into the\nNIGRAM\nmitochondrial matrix. [1]\n(b) Identify protein D and the group of proteins represented by A, B and C and describe their SIHT\nroles in oxidative phosphorylation.\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n............................................................................................................................................. [7]\nTON\nNIGRAM\n(c) The matrix of a mitochondrion contains small circular DNA. In the cells of humans, the\nSIHT\nmitochondria have circular DNA that contains 37 genes.\nNI\nETIRW\nSuggest functions for these mitochondrial genes.\nTON ......................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-mj-42-q06-d77e90bbb2.pdf"
    },
    {
      "id": "9700-2024-mj-42-q07",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-42-q07-p01-4232f53b28.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-42-q07-p02-6d3b4a9279.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:3)(cid:2)(cid:7)(cid:44)\n7 (a) A student investigated the effect of temperature on the rate of photosynthesis at two different\nSIHT\nlight intensities. The rate of photosynthesis was determined by measuring the rate of oxygen\nNI production of the plant.\nETIRW\nFig. 7.1 shows the result of the investigation.\nTON\n4\nOD\nhigh light\nintensity\n3 NIGRAM\nSIHT rate of\noxygen\nNI 2\nproduction ETIRW\nmm3 h–1 /\nTON\nOD 1\nNIGRAM low light\n0\n40 50 0 10 20 30 intensity\n/ °C temperature\nSIHT\nNI\nFig. 7.1\nETIRW\n(i) With reference to Fig. 7.1, describe and explain the results of the investigation for the\nTON\nplant in low light intensity.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n..................................................................................................................................... [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) With reference to Fig. 7.1, describe and explain the results of the investigation for the\nSIHT\nplant in high light intensity above 30 °C.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ...........................................................................................................................................\nNI\n...........................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-mj-42-q07-d9b8483a44.pdf"
    },
    {
      "id": "9700-2024-mj-42-q08",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-42-q08-p01-90d4792906.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-42-q08-p02-6620c5eb96.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-42-q08-p03-adbd5697d2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:3)(cid:2)(cid:9)(cid:44)\n8 (a) Fig. 8.1 is a diagram of a sensory neurone.\nSIHT\nA NI\ncell body\nETIRW\nB\nTON\nOD\nmyelin sheath\nNIGRAM\nFig. 8.1\nSIHT\n(i) Name the type of cell represented by cell A.\nNI\n..................................................................................................................................... [1] ETIRW\n(ii) Name a type of cell that forms a synapse with structure B. TON\nOD\n..................................................................................................................................... [1]\n(iii) Name the cells that form the myelin sheath.\nNIGRAM\n..................................................................................................................................... [1]\n(b) State and explain the dif ferences in the transmission of impulses between myelinated and\nSIHT\nunmyelinated neurones.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Opioids are therapeutic drugs that are used to relieve pain. Morphine is an example of an\nSIHT\nopioid. It affects the functioning of synapses.\nNI\nETIRW\nFig. 8.2 is a diagram of a cholinergic synapse.\npain TON\nstimulus\nOD\npresynaptic\nNIGRAM\nneurone opioid\nreceptor\nSIHT\nion channel B\nNI\nion channel A\nETIRW\nopioid\nTON\nreceptor\nOD\npostsynaptic\nion channel C\nneurone\nNIGRAM\nFig. 8.2\nSIHT\n(i) With reference to Fig. 8.2, describe the dif ferences between ion channel A and ion\nNI\nchannel B. ETIRW\n.........................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-mj-42-q08-a853f7409d.pdf"
    },
    {
      "id": "9700-2024-mj-42-q09",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-42-q09-p01-c858fa7216.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-42-q09-p02-f625a10aac.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:3)(cid:3)(cid:2)(cid:44)\n9 (a) The Asian common toad, Duttaphrynus melanostictus, arrived accidentally in eastern\nSIHT\nMadagascar in 2010 on a ship.\nNI\nETIRW\nFig. 9.1 shows an Asian common toad.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 9.1\nTON\nThe Asian common toad is now regarded as an invasive alien species because:\nOD\n• it breeds fast in the warm, wet conditions of eastern Madagascar\n• it can travel up to 2.5 km per year\n• its skin contains poison, which is highly toxic to humans and predators.\nNIGRAM\nSuggest ways in which the Asian common toad may af fect the ecosystem of eastern\nMadagascar.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [4]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Zoos work with local groups around the world to help conserve endangered species.\nSIHT\nOne example of this is the use of assisted reproduction, such as IVF , with endangered\nNI mammals such as the eastern black rhino.\nETIRW\nOutline the process of IVF .\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n.................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-mj-42-q09-e87669e01c.pdf"
    },
    {
      "id": "9700-2024-mj-42-q10",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_42.pdf?download=true",
      "source_pages": [
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-mj-42-q10-p01-c3cc14ba3a.png",
        "9700-topic-14-homeostasis/assets/9700-2024-mj-42-q10-p02-51468f71cb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-42-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:5)(cid:10)(cid:10)(cid:4)(cid:3)(cid:3)(cid:4)(cid:44)\n10 (a) In mammals, blood glucose concentration is maintained around a set point so that the\nSIHT\nmammal can function efficiently. This is an example of homeostasis.\nNI\nETIRW\nName the mechanism that maintains blood glucose concentration around a set point.\nTON .............................................................................................................................................. [1]\nOD\n(b) Fig. 10.1 shows how the blood glucose concentration varies before, during and after a period\nof exercise.\n7\nNIGRAM period of\nexercise\nSIHT\n6\nNI\nblood glucose ETIRW\nconcentration\ndm–3 mmol / TON\n5\nOD\n4 NIGRAM\n80 100 0 10 30 40 50 60 70 90 20\n/ min time\nSIHT\nNI Fig. 10.1\nETIRW\nExplain the results shown in Fig. 10.1 between 15 minutes and 70 minutes.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Type 1 diabetes in humans is a condition where the pancreas does not produce enough\nSIHT\ninsulin to control blood glucose concentration.\nNI\nETIRW\nT est strips are used to measure the concentration of glucose in a sample of urine. Two\nreactions take place on the test strip and a colour change occurs if glucose is present in\nTON the urine.\nOD\nThe equations for these reactions are:\nglucose oxidase\nA + hydrogen peroxide glucose + oxygen\nNIGRAM\nperoxidase\nSIHT oxidised B + water hydrogen peroxide + B\nNI\n(i) Identify compound A and the type of substrate represented by B.\nETIRW\nA ........................................................................................................................................\n",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-mj-42-q10-15a8cb6c3f.pdf"
    },
    {
      "id": "9700-2024-mj-43-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-mj-43-q01-p01-4bee932146.png",
        "9700-topic-14-homeostasis/assets/9700-2024-mj-43-q01-p02-b14756c8c9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Fig. 1.1 outlines the effect of antidiuretic hormone (ADH) on the cells of the collecting duct. The 1\ncell-signalling mechanism of ADH is similar to that of glucagon on liver cells.\ncollecting duct cell\nreceptor\nprotein\nlumen of blood\ncollecting\nduct\nADH\nA\ncell\nATP surface\nenzyme membrane\nadenylyl\ncascade\ncyclase water\nwater\nC water\nB\nP\nFig. 1.1\nName structures A, and C. (a) B\n............................................................................................................................................... A\n............................................................................................................................................... B\n............................................................................................................................................... C\n[3]\nADH is secreted by the posterior pituitary gland when the water potential of the blood (b)\ndecreases.\nSuggest reasons why the water potential of the blood may decrease.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDiabetes insipidus is a condition affecting osmoregulation by the kidney. One form of diabetes (c)\ninsipidus is caused by a tumour in the pituitary gland, which results in a decreased secretion\nof ADH.\nSuggest the symptoms that would occur in a person with diabetes insipidus.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNeurogenic diabetes insipidus (NDI) is another form of diabetes insipidus. In NDI, ADH (d)\nmolecules cannot bind to the receptor proteins located in the cell surface membranes of the\ncells of the collecting duct.\nWith reference to Fig. 1.1, explain the eff",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-mj-43-q01-8d40501486.pdf"
    },
    {
      "id": "9700-2024-mj-43-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        5
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-43-q02-p01-e9da05525d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Phenotypic variation exists in many forms. 2\nSome examples of phenotypic variation in plants and animals are described in Table 2.1. (a)\nComplete Table 2.1 by stating whether the cause of variation for each described example is\nlikely to be due to:\n• genetic factors, V\nG\n• environmental factors, V\nE\n• a combination of genetic and environmental factors, V + V .\nG E\nTable 2.1\ndescription of phenotypic variation cause of variation\nTomato plants grown in a glasshouse and grown outside vary in\nthe yield of tomatoes they produce. Seventeen genes associated\nwith tomato yield have been identified.\nNew strawberry plants from the variety called Sweet Ann are\nmade by asexual reproduction. The new plants grow to different\nsizes and produce different numbers of fruit.\nThe domestic cat has a blood group system with three possible\nblood types: A, B and AB. The blood types are determined by\nantigens present on the cell surface membrane of red blood cells.\nOver 50 genes have variants that are associated with excessive\nweight gain in humans. Other risk factors for excessive weight gain\ninclude diet and exercise.\nResting heart rate in humans varies between different individuals.\nSome factors that influence resting heart rate include: biological\nsex, family history of heart disease, number of cigarettes smoked,\nmedication taken.\n[3]\nName a spontaneous, random event occurring in cells that can be a source of phenotypic (b)\nvariation.\n............................................................................................................................................. [1]\nOther than the event named in (b), describe the features of sexual reproduction that contribute (c)\nto the production of genetically different offspring.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\n[Total: 7]",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-43-q02-89999f0a9e.pdf"
    },
    {
      "id": "9700-2024-mj-43-q03",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-mj-43-q03-p01-0b10913708.png",
        "9700-topic-16-inheritance/assets/9700-2024-mj-43-q03-p02-d9a1286a17.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "In plants and humans, the phenotype of an organism is determined by the genotype and the 3\nenvironment.\nPlants from the genus Primula have different petal colours. The presence of the pigment (a)\nmalvidin results in blue petals.\nThe metabolic pathway for malvidin synthesis is controlled by gene T/t. The presence of the\ndominant allele results in blue petals. T\nAnother gene, gene D/d, at a different locus, also influences the malvidin synthesis pathway.\nWhen the dominant allele is present, its gene product suppresses the malvidin synthesis D\npathway. This is summarised in Fig. 3.1.\nallele T\nprecursor malvidin blue petals\nprecursor no malvidin non-blue petals\nallele D\nFig. 3.1\nA genetic cross was carried out between two plants heterozygous at both gene loci. The\nresulting offspring genotypes are shown in a Punnett square in Fig. 3.2.\ntd tD Td TD\nTtDd TtDD TTDd TTDD TD\nTtdd TtDd TTdd TTDd Td\nttDd ttDD TtDd TtDD tD\nttdd ttDd Ttdd TtDd td\nFig. 3.2\nState which of the genotypes shown in Fig. 3.2 have blue petals. (i)\n...........................................................................................................................................\n..................................................................................................................................... [1]\nState the ratio of non-blue to blue petals for the cross shown in Fig. 3.2. (ii)\n..................................................................................................................................... [1]\nName the type of gene interaction that has caused the offspring ratio you have stated (iii)\nin (a)(ii).\n..................................................................................................................................... [1]\nGene T/t and gene D/d code for proteins that are involved in the control of the production (iv)\nof malvidin.\nDiscuss the possible roles of the proteins coded for by gene T/t and gene D/d in the\ncontrol of the production of malvidin.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-mj-43-q03-57e39bec72.pdf"
    },
    {
      "id": "9700-2024-mj-43-q04",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-43-q04-p01-813093b284.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-43-q04-p02-93ef4332b1.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-43-q04-p03-8278ae4c1a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genetic engineering is a technique used to modify the genetic material of a specific organism to 4\nchange a characteristic.\nGenetic engineering uses specific enzymes and commonly involves the use of plasmids (a) (i)\nfor the transfer of genes into an organism.\nFour enzymes that are used in genetic engineering techniques involving plasmids are:\n• restriction endonuclease\n• DNA ligase\n• DNA polymerase\n• reverse transcriptase.\nOutline the role of these enzymes in genetic engineering involving plasmids.\nrestriction endonuclease ...................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\nDNA ligase ........................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\nDNA polymerase ...............................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\nreverse transcriptase .........................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n[4]\nExplain why a promoter, as well as the desired gene, is often transferred into an organism. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...............................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-mj-43-q04-2b087d72f1.pdf"
    },
    {
      "id": "9700-2024-mj-43-q05",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-43-q05-p01-0695ee60f4.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-mj-43-q05-p02-da58468eb7.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "BRCA2 is a tumour-suppressor gene. Its gene product, BRCA2, is involved in DNA repair. If the 5\nDNA cannot be repaired, BRCA2 has a role in causing the cell to die. BRCA2 is found in cells of\nbreast tissue.\nWhen a mutation occurs in BRCA2, damaged DNA may not be repaired and this increases the\nrisk of breast cancer.\nWhen a double-stranded piece of DNA breaks, BRCA2 binds to the damaged DNA directly and\ninteracts with the enzyme RAD51 to repair the damage.\nRepairing DNA prevents other mutations and gene rearrangements from occurring which could\notherwise lead to breast cancer.\nDouble-stranded DNA breaks occur naturally during meiosis. (a)\nState the event that is initiated as a result of double-stranded breaks during meiosis.\n............................................................................................................................................. [1]\nScientists have identified hundreds of mutations in BRCA2, but not all of these mutations will (b)\nincrease the risk of cancer. One specific mutation in BRCA2, known as 999del5, is found in\n0.6% of the general global population.\nIceland is an island country in the North Atlantic Ocean. The ancestors of most of the current\npopulation are people who arrived to settle in Iceland in AD 874.\nIn the Icelandic population, mutation 999del5 is the cause of 7–8% of breast cancer cases in\nwomen and 40% of breast cancer cases in men. This is much higher than the percentage of\nbreast cancer cases in the general global population.\nSuggest explain how mutation 999del5 accounts for a very high percentage of (i) and\nbreast cancer cases in Iceland compared with the general global population.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nOne of the largest global genetic screening programmes for breast cancer involves (ii)\nidentifying people with mutations in BRCA2.\nOutline the advantages of genetic screening for mutations in BRCA2.\n..................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-mj-43-q05-edfb270b15.pdf"
    },
    {
      "id": "9700-2024-mj-43-q06",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        15
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-mj-43-q06-p01-3126ca974b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "In plants, stomata open and close in response to changes in environmental conditions. 6 (a)\nExplain why stomata need to open and close according to environmental conditions.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nDescribe the mechanism occurring in guard cells that leads to the opening of a stoma. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..........................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-mj-43-q06-51bfb8c5ab.pdf"
    },
    {
      "id": "9700-2024-mj-43-q07",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-43-q07-p01-271bbce5ad.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-43-q07-p02-3bef6832e6.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-mj-43-q07-p03-04fe909178.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Experiments were carried out to determine the effect of light intensity on the rate of 7 (a)\nphotosynthesis of a species of the unicellular protoctist, Chlorella. A cell suspension of\nChlorella was used.\nCarbon dioxide uptake was used as a measure of the rate of photosynthesis.\n• The suspension of Chlorella was illuminated at a light intensity of 3 lux for 20 seconds.\n• The carbon dioxide uptake by Chlorella was measured at the end of the 20 second\nperiod of illumination.\n• The experiment was repeated at 6 lux, 9 lux, 12 lux and in a dark room.\n• °C. The suspension was maintained at a temperature of 20\nTable 7.1 shows the results of the experiments.\nTable 7.1\ntotal CO uptake after\n2 rate of photosynthesis light intensity\n20 seconds\n–1 μμmol / s / lux\nμμmol /\n0 0 0.0\n3 20 1.0\n6 44\n…………….\n9 72 3.6\n12 80 4.0\nUse Table 7.1 to calculate the rate of photosynthesis at a light intensity of 6 lux. (i)\nComplete Table 7.1 by writing your calculated value in the space provided. [1]\nPlot a graph of the data in Table 7.1 on the grid in Fig. 7.1 to show the effect of light (ii)\nintensity on the rate of photosynthesis.\nDraw a curve extend your curve to show what would happen to the rate of and\nphotosynthesis if the experiment is carried out at 18 lux.\n5\n4\nrate of 3\nphotosynthesis\ns−1 μmol / 2\n1\n0\n0 2 4 6 8 10 12 14 16 18\n/ lux light intensity\nFig. 7.1\n[3]\nSuggest an explanation for the shape of your curve from 12 lux to 18 lux. (iii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nIn photophosphorylation, photoactivation of chlorophyll results in the synthesis of ATP. (b)\nDescribe how photoactivation of chlorophyll results in the synthesis of ATP in\nphotophosphorylation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-mj-43-q07-b451cc301f.pdf"
    },
    {
      "id": "9700-2024-mj-43-q08",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-43-q08-p01-f73c9de6c3.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-mj-43-q08-p02-7460f051eb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q08",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "Fig. 8.1 is a diagram of a mitochondrion. 8 (a)\ninner outer membrane\nmembrane\nFig. 8.1\nOutline the roles played by the mitochondrial membranes in respiration.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nThe shapes and numbers of mitochondria are continually changing due to fission. Fission (b)\noccurs when one mitochondrion splits to form two mitochondria.\nSuggest reasons why mitochondria carry out fission. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nGuanosine ",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-mj-43-q08-dfca35fd47.pdf"
    },
    {
      "id": "9700-2024-mj-43-q09",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-43-q09-p01-358f0446cf.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-mj-43-q09-p02-2f5a75b7c0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Fig. 9.1 shows how the mean global atmospheric carbon dioxide concentration has changed 9 (a)\n×103) years leading up to the year 2020. over the 800 000 (800\n810\n720\n630 mean global\natmospheric\nA\ncarbon dioxide 540\nconcentration\nm−3 mg / 450\n360\n270\n800 700 600 500 400 300 200 100 0\n×103 years before 2020\nthe year 2020\nFig. 9.1\nCalculate the percentage increase in carbon dioxide concentration between point and (i) A\nthe year 2020.\nShow your working. Write your answer to one decimal place.\npercentage increase = ......................................................% [2]\nSuggest how the changes in carbon dioxide concentration between and the year 2020 (ii) A\nmay have affected the environment and biodiversity.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [4]\nOutline reasons for maintaining biodiversity. (b) plant\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-mj-43-q09-e4d7188241.pdf"
    },
    {
      "id": "9700-2024-mj-43-q10",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_43.pdf?download=true",
      "source_pages": [
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-43-q10-p01-a9de3494f9.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-mj-43-q10-p02-515dbce20e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-43-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 10.1 is a diagram of part of a neurone membrane while the resting potential is maintained. 10 (a)\ntissue fluid\ncell\nsurface\nmembrane\naxoplasm\nFig. 10.1\nOn Fig. 10.1, use label lines and letters to label:\n– potassium ions K\n– ATP. A\n[2]\nDescribe the sequence of events that occur during an action potential. (b)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nTable 10.1 shows the axon diameter, myelination and transmission speed of impulses of (c)\nmotor neurones for three animals: squid, cockroach and cat.\nTable 10.1\naxon diameter transmission speed\nanimal myelination\n–1 / mm / m s\nsquid 1.5 no 30\ncockroach 0.05 no 10\ncat 0.02 yes 100\nDescribe suggest explanations for the results shown in Table 10.1. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-mj-43-q10-d0a5a48f89.pdf"
    },
    {
      "id": "9700-2024-mj-51-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-51-q01-p01-74fd015ccb.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q01-p02-f6889ed024.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q01-p03-03e59e2ef5.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q01-p04-16753dd7f9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "A student investigated the effect of wind speed on the rate of transpiration. The student used the 1\nflowering plant Japanese spiraea, Spiraea japonica.\nThe student used the apparatus shown in Fig. 1.1 to measure the rate of transpiration.\nJapanese spiraea plant\nairtight seal\nfan\nplastic bag\ncontainer\nsoil\n00 : 00\nbalance\n740.05 g\nstop-clock\nFig. 1.1\nTo set up the apparatus the student:\n• obtained a Japanese spiraea plant growing in a container of soil\ncm3 • added 200 of water to the soil\n• placed a plastic bag around the container of soil to prevent water loss from the soil\n• placed the plant and container on the balance\n• switched on the fan to a low setting.\nThe roots of the Japanese spiraea plant absorbed water from the soil. Water was carried in the\nxylem and water vapour was lost by transpiration from the leaves of the plant. This caused the\nreading on the balance to decrease during the investigation.\nIdentify the variable in this investigation. (a) (i) independent\n...........................................................................................................................................\n..................................................................................................................................... [1]\nThe student carried out the investigation in a laboratory with standard laboratory apparatus. (ii)\nDescribe a method, using the apparatus shown in Fig. 1.1, that the student could use to\ninvestigate the effect of wind speed on the rate of transpiration by Japanese spiraea.\nYour method should be set out in a logical order and be detailed enough to allow another\nperson to follow it.\nDetails of how to set up the apparatus shown in Fig. 1.1 should be included. not\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-51-q01-6942bef746.pdf"
    },
    {
      "id": "9700-2024-mj-51-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_51.pdf?download=true",
      "source_pages": [
        7,
        8,
        9,
        10,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p01-5f11f3da13.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p02-c20f244079.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p03-3f386547dd.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p04-573af7463e.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p05-af458c11a3.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q02-p06-6ad4990078.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "The responses of plant species to water stress can be classified as either isohydric or anisohydric. 2\n• Isohydric plant species close stomata during times of water stress. This behaviour minimises\nwater loss by transpiration but also reduces carbon dioxide uptake for photosynthesis.\n• Anisohydric plant species do not close stomata during times of water stress. This behaviour\nmaximises carbon dioxide uptake for photosynthesis but also increases water loss\nby transpiration.\nA biologist studied 10 tree species from Australia. The biologist studied the effect of water\nstress high environmental temperatures on five isohydric tree species and five anisohydric and\ntree species.\nFor each tree species studied:\n• The biologist obtained 20 young trees.\n• The young trees were grown in containers of soil in controlled conditions in a glasshouse.\n• The environmental conditions in the glasshouse were chosen to represent summer conditions\n°C. in Australia. The mean glasshouse temperature was 28\n• All the young trees were given a good supply of water for 10 weeks, so that the young trees\nacclimatised to the environmental conditions in the glasshouse.\nState environmental conditions in the glasshouse that should be standardised in the (a) three\n10-week period of acclimatisation, the temperature of the glasshouse. other than\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nAfter the 10-week period of acclimatisation, the biologist divided the young trees from each\nspecies into four groups of five trees.\nTable 2.1 shows the experimental conditions used by the biologist for the next five weeks.\nTable 2.1\ngroup experimental conditions\nThe young trees were given a good supply of water for five weeks. 1\n°C. The mean glasshouse temperature during weeks 1 to 5 was 28\nThe young trees were given a good supply of water for five weeks. 2\n°C. The mean glasshouse temperature during weeks 1 to 4 was 28\n°C. During week 5, the mean glasshouse temperature was increased to 35\nThe young trees were given a reduced supp",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-51-q02-989cf5b109.pdf"
    },
    {
      "id": "9700-2024-mj-51-q03",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_51.pdf?download=true",
      "source_pages": [
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-51-q03-p01-6178d53317.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q03-p02-640c490817.png",
        "9700-experimental-skills/assets/9700-2024-mj-51-q03-p03-5874b4fbe0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-51-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Female mosquitoes feed on human blood. Some species of mosquitoes are vectors of human 3\npathogens. For example, female mosquitoes of the species Aedes aegypti transmit the pathogen\nthat causes the disease yellow fever.\nMosquitoes have sensory receptors that can detect chemicals in the air. Mosquito repellents\ncontain chemicals that are sprayed onto the skin of humans to prevent mosquitoes taking\nblood meals.\nSome scientists investigated the effectiveness of different mosquito repellents using a human (a)\nvolunteer. The scientists carried out the investigation on this person in a laboratory.\nFig. 3.1 is a diagram of the experiment before the start of the investigation.\nfan\ns–1) wind speed (2 m\nfront back\nsection 1 section 2 section 3\nhuman\n1 m distance\nvolunteer\nmosquito cage with\nfine net walls\nwide net to allow\nmosquitoes to move\nbetween sections\nFig. 3.1\nThe same procedure was used for each mosquito repellent studied.\n• A human volunteer sat 1 m from a mosquito cage, as shown in Fig. 3.1.\n• The person sprayed some mosquito repellent onto the skin of both arms.\n• A fan was turned on so that air moved from the person towards the mosquito cage.\nThe moving air carried chemicals, including the mosquito repellent, from the person into\nthe mosquito cage.\n• The scientists added 100 female mosquitoes of A. aegypti to section 2 of the\nmosquito cage.\n• The mosquitoes were left in the mosquito cage for 15 minutes. The walls of the mosquito\ncage were made of fine net to prevent mosquitoes leaving the cage. The mosquitoes\nmoved freely between sections 1, 2 and 3.\n• After 15 minutes, the scientists counted the number of mosquitoes in each section of\nthe mosquito cage. The percentage of mosquitoes in section 1 of the mosquito cage\nwas calculated.\n• This procedure was then repeated three times on different days, using the same person.\nState a suitable control for this investigation. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSuggest risk to the when carrying out this investigation state a (ii) one scientists and\nsuitable precaution they should take.\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nTable 3.1 shows the results of the investigation. (b)\nTable 3.1\nchemical in mosquito percentage percentage of mosquitoes in\nrepelle",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-51-q03-11da8dc95f.pdf"
    },
    {
      "id": "9700-2024-mj-52-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 19,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p01-0df8828101.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p02-e9f3728a2f.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p03-4f48d49138.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p04-0882ad158f.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p05-a96bda3eda.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q01-p06-35047f5661.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:6)(cid:5)(cid:3)(cid:3)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Tree plantations are areas where trees are planted for a particular purpose. Some tree plantations\nSIHT\nincrease the supply of wood for construction and fuel. In many plantations, fast‑growing, alien tree\nNI species are planted.\nETIRW\nUndergrowth is found in and around plantations, increasing plant biodiversity . Undergrowth is\nTON mainly made up of indigenous (native) small plants and shrubs.\nOD\nFig. 1.1 shows part of a plantation of eucalyptus trees.\nNIGRAM\neucalyptus trees\nSIHT\nNI\nETIRW\nTON\nsmall plants\nOD and shrubs in\nundergrowth\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 1.1\nOD\nBangladesh has many tree plantations as part of a national tree‑planting programme.\nTwo fast‑growing, alien tree species planted in Bangladesh are acacia, Acacia auriculiformis, and\nNIGRAM\neucalyptus, Eucalyptus camaldulensis.\nAcacia and eucalyptus are considered to be invasive species as they outcompete indigenous tree\nSIHT\nspecies such as sal tree, Shorea robusta, and mango, Mangifera indica.\nNI\nETIRW\nScientists carried out an investigation into the biodiversity of plant species in the undergrowth in\nplantations of alien trees compared with the undergrowth in plantations of indigenous trees.\nTON\nThe hypothesis that the scientists tested was stated as:\nOD\nThe undergrowth in plantations of alien tree species will have a lower biodiversity of plant\nspecies than the undergrowth in plantations of indigenous tree species.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) (i) Identify the independent variable in this investigation.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n..................................................................................................................................... [1]\nTON\nFor the investigation, plantations were selected within the same region of Bangladesh with the\nOD\nsame environmental conditions.\n• Three plantations of each tree species were selected: acacia, eucalyptus, sal tree andm ango.\n• A plot measuring 36 m by 36 m was studied within each plantation.\nNIGRAM\n• For each plot, belt transects were used to collect the data needed to determine Simpson’ s\nindex of diversity (D).\nSIHT\nFor each type of plantation, Simpson’s index of diversity (D) was calculated for the plant species\nNI\nin the undergrowth.\nETIRW\nEach plot was studied in April, July and November.\nTON\nOD (ii) Identify two variables the scientists standardised in this investigation.\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-52-q01-69e1cc3ce4.pdf"
    },
    {
      "id": "9700-2024-mj-52-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-52-q02-p01-2f3baccb05.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q02-p02-a9ff2c62aa.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q02-p03-421796ec4f.png",
        "9700-experimental-skills/assets/9700-2024-mj-52-q02-p04-db5e912ffc.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:2)(cid:10)(cid:7)(cid:6)(cid:6)(cid:5)(cid:3)(cid:3)(cid:2)(cid:1)(cid:9)(cid:44)\n2 Tumours may be described as benign or malignant. Malignant tumours can lead to greater\nSIHT\ncomplications for the person with the tumour.\nNI\nETIRW\nEarly identification of tumours, particularly malignant tumours, is important for effective treatment.\nBenign tumour cells and malignant tumour cells can look similar when viewed using a light\nTON microscope.\nOD\nSome scientists wanted to develop a diagnostic test to identify tumour cells as benign or malignant.\nThe scientists investigated whether the diameter of the cell nucleus could be used to identify the\ntype of tumour cell as benign or malignant.\nNIGRAM\nThe scientists used a light microscope with a calibrated eyepiece graticule to measure the\ndiameter of the nuclei of stained tumour cells.\nSIHT\nFig. 2.1 shows a photomicrograph of stained tumour cells viewed using a light microscope.\nNI\nETIRW\nAn eyepiece graticule was placed across the nucleus of one of the tumour cells.\nTON\n0 OD\n0\n1\n0\n9\n0\n8\nNIGRAM\n0\n7\ncell surface\n0 SIHT\n6 membrane\n0 NI\n5\nETIRW\nnucleus 0\n4\nTON\n0\n3\ncell X\nOD\n0\n2\n0 eyepiece\n1\ngraticule\n0 NIGRAM\nimage\nSIHT magnification\n×400\nNI\nETIRW\nFig. 2.1\nTON\nThe calibration of the eyepiece graticule scale is:\nOD\none eyepiece graticule division = 320 nm\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) Use the calibration of the eyepiece graticule scale to calculate the actual diameter of the\nSIHT\nnucleus of cell X, shown in Fig. 2.1.\nNI\nETIRW\nShow your working and state your answer in µm.\nTON\nOD\nNIGRAM\nactual diameter of the nucleus of cell X = ................................................... µm [3]\nSIHT\nNI (b) For the investigation into whether the diameter of the nucleus could be used to identify the\nETIRW\ntype of tumour cell, tumours of the thyroid gland (an endocrine gland) were used.\nTON For the diagnostic test the scientists carried out a procedure using two dif ferent stains.\nOD\nstep 1 5 0 people diagnosed with benign tumours and 24 people diagnosed with malignant\ntumours were selected.\nstep 2 A sample of tumour cells was removed from the thyroid gland of each person.\nNIGRAM\nstep 3 T he cells were stained with either Papanicolaou stain (Pap) or haematoxylin and\nSIHT eosin stain (H&E).\nNI\nstep 4 S amples were viewed using a light microscope with a magnification of ×400.\nETIRW\nstep 5 I n each sample, the diameters of 100 nuclei were measured.\nTON\nOD step 6 T he measurements were made by one of the scientists who did not know the type of\ntumour cell they were measuring.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nT able 2.1 shows the results of the investigation.\nSIHT\nNI Table 2.1\nETIRW\nstandard standard category mean nuclear\nTON\n/µµ m deviation /µµ m error /µµ m diameter\nOD\n7.3 total benign cells (50 people) 5000 cells: 0.8 ± ± 0.011\n• Pap‑stained cells (41 people) 4100 cells 7.3 ± 0.7 ± 0.011\nNIGRAM\n• H&E‑stained cells (9 people) 900 cells 7.2 ± 1.0 ± 0.033\n9.0 ± 0.6 ±",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-52-q02-8a2a10e58d.pdf"
    },
    {
      "id": "9700-2024-mj-53-q01",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 19,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-53-q01-p01-8614862e02.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q01-p02-3559e86bde.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q01-p03-a72ce8a07d.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q01-p04-ce4d33335a.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q01-p05-e20c1ea1a9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Blood vessels must be able to withstand and maintain varying blood pressures. 1\nThere are several methods that could be used to determine the strength of a blood vessel. One\nmethod is called the circumferential tensile strength (CTS) test, as shown in Fig. 1.1.\nclamp stand\nhook\nring of blood vessel\nmass hanger\n10 g masses\nFig. 1.1\nThe ring of blood vessel, shown in Fig. 1.1, was cut from a length of artery or vein that was\nprepared by removing surrounding tissues.\nA student wanted to use the apparatus in Fig. 1.1 to determine the mass needed to break an\nartery and the mass needed to break a vein.\nIdentify the independent variable the dependent variable in this investigation. (a) (i) and\nindependent variable .........................................................................................................\n...........................................................................................................................................\ndependent variable ............................................................................................................\n...........................................................................................................................................\n[1]\nThe student was provided with lengths of a large artery, lengths of a large vein and (ii)\nstandard laboratory apparatus.\nDescribe how the student could use the CTS test to determine the mass needed to break\nan artery and the mass needed to break a vein.\nYour method should be set out in a logical order and be detailed enough to let another\nperson follow it.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n......",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-53-q01-c1314d419a.pdf"
    },
    {
      "id": "9700-2024-mj-53-q02",
      "subject": "9700",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-may-june/9700_s24_qp_53.pdf?download=true",
      "source_pages": [
        7,
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-May-June/qp/9700_s24_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-mj-53-q02-p01-c8184036f7.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q02-p02-6e825cb501.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q02-p03-17c46588a9.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q02-p04-cce9359e41.png",
        "9700-experimental-skills/assets/9700-2024-mj-53-q02-p05-ee7fb6011a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "Gibberellins are a group of plant hormones that are involved in the elongation of plant stems. 2\n, A student investigated the effect of two different concentrations of a gibberellin, known as GA\n3\non stem elongation of 10-day old pea seedlings.\n10–4 dm–3. × with a concentration of 2.85 mol The student was given a stock solution of GA\n3\nwith a concentration of Describe a method the student could use to make a solution of GA (a)\n3\n10–6 dm–3 × mol state the dilution factor used. 1.90 and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\nIn the investigation, the student: (b)\n10–4 dm–3 • × used the two concentrations of GA : 2.85 mol (high concentration GA ) and\n3 3\n10–6 dm–3 × mol (low concentration GA ) 1.90\n3\n• applied the high concentration GA to one batch of 10-day-old pea seedlings\n3\n• applied the low concentration GA to another batch of 10-day-old pea seedlings\n3\n• applied distilled water to a third batch of 10-day-old pea seedlings, as a control\n• standardised all other variables\n• measured the length of the stem of each seedling every two days until the seedlings\nwere 20 days old\nday–1. • calculated the rate of stem elongation in cm\nState way in which the student could standardise the measuring of stem length. one\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nThe results of the investigation are shown in Table 2.1. (c)\nTable 2.1\nage of seedling mean stem length mean stem mean stem length\n/ days with high length with low with distilled water\nconcentration GA concentration GA added / cm\n3 3\nadded / cm added / cm\n10 2 2 2\n12 4 2 3\n14 7 5 4\n16 6 12 7\n18 20 15 8\n20 25 21 8\nrate of stem\n2.3 0.6 elongation\n–1 ………………… / cm day\nday–1 for the Complete Table 2.1 by calculating the rate of stem elongation in cm (i)\nadded. seedlings with low concentration GA\n3\n[1]\nA scientist stated that they did not have enough confidence in the results in Table 2.1 to (ii)\nmake any conclusions.\nState reaso",
      "pdf_path": "9700-experimental-skills/print/9700-2024-mj-53-q02-caead0e6ff.pdf"
    },
    {
      "id": "9700-2024-on-41-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-41-q01-p01-5ae5f3340e.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-41-q01-p02-3ff8b7d6f1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Different species of animal have neurones with different characteristics.\nSIHT\nNI (a) Fig. 1.1 is a diagram of a motor neurone of a rat and a motor neurone of a snail.\nETIRW\naxon (diameter of 7 μm)\nTON\nOD\nrat\nmotor\nneurone\nNIGRAM\nSchwann cell\nSIHT\nC\nNI\nB cell body ETIRW A\nTON\nOD\nsnail\nmotor\nneurone\nNIGRAM\nμm) axon (diameter of 40\nSIHT\nNI\nFig. 1.1\nETIRW\n(i) Name the structures labelled A, B and C on Fig. 1.1.\nTON\nA ........................................................................................................................................ OD\nB ........................................................................................................................................\nC ........................................................................................................................................ NIGRAM\n[3]\nSIHT s–1. The snail motor (ii) The rat motor neurone has an impulse transmission speed of 50 m\ns–1. neurone has an impulse transmission speed of 8 m\nNI\nETIRW\nExplain why the rat motor neurone has a faster impulse transmission speed than the\nsnail motor neurone.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n..................................................................................................................................... [2]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Fig. 1.2 shows an action potential in a rat neurone and Fig. 1.3 shows an action potential in a\nSIHT\nsnail neurone.\nNI\nETIRW rat snail\n60 60\nTON\n40 40\nOD\n20 20\nNIGRAM 0 membrane membrane 0\npotential potential\n/ mV mV / −20 −20\nSIHT\nNI −40 −40\nETIRW\n−60 −60\nTON\nOD −80 −80\n5 15 20 25 30 0 5 10 15 20 25 30 0 10\n/ ms / ms time time\nNIGRAM\nFig. 1.2 Fig. 1.3\nContrast the two action potentials shown in Fig. 1.2 and Fig. 1.3. SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n..........................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-on-41-q01-e9f6c60f01.pdf"
    },
    {
      "id": "9700-2024-on-41-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-41-q02-p01-0799636912.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-41-q02-p02-7fc8a943a1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q02",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 The leaves of many plants have stomata that show a regular daily rhythm of opening and closing\nSIHT\nover a period of 24 hours.\nNI\nETIRW\n(a) Explain why it is important for plants to open and close their stomata in a daily rhythm.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [3]\nETIRW\n(b) Fig. 2.1 shows the results of an experiment to monitor this rhythm over three days and nights\nTON\nfor the plant Arabidopsis thaliana. The percentage of open stomata is shown. Each day\nOD consisted of 14 hours of light (white bar) and each night consisted of 10 hours of darkness\n(black bar).\n100\nNIGRAM Key\nday\nopen night\nSIHT\nstomata 50\n/ % NI\nETIRW\nTON\n0\n24 0 48 72 OD\ntime / hours\nFig. 2.1\nNIGRAM\nFig. 2.1 shows that the percentage of open stomata increases in the first seven hours of the\nexperiment. Describe the sequence of changes that occurs in the guard cells that leads to the\nSIHT\nstomata opening.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n.....................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-on-41-q02-02e317f4a6.pdf"
    },
    {
      "id": "9700-2024-on-41-q03",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-41-q03-p01-8c9128fad3.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-41-q03-p02-bac94a580f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 During aerobic respiration, cells respire substrates such as glucose to produce ATP.\nSIHT\nNI Some events that occur during aerobic respiration are:\nETIRW\n• The respiratory substrate breaks down into smaller and smaller molecules. These series of\nTON reactions are described as catabolism.\n• Coenzymes take part in various reactions. In some reactions, coenzymes are reduced or\nOD\noxidised.\n• Carbon dioxide is released.\n(a) Aerobic respiration occurs in four successive stages: glycolysis (G), link reaction (LR), Krebs\nNIGRAM\ncycle (KC) and oxidative phosphorylation (OP).\nComplete Table 3.1 to show which events occur in each stage of aerobic respiration. Use SIHT\na tick (✓) to show that the event does occur or a cross (✗) to show that the event does\nNI\nnot occur.\nETIRW\nTable 3.1\nTON\nOD\nstage\nevent\nG LR KC OP\nNIGRAM catabolism\nSIHT coenzyme is reduced or oxidised\nNI\na coenzyme forms a covalent bond with a ETIRW\nrespiratory intermediate\nTON\ncarbon dioxide is released\nOD\n[4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) A new hand‑held technological device shows the main type of respiratory substrate being\nSIHT\nused in the cells of a person.\nNI\nETIRW\nThe device consists of a carbon dioxide sensor and air‑flow meter . The person inhales\nthrough the device for a fixed time and then exhales into it.\nTON\nThe device calculates the respiratory quotient (RQ) value to show whether the cells are\nOD\nmainly respiring carbohydrates or lipids.\n(i) Explain how the device calculates the RQ value and how this shows whether the cells\nare mainly respiring carbohydrates or lipids.\nNIGRAM\n… ........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n… ........................................................................................................................................\nTON\n… ........................................................................................................................................\nOD\n… ........................................................................................................................................\n… ........................................................................................................................................\nNIGRAM\n… ........................................................................................................................................\nSIHT\n..................................................................................................................................... [4]\nNI\nETIRW (ii) State the dif ference in the relative energy values of carbohydrates and lipids as\nrespiratory substrates, and explain",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-on-41-q03-baa349fc66.pdf"
    },
    {
      "id": "9700-2024-on-41-q04",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-on-41-q04-p01-7c8a8d806b.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-on-41-q04-p02-81ce24efb4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q04",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 The leaves of Mim osa pudica plants are made of a number of structures known as pinnae. The\nSIHT\npinnae fold when the leaf is touched. This closes the leaf.\nNI\nETIRW\nFig. 4.1 shows an open leaf of M. pudica before it is touched. Fig. 4.2 shows the same leaf that\nhas closed after being touched.\nTON\nOD\nNIGRAM\nSIHT\npinnae NI\nETIRW\nTON\nA open (not touched) closed (touched) B\nOD\nFig. 4.1 Fig. 4.2\n(a) A touch stimulus to an M. pudica leaf causes an action potential to be generated.\nNIGRAM\nThe action potential results in changes in cells, which cause the leaf to close.\nSIHT\nFig. 4.3 shows the mechanism in M . pudica cells that causes the leaf to close.\nNI\nETIRW\n− K+ cells water moves Cl and protons action\nbecome out of cells ions move out pumped into potential after\nTON flaccid by osmosis of cells extensor cells touch stimulus\nOD\nFig. 4.3\nThe leaves of M . pudica and the leaves of Venus fly traps move in response to touch stimuli,\nNIGRAM but the mechanisms that cause the responses are different.\nDescribe the dif ferences between the mechanism shown in Fig. 4.3 and the mechanism that\nSIHT\ncauses the closure of the modified leaves in Venus fly traps.\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [3]\nETIRW\nTON\nNIGRAM\n(b) The rate of photosynthesis decreases by 40% when the leaves of M. pudica close.\nSIHT\nNI Explain why the rate of photosynthesis decreases when the leaves of M. pudica close.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n.......",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-on-41-q04-cd35ab77f2.pdf"
    },
    {
      "id": "9700-2024-on-41-q05",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-41-q05-p01-0ed56e73cb.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-41-q05-p02-fdd15b4964.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 Lunularia cruciata is a primitive plant. Its body consists of a flattened sheet of photosynthetic\nSIHT\ntissue called a thallus.\nNI\nETIRW\nFig. 5.1 shows L. cruciata with two dif ferent types of reproductive structure, labelled A and B, on\nits surface.\nTON\nOD\nNIGRAM\nSIHT A\nNI\nETIRW\nTON\nB\nOD\nNIGRAM\nFig. 5.1\n(a) Structures A and B and the thallus of L. cruciata are haploid.\nSIHT\nNI Explain what is meant by haploid.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\n(b) Structure A contains pale discs of tissue, C, that can germinate into new L. cruciata. The new\nSIHT plants that develop from C are genetically identical to the parent plant in Fig. 5.1.\nNI\nStructure B contains male sperm that are chemically attracted to swim to female eggs on a\nETIRW\nneighbouring parent plant. When the egg and sperm fuse, they form structure D. Structure D\ndevelops to produce spores that grow into new plants that are genetically dif ferent from the\nTON\ntwo parent plants.\nOD\nIdentify which of the structures A, B, C and D are:\nassociated with sexual reproduction …………………………………………\nNIGRAM\nproduced by mitosis …………………………………………\nSIHT\nthe site of meiosis …………………………………………\nNI\nETIRW Each letter may be used once, more than once, or not at all. [3]\nTON\nNIGRAM\n(c) Fig. 5.2 shows a horse, Equus caballus. Horses are diploid animals that reproduce sex ually.\nSIHT\nMale and female horses produce gametes, which fuse to form genetically different offspring.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 5.2\nOD\nExplain the need for a reduction division during meiosis in the production of gametes in\nanimals such as horses.\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n.........................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-41-q05-8e0d6b0e38.pdf"
    },
    {
      "id": "9700-2024-on-41-q06",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-on-41-q06-p01-e972c74eaf.png",
        "9700-topic-14-homeostasis/assets/9700-2024-on-41-q06-p02-c874034197.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n6 The mammalian kidney is responsible for:\nSIHT\n• NI the excretion of urea\nETIRW • osmoregulation (the homeostatic control of the water potential of the blood).\nTON (a) (i) Outline how and where the excretory product urea is made in the body.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\n(ii) Homeostatic control of the water potential of blood includes receptors, ef fectors and SIHT\ntarget cells.\nNI\nETIRW\nIdentify the names and locations of these components of homeostatic control in\nosmoregulation.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n..................................................................................................................................... [3]\nTON\n(b) The glomerulus and Bowman’s capsule of the nephron are important in the formation of urine.\nOD\nOutline the role of the glomerulus and Bowman’ s capsule in the formation of urine.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n..............................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-on-41-q06-c079e5ce8b.pdf"
    },
    {
      "id": "9700-2024-on-41-q07",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        14,
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-41-q07-p01-1ff95d78eb.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-41-q07-p02-36279164e9.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-41-q07-p03-05c04562a8.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-41-q07-p04-ee7f7b9741.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q07",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n7 Spea multiplicata is one of several species of American spadefoot toad.\nSIHT\nNI (a) Young spadefoot toads are called tadpoles and live in water in ponds.\nETIRW\nS. multiplicata tadpoles show three dif ferent phenotypes due to genetic variation. The three\nTON phenotypes are: detritus feeder, intermediate and carnivore.\nOD\nDetritus feeders are small, and carnivores are large. Intermediates vary in size between the\ntwo extremes.\nA detritus feeder and a carnivore are shown in Fig. 7.1.\nNIGRAM\nSIHT\nNI\nETIRW\ncarnivore\nTON\nOD\ndetritus feeder\nNIGRAM fairy shrimp\nSIHT\nNI\nETIRW Fig. 7.1\nTON Detritus feeders:\nOD\n• eat detritus (small pieces of dead organic matter) and algae (photosynthetic protoctists)\n• have smooth mouthparts, small jaw muscles and long intestines.\nIntermediates:\nNIGRAM\n• can eat all available food (detritus, algae and fairy shrimps)\n• have teeth‑like mouthparts, medium‑sized jaw muscles and medium‑sized intestines. SIHT\nNI\nCarnivores:\nETIRW\n• eat fairy shrimps and other small animals\nTON\n• have teeth‑like mouthparts, large jaw muscles and short intestines.\nOD\nScientists counted the number of each type of tadpole in two dif ferent ponds: pond 1 and\npond 2.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) In pond 1, the scientists observed:\nSIHT\n• NI a high density of tadpoles\nETIRW • a low abundance of food\n• that most of the tadpoles they counted were either detritus feeders or carnivores,\nTON with very few intermediates present.\nOD\nDescribe and suggest explanations for the type of natural selection that appears to be\nacting in pond 1.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [3]\nNIGRAM\n(ii) In pond 2, the scientists observed:\nSIHT\n• a low density of tadpoles\n• sufficient food availability for all tadpoles NI\n• ETIRW that most of the tadpoles they counted were intermediates, with fewer detritus\nfeeders or carnivores.\nTON\nDescribe and suggest explanations for the type of natural selection that ap",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-on-41-q07-166c98d8bc.pdf"
    },
    {
      "id": "9700-2024-on-41-q08",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-on-41-q08-p01-37484094f6.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-on-41-q08-p02-e10975aa2d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 Scientists use many different techniques in genetic engineering.\nSIHT\nNI (a) Sometimes the gene for genetic engineering cannot be extracted from the donor organism.\nETIRW\nInstead, the gene is synthesised using one of two different methods.\nTON Outline the two methods for synthesising a gene for use in genetic engineering.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ............................................................................................................................................. [3]\n(b) DNA ligase and DNA polymerase are two enzymes that are used in genetic engineering.\nNIGRAM Complete Table 8.1 to show the roles of DNA ligase and DNA polymerase in genetic\nengineering.\nSIHT\nUse a tick (✓) if the enzyme has the role or a cross (✗) if the enzyme does not have the role.\nNI\nETIRW Table 8.1\nTON\nrole in genetic engineering DNA ligase DNA polymerase\nOD\njoins two sections of sugar\nphosphate backbone in DNA\nadds a gene to a plasmid\nNIGRAM\nadds free activated DNA\nnucleotides to a polynucleotide SIHT\n[3]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The polymerase chain reaction (PCR) is used to make many copies of a gene.\nSIHT\nNI Three temperatures are used in a PCR cycle.\nETIRW\nState the three temperatures that are used, and outline what happens at each temperature\nTON during a PCR cycle.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n..................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-on-41-q08-97d2bc3183.pdf"
    },
    {
      "id": "9700-2024-on-41-q09",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-41-q09-p01-a27fc2aa65.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-41-q09-p02-1190a84d8d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 Lichens are found growing on trees, walls, rocks and soil.\nSIHT\nNI Fig. 9.1 shows a lichen of the genus Usnea. Usnea can tolerate only low concentrations of\nETIRW\nsulfur dioxide and does not grow in places where the air is polluted with sulfur dioxide.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 9.1\nOD\nUsnea is composed of a mixture of two types of cell:\n• photosynthetic cells that are classified in the kingdom Protoctista\n• fungal cells that are classified in the kingdom Fungi. NIGRAM\n(a) Outline the characteristic features of the kingdoms Protoctista and Fungi.\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Xanthoria is a lichen that can grow in places where there is a high concentration of\nSIHT\nsulfur dioxide in the air, for example in towns where homes, factories and vehicles burn fuels.\nNI\nETIRW\nFig. 9.2 shows a lichen of the genus Xanthoria.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON Fig. 9.2\nOD\n(i) A student planned a method to measure the relative abundance of Usnea and Xanthoria\non trees along a transect from the town centre at 0 km to unpolluted countryside at 4 km.\nSuggest why measuring the relative abundance of the two types of lichen gives\nNIGRAM\ninformation that is useful for conservation.\n… ........................................................................................................................................ SIHT\nNI\n… .............",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-on-41-q09-c3d1082c4f.pdf"
    },
    {
      "id": "9700-2024-on-41-q10",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-41-q10-p01-d1be73f805.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-41-q10-p02-1ecde434fb.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-41-q10-p03-710bc6dfd6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-41-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 Populations of the moth Biston betularia live in Europe and in North America. The most common\nSIHT\nphenotype on both continents is a pale wing colour with light‑grey shading (the typical form).\nNI\nETIRW\nA moth phenotype with dark wing colour (the melanic form) also occurs on both continents.\nTON Fig. 10.1 shows the typical form of the moth.\nOD\nFig. 10.2 shows the melanic form of the moth.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 10.1 Fig. 10.2\nOD\n(a) Two melanic European moths were crossed together. The wing colours of the offspring were\n15 typical and 41 melanic.\nNIGRAM Construct a genetic diagram to explain these results. You may use the symbols A and a to\nrepresent the alleles.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\n[3]\nTON\nOD (b) In a similar experiment, two melanic North American moths were crossed together . The\ncolours of the offspring were 10 typical and 31 melanic.\nWhat can be concluded about the allele that causes the melanic form in the moth populations\nNIGRAM in both continents?\n……………… ....................................................................................................................... [1]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Researchers did not know if the allele causing the melanic form in European moths occurred\nSIHT\nat the same locus as the allele causing the melanic form in North American moths. To find\nNI out, they carried out the following crosses:\nETIRW\n• Cross 1: European moths that were heterozygous at the European melanic locus only\nTON were crossed with North American moths that were heterozygous at the North American\nmelanic locus only.\nOD\n• Cross 2: The melanic and the typical offspring of cross 1 were mated together.\n(i) Explain why cross 2 is a test cross.\nNIGRAM\n… ........................................................................................................................................\n… ........................................................................................................................................ SIHT\nNI\n..................................................................................................................................... [1]\nETIRW\n(ii) Complete Table 10.1 to show the predicted results if:\nTON\n• OD the European and North American melanic alleles are on the same locus (A/a)\n• the European and North American melanic alleles are on two different loci (A/a\nand B/b).\nNIGRAM Table 10.1\nsame locus different loci\nSIHT\n(A/a) (A/a and B/b)\nNI\nETIRW\ngenotypes of melanic moths from\nTON\ncross 1\nOD\nproportion of test crosses (cross 2)\ngiving 100% melanic offspring\nNIGRAM\n[3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(d) A light trap was used to estimate the total size of a population of B. betularia in a woodland.\nSIHT\nOn night one, 24 moths were captured. These were marked with a small spot of harmless\nNI paint. On night two, 29 moths were captured, ",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-41-q10-81da5a71fc.pdf"
    },
    {
      "id": "9700-2024-on-42-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-on-42-q01-p01-0faba10213.png",
        "9700-topic-14-homeostasis/assets/9700-2024-on-42-q01-p02-7b8bdef723.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) Fig. 1.1 is a diagram of part of a Bowman’s capsule and a glomerular capillary.\nSIHT\nNI\nA\nETIRW\nTON\nOD\nB\nNIGRAM\nFig. 1.1\nSIHT\n(i) Identify structures A and B.\nNI\nETIRW\nA ........................................................................................................................................\nTON\nB ........................................................................................................................................\n[2] OD\n(ii) The glomerular filtrate is produced in the Bowman’ s capsule by the process of\nultrafiltration.\nNIGRAM\nState the conditions required for ultrafiltration.\nSIHT ...........................................................................................................................................\nNI\n........................................................................................................................................... ETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) As the newly formed glomerular filtrate passes along the proximal convoluted tubule, selective\nSIHT\nreabsorption takes place. The fluid remaining in the proximal convoluted tubule eventually\nNI forms urine.\nETIRW\nT able 1.1 lists three of the components of blood plasma that enter the glomerulus in a healthy\nTON person.\nOD\nComplete Table 1.1 by writing:\n• increased, if the component is present and is higher in concentration than in blood\nplasma\nNIGRAM\n• decreased, if the component is present and is lower in concentration than in blood\nplasma\n• same, if the component is present and is of the same concentration as blood plasma SIHT\n• not present, if the component is absent.\nNI\nETIRW\nY ou may use each response once, more than once or not at all.\nTON\nTable 1.1\nOD\ncomponent of blood plasma component in newly\ncomponent in urine\nentering glomerulus formed glomerular filtrate\nNIGRAM\nglucose\nSIHT\nlarge plasma proteins\nNI\nETIRW\nurea\nTON\nOD [3]\n(c) Sweating during exercise can lead to a response in the posterior pituitary gland and in the\nkidney.\nNIGRAM\nState the response to sweating that occurs in the posterior pituitary gland and the response\nthat occurs in the kidney.\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-on-42-q01-8b677fc378.pdf"
    },
    {
      "id": "9700-2024-on-42-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-42-q02-p01-40af60b570.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 Several different processes can affect allele frequencies in populations.\nSIHT\nNI (a) Outline the processes that may affect allele frequencies in wildlife populations.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n............................................................................................................................................. [6]\n(b) The process of selective breeding changes allele frequencies in a population. Plant breeders\nuse selective breeding to improve crop plants such as maize.\nNIGRAM\nExplain how selective breeding can be used to obtain a variety of maize where the plants are\nvigorous and of uniform height.\nSIHT\nNI ..................................................................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-on-42-q02-2dbab8c349.pdf"
    },
    {
      "id": "9700-2024-on-42-q03",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-on-42-q03-p01-b3397dda21.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-on-42-q03-p02-389e801750.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-on-42-q03-p03-01100bfed4.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-on-42-q03-p04-4516629341.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:6)(cid:44)\n3 The treatment of many diseases has been improved by the use of genetic technology in medicine.\nSIHT\nNI (a) Explain what is meant by genetic engineering.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [3]\n(b) (i) Leber Congenital Amaurosis (LCA) is an inherited eye disease. In LCA, the photoreceptor\nNIGRAM cells in the retina die at an early age. This causes impaired vision (reduced eyesight) in\nchildren, which can progress to blindness.\nSIHT\nMutations in dif ferent genes cause dif ferent forms of LCA. One form of this disease,\nLCA2, is caused by a mutation in the RPE65 gene. Gene therapy has been used to treat NI\nETIRW LCA2.\nOutline how an inherited eye disease, such as LCA2, is treated with gene therapy. TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ...........................................................................................................................................\nETIRW\n........................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-on-42-q03-9822aff18d.pdf"
    },
    {
      "id": "9700-2024-on-42-q04",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        10
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-42-q04-p01-4c0f8a1a37.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q04",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 Genetic crosses can be used to investigate patterns of inheritance.\nSIHT\nNI (a) A mutation in a gene involved in fruit colour in tomato plants, Solanum lycopersicum, results\nETIRW\nin the production of yellow fruits instead of red fruits.\nTON A genetic cross was carried out between a pure breeding plant with a red fruit and a pure\nbreeding plant with a yellow fruit to produce the F1 generation. All offspring plants produced\nOD\nred fruits.\nThe F1 plants were then crossed with each other and the seeds produced were planted to\nobtain the F2 generation.\nNIGRAM\nConstruct a genetic diagram to show the cross of the F1 generation that produced this\nF2 generation. SIHT\nNI\nUse the symbols R and r for the alleles.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW [3]\n(b) A theoretical dihybrid cross involves two genes located on dif ferent autosomes. Each gene TON\nhas two alleles, one dominant and one recessive.\nOD\nA parent, homozygous dominant for both genes, is crossed with a parent that is homozygous\nrecessive for both genes. This produces F1 individuals that are then crossed to produce the\nF2 generation.\nNIGRAM\nState the phenotypic ratio of this dihybrid F2 generation and explain why some of these\noffspring phenotypes are different from the original parental phenotypes.\nSIHT\nNI ratio ...........................................................................................................................................\nETIRW\nexplanation ...............................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [3]\nNI\nETIRW\n[Total: 6]\nTON",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-42-q04-1760cf6a3a.pdf"
    },
    {
      "id": "9700-2024-on-42-q05",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-42-q05-p01-2b10c9ea9d.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-42-q05-p02-d44120918a.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-42-q05-p03-e08987aa58.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:2)(cid:44)\n5 Phenotypic variation exists in natural populations. There are many causes of variation. Natural\nSIHT\nselection determines which phenotypes are advantageous.\nNI\nETIRW\n(a) Variation in a particular characteristic can be described as either discontinuouso r continuous.\nTON T able 5.1 contains a list of statements that apply to discontinuous variation, continuous\nvariation or both.\nOD\n(✓) if the statement applies and Complete both columns of Table 5.1. Put a tick in the box\nleave the box empty if the statement does not apply.\nNIGRAM\nTable 5.1\nSIHT\nstatement discontinuous continuous\nvariation variation NI\nETIRW\noften involves one gene only\nTON\nenvironmental factors may affect\nOD\ngene expression\nthere is an additive effect of genes\nthat contributes to the phenotype\nNIGRAM\nthere are distinct differences between\nthe various forms of a characteristic\nSIHT\n[4]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) There is variation in the quantity of vitamin D stored in the body.\nSIHT\nNI V itamin D has an important role in keeping bones healthy. The main storage form of vitamin D\nETIRW\nin the body is serum 25‑hydroxyvitamin D (serum 25‑OHD).\nTON A study was carried out on 262 healthy women to investigate if the concentration of\nserum 25‑OHD varied between summer and winter . The women had taken no vitamin D\nOD\nsupplements. The age range of the women in the sample was 40 to 72 years old.\nT able 5.2 shows the results of the study.\nNIGRAM\nTable 5.2\nSIHT\nstandard group mean concentration of\ndeviation serum 25‑OHD NI\ncm–3 ETIRW / ng\nsampled during summer n = 138 32.7 7.6\nTON\nsampled during winter n = 124 28.5 8.3\nOD\nwhole sample n = 262 30.7 8.2\nAdditional analysis showed that there was no significant correlation between age and\nNIGRAM\nserum 25‑OHD concentration.\n(i) Explain what is meant by standard deviation. SIHT\nNI ...........................................................................................................................................\nETIRW\n..................................................................................................................................... [1]\nTON\n(ii) The t‑test was used to compare the mean concentration of serum 25‑OHD when sampled OD\nduring the summer with the mean concentration of serum 25‑OHD when sampled during\nthe winter, as shown in Table 5.2.\nCalculate the value of t using the formula provided. NIGRAM\nkey to symbols: -x x\nt = 1 2 SIHT x = mean ¯\ns2 s2\ns = sample standardf d1ev iat2iopn\n+ NI\nn n\nn = sample size (number of observations) ETIRW 2 1\nGive your answer to four significant figures.\nTON\nOD There is space for your working.\nNIGRAM\nSIHT\nNI\nETIRW\nTON t‑test value ......................................................... [3]\nNIGRAM\n(iii) The critical value at the 0.0001 probability level is 3.773.\nSIHT\nNI State the conclusion that can be made about the res",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-42-q05-b684132fc6.pdf"
    },
    {
      "id": "9700-2024-on-42-q06",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-42-q06-p01-2d7d3a95df.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-42-q06-p02-460a96b844.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q06",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 (a) DCPIP is an indicator and can be used to determine the rate of respiration of organisms such\nSIHT\nas yeast.\nNI\nETIRW\n(i) State the category of indicators to which DCPIP belongs.\nTON ..................................................................................................................................... [1]\nOD\n(ii) Describe the colour change that occurs in DCPIP during experiments to determine the\nrate of respiration.\n..................................................................................................................................... [1]\nNIGRAM\n(b) An investigation was carried out to determine the ef fect of temperature on the rate of\nrespiration of yeast. SIHT\nNI\n• A suspension of yeast cells was added to a test‑tube containing glucose solution.\nETIRW\n• A further four test‑tubes were set up in the same way.\n• One test‑tube was placed in a water‑bath at 10 °C for 5 minutes.\nTON\n• DCPIP was added to the test‑tube and the time taken for the DCPIP to change colour\nOD was measured.\n• The experiment was repeated using the other test‑tubes at 20° C, 30 °C, 40 °C and 50 °C.\nThe results are shown in Fig. 6.1.\nNIGRAM\n25\nSIHT\n20\nNI\nETIRW time taken for DCPIP 15\nto change colour\nTON / minutes\n10\nOD\n5\n0\n0 10 20 30 40 50 NIGRAM\n/ °C temperature\nSIHT\nFig. 6.1\nNI\nETIRW\n(i) State two variables that need to be kept constant in this experiment.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Explain the results shown in Fig. 6.1.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n.................................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-on-42-q06-e7675b0c51.pdf"
    },
    {
      "id": "9700-2024-on-42-q07",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-on-42-q07-p01-2e83a2ee64.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-on-42-q07-p02-35147b9fdb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 (a) Paper chromatography is a technique that can be used to separate and identify dif ferent\nSIHT\nchloroplast pigments.\nNI\nETIRW\nDescribe how the results of paper chromatography can be used to identify chloroplast\npigments.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [3]\nOD\n(b) Carotene and xanthophyll are chloroplast pigments.\nDescribe the role played by these pigments in photosynthesis.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The light‑dependent stage of photosynthesis produces ATP and reduced NADP , which are\nSIHT\nused in the light‑independent stage.\nNI\nETIRW\nDescribe the light‑independent stage of photosynthesis.\nTON ...................................................................................................................................................\nOD",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-on-42-q07-99d2ede295.pdf"
    },
    {
      "id": "9700-2024-on-42-q08",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-42-q08-p01-4f45b00573.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-42-q08-p02-7b82cbe81a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 (a) Fig. 8.1 is a diagram of a motor neurone.\nSIHT\nNI\nETIRW\nTON\nOD\nX\nY\nNIGRAM\nZ\nSIHT Fig. 8.1\nNI\n(i) Name cell X and part Y.\nETIRW\nX ........................................................................................................................................\nTON\nOD Y ........................................................................................................................................\n[2]\n(ii) Name a type of cell that forms a synapse with structure Z.\nNIGRAM\n..................................................................................................................................... [1]\nSIHT\n(b) Table 8.1 shows some of the events that occur during muscle contraction.\nNI\nETIRW They are not listed in the correct order .\nTable 8.1 TON\nOD\nevent description of event\nA calcium ions diffuse out of sarcoplasmic reticulum\nNIGRAM B myosin heads bind to actin\nC sarcolemma depolarised\nSIHT\nD sarcomere shortens\nNI\nE calcium ions bind to troponin ETIRW\nF T‑tubule system membranes depolarised\nTON\nG binding sites on actin exposed\nOD\nH myosin heads tilt\nI tropomyosin moves\nNIGRAM J troponin changes shape\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nComplete Table 8.2 to show the correct order of the events that occur during muscle\nSIHT\ncontraction. Two of the events have been completed for you.\nNI\nTable 8.2 ETIRW\nTON correct order letter of event\nOD\n1\n.......\n2\n.......\nNIGRAM\n3\n.......\nSIHT 4\n.......\nNI\n5 J ETIRW\n6 TON .......\nOD\n7\n.......\n8\n.......\nNIGRAM\n9\n.......\nSIHT\n10 D\nNI\n[4]\nETIRW\n(c) Lambert‑Eaton myasthenic syndrome (LEMS) is a rare disorder of the neuromuscular\nTON\njunction. A person with LEMS produces antibodies that bind to the voltage‑gated calcium\nchannels on the presynaptic knob. One symptom of LEMS is weaker muscle contraction. OD\nSuggest and explain why LEMS leads to weaker muscle contraction.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n..............................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-on-42-q08-294cf3def9.pdf"
    },
    {
      "id": "9700-2024-on-42-q09",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-42-q09-p01-caa2842e35.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-42-q09-p02-a3dabb264c.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-42-q09-p03-10e2012130.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 The red ruf fed lemur, Varecia rubra, is a mammal found only in the rainforests of the Masoala\nSIHT\nregion in north east Madagascar.\nNI\nETIRW\nFig. 9.1 shows a red ruffed lemur.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFig. 9.1 NI\nETIRW\nSpeciesTM (a) The International Union for Conservation of Nature (IUCN) Red List of Threatened\nstates that the red ruffed lemur is critically endangered. TON\nOD\nSuggest why the red ruf fed lemur has become critically endangered.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n............................................................................................................................................. [3] NIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Many zoos around the world operate captive breeding programmes for the red ruffed lemur.\nSIHT\nNI Fig. 9.2 shows the numbers of captive‑born red ruf fed lemurs in North American zoos from\nETIRW\n1970 to 2020.\n200 TON\nOD\n150\nNIGRAM\nnumber of\nSIHT\n100 captive-born\nNI red ruffed lemurs\nETIRW\nTON\n50\nOD\n0\nNIGRAM 1980 1990 2000 2010 2020 1970\nyear\nSIHT\nFig. 9.2\nNI\nETIRW\nDescribe the results shown in Fig. 9.2.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-on-42-q09-395e6c0486.pdf"
    },
    {
      "id": "9700-2024-on-42-q10",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_42.pdf?download=true",
      "source_pages": [
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-on-42-q10-p01-bca48ad532.png",
        "9700-topic-14-homeostasis/assets/9700-2024-on-42-q10-p02-30cfd66ea5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-42-q10",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:4)(cid:44)\n10 (a) Blood glucose concentration is maintained around a set point by homeostasis.\nSIHT\nNI Explain the principles of homeostasis.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n(b) Glycogen phosphorylase catalyses the conversion of glycogen to glucose in liver cells. The\nproduction of glycogen phosphorylase is coded for by the gene PYGL.\nSIHT\nA mutation in PYGL leads to a condition called glycogen storage disease type VI (GSDVI), in NI\nETIRW which glycogen is not broken down efficiently.\nSuggest and explain why cell signalling by glucagon is likely to be affected in the liver cells of TON\na person with GSDVI.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n....................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-on-42-q10-63daae889d.pdf"
    },
    {
      "id": "9700-2024-on-43-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-43-q01-p01-5ae5f3340e.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-43-q01-p02-3ff8b7d6f1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Different species of animal have neurones with different characteristics.\nSIHT\nNI (a) Fig. 1.1 is a diagram of a motor neurone of a rat and a motor neurone of a snail.\nETIRW\naxon (diameter of 7 μm)\nTON\nOD\nrat\nmotor\nneurone\nNIGRAM\nSchwann cell\nSIHT\nC\nNI\nB cell body ETIRW A\nTON\nOD\nsnail\nmotor\nneurone\nNIGRAM\nμm) axon (diameter of 40\nSIHT\nNI\nFig. 1.1\nETIRW\n(i) Name the structures labelled A, B and C on Fig. 1.1.\nTON\nA ........................................................................................................................................ OD\nB ........................................................................................................................................\nC ........................................................................................................................................ NIGRAM\n[3]\nSIHT s–1. The snail motor (ii) The rat motor neurone has an impulse transmission speed of 50 m\ns–1. neurone has an impulse transmission speed of 8 m\nNI\nETIRW\nExplain why the rat motor neurone has a faster impulse transmission speed than the\nsnail motor neurone.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n..................................................................................................................................... [2]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Fig. 1.2 shows an action potential in a rat neurone and Fig. 1.3 shows an action potential in a\nSIHT\nsnail neurone.\nNI\nETIRW rat snail\n60 60\nTON\n40 40\nOD\n20 20\nNIGRAM 0 membrane membrane 0\npotential potential\n/ mV mV / −20 −20\nSIHT\nNI −40 −40\nETIRW\n−60 −60\nTON\nOD −80 −80\n5 15 20 25 30 0 5 10 15 20 25 30 0 10\n/ ms / ms time time\nNIGRAM\nFig. 1.2 Fig. 1.3\nContrast the two action potentials shown in Fig. 1.2 and Fig. 1.3. SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n..........................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-on-43-q01-556dd2ce8f.pdf"
    },
    {
      "id": "9700-2024-on-43-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-43-q02-p01-0799636912.png",
        "9700-topic-15-control-and-coordination/assets/9700-2024-on-43-q02-p02-7fc8a943a1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q02",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 The leaves of many plants have stomata that show a regular daily rhythm of opening and closing\nSIHT\nover a period of 24 hours.\nNI\nETIRW\n(a) Explain why it is important for plants to open and close their stomata in a daily rhythm.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [3]\nETIRW\n(b) Fig. 2.1 shows the results of an experiment to monitor this rhythm over three days and nights\nTON\nfor the plant Arabidopsis thaliana. The percentage of open stomata is shown. Each day\nOD consisted of 14 hours of light (white bar) and each night consisted of 10 hours of darkness\n(black bar).\n100\nNIGRAM Key\nday\nopen night\nSIHT\nstomata 50\n/ % NI\nETIRW\nTON\n0\n24 0 48 72 OD\ntime / hours\nFig. 2.1\nNIGRAM\nFig. 2.1 shows that the percentage of open stomata increases in the first seven hours of the\nexperiment. Describe the sequence of changes that occurs in the guard cells that leads to the\nSIHT\nstomata opening.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n.....................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2024-on-43-q02-0f245e1bbf.pdf"
    },
    {
      "id": "9700-2024-on-43-q03",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-43-q03-p01-8c9128fad3.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2024-on-43-q03-p02-bac94a580f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q03",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 During aerobic respiration, cells respire substrates such as glucose to produce ATP.\nSIHT\nNI Some events that occur during aerobic respiration are:\nETIRW\n• The respiratory substrate breaks down into smaller and smaller molecules. These series of\nTON reactions are described as catabolism.\n• Coenzymes take part in various reactions. In some reactions, coenzymes are reduced or\nOD\noxidised.\n• Carbon dioxide is released.\n(a) Aerobic respiration occurs in four successive stages: glycolysis (G), link reaction (LR), Krebs\nNIGRAM\ncycle (KC) and oxidative phosphorylation (OP).\nComplete Table 3.1 to show which events occur in each stage of aerobic respiration. Use SIHT\na tick (✓) to show that the event does occur or a cross (✗) to show that the event does\nNI\nnot occur.\nETIRW\nTable 3.1\nTON\nOD\nstage\nevent\nG LR KC OP\nNIGRAM catabolism\nSIHT coenzyme is reduced or oxidised\nNI\na coenzyme forms a covalent bond with a ETIRW\nrespiratory intermediate\nTON\ncarbon dioxide is released\nOD\n[4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) A new hand‑held technological device shows the main type of respiratory substrate being\nSIHT\nused in the cells of a person.\nNI\nETIRW\nThe device consists of a carbon dioxide sensor and air‑flow meter . The person inhales\nthrough the device for a fixed time and then exhales into it.\nTON\nThe device calculates the respiratory quotient (RQ) value to show whether the cells are\nOD\nmainly respiring carbohydrates or lipids.\n(i) Explain how the device calculates the RQ value and how this shows whether the cells\nare mainly respiring carbohydrates or lipids.\nNIGRAM\n… ........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n… ........................................................................................................................................\nTON\n… ........................................................................................................................................\nOD\n… ........................................................................................................................................\n… ........................................................................................................................................\nNIGRAM\n… ........................................................................................................................................\nSIHT\n..................................................................................................................................... [4]\nNI\nETIRW (ii) State the dif ference in the relative energy values of carbohydrates and lipids as\nrespiratory substrates, and explain",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2024-on-43-q03-c15e6ae6ee.pdf"
    },
    {
      "id": "9700-2024-on-43-q04",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2024-on-43-q04-p01-7c8a8d806b.png",
        "9700-topic-13-photosynthesis/assets/9700-2024-on-43-q04-p02-81ce24efb4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q04",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 The leaves of Mim osa pudica plants are made of a number of structures known as pinnae. The\nSIHT\npinnae fold when the leaf is touched. This closes the leaf.\nNI\nETIRW\nFig. 4.1 shows an open leaf of M. pudica before it is touched. Fig. 4.2 shows the same leaf that\nhas closed after being touched.\nTON\nOD\nNIGRAM\nSIHT\npinnae NI\nETIRW\nTON\nA open (not touched) closed (touched) B\nOD\nFig. 4.1 Fig. 4.2\n(a) A touch stimulus to an M. pudica leaf causes an action potential to be generated.\nNIGRAM\nThe action potential results in changes in cells, which cause the leaf to close.\nSIHT\nFig. 4.3 shows the mechanism in M . pudica cells that causes the leaf to close.\nNI\nETIRW\n− K+ cells water moves Cl and protons action\nbecome out of cells ions move out pumped into potential after\nTON flaccid by osmosis of cells extensor cells touch stimulus\nOD\nFig. 4.3\nThe leaves of M . pudica and the leaves of Venus fly traps move in response to touch stimuli,\nNIGRAM but the mechanisms that cause the responses are different.\nDescribe the dif ferences between the mechanism shown in Fig. 4.3 and the mechanism that\nSIHT\ncauses the closure of the modified leaves in Venus fly traps.\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [3]\nETIRW\nTON\nNIGRAM\n(b) The rate of photosynthesis decreases by 40% when the leaves of M. pudica close.\nSIHT\nNI Explain why the rate of photosynthesis decreases when the leaves of M. pudica close.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n.......",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2024-on-43-q04-f15cae030d.pdf"
    },
    {
      "id": "9700-2024-on-43-q05",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-43-q05-p01-0ed56e73cb.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-43-q05-p02-fdd15b4964.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 Lunularia cruciata is a primitive plant. Its body consists of a flattened sheet of photosynthetic\nSIHT\ntissue called a thallus.\nNI\nETIRW\nFig. 5.1 shows L. cruciata with two dif ferent types of reproductive structure, labelled A and B, on\nits surface.\nTON\nOD\nNIGRAM\nSIHT A\nNI\nETIRW\nTON\nB\nOD\nNIGRAM\nFig. 5.1\n(a) Structures A and B and the thallus of L. cruciata are haploid.\nSIHT\nNI Explain what is meant by haploid.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\n(b) Structure A contains pale discs of tissue, C, that can germinate into new L. cruciata. The new\nSIHT plants that develop from C are genetically identical to the parent plant in Fig. 5.1.\nNI\nStructure B contains male sperm that are chemically attracted to swim to female eggs on a\nETIRW\nneighbouring parent plant. When the egg and sperm fuse, they form structure D. Structure D\ndevelops to produce spores that grow into new plants that are genetically dif ferent from the\nTON\ntwo parent plants.\nOD\nIdentify which of the structures A, B, C and D are:\nassociated with sexual reproduction …………………………………………\nNIGRAM\nproduced by mitosis …………………………………………\nSIHT\nthe site of meiosis …………………………………………\nNI\nETIRW Each letter may be used once, more than once, or not at all. [3]\nTON\nNIGRAM\n(c) Fig. 5.2 shows a horse, Equus caballus. Horses are diploid animals that reproduce sex ually.\nSIHT\nMale and female horses produce gametes, which fuse to form genetically different offspring.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 5.2\nOD\nExplain the need for a reduction division during meiosis in the production of gametes in\nanimals such as horses.\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n.........................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-43-q05-49ec6c6574.pdf"
    },
    {
      "id": "9700-2024-on-43-q06",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2024-on-43-q06-p01-e972c74eaf.png",
        "9700-topic-14-homeostasis/assets/9700-2024-on-43-q06-p02-c874034197.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n6 The mammalian kidney is responsible for:\nSIHT\n• NI the excretion of urea\nETIRW • osmoregulation (the homeostatic control of the water potential of the blood).\nTON (a) (i) Outline how and where the excretory product urea is made in the body.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\n(ii) Homeostatic control of the water potential of blood includes receptors, ef fectors and SIHT\ntarget cells.\nNI\nETIRW\nIdentify the names and locations of these components of homeostatic control in\nosmoregulation.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n..................................................................................................................................... [3]\nTON\n(b) The glomerulus and Bowman’s capsule of the nephron are important in the formation of urine.\nOD\nOutline the role of the glomerulus and Bowman’ s capsule in the formation of urine.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n..............................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2024-on-43-q06-308dee8819.pdf"
    },
    {
      "id": "9700-2024-on-43-q07",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-43-q07-p01-1ff95d78eb.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-43-q07-p02-36279164e9.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-43-q07-p03-05c04562a8.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2024-on-43-q07-p04-ee7f7b9741.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q07",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n7 Spea multiplicata is one of several species of American spadefoot toad.\nSIHT\nNI (a) Young spadefoot toads are called tadpoles and live in water in ponds.\nETIRW\nS. multiplicata tadpoles show three dif ferent phenotypes due to genetic variation. The three\nTON phenotypes are: detritus feeder, intermediate and carnivore.\nOD\nDetritus feeders are small, and carnivores are large. Intermediates vary in size between the\ntwo extremes.\nA detritus feeder and a carnivore are shown in Fig. 7.1.\nNIGRAM\nSIHT\nNI\nETIRW\ncarnivore\nTON\nOD\ndetritus feeder\nNIGRAM fairy shrimp\nSIHT\nNI\nETIRW Fig. 7.1\nTON Detritus feeders:\nOD\n• eat detritus (small pieces of dead organic matter) and algae (photosynthetic protoctists)\n• have smooth mouthparts, small jaw muscles and long intestines.\nIntermediates:\nNIGRAM\n• can eat all available food (detritus, algae and fairy shrimps)\n• have teeth‑like mouthparts, medium‑sized jaw muscles and medium‑sized intestines. SIHT\nNI\nCarnivores:\nETIRW\n• eat fairy shrimps and other small animals\nTON\n• have teeth‑like mouthparts, large jaw muscles and short intestines.\nOD\nScientists counted the number of each type of tadpole in two dif ferent ponds: pond 1 and\npond 2.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) In pond 1, the scientists observed:\nSIHT\n• NI a high density of tadpoles\nETIRW • a low abundance of food\n• that most of the tadpoles they counted were either detritus feeders or carnivores,\nTON with very few intermediates present.\nOD\nDescribe and suggest explanations for the type of natural selection that appears to be\nacting in pond 1.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [3]\nNIGRAM\n(ii) In pond 2, the scientists observed:\nSIHT\n• a low density of tadpoles\n• sufficient food availability for all tadpoles NI\n• ETIRW that most of the tadpoles they counted were intermediates, with fewer detritus\nfeeders or carnivores.\nTON\nDescribe and suggest explanations for the type of natural selection that ap",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2024-on-43-q07-c9901ecd2e.pdf"
    },
    {
      "id": "9700-2024-on-43-q08",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2024-on-43-q08-p01-37484094f6.png",
        "9700-topic-19-genetic-technology/assets/9700-2024-on-43-q08-p02-e10975aa2d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 Scientists use many different techniques in genetic engineering.\nSIHT\nNI (a) Sometimes the gene for genetic engineering cannot be extracted from the donor organism.\nETIRW\nInstead, the gene is synthesised using one of two different methods.\nTON Outline the two methods for synthesising a gene for use in genetic engineering.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ............................................................................................................................................. [3]\n(b) DNA ligase and DNA polymerase are two enzymes that are used in genetic engineering.\nNIGRAM Complete Table 8.1 to show the roles of DNA ligase and DNA polymerase in genetic\nengineering.\nSIHT\nUse a tick (✓) if the enzyme has the role or a cross (✗) if the enzyme does not have the role.\nNI\nETIRW Table 8.1\nTON\nrole in genetic engineering DNA ligase DNA polymerase\nOD\njoins two sections of sugar\nphosphate backbone in DNA\nadds a gene to a plasmid\nNIGRAM\nadds free activated DNA\nnucleotides to a polynucleotide SIHT\n[3]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The polymerase chain reaction (PCR) is used to make many copies of a gene.\nSIHT\nNI Three temperatures are used in a PCR cycle.\nETIRW\nState the three temperatures that are used, and outline what happens at each temperature\nTON during a PCR cycle.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n..................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2024-on-43-q08-16209d46b7.pdf"
    },
    {
      "id": "9700-2024-on-43-q09",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-43-q09-p01-a27fc2aa65.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2024-on-43-q09-p02-1190a84d8d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 Lichens are found growing on trees, walls, rocks and soil.\nSIHT\nNI Fig. 9.1 shows a lichen of the genus Usnea. Usnea can tolerate only low concentrations of\nETIRW\nsulfur dioxide and does not grow in places where the air is polluted with sulfur dioxide.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 9.1\nOD\nUsnea is composed of a mixture of two types of cell:\n• photosynthetic cells that are classified in the kingdom Protoctista\n• fungal cells that are classified in the kingdom Fungi. NIGRAM\n(a) Outline the characteristic features of the kingdoms Protoctista and Fungi.\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Xanthoria is a lichen that can grow in places where there is a high concentration of\nSIHT\nsulfur dioxide in the air, for example in towns where homes, factories and vehicles burn fuels.\nNI\nETIRW\nFig. 9.2 shows a lichen of the genus Xanthoria.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON Fig. 9.2\nOD\n(i) A student planned a method to measure the relative abundance of Usnea and Xanthoria\non trees along a transect from the town centre at 0 km to unpolluted countryside at 4 km.\nSuggest why measuring the relative abundance of the two types of lichen gives\nNIGRAM\ninformation that is useful for conservation.\n… ........................................................................................................................................ SIHT\nNI\n… .............",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2024-on-43-q09-79a32cdac9.pdf"
    },
    {
      "id": "9700-2024-on-43-q10",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_43.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2024-on-43-q10-p01-d1be73f805.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-43-q10-p02-1ecde434fb.png",
        "9700-topic-16-inheritance/assets/9700-2024-on-43-q10-p03-710bc6dfd6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-43-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 Populations of the moth Biston betularia live in Europe and in North America. The most common\nSIHT\nphenotype on both continents is a pale wing colour with light‑grey shading (the typical form).\nNI\nETIRW\nA moth phenotype with dark wing colour (the melanic form) also occurs on both continents.\nTON Fig. 10.1 shows the typical form of the moth.\nOD\nFig. 10.2 shows the melanic form of the moth.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 10.1 Fig. 10.2\nOD\n(a) Two melanic European moths were crossed together. The wing colours of the offspring were\n15 typical and 41 melanic.\nNIGRAM Construct a genetic diagram to explain these results. You may use the symbols A and a to\nrepresent the alleles.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\n[3]\nTON\nOD (b) In a similar experiment, two melanic North American moths were crossed together . The\ncolours of the offspring were 10 typical and 31 melanic.\nWhat can be concluded about the allele that causes the melanic form in the moth populations\nNIGRAM in both continents?\n……………… ....................................................................................................................... [1]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Researchers did not know if the allele causing the melanic form in European moths occurred\nSIHT\nat the same locus as the allele causing the melanic form in North American moths. To find\nNI out, they carried out the following crosses:\nETIRW\n• Cross 1: European moths that were heterozygous at the European melanic locus only\nTON were crossed with North American moths that were heterozygous at the North American\nmelanic locus only.\nOD\n• Cross 2: The melanic and the typical offspring of cross 1 were mated together.\n(i) Explain why cross 2 is a test cross.\nNIGRAM\n… ........................................................................................................................................\n… ........................................................................................................................................ SIHT\nNI\n..................................................................................................................................... [1]\nETIRW\n(ii) Complete Table 10.1 to show the predicted results if:\nTON\n• OD the European and North American melanic alleles are on the same locus (A/a)\n• the European and North American melanic alleles are on two different loci (A/a\nand B/b).\nNIGRAM Table 10.1\nsame locus different loci\nSIHT\n(A/a) (A/a and B/b)\nNI\nETIRW\ngenotypes of melanic moths from\nTON\ncross 1\nOD\nproportion of test crosses (cross 2)\ngiving 100% melanic offspring\nNIGRAM\n[3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(d) A light trap was used to estimate the total size of a population of B. betularia in a woodland.\nSIHT\nOn night one, 24 moths were captured. These were marked with a small spot of harmless\nNI paint. On night two, 29 moths were captured, ",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2024-on-43-q10-1e52b3290a.pdf"
    },
    {
      "id": "9700-2024-on-51-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 21,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_51.pdf?download=true",
      "source_pages": [
        3,
        4,
        5,
        6,
        7,
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p01-9f838bdf3a.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p02-caf2c6f8cd.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p03-9b890f4215.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p04-c5395b18fb.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p05-017ae6e7f1.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p06-e873429ac0.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p07-6fe08b97b7.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p08-6aa203dcca.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q01-p09-5f653818e8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44)\n1 Algae are a diverse group of photosynthetic eukaryotes in the kingdom Protoctista.\nSIHT\nNI Rhodomonas salina and Skeletonema costatum are two species of alga that live in seawater.\nETIRW\nFig. 1.1 shows a scanning electron micrograph of R. salina, which is a unicellular organism.\nTON R. salina is red in colour.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\n×2200\nOD\nFig. 1.1\n1.2 shows a scanning electron micrograph of S. costatum, which is also a unicellular organism. Fig.\nNIGRAM\nThe individual cells can group together in a chain. S. costatum is yellow‑brown in colour.\nSIHT\nNI\nETIRW\nTON\none cell of S. costatum\nOD\nNIGRAM\n×450 SIHT\nNI\nFig. 1.2 ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) A student prepared extracts of the photosynthetic pigments from R. salina and S. costatum.\nSIHT\nThe student carried out chromatography to identify and compare the pigm ents in the\nNI two species.\nETIRW\nThe chromatograms are shown in Fig. 1.3.\nTON\nR. salina S. costatum\nOD\nsolvent\nNIGRAM\nfront\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM alloxanthin\nSIHT\nNI\nETIRW\nTON\nOD\norigin\nFig. 1.3\nNIGRAM\nvalue of alloxanthin from R. salina. (i) Use Fig. 1.3 to calculate the R\nf\nSIHT\nNI\nETIRW\nTON\nOD of alloxanthin = ......................................................... [1] R\nf\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Table 1.1 shows the R values of some photosynthetic pigments.\nSIHT f\nNI Table 1.1\nETIRW\nphotosynthetic pigment R value\nf TON\nβ‑carotene 0.94\nOD\nchlorophyll a 0.68\nchlorophyll b 0.54\nNIGRAM\nchlorophyll c 0.15\nSIHT diadinoxanthin 0.32\nNI\nfucoxanthin 0.43\nETIRW\nphycocyanin 0.35\nTON\nOD R. salina lacks two photosynthetic pigments that are present in S. costatum.\nUse Fig. 1.3 and Table 1.1 to identify the two photosynthetic pigments that are present in\nonly S. costatum.\nNIGRAM\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [2]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The extract of the photosynthetic pigments from each of the two species was used to obtain\nSIHT\nabsorption spectra as shown in Fig. 1.4.\nNI\nETIRW Key\nR. salina\nTON\nS. costatum\nOD\n1.8\nNIGRAM\n1.5\nSIHT\nNI\nETIRW 1.2\nTON\nOD\nabsorbance 0.9\nNIGRAM\n0.6\nSIHT\nNI\nETIRW\n0.3\nTON\nOD\n0.0\n400 450 500 550 600 650 700\n/ nm wavelength\nNIGRAM\nFig. 1.4\nSIHT\nThe student planned to investigate the ef fect of light wavelength on the rate of photosynthesis\nNI in R. salina and S. costatum.\nETIRW\nThe student planned to determine the rate of photosynthesis in the two species when they\nTON\nwere exposed to three different colours of light:\nOD\n• blue light, peak wavelength of 455 nm\n• green light, peak wavelength of 550 nm\n• red light, peak wavelength of 670 nm.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRA",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-51-q01-10d15ed21d.pdf"
    },
    {
      "id": "9700-2024-on-51-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_51.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-51-q02-p01-2149ea5f2a.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q02-p02-860f3c9091.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q02-p03-a8d009d5b4.png",
        "9700-experimental-skills/assets/9700-2024-on-51-q02-p04-ee81e56d9f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n2 A group of ecologists investigated the biodiversity of two peat bog ecosystems, A and B.\nSIHT\nNI Peat bogs are wetland ecosystems that often contain rare species.\nETIRW\nFig. 2.1 shows an example of a peat bog.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 2.1\n(a) The ecologists sampled the plant species in peat bog A and peat bog B.\nNIGRAM\nThe ecologists decided to use two suitable indices of biodiversity:\n• Simpson’s index of diversity SIHT\n• Shannon diversity index.\nNI\nETIRW\nThe ecologists had read that the two indices can lead to dif ferent conclusions.\nTON\nThe Shannon diversity index gives values in the range 1.5–3.5, where 1.5 is low biodiversity\nand 3.5 is high biodiversity. OD\nThe results are shown in Table 2.1.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nTable 2.1\nSIHT\nNI\npeat bog A peat bog B\nETIRW\nspecies number of individuals species number of individuals\nTON\nL 34 L 30\nOD\nM 10 M 15\nN 11 N 18\nNIGRAM\nO 15 O 24\nP 8 P 16 SIHT\nNI Q 7 Q 8\nETIRW\nR 4 R 20\nTON\nS 1 S 4\nOD\nT 3 T 0\nU 3 U 0\nNIGRAM\nV 2 V 0\nW 1 W 0\nSIHT\nSimpson’s index (D) = 0.82 Simpson’s index (D) = 0.85\nNI\nETIRW\nShannon index = 2.03 Shannon index = 1.96\nTON\nThe ecologists concluded that one index indicated peat bog A had more biodiversity , but the\nOD\nother index indicated peat bog B had more biodiversity.\nSuggest how the dif ferences in the data in the two samples, shown inT able 2.1, may have led\nto the different conclusions.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n............................................................................................................................................. [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The ecologists sampled the invertebrate species in peat bogs A and B and calculated\nSIHT\ninvertebrate biodiversity using Simpson’s index of diversity.\nNI\nETIRW\n(i) The ecologists sampled invertebrate species in peat bog A:\nTON • between 13:00 and 16:00 on three different days within a 30‑day period\n• at 10 random sites along the edge of a path in the peat bog\nOD\nm2 • by sampling the invertebrates within a 1 quadrat placed at each site\n• by using an identification key to identify the species\n• by counting the number of individuals of each species.\nNIGRAM\nDescribe how this sampling method could be improved to make sure the results are\nmore represen",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-51-q02-2ace1ae4aa.pdf"
    },
    {
      "id": "9700-2024-on-52-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p01-b1732d7fd7.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p02-20dbe8cb08.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p03-97287f5ab2.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p04-f15c66b80d.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p05-77f3a10154.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q01-p06-068740a1e2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 In the light‑dependent stage of photosynthesis, electrons and hydrogen ions are accepted by the\nSIHT\ncoenzyme NADP, which becomes reduced.\nNI\nETIRW\nDCPIP is a dye that can act as an electron and hydrogen ion acceptor . The dye is blue when\noxidised and colourless when reduced. In laboratory experiments, DCPIP can be used to follow\nTON the progress of the light‑dependent stage because it can replace NADP as the acceptor molecule\nfor electrons and hydrogen ions, as shown in Fig. 1.1.\nOD\nreduced DCPIP oxidised DCPIP + electrons + hydrogen ions\n(blue) (colourless)\nNIGRAM\nFig. 1.1\nThe effects of various factors on the light‑dependent stage of photosynthesis can be investigated SIHT\nby using suspensions of isolated chloroplasts (chloroplast suspensions) and DCPIP.\nNI\nETIRW\nA student used DCPIP to investigate the ef fect of temperature on the rate of the light‑dependent\nstage of photosynthesis in spinach, Spinacia oleracea.\nTON\nOD The student prepared a leaf extract to make a stock chloroplast suspension and then carried out\na preliminary experiment to determine a suitable concentration of chloroplast suspension to use in\nthe investigation.\nNIGRAM To carry out the preliminary experiment, the student followed a set of instructions, steps 1 to 11.\n1 Cut spinach leaves into small pieces and place these pieces in a blender containing ice‑cold\nSIHT\n10% sucrose solution buffered at pH 7.0.\nNI\nETIRW 2 Turn on the blender for 15 seconds and then filter the extract to remove all the small pieces of\nleaf.\nTON\n3 Place the leaf extract in a centrifuge and spin at low speed.\nOD\n4 Pour of f the supernatant that contains the chloroplasts. Keep this stock chloroplast\nsuspension ice cold and in the dark.\nNIGRAM\n5 Prepare 5 different concentrations of the chloroplast suspension using 10% sucrose solution.\nThe percentage concentrations are 10%, 20%, 30%, 40% and 50% of the stock chloroplast\nsuspension.\nSIHT\nNI 6 Wrap 5 flat‑bottomed tubes in black plastic film to prevent light entering.\nETIRW\ncm3 of each concentration of stock chloroplast suspension into a flat‑bottomed tube, 7 Put 10\nTON cm3 of DCPIP solution to each tube. The chloroplast suspension is now blue‑green and add 1\nin colour. OD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n8 Place 1 of the tubes beneath a light source as shown in Fig. 1.2.\nSIHT\nNI\nETIRW\nTON min s\nOD\nchloroplast\nsuspension\nNIGRAM\nSIHT\nFig. 1.2 NI\nETIRW\n9 Start a timer . Remove the black plastic film from the tube. Record the time taken for the\nDCPIP to decolourise so that the chloroplast suspension is green. TON\nOD\n10 Calculate the rate of the light‑dependent stage of photosynthesis by using the formula:\n1000\nrate =\nt\nNIGRAM\nt = time taken in seconds for the chloroplast suspension to reach a green colour when all the\nDCPIP is decolourised.\nSIHT\n11 Repeat step 8 to step 10 for the other tubes.\nNI\nETIRW\n(a) Suggest two suitable control experim",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-52-q01-c53d292899.pdf"
    },
    {
      "id": "9700-2024-on-52-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_52.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12,
        13,
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p01-c5e4d76038.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p02-3217e3a390.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p03-19290fa843.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p04-597b98caeb.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p05-d5931e66f4.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p06-c2f4455aeb.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p07-04996ff042.png",
        "9700-experimental-skills/assets/9700-2024-on-52-q02-p08-756e65a300.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:10)(cid:44)\n2 The two‑spot ladybird, Adalia bipunctata, is a species of small, flying beetle that is found in\nSIHT\nnorthern Europe and other parts of the world. The wings of these beetles are covered by two\nNI tough structures known as elytra, as shown in Fig. 2.1.\nETIRW\nTON\nOD\nprotective elytra\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 2.1 TON\nOD\nA. bipunctata is an important predator of insect pests such as aphids. These insect pests feed on\nplants, including many crop species. Population numbers of A. bipunctata fluctuate in response to\nchanges in the population numbers of their prey.\nNIGRAM\nThe elytra of A. bipunctata show phenotypic variation. The three most common morphs (forms)\nfound in northern Europe are known as typica (T), quadrimaculata (Q) and sexpustulata (S).\nSIHT\n• The typica morph, shown in Fig. 2.1, is mostly red with two black spots and is described as\nNI\nnon‑melanic.\nETIRW\n• The other two morphs are described as melanic as they are mainly black with some red.\n• The distribution of the colours in these three morphs is shown in Fig. 2.2.\nTON\nnon-melanic melanic OD\nred\nblack\nNIGRAM\nred\nblack SIHT\nNI\nETIRW\ntypica (T) quadrimaculata (Q) sexpustulata (S)\nTON\nOD Fig. 2.2\nIn northern Europe, the populations of A. bipunctata hibernate (are inactive) during the winter. As\nthe temperature increases in early spring, the populations become active. The populations remain\nNIGRAM active and produce three generations before the next winter begins.\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) It was observed that in some areas in populations of A. bipunctata, the proportion of\nSIHT\nnon‑melanic to melanic phenotypes in the early spring was different to the proportion later in\nNI the year in the autumn.\nETIRW\nResearchers carried out a study in one area of Germany to compare the proportion of\nTON non‑melanic to melanic phenotypes in early spring with the proportion in autumn after three\ngenerations had been produced. The data were collected over a period of 12 years.\nOD\nThe results are shown in Fig. 2.3.\n100\nNIGRAM\n90\nSIHT 80\nNI 70 percentage\nETIRW\nfrequency in\n60\nsampled\nTON populations of 50\nND ND non-melanic\nOD\n40\nand melanic\nphenotypes of 30\nA. bipunctata\n20\nNIGRAM\n10\n0 SIHT\nnmutua nmutua nmutua nmutua nmutua nmutua nmutua nmutua nmutua nmutua nmutua nmutua gnirps gnirps gnirps gnirps gnirps gnirps gnirps gnirps gnirps gnirps gnirps gnirps\nNI\nETIRW\n1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 1940\nTON\nsampling times\nOD\nKey\nnon-melanic\nmelanic\nNIGRAM\nND no data\nSIHT\nFig. 2.3\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Outline a method that could be followed to collect the data shown in Fig. 2.3.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...............................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-52-q02-fbcf067823.pdf"
    },
    {
      "id": "9700-2024-on-53-q01",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 21,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_53.pdf?download=true",
      "source_pages": [
        3,
        4,
        5,
        6,
        7,
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p01-7e6595262d.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p02-ea894d8c36.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p03-82ce994b66.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p04-a832028337.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p05-017ae6e7f1.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p06-54901447ae.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p07-e43670d594.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p08-f1bcde24e5.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q01-p09-e99b6dcdcd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44)\n1 Algae are a diverse group of photosynthetic eukaryotes in the kingdom Protoctista.\nSIHT\nNI Rhodomonas salina and Skeletonema costatum are two species of alga that live in seawater.\nETIRW\nFig. 1.1 shows a scanning electron micrograph of R. salina, which is a unicellular organism.\nTON R. salina is red in colour.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\n×2200\nOD\nFig. 1.1\n1.2 shows a scanning electron micrograph of S. costatum, which is also a unicellular organism. Fig.\nNIGRAM\nThe individual cells can group together in a chain. S. costatum is yellow‑brown in colour.\nSIHT\nNI\nETIRW\nTON\none cell of S. costatum\nOD\nNIGRAM\n×450 SIHT\nNI\nFig. 1.2 ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) A student prepared extracts of the photosynthetic pigments from R. salina and S. costatum.\nSIHT\nThe student carried out chromatography to identify and compare the pigm ents in the\nNI two species.\nETIRW\nThe chromatograms are shown in Fig. 1.3.\nTON\nR. salina S. costatum\nOD\nsolvent\nNIGRAM\nfront\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM alloxanthin\nSIHT\nNI\nETIRW\nTON\nOD\norigin\nFig. 1.3\nNIGRAM\nvalue of alloxanthin from R. salina. (i) Use Fig. 1.3 to calculate the R\nf\nSIHT\nNI\nETIRW\nTON\nOD of alloxanthin = ......................................................... [1] R\nf\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Table 1.1 shows the R values of some photosynthetic pigments.\nSIHT f\nNI Table 1.1\nETIRW\nphotosynthetic pigment R value\nf TON\nβ‑carotene 0.94\nOD\nchlorophyll a 0.68\nchlorophyll b 0.54\nNIGRAM\nchlorophyll c 0.15\nSIHT diadinoxanthin 0.32\nNI\nfucoxanthin 0.43\nETIRW\nphycocyanin 0.35\nTON\nOD R. salina lacks two photosynthetic pigments that are present in S. costatum.\nUse Fig. 1.3 and Table 1.1 to identify the two photosynthetic pigments that are present in\nonly S. costatum.\nNIGRAM\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [2]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The extract of the photosynthetic pigments from each of the two species was used to obtain\nSIHT\nabsorption spectra as shown in Fig. 1.4.\nNI\nETIRW Key\nR. salina\nTON\nS. costatum\nOD\n1.8\nNIGRAM\n1.5\nSIHT\nNI\nETIRW 1.2\nTON\nOD\nabsorbance 0.9\nNIGRAM\n0.6\nSIHT\nNI\nETIRW\n0.3\nTON\nOD\n0.0\n400 450 500 550 600 650 700\n/ nm wavelength\nNIGRAM\nFig. 1.4\nSIHT\nThe student planned to investigate the ef fect of light wavelength on the rate of photosynthesis\nNI in R. salina and S. costatum.\nETIRW\nThe student planned to determine the rate of photosynthesis in the two species when they\nTON\nwere exposed to three different colours of light:\nOD\n• blue light, peak wavelength of 455 nm\n• green light, peak wavelength of 550 nm\n• red light, peak wavelength of 670 nm.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRA",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-53-q01-54fcd261e3.pdf"
    },
    {
      "id": "9700-2024-on-53-q02",
      "subject": "9700",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2024-oct-nov/9700_w24_qp_53.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2024-Oct-Nov/qp/9700_w24_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2024-on-53-q02-p01-6c5cef24ed.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q02-p02-e1614df4ed.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q02-p03-b7ba2e82f7.png",
        "9700-experimental-skills/assets/9700-2024-on-53-q02-p04-d35b6604e0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2024-on-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n2 A group of ecologists investigated the biodiversity of two peat bog ecosystems, A and B.\nSIHT\nNI Peat bogs are wetland ecosystems that often contain rare species.\nETIRW\nFig. 2.1 shows an example of a peat bog.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 2.1\n(a) The ecologists sampled the plant species in peat bog A and peat bog B.\nNIGRAM\nThe ecologists decided to use two suitable indices of biodiversity:\n• Simpson’s index of diversity SIHT\n• Shannon diversity index.\nNI\nETIRW\nThe ecologists had read that the two indices can lead to dif ferent conclusions.\nTON\nThe Shannon diversity index gives values in the range 1.5–3.5, where 1.5 is low biodiversity\nand 3.5 is high biodiversity. OD\nThe results are shown in Table 2.1.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nTable 2.1\nSIHT\nNI\npeat bog A peat bog B\nETIRW\nspecies number of individuals species number of individuals\nTON\nL 34 L 30\nOD\nM 10 M 15\nN 11 N 18\nNIGRAM\nO 15 O 24\nP 8 P 16 SIHT\nNI Q 7 Q 8\nETIRW\nR 4 R 20\nTON\nS 1 S 4\nOD\nT 3 T 0\nU 3 U 0\nNIGRAM\nV 2 V 0\nW 1 W 0\nSIHT\nSimpson’s index (D) = 0.82 Simpson’s index (D) = 0.85\nNI\nETIRW\nShannon index = 2.03 Shannon index = 1.96\nTON\nThe ecologists concluded that one index indicated peat bog A had more biodiversity , but the\nOD\nother index indicated peat bog B had more biodiversity.\nSuggest how the dif ferences in the data in the two samples, shown inT able 2.1, may have led\nto the different conclusions.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n............................................................................................................................................. [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The ecologists sampled the invertebrate species in peat bogs A and B and calculated\nSIHT\ninvertebrate biodiversity using Simpson’s index of diversity.\nNI\nETIRW\n(i) The ecologists sampled invertebrate species in peat bog A:\nTON • between 13:00 and 16:00 on three different days within a 30‑day period\n• at 10 random sites along the edge of a path in the peat bog\nOD\nm2 • by sampling the invertebrates within a 1 quadrat placed at each site\n• by using an identification key to identify the species\n• by counting the number of individuals of each species.\nNIGRAM\nDescribe how this sampling method could be improved to make sure the results are\nmore represen",
      "pdf_path": "9700-experimental-skills/print/9700-2024-on-53-q02-2db2611eb1.pdf"
    },
    {
      "id": "9700-2025-m-42-q01",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q01-p01-04fe0911c9.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q01-p02-42a6e2fce5.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q01-p03-52a715133e.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q01-p04-cab0e8bce9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) Different types of respiratory substrate can have different energy values and therefore release\nSIHT\ndifferent quantities of energy when they are respired.\nNI\nETIRW\nComplete Table 1.1 to show the energy value of each of the three main types of respiratory\nsubstrate.\nTON\nUse one tick (3) to identify which of the two possible energy values is correct for each\nOD\nrespiratory substrate.\nTable 1.1\nNIGRAM\ng–1 energy value / kJ\ntype of respiratory\nsubstrate\napproximately 17 approximately 37 SIHT\nNI\ncarbohydrate\nETIRW\nlipid TON\nOD\nprotein\n[1]\nNIGRAM\n(b) Determining the respiratory quotient (RQ) of an organism can be used to indicate the main\ntype of respiratory substrate that is being metabolised in respiration. This is because the\ndifferent types of respiratory substrate have different RQ values. Table 1.2 shows typical RQ SIHT\nvalues for carbohydrate, lipid and protein.\nNI\nETIRW\nTable 1.2\nTON\ntype of RQ\nOD respiratory substrate value\ncarbohydrate 1.0\nlipid 0.7\nNIGRAM\nprotein 0.8\nSIHT\n(i) State the name of the laboratory apparatus that can be used to determine the RQ value\nNI of organisms such as blowfly larvae.\nETIRW\n..................................................................................................................................... [1]\nTON\n(ii) When determining RQ values using the laboratory apparatus stated in (b)(i), chemicals\nOD\nsuch as soda lime or potassium hydroxide solution are used.\nState the reason for using chemicals such as soda lime or potassium hydroxide solution\nwhen measuring RQ values.\nNIGRAM\n..................................................................................................................................... [1]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Organic acids such as malic acid can also act as respiratory substrates. When respired\nSIHT\naerobically, their RQ values may be dif ferent to the RQ values of the m ain respiratory\nNI substrates.\nETIRW\nFig. 1.1 shows the formula that is used to calculate RQ values.\nTON\nnumber of molecules of carbon dioxide produced\nRQ =\nOD number of molecules of oxygen taken in\nFig. 1.1\nWhen malic acid is respired aerobically, the equation is:\nNIGRAM\nH O + .......... O 4CO + 3H O + energy C\n4 6 5 2 2 2\nSIHT\n(i) Calculate how many molecules of oxygen are taken in when one molecule of malic acid\nNI\nis respired aerobically.\nETIRW\nTON\nnumber of molecules of oxygen = ......................................................... [1]\nOD\n(ii) Calculate the RQ for malic acid.\nGive your answer to two decimal places.\nNIGRAM\nSIHT\nRQ = ......................................................... [1]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(d) The deer mouse, Peromyscus maniculatus, lives in forests in North America.\nSIHT\nNI Fig. 1.2 shows a deer mouse.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nFig. 1.2\nOD\nThe deer mouse is active throughout the year and ",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-m-42-q01-17446ff59d.pdf"
    },
    {
      "id": "9700-2025-m-42-q02",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-m-42-q02-p01-9471206a69.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-m-42-q02-p02-519c50293c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-m-42-q02-p03-b357b62f11.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-m-42-q02-p04-2a45e97f22.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-m-42-q02-p05-7804498e5a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n2 The orca, Orcinus orca, has the largest distribution of all aquatic mammals and is found in nearly\nSIHT\nall seas and oceans. Orca are social mammals that usually live in groups. These groups can vary\nNI in size.\nETIRW\nFig. 2.1 shows an orca.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 2.1\nTON\nThere are a number of distinct types of orca.T hese distinct types of orca are classified as members OD\nof the same species. However, there is evidence that sympatric speciation is occurring.\n(a) (i) There are two distinct types of orca in the Northeast Atlantic Ocean: Type 1 and Type 2.\nType 1 orca feed mainly on fish. Type 2 orca feed mainly on aquatic mammals, such as NIGRAM\nseals.\nSIHT\nFig. 2.2 shows the locations in the Northeast Atlantic Ocean where Type 1 orca and\nType 2 orca have been observed. Orca do not occur only in these areas and some\nNI\ngroups of orca travel great distances. ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNortheast NI\nETIRW Atlantic\nOcean\nTON\nOD\nkey\nType 1 orca\nNIGRAM\nType 2 orca km 400\nSIHT\nNI\nFig. 2.2\nETIRW\nTON\nNIGRAM\nWith reference to Fig. 2.2, explain why the type of speciation that is occurring in the orca\nSIHT\nis described as sympatric speciation.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n..................................................................................................................................... [1]\n(ii) Suggest examples of behavioural separation that would contribute to sympatric\nspeciation of Type 1 orca and Type 2 orca.\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) In the Southern Ocean, which surrounds Antarctica, there are three ",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-m-42-q02-bd67d5f0f0.pdf"
    },
    {
      "id": "9700-2025-m-42-q03",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-m-42-q03-p01-e3c4196e0b.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-m-42-q03-p02-23bd778c49.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n3 Gentamicin is an antibiotic used to treat severe bacterial infections in children.\nSIHT\n(a) Some children have a genetic mutation in the gene M T-RNR1. If gentamicin is given to NI\nETIRW children with this genetic mutation, it can cause deafness.\nBefore gentamicin can be given to a child with a severe bacterial infection, PCR (polymerase TON\nchain reaction) and electrophoresis are used to test whether the child has this mutation. If the\nOD\nmutation is found, a different antibiotic must be given.\n(i) Describe and explain the role of Taq polymerase in PCR.\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [4]\nNI\nETIRW\n(ii) PCR with primers specific to the MT -RNR1 gene is used to amplify DNA from the child\nthat is being tested.\nTON\nThe PCR primers are designed so that the amplified product of the normal allele of OD\nMT-RNR1 is longer than the amplified product of the mutant allele.\nGel electrophoresis is used to separate the PCR products.\nNIGRAM\nFig. 3.1 shows the results of gel electrophoresis after using this method of PCR on DNA\nsamples collected from three children.\nSIHT\nchild 1 child 2 child 3\nNI\nETIRW\nTON\nOD\ndirection\nof DNA\nmovement\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 3.1 TON\nNIGRAM\nState which of the children in Fig. 3.1 cannot receive the antibiotic gentamicin to treat a\nSIHT\nsevere bacterial infection.\nNI\nETIRW\n..................................................................................................................................... [1]\nTON (iii) Explain how gel electrophoresis produces the pattern of results shown in Fig. 3.1 from\nthe PCR products of the MT-RNR1 gene.\nOD\n................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-m-42-q03-551ec99db8.pdf"
    },
    {
      "id": "9700-2025-m-42-q04",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        14,
        15,
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q04-p01-b8bbb465e3.png",
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q04-p02-8a9ce70038.png",
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q04-p03-5e1e753174.png",
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q04-p04-27f91b3f73.png",
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q04-p05-5aee4f4538.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q04",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n4 The HFE gene codes for the HFE protein, which has a role in the regulation of iron absorption by\nSIHT\nthe body. Iron is an essential mineral that can be obtained only from the diet.\nNI\nETIRW\nA mutation of the HFE gene known as C282Y causes hereditary haemochromatosis, which is an\nautosomal recessive disease. The mutant allele codes for a non-functioning protein. People who\nTON are homozygous for the mutant allele produce no functioning HFE protein and this results in an\nexcess of iron being absorbed by the body . The accumulation (build-up) of iron in body organs\nOD\nover many years can cause organ damage.\nPeople that are heterozygous for the HFE gene do not have hereditary haemochromatosis. They\ndo absorb more iron from their diet than people who do not have the mutation, but this does not\nNIGRAM\nusually have any health effects.\n(a) Construct a genetic diagram of a monohybrid cross to show how two parents who do not SIHT\nhave hereditary haemochromatosis can produce a child with the disease.\nNI\nETIRW\nUse the following symbols:\nH = normal HFE allele\nTON\nh = mutant HFE allele.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\n[3] NIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Some scientists believe that the C282Y mutation may have first occurred in Ireland.\nSIHT\nNI Scientists sequenced DNA obtained from two human fossil skeletons in Ireland. One of the\nETIRW\nfossils was 5200 years old and the other was 4000 years old.\nTON The scientists concluded that:\n• the human living 4000 years ago did have the C282Y mutation\nOD\n• the human living 5200 years ago did not have the C282Y mutation.\nExplain how analysis of the results of these DNA sequencing studies could have been carried\nout to allow the scientists to make these conclusions.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) At about the same time as the C282Y allele is thought to have first occurred in Ireland, the\nSIHT\nlifestyle of people in Europe began to change from hunter-gatherers to farmers.\nNI\nETIRW\nHunter-gatherers ate mainly meat with some wild plant food. Their diets had high quant",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-m-42-q04-ba39e9ebd2.pdf"
    },
    {
      "id": "9700-2025-m-42-q05",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-m-42-q05-p01-7f6cfb1a06.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:10)(cid:44)\n5 Fig. 5.1 shows a photomicrograph of a single plant cell in a stage of meiosis.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 5.1\nDescribe the stage of meiosis shown in Fig. 5.1. SIHT\nNI\n..........................................................................................................................................................\nETIRW\n..........................................................................................................................................................\nTON\n.......................................................................................................................................................... OD\n..........................................................................................................................................................\n.......................................................................................................................................................... NIGRAM\n..........................................................................................................................................................\nSIHT\n..........................................................................................................................................................\nNI\nETIRW\n..........................................................................................................................................................\nTON\n..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n.................................................................................................................................................... [5]\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-m-42-q05-eb70e67df0.pdf"
    },
    {
      "id": "9700-2025-m-42-q06",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-m-42-q06-p01-d18a5783cd.png",
        "9700-topic-14-homeostasis/assets/9700-2025-m-42-q06-p02-846bade7b2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n6 (a) Fig. 6.1 outlines part of the control mechanism that regulates blood glucose concentration.\nSIHT\nNI\nETIRW meal rich in\ncarbohydrates\nTON\nOD\nincrease in blood\nglucose concentration\nabove the set point\nNIGRAM\npancreas\nSIHT\nreleases P\nNI\nETIRW\nP stimulates glucose P stimulates liver\nTON\nuptake in fat cells cells to convert\nand Q cells glucose to R OD\ndecrease in blood\nglucose NIGRAM\nconcentration\nSIHT\nFig. 6.1\nNI\nETIRW\n(i) Identify P, Q and R.\nTON\nP ....................................................\nOD\nQ ...................................................\nR ....................................................\n[3] NIGRAM\n(ii) State the type of homeostatic control mechanism operating in Fig. 6.1.\nSIHT\n..................................................................................................................................... [1]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) When the blood glucose concentration decreases below the set point, a hormone is released\nSIHT\nfrom the pancreas.\nNI\nETIRW\nDescribe how the release of this hormone leads to the blood glucose concentration returning\nto the set point.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-m-42-q06-cde93ca184.pdf"
    },
    {
      "id": "9700-2025-m-42-q07",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-m-42-q07-p01-7b482ef543.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-m-42-q07-p02-8908a55c1b.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-m-42-q07-p03-65d45a52f7.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q07",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n7 The Sumatran tiger , Panthera tigris sumatrae, is classified as critically endangered on the\nSIHT\nSpeciesTM. International Union for Conservation of Nature (IUCN) Red List of Threatened\nNI\nETIRW\nFig. 7.1 shows a Sumatran tiger.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 7.1\n(a) Fig. 7.2 shows the number of wild Sumatran tigers in the world between 1970 and 2020.\nNIGRAM\n1200\nSIHT\nNI\nETIRW\n1000\nTON\nOD\n800\nNIGRAM\nnumber of\nSumatran 600\ntigers SIHT\nNI\nETIRW\n400\nTON\nOD\n200\nNIGRAM\n0\nSIHT\n1970 1980 1990 2000 2010 2020\nyear\nNI\nETIRW\nFig. 7.2\nTON\nNIGRAM\n(i) Calculate the mean rate of decrease in the Sumatran tiger population between 1970 and\nSIHT\n2020.\nNI\nETIRW\nTON\nOD\n–1 mean rate of decrease = .............................................. year [2]\nNIGRAM\n(ii) There are a number of dif ferent ways to help conserve Sumatran tigers. For example,\nsome zoos have captive breeding programmes. SIHT\nNI\nOutline ways in which Sumatran tigers may be conserved, other than captive breeding\nETIRW\nprogrammes.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n..................................................................................................................................... [3] TON\nOD\n(b) Outline reasons for maintaining animal biodiversity.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n.................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-m-42-q07-d3d1f59e38.pdf"
    },
    {
      "id": "9700-2025-m-42-q08",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q08-p01-48d44bd797.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-m-42-q08-p02-91bb1b116e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q08",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:7)(cid:44)\n8 (a) In the link reaction, a two-carbon acetyl group is produced from pyruvate. The acetyl group is\nSIHT\ntransferred to coenzyme A to form acetyl coenzyme A (acetyl-coA).\nNI\nETIRW\nState the terms used to summarise the two chemical changes that occur in the link reaction\nto produce an acetyl group from pyruvate.\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [2]\n(b) Acetyl-coA combines with oxaloacetate in the Krebs cycle to form citrate. This reaction is\nNIGRAM\ncatalysed by the enzyme citrate synthase.\nAcetyl-coA has a similar shape to succinyl-coA, one of the compounds made in a later part of SIHT\nthe Krebs cycle.\nNI\nETIRW\nAn experiment was carried out to investigate the ef fect of increasing the concentration of\nacetyl-coA on the activity of citrate synthase. The experiment was repeated, this time adding\nTON\na solution of succinyl-coA.\nOD\nFig. 8.1 shows the results of these experiments.\n1.0\nNIGRAM\nSIHT\n0.8\nNI\nETIRW\nwithout\nsuccinyl-coA\nTON\n0.6\nactivity of OD\ncitrate synthase\n/ arbitrary units\n0.4\nNIGRAM\nwith SIHT\nsuccinyl-coA 0.2\nNI\nETIRW\nTON\n0.0\nOD\n0 20 40 60 80 100 120\nconcentration of acetyl-coA / arbitrary units\nNIGRAM Fig. 8.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nWith reference to Fig. 8.1, explain how succinyl-coA could help to regulate the Krebs cycle.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n......................................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-m-42-q08-6aea8ed17a.pdf"
    },
    {
      "id": "9700-2025-m-42-q09",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        28,
        29
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-m-42-q09-p01-dcc0fde34e.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-m-42-q09-p02-7fa7d46988.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:9)(cid:44)\n9 The chloroplasts of leaves of tobacco plants, Nicotiana sp., contain chlorophyll a and chlorophyll b.\nSIHT\nNI (a) Describe the role of chlorophyll b in photosynthesis.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [2]\nSIHT\n(b) A mutant tobacco plant was found to contain more chlorophyll b than normal tobacco plants.\nNI\nETIRW\nAn investigation was carried out to measure the rate of photosynthesis of normal and mutant\ntobacco plants at increasing light intensities. The rate of production of oxygen was used as a\nTON\nmeasure of the rate of photosynthesis.\nOD\nOther variables were kept constant.\nFig. 9.1 shows the results of this investigation.\nNIGRAM\n300\nmutant plants\nSIHT\nNI\nETIRW\n200\nrate of production normal plants\nTON of oxygen\nmg–1 hour–1 / mmol\nOD\n100\nNIGRAM\n0\n0 200 400 600 800 1000\nSIHT\nlight intensity / lux\nNI\nETIRW\nFig. 9.1\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Describe the results shown in Fig. 9.1.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [3]\nNI\nETIRW\n(ii) It was observed that the mutant tobacco plants had a faster growth rate than the normal\ntobacco plants.\nTON\nOD Suggest explanations for this observation.\n..........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-m-42-q09-7c6ee495d4.pdf"
    },
    {
      "id": "9700-2025-m-42-q10",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_42.pdf?download=true",
      "source_pages": [
        30,
        31
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-m-42-q10-p01-a188881b6e.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-m-42-q10-p02-d0bba0543c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-42-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:1)(cid:44)\n10 (a) Fig. 10.1 shows chemoreceptor cells in a taste bud. Two of the chemoreceptor cells have\nSIHT\nformed synapses with sensory neurone dendrites.\nNI\nETIRW microvilli\nTON\nsupport cell\nOD\nNIGRAM\nchemoreceptor cell\nSIHT\nNI\nETIRW\nTON\nOD\nsensory neurone\ndendrite\nNIGRAM\nSIHT\nFig. 10.1\nNI\nETIRW\nDescribe how the contact of sodium ions with the microvilli of the chemoreceptor cell can lead\nto the release of a neurotransmitter by the cell.\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Some of the neurotransmitters in the brain are produced through a series of reactions\nSIHT\n(reaction pathway) from a chemical called DOPA.\nNI\nETIRW\nDOP A is also involved in other reaction pathways. For example, in the skin and eyes, DOP A\nis part of a different reaction pathway that depends on the TYR gene.\nTON\nDescribe and explain the phenotypic consequences for the skin and eyes of a person who is\nOD\nhomozygous for a mutated, non-functional allele of the TYR gene.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n.....................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-m-42-q10-a9eefdba54.pdf"
    },
    {
      "id": "9700-2025-m-52-q01",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-m-52-q01-p01-98fb3f0302.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q01-p02-f022cdffbd.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q01-p03-f0b5edd669.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q01-p04-e7ae9c27ce.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Sand dunes are important ecosystems that are found in many regions of the world. Sand dune\nSIHT\necosystems can be damaged by natural events, such as extreme weather conditions, and by\nNI human activities including farming and tourism.\nETIRW\nFig. 1.1 is a map of part of a national park in a coastal area of northern Europe. The map shows\nTON the range of ecosystems present, including two types of sand dune ecosystem: yellow dune and\ngrey dune.\nOD\nkey\nyellow dune\nNIGRAM\nroad\ngrey dune\nSIHT\nNI\nsalt marsh ETIRW\nTON\nrocky shore\nOD\nsea farmland\n0 1\nNIGRAM sand km\nSIHT\nFig. 1.1\nNI\nETIRW A biologist decided to study the grey dune ecosystem. The biologist visited the ecosystem in the\nsummer, in the month of June.\nTON\n(a) The grey dune ecosystem is a habitat for the grey bush cricket, Platycleis albopunctata.\nOD\nFig. 1.2 shows a grey bush cricket.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nFig. 1.2\nETIRW\nTON\nNIGRAM\nThe biologist wanted to estimate the population size of the grey bush cricket in the grey dune\nSIHT\necosystem.\nNI\nETIRW\nThe biologist used a sweep net to catch grey bush crickets. The biologist moved the sweep\nnet through the plants growing in the grey dune ecosystem. Any grey bush crickets on the\nTON plants were caught in the net.\nOD\nFig. 1.3 shows a biologist using a sweep net.\nNIGRAM\nSIHT\nNI\nsweep net\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW Fig. 1.3\nDescribe the mark-release-recapture method that the biologist could use to estimate the TON\npopulation size of the grey bush cricket.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The biologist also estimated the species diversity of the plants in the grey dune ecosystem.\nSIHT\nNI The biologist counted the number of individual plants of each species in a small area and\nETIRW\ncalculated Simpson’s index of diversity (D).\nTON The formula for Simpson’ s index ",
      "pdf_path": "9700-experimental-skills/print/9700-2025-m-52-q01-c9988ff403.pdf"
    },
    {
      "id": "9700-2025-m-52-q02",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_52.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-m-52-q02-p01-f42611c5f6.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q02-p02-31af45d0a1.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q02-p03-451ad68ffe.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q02-p04-ca96d8e045.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n2 Nagarahole National Park in southern India is a forest ecosystem that is important for the\nSIHT\nconservation of large mammals such as the Asian elephant, Elephas maximus. Conflicts between\nNI people and elephants in the area surrounding the national park can threaten conservation work.\nETIRW\nFig. 2.1 shows a group of Asian elephants.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 2.1\nSIHT\nSeveral coffee farms (farms where cof fee is the crop) are found close to Nagarahole National\nNI Park. Asian elephants sometimes leave the national park and enter the cof fee farms. Elephants\nETIRW\ndamage crops and trees, and sometimes injure people who work on the farms.\nTON December trees, Erythrina subumbrans, are planted on cof fee farms to provide shade and are\noften damaged when elephants enter the farms.\nOD\nSome scientists wanted to investigate the reasons why elephants enter cof fee farms close to\nNagarahole National Park.\nNIGRAM\n20 coffee farms in an area to the south of Nagarahole National Park were chosen at random.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nIn March 2008, the scientists measured four variables at each coffee farm, as shown in Table 2.1.\nSIHT\nNI Table 2.1\nETIRW\nhow the variable was determined for each\nvariable TON\ncoffee farm\nOD\nwater availability counting the number of ponds, lakes and water\nholes visible in satellite images\ndensity of December trees counting the number of December trees in ten\nNIGRAM circular areas, each with a radius of 10 m\ngrass biomass drying and measuring the dry mass of all the\nSIHT grass leaves collected in 50 randomly selected\nm2 areas of 0.25\nNI\nETIRW\ndistance from Nagarahole using GPS to measure the distance between\nNational Park the boundary of Nagarahole National Park and\nTON\nthe boundary of the farm\nOD\nFarmers from each coffee farm completed a questionnaire to record when elephants entered the\nfarm.\nNIGRAM The scientists decided to use Spearman’s rank correlation to investigate the relationship between\neach of the four variables in Table 2.1 and the number of months that elephants entered each\ncoffee farm per year.\nSIHT\n(a) State two conditions that allow Spearman’s rank correlation to be used in this investigation. NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-m-52-q02-b8aed58457.pdf"
    },
    {
      "id": "9700-2025-m-52-q03",
      "subject": "9700",
      "year": 2025,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-march/9700_m25_qp_52.pdf?download=true",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-March/qp/9700_m25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-m-52-q03-p01-f9bae471c4.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q03-p02-c6022ba4c6.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q03-p03-721e33e44f.png",
        "9700-experimental-skills/assets/9700-2025-m-52-q03-p04-b8bd28787c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-m-52-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n3 Type 1 diabetes is a disease that occurs when the pancreas stops producing insulin. Type 1\nSIHT\ndiabetes occurs most commonly in children and is often the result of the immune system destroying\nNI beta cells in the pancreas. Beta cells are the cells in the pancreas that make insulin.\nETIRW\nRotavirus is a genus of RNA viruses that can cause infections in people throughout most of the\nTON world. Infection by Rotavirus causes sickness and diarrhoea in babies and young children, and\nmay damage the pancreas. Rotavirus infections are also associated with the development of\nOD\ntype 1 diabetes in children.\nIn May 2007, a vaccination against Rotavirus was introduced in Australia. All babies aged from\n2 months to 4 months were offered the Rotavirus vaccine.\nNIGRAM\nSome scientists decided to investigate whether vaccination against Rotavirus had reduced the\nnumber of children with type 1 diabetes in Australia. SIHT\nNI\n(a) Identify the independent variable and the dependent variable in this investigation.\nETIRW\nindependent variable ................................................................................................................\nTON\nOD dependent variable ...................................................................................................................\n[2]\n(b) The scientists used data that had been collected from 2000 to 2015. For each year from 2000\nNIGRAM to 2015, the scientists used the data to:\n• find the number of children aged from 0 years to 4 years\nSIHT\n• find the number of children aged from 0 years to 4 years with type 1 diabetes NI\nETIRW\n• calculate the number of children aged from 0 years to 4 years with type 1 diabetes per\n100 000 children that were aged from 0 to 4 years. TON\nOD\nThe scientists repeated this analysis for children aged from 10 years to 14 years.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Fig. 3.1 shows the number of children aged from 0 years to 4 years with type 1 diabetes\nSIHT\nper 100 000 for each year from 2000 to 2015.\nNI\nETIRW\nThe predicted number of children aged from 0 years to 4 years with type 1 diabetes per\n100 000, if the Rotavirus vaccine had not been introduced, is also shown from 2008 to\nTON 2015.\nOD\n18\nNIGRAM\n16 number of\nchildren with\nSIHT\ntype 1\ndiabetes per NI\nETIRW 100 000\n14\nRRoottaavviirruuss\nvvaacccciinnee\nTON iinnttrroodduucceedd\nOD\n12\n2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015\nyear\nNIGRAM\nkey\nnumber of children with type 1 diabetes\nSIHT predicted number of children with type 1 diabetes\nNI\nETIRW Fig. 3.1\nState two conclusions that can be made from the data shown in Fig. 3.1. TON\nOD\n...........................................................................................................................................\n.........................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-m-52-q03-12da9f9996.pdf"
    },
    {
      "id": "9700-2025-mj-41-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-41-q01-p01-8056c59d86.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-41-q01-p02-e50c28f761.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Organisms need a source of energy for many cellular processes. Respiration involves the release\nSIHT\nof energy from energy-rich molecules for the synthesis of ATP molecules. ATP is described as the\nNI energy currency of cells.\nETIRW\n(a) Name two energy-rich molecules that enter the respiration pathway to synthesise ATP.\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [2]\n(b) Fig. 1.1 shows a molecule of ATP.\nNIGRAM\nNH\n2\nSIHT\nN\nNI N O O O\nETIRW\nP O HO P O P O\nTON N\nN\nO\nOD O– O– O–\nNIGRAM\nOH OH\nSIHT\nFig. 1.1\nNI\nETIRW Explain the features of ATP that make it suitable to be the universal energy currency of cells.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Identify the type of phosphorylation reaction to synthesise ATP that occurs during glycolysis\nSIHT\nand the Krebs cycle.\nNI\nETIRW\n............................................................................................................................................. [1]\nTON (d) (i) Pyruvate moves into the matrix of the mitochondrion only when a particular inorganic\nmolecule is present.\nOD\nName the inorganic molecule that must be present in the cell for pyruvate to enter the\nmatrix of the mitochondrion.\nNIGRAM\n.................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-41-q01-41e911d301.pdf"
    },
    {
      "id": "9700-2025-mj-41-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-41-q02-p01-1b774c818e.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-41-q02-p02-7bf9ebab0f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q02",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 A respirometer can be used to investigate how temperature af fects the rate of respiration\nSIHT\nof woodlice.\nNI\nETIRW\nWoodlice are small invertebrate animals. Fig. 2.1 shows a single woodlouse.\nTON\nOD\nNIGRAM\nFig. 2.1\nFig. 2.2 shows a simple respirometer.\nSIHT\nNI sodium\nETIRW hydroxide drop of\npellets boiling tube liquid\nTON\nOD\ngraduated wire mesh woodlouse\ncapillary tube\nNIGRAM\nFig. 2.2\nSIHT\n(a) Explain how the experimental set-up in Fig. 2.2 allows the rate of respirationo f woodlice to be\nNI\ndetermined. ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [4]\nNI\nETIRW\nTON\nNIGRAM\n(b) A student determined the rate of respiration of woodlice at a temperature of 5 °C and at a\nSIHT\ntemperature of 20 °C.\nNI\nETIRW\nPredict and explain how the rate of respiration of woodlice dif fers at 5 °C compared to 20 °C.\nTON ...................................................................................................................................................\nOD\n...............................................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-41-q02-e16fff1773.pdf"
    },
    {
      "id": "9700-2025-mj-41-q03",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-41-q03-p01-d4f2875faa.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-41-q03-p02-c243523a28.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-41-q03-p03-964ef8737f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 Porphyria is a group of rare genetic diseases in which molecules called porphyrins accumulate in\nSIHT\nthe body.\nNI\nETIRW\nThree examples of porphyria are:\nTON • X-linked protoporphyria\n• variegate porphyria\nOD\n• congenital erythropoietic porphyria (CEP).\n(a) X-linked protoporphyria is caused by a mutant allele located on the X chromosome.\nNIGRAM\nFig. 3.1 shows the pattern of inheritance of X-linked protoporphyria in one family .\nkey SIHT\nNI female without the disease\n1 2 ETIRW\nmale without the disease\nTON\nOD\nfemale with the disease\n3 4 5 6\nmale with the disease\nNIGRAM\nFig. 3.1\nSIHT\nX-linked protoporphyria is described as a sex -linked disease.\nNI\nETIRW\nUse the information in Fig. 3.1 to make one other conclusion about the pattern of inheritance\nof X-linked protoporphyria.\nTON\nOD Give evidence to support your conclusion.\nconclusion .................................................................................................................................\nNIGRAM evidence ...................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n[2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Variegate porphyria occurs in 1 in 100 000 people in Europe.\nSIHT\nNI In the 1600s, approximately 100 Dutch people migrated from Europe to South Africa.\nETIRW\nThe population of people of Dutch descent in South Africa is now 2.5 million. V ariegate\nTON porphyria occurs in 1 in 1000 people in this population.\nOD\nSuggest and explain why variegate porphyria is more common among people of Dutch\ndescent in South Africa than among people in Europe.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-mj-41-q03-014093c355.pdf"
    },
    {
      "id": "9700-2025-mj-41-q04",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-41-q04-p01-d6b5d2e893.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-41-q04-p02-9588b267a5.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-41-q04-p03-556b7d2260.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q04",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:10)(cid:44)\n4 Scientists have invented a way of producing food by artificial photosynthesis.\nSIHT\n• NI Solar panels convert sunlight energy into electricity.\nETIRW • The electricity powers an electrolysis reaction between carbon dioxide gas and water to form\nthe organic product acetate.\nTON • The single-celled alga Chlamydomonas, a protoctist, can use acetate to grow and reproduce\nin the dark, instead of photosynthesising in the light.\nOD\n• The algae can be processed to make a food product.\nFig. 4.1 shows an outline of the artificial photosynthesis process.\nNIGRAM\nelectrolyser\nO O H\n2 2\nSIHT\nacetate CO\n2 NI\nfood algae ETIRW\nelectron\nTON flow\nsolar panel OD\nFig. 4.1\nNIGRAM (a) Identify three similarities between the artificial photosynthesis process shown in Fig. 4.1 and\nthe normal process of photosynthesis.\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Fig. 4.2 shows the single-celled alga Chlamydomonas. It contains a large, cup-shaped\nSIHT\nchloroplast for photosynthesis.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nB\nNI\nETIRW\nTON\nOD\nA\nNIGRAM\nSIHT Fig. 4.2\nNI\nStructures A and B in Fig. 4.2 are found within the chloroplast.\nETIRW\nName structures A and B.\nTON\nOD A ...............................................................\nB ...............................................................\n[2]\nNIGRAM\n(c) Thylakoid membranes are the site of the light-dependent stage of photosynthesis.\nSIHT\n(i) Name the products of the light-dependent stage of photosynthesis.\nNI\nETIRW ..................................................................................................................................... [1]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Some of the protein components of the thylakoid membrane have a role in the\nSIHT\nlight-dependent stage of photosynthesis.\nNI\nETIRW\nExplain the roles of the dif ferent proteins that function in the light-dependent stage\nof photosynthesis.\nTON\n.........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-mj-41-q04-b8cd746266.pdf"
    },
    {
      "id": "9700-2025-mj-41-q05",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q05-p01-94cc210a2f.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q05-p02-ff3c5c90a5.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q05-p03-6b37094c0d.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q05-p04-fb1f12974a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q05",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 The Galápagos is a group of islands with a high biodiversity.\nSIHT\nNI (a) Geospiza fortis is one of the many species of finch that live on the Galápagos islands. On one\nETIRW\nof the Galápagos islands, Daphne Major, scientists measured the bill size (length, width and\ndepth) of individuals in a population of G. fortis over several years.\nTON\nFig. 5.1 shows a G. fortis female.\nOD\nNIGRAM\nbill\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 5.1\nG. fortis feed on seeds. Seed size and bill size vary . It is easier for G. fortis with smaller bills\nNIGRAM to eat small seeds and for G. fortis with larger bills to eat large seeds.\nIn 1977, a drought occurred on Daphne Major , which resulted in a large decrease in the\nSIHT\navailability of seeds, particularly small seeds.\nNI\nETIRW Scientists observed an increase in the mean bill size in the G. fortis population on\nDaphne Major after the drought.\nTON\n(i) State the type of natural selection that occurred in the G. fortis population on\nOD\nDaphne Major as a result of the drought.\n..................................................................................................................................... [1]\nNIGRAM\n(ii) Fig. 5.2 shows the distribution of bill size in the G. fortis population before the drought\nin 1977.\nSIHT\nNI\nETIRW\nnumber of\nTON\nG. fortis\nOD\nNIGRAM\nbill size\nSIHT Fig. 5.2\nNI\nSketch a new curve on Fig. 5.2 to show the distribution of bill size in the G. fortis\nETIRW\npopulation after the drought.\n[1]\nTON\nNIGRAM\n(b) G. fortis also lives on Santa Cruz, another island in the Galápagos.\nSIHT\nNI Most of the seeds available for G. fortis to eat on Santa Cruz are either small or large. There\nETIRW\nare few intermediate sizes of seed.\nTON Fig. 5.3 shows the distribution of bill size in the G. fortis population on Santa Cruz.\nOD\nnumber of NIGRAM\nG. fortis\nSIHT\nNI\nETIRW\nbill size\nTON\nFig. 5.3 OD\nExplain how natural selection in the G. fortis population on Santa Cruz has resulted in the\ndistribution in bill size shown in Fig. 5.3.\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n..............................................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-41-q05-f064507ae2.pdf"
    },
    {
      "id": "9700-2025-mj-41-q06",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-mj-41-q06-p01-928a67a6d8.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-41-q06-p02-4a74505b19.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n6 (a) In the kidney, various molecules move between the blood and the kidney nephron.\nSIHT\nNI T able 6.1 describes the movements of molecules at three regions of the nephron, P,Q and R.\nETIRW\nTable 6.1\nTON\nOD region of\nmolecule(s) direction of movement\nnephron\namino acids, glucose\nP from nephron to blood\nand water\nNIGRAM\nfrom nephron to blood in the\nQ water presence of antidiuretic hormone\nSIHT\n(ADH)\nNI\nwater and molecules\nETIRW\nR with a molecular mass from blood to nephron\nof less than 68,000\nTON\nOD (i) Name the processes, other than dif fusion, osmosis and active transport, that are\noccurring at P, Q and R.\nP ...............................................................\nNIGRAM\nQ ...............................................................\nSIHT\nR ...............................................................\n[3] NI\nETIRW\n(ii) Identify the regions of the nephron represented by P, Q and R.\nTON\nP ...............................................................\nOD\nQ ...............................................................\nR ...............................................................\nNIGRAM\n[3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) A student cut a fresh kidney lengthways and placed one half in the freezer . After 24 hours,\nSIHT\nthe student examined the kidney section and tested its firmness with a mounted needle.\nNI\nETIRW\nSodium chloride concentration af fects the freezing point of a solution.\nTON The student drew the diagram in Fig. 6.1 to show an area that had frozen hard and an area\nthat was softer and less frozen.\nOD\nA: frozen hard\nNIGRAM\nB: softer and\nSIHT\nless frozen\nNI\nETIRW\nTON\nOD\nFig. 6.1\nSuggest an explanation for the observations in Fig. 6.1 made by the student.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\n[Total: 8]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-mj-41-q06-b777cdf497.pdf"
    },
    {
      "id": "9700-2025-mj-41-q07",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-mj-41-q07-p01-8690a6a676.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-41-q07-p02-65f028c731.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q07",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n7 Insulin is an example of a cell-signalling molecule of the endocrine system.\nSIHT\nNI (a) Outline why insulin can be described as an example of a cell-signalling molecule of the\nETIRW\nendocrine system.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n............................................................................................................................................. [4]\nTON\nOD (b) Researchers compared the effect of two types of meal on thme ean blood glucose concentration\nand the mean blood insulin concentration of twelve volunteers. The measurements were\nmade for three hours after the meal.\nNIGRAM Meal A contained 55% carbohydrate, 27% fat and 18% protein.\nMeal B contained 3% carbohydrate, 90% fat and 7% protein.\nSIHT\nFig. 7.1 shows the results.\nNI\n600 ETIRW\n120 TON\nOD\n110 400\nmean blood\ninsulin\nmean blood 100\nconcentration\nglucose NIGRAM dm–3 pmol /\nconcentration 90 200\ncm–3 / mg 100\nSIHT\n80\nNI\nETIRW\n0 0\n0 0 30 60 90 120 150 180 30 60 90 120 150 180\nTON time / min time / min\nOD\nkey\nmeal A\nmeal B\nNIGRAM\nFig. 7.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) With reference to Fig. 7.1, explain how negative feedback operates to control mean\nSIHT\nblood glucose concentration in the volunteers who ate meal A.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT .........................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-mj-41-q07-2dfabaf834.pdf"
    },
    {
      "id": "9700-2025-mj-41-q08",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        20
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-41-q08-p01-146e34967c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n8 The presence of gibberellins in a plant cell leads to the expression of genes involved in\nSIHT\nstem elongation.\nNI\nETIRW\nDescribe how gibberellin causes stem elongation in plants.\nTON ..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM\n..........................................................................................................................................................\n.......................................................................................................................................................... SIHT\nNI\n..........................................................................................................................................................\nETIRW\n..........................................................................................................................................................\nTON\nOD ..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM ..........................................................................................................................................................\n..........................................................................................................................................................\nSIHT\n.......................................................................................................................................................... NI\nETIRW\n..........................................................................................................................................................\nTON\n..........................................................................................................................................................\nOD\n[Total: 6]\nNIGRAM",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-mj-41-q08-eee4121502.pdf"
    },
    {
      "id": "9700-2025-mj-41-q09",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q09-p01-5068ab57c5.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-41-q09-p02-e8ce3670ad.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n9 A species is classified into one of three domains.\n(a) Two of the domains contain only prokaryotic species.\nSIHT\nNI With reference to the two prokaryotic domains, describe the features used to classify\nETIRW\nprokaryotic species into two different domains.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n............................................................................................................................................. [3]\nTON\nNIGRAM\n(b) Tigers are large mammals that live in many parts of Asia.\nSIHT\nNI Fig. 9.1 shows a tiger , Panthera tigris.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 9.1\nTON\nComplete Table 9.1 by writing the domain, kingdom and genus in which tigers are classified.\nOD\nTable 9.1\nNIGRAM\ntaxon name\ndomain\nSIHT\nkingdom NI\nETIRW\nphylum Chordata\nTON\nclass Mammalia\nOD\norder Carnivora\nfamily Felidae\nNIGRAM\ngenus\nSIHT [3]\nNI\n[Total: 6] ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-41-q09-e5577eb8a5.pdf"
    },
    {
      "id": "9700-2025-mj-41-q10",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_41.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-41-q10-p01-aa46b04eaf.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-41-q10-p02-5cc61e64ee.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-41-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 During the course of an action potential, the potential dif ference across a neurone\nSIHT\nmembrane changes.\nNI\nETIRW\nTable 10.1 shows the potential difference across the neurone membrane at three sequential time\npoints, X, Y and Z, just before and during an action potential.\nTON\nTable 10.1\nOD\ntime potential difference\npoint / mV\nNIGRAM\nX –70\nY +40\nSIHT\nZ –90\nNI\nETIRW\n(a) Describe the events that cause each of the potential differences at points X, Y and Z.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n..........................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-mj-41-q10-7949ecb9e6.pdf"
    },
    {
      "id": "9700-2025-mj-42-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-42-q01-p01-5abbc5a14c.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-42-q01-p02-aeccf2b983.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Fig. 1.1 is a diagram of part of the inner membrane of a mitochondrion.\nSIHT\nintermembrane space NI\nETIRW\ncarriers\nTON\nOD\n4 inner 3 1\nmembrane\n2\nNIGRAM\n5\nSIHT\nmatrix\nNI\nETIRW Fig. 1.1\nTON (a) (i) The coenzymes NAD and F AD deliver hydrogen atoms to the electron transport\nchain (ETC).\nOD\nWith reference to Fig. 1.1, state which of the four carriers receives hydrogen atoms from\nreduced FAD.\nNIGRAM\n…………………………………………….. [1]\n(ii) Hydrogen atoms that are delivered to the ETC by reduced NAD and reduced F AD split SIHT\ninto protons and electrons. Energy is released as electrons pass along the ETC.\nNI\nETIRW\nDescribe the events that occur as a result of this release of energy.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\n(iii) Describe how structure 5 in Fig. 1.1 is involved in oxidative phosphorylation.\nSIHT\n...........................................................................................................................................\nNI\n........................................................................................................................................... ETIRW\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ..................................................................................................................................... [3]\nETIRW\nTON\nNIGRAM\n(CN–) (b) Cyanide ions are highly toxic.\nSIHT\nNI Cyanide ions bind to carrier 4 in Fig. 1.1 and inactivate the carrier .\nETIRW\nSuggest and explain how the binding of cyanide ions to carrier 4 can have an ef fect on\nTON respiration.\nOD\n...................................................................................................................................................\n..........................................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-42-q01-9eec131abb.pdf"
    },
    {
      "id": "9700-2025-mj-42-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-42-q02-p01-6e4376aacb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 When organisms reproduce, they pass on their alleles to the next generation. There are many\nSIHT\nfactors that can affect how allele frequencies change over time in a population.\nNI\nETIRW\nExplain how genetic drift and the founder effect may affect allele frequencies in populations.\nTON ..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM\n..........................................................................................................................................................\n.......................................................................................................................................................... SIHT\nNI\n..........................................................................................................................................................\nETIRW\n..........................................................................................................................................................\nTON\nOD ..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM ..........................................................................................................................................................\n..........................................................................................................................................................\nSIHT\n.......................................................................................................................................................... NI\nETIRW\n..........................................................................................................................................................\nTON\n..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM\n[Total: 6]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIR",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-42-q02-b63334055c.pdf"
    },
    {
      "id": "9700-2025-mj-42-q03",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-mj-42-q03-p01-5de8ba28ad.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-42-q03-p02-2e565923c5.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-42-q03-p03-5c1cfb3a9e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:6)(cid:44)\n3 Rice, Oryza sativa, is an important grain crop. A rice grain is a seed and can have a structure\nSIHT\nknown as an awn, which projects from the tip of the grain.\nNI\nETIRW Fig. 3.1 shows a rice grain with an awn present and a rice grain with no awn present.\nTON\nawn\nOD\nno awn present\nNIGRAM\ngrain\nSIHT\nFig. 3.1\nNI\nETIRW\n(a) (i) The development of awns is controlled by two genes: gene A/a and gene B/b. The\ngenes are present on dif ferent autosomes. The presence of either dominant allele A or\nTON\ndominant allele B results in rice plants producing grains with awns.\nOD\nA cross was carried out between a rice plant that is homozygous dominant for the two\ngenes (double homozygous dominant) and a rice plant that is homozygous recessive for\nthe two genes (double homozygous recessive). All F1 offspring plants produced grains\nwith awns. NIGRAM\nConstruct a genetic diagram, including a Punnett square, to show the cross that produces\nSIHT the F2 generation, including phenotypes.\nNI\nState the ratio of the offspring phenotypes produced.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\n[5]\nTON\nNIGRAM\n(ii) Deduce the type of inheritance shown by the ratio of the of fspring phenotypes stated in\nSIHT\n3(a)(i).\nNI\nETIRW ..................................................................................................................................... [1]\n(b) It is common for wild rice plants to have grains with awns present. TON\nOD\nSuggest a selective advantage to wild rice of having awns present on their grains.\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [1]\n(c) One of the changes that occurred during the domestication of wild rice to cultivated rice was\nSIHT\nthe loss of the awns from rice grains.\nNI\nETIRW\nFarmers found that long awns made storing and processing rice grains more dif ficult.\nTON\nIt was also observed that rice plants that have grains with no awns have an increased\ngrain yield. OD\n(i) Explain the principles used by farmers to produce rice plant grains with no awns.\n........................................................................................................................................... NIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n..........................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-mj-42-q03-8b3c2b322a.pdf"
    },
    {
      "id": "9700-2025-mj-42-q04",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-mj-42-q04-p01-ff6dd263c3.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-42-q04-p02-31ba0dcbcb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q04",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 Sexual reproduction in plants and animals involves:\nSIHT\n• NI the formation of gametes as a result of meiosis\nETIRW • the process of fertilisation.\nTON (a) In most species of plants and animals, the cell that is formed as a result of fertilisation is\ndiploid and contains homologous chromosomes.\nOD\nExplain why the cell that is formed as a result of fertilisation is a diploid cell and contains\nhomologous chromosomes.\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [3]\nOD\n(b) State the name of the stage in meiosis when reduction division occurs and ex plain a reason\nfor your choice.\nNIGRAM\nstage in meiosis ........................................................................................................................\nreason .......................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n[2]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Microscope slides can be",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-mj-42-q04-60bffb7d20.pdf"
    },
    {
      "id": "9700-2025-mj-42-q05",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-42-q05-p01-11fd58f26a.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-42-q05-p02-1f39587769.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 A number of diseases in humans can be treated using recombinant human proteins. These are\nSIHT\nproduced by recombinant DNA technology.\nNI\nETIRW\n(a) To produce a human protein for treatment of a disease, recombinant DNA technology needs\na gene coding for the particular human protein.\nTON\nOutline the dif ferent ways that can be used to obtain a gene that codes for a human protein.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ............................................................................................................................................. [4]\n(b) Diabetes mellitus is a disease in which the blood glucose concentration cannot be controlled. TON\nMany people with diabetes mellitus use recombinant human insulin to help control their blood\nOD\nglucose concentration.\nBefore recombinant human insulin became available, animals were the main source of insulin.\nNIGRAM\nExplain the advantages of using recombinant human insulin to treat diabetes.\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n....................................................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-mj-42-q05-f394af2111.pdf"
    },
    {
      "id": "9700-2025-mj-42-q06",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-mj-42-q06-p01-01bd3987c4.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-42-q06-p02-375cf0340e.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-42-q06-p03-e41d3deacf.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n6 Guard cells are located in the epidermis of the leaves of most plants. When environmental\nSIHT\nconditions change, this causes changes in guard cells that control the opening and closing\nNI of stomata.\nETIRW\n(a) Describe the structure of guard cells.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n(b) Stomata have daily rhythms of opening and closing.\nSIHT\nFig. 6.1 shows the percentage of stomata open at dif ferent times of the day, over a period of\nthree days, in the thale cress plant, Arabidopsis thaliana. NI\nETIRW\n100 Key\nTON\n12:00 = midday\n90\nOD\n24:00 = midnight\n= darkness 80\n70 NIGRAM\n60\nSIHT\npercentage\nof stomata 50 NI\nETIRW open\n40\nTON\n30 OD\n20\nNIGRAM\n10\n0 SIHT\n24:00 12:00 24:00 12:00 24:00 12:00 24:00\ntime of day NI\nETIRW\nFig. 6.1\nTON\nNIGRAM\n(i) With reference to Fig. 6.1, describe the rhythm of stomatal opening and closing.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... ",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-mj-42-q06-f1f0c6f226.pdf"
    },
    {
      "id": "9700-2025-mj-42-q07",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-42-q07-p01-2c069bacce.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-42-q07-p02-a47f65f2cc.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-42-q07-p03-2ffe29e602.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:6)(cid:44)\n7 (a) An axon membrane is described as being at its resting potential when an action potential is\nSIHT\nnot occurring.\nNI\nETIRW\nDescribe and explain how a resting potential of an ax on membrane is maintained.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [6]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n+) (b) Hypokalaemia is a condition in which there is a low concentration of potassium ions (K in\nSIHT\nthe body. This can affect nervous coordination.\nNI\nETIRW\nFig. 7.1 shows a normal action potential and Fig. 7.2 shows an action potential of a person\nwith hypokalaemia.\nTON\n+40\nOD\n+20\n0 NIGRAM\n–20\nSIHT\nmembrane\nNI –40 potential\nETIRW\n/",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-mj-42-q07-b11340b012.pdf"
    },
    {
      "id": "9700-2025-mj-42-q08",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-42-q08-p01-730fd23e9c.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-42-q08-p02-2f4dc9871f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q08",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 A lichen describes a mutually beneficial association of a fungus with an organism termed\nSIHT\na photobiont. An example of a photobiont is the green alga, T rebouxia sp., which is a\nNI photosynthetic protoctist.\nETIRW\nFig. 8.1 shows lichen attached to a tree.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 8.1\nNIGRAM\n(a) Outline the characteristic features of the kingdom Fungi.\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n............................................................................................................................................. [4]\nTON\n(b) Suggest how a fungus and a green alga benefit from the relationship shown by a lichen.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [2]\nNI\nETIRW\nTON\nNIGRAM\n(c) A suspension of Trebouxia in water was used to investigate the ef fect of the intensity of light\nSIHT\non the rate of photosynthesis. The volume of oxygen released over a set period of time was\nNI used as a measure of the rate of photosynthesis at each different light intensity.\nETIRW\nAll other conditions were kept constant.\nTON\nFig. 8.2 shows how light intensity af fected the volume of oxygen released by Treboux",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-mj-42-q08-de281f5f33.pdf"
    },
    {
      "id": "9700-2025-mj-42-q09",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-42-q09-p01-1ad218e73b.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-42-q09-p02-1ffb6eb0eb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q09",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 (a) The biodiversity of an area can be assessed using a variety of sampling methods.\nSIHT\nNI Outline how a frame quadrat could be used to assess the biodiversity of plants in a field.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [4]\nSIHT\n(b) A student investigated whether the height of the soft rush plant, Juncus effusus, decreases\nwith an increase in altitude on a hillside in the United Kingdom. NI\nETIRW\n• 12 sites were chosen at increasing altitudes.\n• The mean height of 10 plants was calculated at each altitude. TON\nOD\nSpearman’ s rank correlation was used to assess the relationship between the height of the\nplants and altitude.\n) is: The equation for Spearman’ s rank correlation (r\ns NIGRAM\nKey to symbols:\n( )\nΣD2 6 ×\nr = 1 –\nSIHT D = difference in rank between each pair of measurements s n3 – n\nn = number of pairs of items in the sample\nNI\nETIRW\nΣD2 was calculated to be 550.\nTON\n(i) Calculate the Spearman’s rank correlation for these data.\nOD\nGive your answer to three decimal places.\nNIGRAM\nSIHT\nNI\nETIRW answer .......................................................... [3]\nTON\nNIGRAM\n(ii) The null hypothesis for this investigation is: there is no correlation between the altitude\nSIHT\nand the height of the soft rush plants.\nNI\nETIRW\nT",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-42-q09-ae035a35b1.pdf"
    },
    {
      "id": "9700-2025-mj-42-q10",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_42.pdf?download=true",
      "source_pages": [
        22,
        23,
        24
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-42-q10-p01-1d5364c90c.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-42-q10-p02-a2341f5602.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-42-q10-p03-2592d1522f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-42-q10",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 (a) The grey seal, Halichoerus grypus, is an aquatic mammal that lives in the North Atlantic\nSIHT\nOcean. It feeds on fish, which it hunts at depths of up to 70 metres.\nNI\nETIRW\nFig. 10.1 shows a grey seal.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 10.1 NIGRAM\nDiving to hunt for fish has an ef fect on the respiration of the grey seal.\nSIHT\nA study was carried out to measure the blood lactate concentration of a grey seal before, NI\nduring and after a dive in deep water. ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nFig. 10.2 shows the results of this study .\nSIHT\n30\nNI\nETIRW\ndive\nTON\n25\nOD\nNIGRAM 20\nSIHT\nblood\nlactate\nNI 15\nconcentration ETIRW\ndm–3 / mmol\nTON\nOD 10\nNIGRAM\n5\nSIHT\nNI\n0 ETIRW\n80 0 20 40 60\ntime / min\nTON\nOD Fig. 10.2\nWith reference to Fig. 10.2, suggest reasons for the change in blood lactate concentration of\nthe seal:\nNIGRAM\n• during the dive\n• after the dive.\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [4]\nETIRW\nTON\nNIGRAM\n(b) Some seal species are classified as endangered on the IUCN Red List of Threatened\nSIHT\nSpeciesTM.\nNI\nETIRW\nSuggest ways in which seal species may be conserved.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n..................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-42-q10-7ce9b58997.pdf"
    },
    {
      "id": "9700-2025-mj-43-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-43-q01-p01-6bec54f58b.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-43-q01-p02-71125027ca.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Fig. 1.1 is a diagram of part of a liver cell.\nSIHT\nNI glucose A\nETIRW\ntissue fluid\ncell surface\nTON\nmembrane\nOD\nglucose\nB\ncytoplasm pyruvate\nouter mitochondrial NIGRAM\nmembrane\nC\nSIHT\nNI inner mitochondrial\nETIRW\nmembrane\nmitochondrial matrix\npyruvate TON\nmitochondrial\npyruvate carrier (MPC) OD\nFig. 1.1\nNIGRAM (a) With reference to Fig. 1.1, name:\nthe type of membrane transport protein represented by A ……………………………\nSIHT\nprocess B ………………………………………………………….. NI\nETIRW\narea C ………………………………………………………………\n[3] TON\nOD\n(b) The mitochondrial pyruvate carrier (MPC), shown in Fig. 1.1, allows the passage of pyruvate\ninto the mitochondrial matrix. When pyruvate enters the mitochondrial matrix, it takes part in\nthe link reaction.\nNIGRAM\nDescribe the link reaction.\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................................................................................................................................. [4]\nTON\nNIGRAM\n(c) Some tumour cells have a greatly reduced ability to transport pyruvate into the matrix of\nSIHT\nthe mitochondrion.\nNI\nETIRW\nSuggest how a reduction in pyruvate transport could af fect respiration in these tumour cells.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-43-q01-a50ef19171.pdf"
    },
    {
      "id": "9700-2025-mj-43-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-43-q02-p01-db3b944e24.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-43-q02-p02-d95f941cd4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q02",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 Natural selection and selective breeding (artificial selection) are processes that result in changes\nSIHT\nin the gene pool of a population.\nNI\nETIRW\nNatural selection and selective breeding have implications for humans.\nTON (a) If a person with a bacterial infection does not finish the course of an antibiotic given, it provides\nthe conditions for a population of bacteria to become resistant to this antibiotic.\nOD\n(i) A mutation in a bacterial gene can give resistance to an antibiotic. Directional selection\ncan occur when the antibiotic is present in the environment.\nNIGRAM\nA bacterium can also gain resistance when it receives genetic material from another\nbacterium in a process known as horizontal gene transfer.\nSIHT\nOutline how directional selection and horizontal gene transfer result in a new population\nNI\nof bacteria that is resistant to an antibiotic.\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [4]\nOD\n(ii) Some bacterial diseases can be treated only with one antibiotic, because the bacterial\npathogens are resistant to all other antibiotics.\nNIGRAM\nA drug is being developed to help treatment.\n• The drug is a small polynucleotide.\n• The drug inhibits translation of the messenger RNA (mRNA) produced by\nSIHT\ntranscription of the gene associated with antibiotic resistance.\nNI • The bacteria are then susceptible to more antibiotics.\nETIRW\nSuggest and explain how the drug could cause bacteria to become susceptible to\nTON\nmore antibiotics.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n.........................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-43-q02-66fb1ea2b1.pdf"
    },
    {
      "id": "9700-2025-mj-43-q03",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-mj-43-q03-p01-7d749fdc5f.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-43-q03-p02-be60cdcb02.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 An operon is a section of DNA found in prokaryotes.\nSIHT\nNI (a) Explain why the enzymes coded for by the lac operon are described as inducible enzymes.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\n(b) An investigation into the induction and action of the lac operon was carried out using the\nbacterium, Escherichia coli, grown in a growth medium containing glucose. SIHT\nNI\nWhen the bacteria had used all the glucose, an excess of lactose was added to the growth\nETIRW\nmedium. The activity of β-galactosidase was measured from this tim e (0 min), as shown\nin Fig. 3.1.\nTON\n1.0 OD\n0.8\nNIGRAM\n0.6 b - galactosidase SIHT\nactivity\nNI\narbitrary units /\nETIRW\n0.4\nTON\nOD 0.2\n0.0\nNIGRAM 20 0 40 60 80 100 120 140 160 180 200\ntime / min\nSIHT\nFig. 3.1\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nWith reference to the lac operon, explain the shape of the curve in Fig. 3.1.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n....................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-mj-43-q03-02b8add526.pdf"
    },
    {
      "id": "9700-2025-mj-43-q04",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-43-q04-p01-c1f96513d0.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-43-q04-p02-35a2f76aa9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 The Grand Canyon is located inA rizona, USA. It is estimated to have formed over five million years\nSIHT\nago as the Colorado River began to create a deep channel (canyon) in the surrounding rocks.\nNI\nETIRW\nBefore the canyon formed, an ancestral species of antelope squirrel lived in the area. An antelope\nsquirrel is a type of rodent and member of the squirrel family, Sciuridae.\nTON\nIt is estimated that around 3.6 million years ago, an ancestral species diverged into the two species\nOD\nthat are present today.\n• Harris’s antelope squirrel, Ammospermophilus harrisii, has its habitat range extending from\nthe south rim of the canyon.\n• The white-tailed antelope squirrel, Ammospermophilus leucurus, has its habitat range\nNIGRAM\nextending from the north rim of the canyon.\nFig. 4.1 shows the location of the Colorado River in the Grand Canyon and the location of these SIHT\nspecies of antelope squirrel.\nNI\nETIRW\nA.harrisii A.leucurus\nTON\nOD\nNIGRAM\nSIHT\nSouth Rim North Rim\nNI\nETIRW\nTON\nOD\nColorado River\nNIGRAM\nFig. 4.1\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) Suggest and explain how A. harrisii and A. leucurus evolved from an ancestral species.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................................................................................................................................. [4]\nTON\n(b) Scientists investigated the evolutionary relationships of the squirrel family , Sciuridae. The\nOD scientists took samples from the current species in the family and carried out DNA sequencing\nand morphological analysis.\nT o compare current species with species from the past:\n• NIGRAM specimens from museums were used to provide the tissue for DNA sequencing\n• teeth and skulls from fossils were compared as part of the morphological analysis.\nSIHT\nDescribe the advantages of using DNA",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-43-q04-87d19c84a5.pdf"
    },
    {
      "id": "9700-2025-mj-43-q05",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-43-q05-p01-db7e74ca22.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-43-q05-p02-5d34839827.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 Recombinant DNA technology is used to make recombinant human proteins.\nSIHT\nNI Two of the available methods to obtain the gene of interest are:\nETIRW • cutting the gene out of genomic DNA using restriction enzymes\n• obtaining messenger RNA (mRNA) from cells that are expressing the gene and then using\nTON reverse transcriptase to make complementary DNA (cDNA).\nOD\nPlasmids can be used as vectors to transfer the gene of interest into a host organism.\n(a) Recombinant human insulin is a protein that is made using recombinant DNA technology.\nNIGRAM\nBacteria can be used as host cells to express the recombinant protein.\nFor the human insulin gene to be successfully expressed in bacteria, one method chosen to SIHT\nobtain the gene is to extract mRNA from β-cells in the pancreas.\nNI\nETIRW\nThe gene coding for insulin is not expresse d in the bacterial host when it has been obtained\nby cutting it out of genomic DNA.\nTON\nOD Suggest and explain how the structural dif ference of cDNA and genomic DNA leads to only\ncDNA being expressed successfully.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Plasmids are cut using a restriction enzyme to create sticky endsT. he plasmids are mixed with\nSIHT\nmany copies of the desired gene and DNA ligase. The gene is inserted into many plasmids.\nNI\nETIRW\n(i) State the role of DNA ligase in the formation of recombinant plasmids.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-mj-43-q05-a4ed2b75a5.pdf"
    },
    {
      "id": "9700-2025-mj-43-q06",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-mj-43-q06-p01-773240da45.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-43-q06-p02-6acb58182d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n6 The kidneys have a role in excretion and in osmoregulation.\nSIHT\nNI Excretion is the removal of the waste products of metabolism or the removal of substances that\nETIRW\nare in excess.\nTON (a) (i) Name the main metabolic waste product excreted by the kidneys.\nOD\n..................................................................................................................................... [1]\n(ii) State how this metabolic product is made.\nNIGRAM\n..................................................................................................................................... [1]\n(iii) State where this metabolic product is made. SIHT\nNI\n..................................................................................................................................... [1]\nETIRW\n(b) Glomerular filtrate is formed by the process of ultrafiltration.\nTON\nOD Describe the process of ultrafiltration.\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-mj-43-q06-f0141c28d6.pdf"
    },
    {
      "id": "9700-2025-mj-43-q07",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-43-q07-p01-fcd558c581.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-43-q07-p02-02784b9e79.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n7 (a) A chloroplast is composed of many structures, each with a dif ferent function. Several\nSIHT\nchloroplast structures are listed.\nNI\nETIRW\nstroma lamellae thylakoid membrane ribosome\nTON thylakoid space starch grain DNA outer membrane\nOD\nFrom the list:\n(i) identify the structures involved in the production of rubisco\nNIGRAM\n……………………………………………….\n………………………………………………. SIHT\n[2]\nNI\nETIRW\n(ii) identify the structure that contains a high concentration of protons in daylight.\nTON\n………………………………………………. [1]\nOD\n(b) Paper chromatography is a technique that can be used to separate a mixture of four common\nvalues. chloroplast pigments. The pigments can be identified by calculating their R\nf\nNIGRAM A student carried out paper chromatography on a solution containing a mixture of\nchloroplast pigments.\nSIHT\nThe results are shown in Table 7.1.\nNI\nETIRW Table 7.1\nTON\ndistance travelled distance travelled\npigment by pigment from by solvent from R OD\nf\nbaseline / cm baseline / cm\n6.5 8.9 0.73\n………………\nNIGRAM\nchlorophyll a 4.6 8.9 0.52\nSIHT\nNI\nETIRW chlorophyll b 8.9 0.38\n……………………..\nTON\ncarotene 8.2 8.9\nOD ………..\nComplete T able 7.1. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Fig. 7.1 shows the absorption spectrum for carotene and for chlorophyll a.\nSIHT\nNI\nETIRW\ncarotene chlorophyll a TON\nOD\nabsorbance\nNIGRAM\nSIHT\nNI\nETIRW\n200 300 400 500 600 700\nTON\n/ nm wavelength\nOD\nFig. 7.1\nWith reference to Fig. 7.1, describe and explain the role of carotene in photosynthesis.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n............................................................................................................................................. [4] ETIRW\n[Total: 10]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-mj-43-q07-2c907cf0a0.pdf"
    },
    {
      "id": "9700-2025-mj-43-q08",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-43-q08-p01-a4c91708e3.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-43-q08-p02-8e651fe92a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n8 (a) The Boelen’s python, Simalia boeleni, is a non-venomous snake found only on the island of\nSIHT\nPapua New Guinea.\nNI\nETIRW\nFig. 8.1 shows a Boelen’ s python.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 8.1 SIHT\nNI\nThe International Union for Conservation of Nature (IUCN) has not assessed the conservation\nETIRW\nstatus of S. boeleni because the python is very hard to detect and locate.\nTON\nS. boeleni is listed in one of the appendices of the Convention on International Trade in\nEndangered Species of Wild Fauna and Flora (CITES). OD\nSuggest the advantages of S. boeleni being listed in CITES, even though it does not have an\nIUCN conservation status.\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) S. boeleni is a member of the kingdom Animalia.\nSIHT\nNI Outline the characteristic features of the kingdom Animalia.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n......................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-43-q08-f785e6a4e3.pdf"
    },
    {
      "id": "9700-2025-mj-43-q09",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-43-q09-p01-89b6450fa2.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-43-q09-p02-807422a108.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n9 (a) Most carnivorous mammals need to move to hunt their prey.\nSIHT\nNI Outline why a carnivorous mammal makes more use of its nervous system, rather than its\nETIRW\nendocrine system, when it hunts.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ............................................................................................................................................. [4]\n(b) Fig. 9.1 is a diagram of a motor neurone.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFig. 9.1 NI\nETIRW\nOn Fig. 9.1, add label lines and the letters R, S and T to label a part of the neurone that:\nTON\n• can become depolarised – use the letter R\nOD\n• contains many mitochondria – use the letter S\n• acts as an insulator – use the letter T.\nNIGRAM\n[3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Fig. 9.2 summarises changes that occur during the contraction of a sarcomere.\nSIHT\n0 5 NI\nETIRW\nwidth of\nsarcomere 1 4 TON\nOD\ncalcium ion 2 3\ncalcium ion width of\nconcentration\nconcentration sarcomere\nin sarcoplasm\narbitrary units arbitrary units / /\n3 2 NIGRAM\nSIHT 4 1\nNI\nETIRW\n5 0\n0 10 20 30 40 50 60 70 80\nTON\n/ ms time\nOD\nFig. 9.2\n(i) Suggest an explanation for the shape of the curve that shows changes in the width of\nNIGRAM\nthe sarcomere.\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n......................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-mj-43-q09-1678e31d18.pdf"
    },
    {
      "id": "9700-2025-mj-43-q10",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_43.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-43-q10-p01-e473f73f51.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-43-q10-p02-f060963eb0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-43-q10",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n10 The golden poison dart frog, Phyllobates terribilis, lives in the Colombian rainforest ecosystem.\nSIHT\nNI Fig. 10.1 shows golden poison dart frogs.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 10.1\n(a) Define the term ecosystem.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) One way of estimating the size of a population of golden poison dart frogs is to use the\nSIHT\nmark-release-recapture method.\nNI\nETIRW\n(i) Suggest the assumptions that must be made for the mark-release-recapture method to\nbe valid.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [3]\nOD\n(ii) A first sample of 27 golden poison dart frogs was captured, marked and released.\nWhen a second sample of 33 frogs was captured, 13 had marks on them.\nNIGRAM Use the Lincoln index to estimate the population size of the frogs.\nn × n\n1 2\nestimate of population size = m\nSIHT 2\nn =",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-43-q10-d310b7cd82.pdf"
    },
    {
      "id": "9700-2025-mj-44-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-mj-44-q01-p01-2532d5ab28.png",
        "9700-topic-14-homeostasis/assets/9700-2025-mj-44-q01-p02-6fa02e84d9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q01",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "Describe how a decrease in the water potential of the blood leads to an increase in the 1 (a)\nconcentration of antidiuretic hormone (ADH) in the blood.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nADH is transported in the blood to the cells of the collecting duct of the kidney. (b)\nFig. 1.1 is a diagram outlining the action of ADH on the cells of the collecting duct.\ncollecting blood\nvesicle\nduct cell capillary\nADH\nreceptor\nADH\nlumen of\ncollecting\nduct\nkinase\ncAMP\nFig. 1.1\nBinding of ADH to the ADH receptor triggers reactions within the cell. One of the first (i)\nreactions is the production of cyclic AMP (cAMP).\nState the term used to describe molecules such as cAMP in cell signalling.\n..................................................................................................................................... [1]\nDescribe the role of the vesicle when stimulated by kinase enzyme. (ii)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nOn Fig. 1.1, draw arrows to show the direction of movement of water molecules after the (iii)\ncells of the collecting duct have responded to ADH. [1]\nSome people have a rare kidney disorder in which ADH is not able to bind to ADH receptors. (c)\nSuggest the effects of this disorder on osmoregulation.\n..........................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-mj-44-q01-7660d9f82c.pdf"
    },
    {
      "id": "9700-2025-mj-44-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-mj-44-q02-p01-f48cee5cfc.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-44-q02-p02-8c1d862ab0.png",
        "9700-topic-16-inheritance/assets/9700-2025-mj-44-q02-p03-1e8cfc4dcc.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "Inherited diseases are caused by genetic mutations. 2\nHuntington’s disease is an inherited genetic disease. (a)\nUsing Huntington’s disease as an example, outline the relationship between genes, proteins\nand phenotype.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nRetinitis pigmentosa is an inherited genetic disease that causes loss of vision. (b)\nThe inheritance of a rare form of retinitis pigmentosa in a family is shown in Fig. 2.1.\nContent removed due to copyright restrictions.\nFig. 2.1\nScientists concluded that the inheritance of this rare form of retinitis pigmentosa is linked to\nthe Y chromosome.\nUsing evidence shown in Fig. 2.1, explain why the scientists reached this conclusion.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.....................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-mj-44-q02-9c4e25c75c.pdf"
    },
    {
      "id": "9700-2025-mj-44-q03",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q03-p01-5f3b227bdb.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q03-p02-4e357b5b7b.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q03-p03-fc11616de4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q03",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The genetics and evolution of the cat family, Felidae, have been studied by scientists. 3\nThe domestic cat, Felis catus, is a popular pet. (a)\nFur length in F. catus is coded for by two alleles, and h: H\n• the allele for short hair, H, is dominant\n• the allele for long hair, h, is recessive.\nThe population of domestic cats in a city was studied: (i)\n• the population consisted of 15 000 cats\n• 6.8% of these cats were long-haired.\nCalculate the percentage of cats in the city that were heterozygous for hair length.\nYou should use the Hardy–Weinberg equations in your calculation.\n2 q2 + 2pq + = 1 p + q = 1 p\nShow your working.\nanswer = ......................................................... [3]\nGive reasons why the domestic cat population does meet the conditions needed (ii) two not\nto apply the Hardy–Weinberg principle.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [2]\nSome domestic cats have no tails (tailless). The tailless phenotype in cats is genetically (b)\ncontrolled.\nFig. 3.1 shows a tailless cat.\nContent removed due to copyright restrictions.\nFig. 3.1\nA small population of domestic cats, including some tailless cats, was introduced to an island\ncalled the Isle of Man in the 1700s.\nAfter several generations, without artificial selection from humans, a high proportion of the cat\npopulation was tailless.\nSuggest why the tailless phenotype became common in the small cat population on the Isle\nof Man.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.....................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-44-q03-2e6f300c18.pdf"
    },
    {
      "id": "9700-2025-mj-44-q04",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-44-q04-p01-172b3e6688.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-mj-44-q04-p02-9147cf963a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q04",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "Organisms are classified in three domains. The domain Eukarya is divided into four kingdoms. 4\nThe four eukaryotic kingdoms are listed in Table 4.1. (a)\nComplete Table 4.1 by writing ‘yes’ or ‘no’ to produce a summary of some of the main features\nof each kingdom.\nTable 4.1\nspecies may be species may have cell species may show\nkingdom\nunicellular walls autotrophic nutrition\nAnimalia no\nFungi no\nPlantae yes\nProtoctista\n[4]\nMeiosis occurs in the kingdom Plantae. (b)\nFig. 4.1 shows drawings of photomicrographs of three stages of meiosis in the lily plant,\nLilium grandiflorum. Individual chromosomes and their structure be seen clearly. cannot\nA B C\nFig. 4.1\nIdentify the three stages of meiosis shown in Fig. 4.1.\n................................................................................... A\n................................................................................... B\n................................................................................... C\n[3]\nThe organisms in the other domains are prokaryotes. (c)\nName the domains containing prokaryotic organisms describe differences between two and\nthese two domains.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nViruses are not cellular organisms, but they are classified based on their characteristics. (d)\nOutline how viruses are classified.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\n[Total: 12]",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-mj-44-q04-898d76618b.pdf"
    },
    {
      "id": "9700-2025-mj-44-q05",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-44-q05-p01-e428fb970b.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-mj-44-q05-p02-00c6e8e4b0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "Genetic engineering may often involve the transfer of a gene into an organism. 5\nThe polymerase chain reaction (PCR) can be used to produce many copies of a gene for (a)\ntransfer.\nDescribe explain the steps involved in PCR. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [5]\nSome species of Anopheles mosquitoes, such as Anopheles stephensi, transmit malaria. (b)\nScientists have created genetically engineered A. stephensi mosquitoes in an attempt to\nreduce the spread of malaria.\nA gene called tTav was constructed by scientists and transferred into the eggs of A. stephensi.\n• tTav consists of DNA sequences from a bacterium, Escherichia coli, and herpes simplex\nvirus, which is a DNA virus.\n• The gene codes for tTav protein, which stops the expression of genes that are essential\nto mosquito development.\n• Genetically engineered A. stephensi do not survive beyond the larval stage and therefore\ndo not develop into adults.\n• A chemical called tetracycline can stop the action of the tTav protein.\nMale genetically engineered A. stephensi were released into the wild to breed with females.\nTheir offspring had the tTav gene.\nDescribe how the tTav gene could have been synthesised and transferred into the eggs (i)\nof A. stephensi.\n...........................................................................................................................................\n...........................................................................................................................................\n.....................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-mj-44-q05-fa9fd0d431.pdf"
    },
    {
      "id": "9700-2025-mj-44-q06",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q06-p01-ea71f50fd6.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q06-p02-7c0bbcd1e6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q06",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "Ensatina eschscholtzii is a species of salamander that lives in woodland ecosystems in 6\nCalifornia, USA.\nFig. 6.1 shows an ensatina salamander.\nFig. 6.1\nDefine the term ecosystem. (a)\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [3]\nThe term species can be defined using different concepts, including the biological species (b)\nconcept and the morphological species concept.\nThe morphological species concept is based on appearance or observable characteristics.\nDescribe what is meant by the biological species concept.\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [1]\nPopulations of E. eschscholtzii are separated by the Central Valley, shown in Fig. 6.2, which (c)\ncontains conditions unsuitable for salamanders.\nSome scientists think that the separated populations of E. eschscholtzii have evolved to form\ndifferent species of salamander.\npopulation B\nCentral Valley\npopulation A\nFig. 6.2\nIndividuals from different salamander populations may not be able to reproduce with each\nother, even if they are able to interact.\nExplain how salamanders from different populations have evolved to be unable to reproduce\nwith each other.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n..............................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-44-q06-9d8ca15a19.pdf"
    },
    {
      "id": "9700-2025-mj-44-q07",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 7,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q07-p01-0a232428b1.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-mj-44-q07-p02-5fa1b5a0e4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q07",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "The grass Oryza sativa is grown as a food crop to produce rice. Rice plants are adapted to grow 7\nwith their roots submerged in water.\nAn experiment was carried out to investigate the development of aerenchyma tissue in the (a)\nroots of rice plants.\n• 10-day old seedlings were grown with their roots submerged in water.\n• Some were grown in water that was kept oxygenated and others were grown in water\nthat did not contain oxygen (deoxygenated).\n• At 12 hour intervals, some seedlings from each group were removed and transverse\nsections of their roots were prepared and examined.\n• The sections were cut at the same distances from the root tip to check for development.\n• The percentage of root tissue that had developed into aerenchyma tissue was calculated.\nThe results are shown in Fig. 7.1.\n10 Key\noxygenated\n8\ndeoxygenated\npercentage of 6\nroot tissue as\naerenchyma\n4\n2\n0\n0 12 24 36 48 60\ntime / hours\nFig. 7.1\nDescribe the results shown in Fig. 7.1. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [3]\nDescribe the structure of aerenchyma explain how this is an adaptation that allows (ii) and\nroots of rice plants to be submerged in water.\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n.................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-mj-44-q07-32386e4bf7.pdf"
    },
    {
      "id": "9700-2025-mj-44-q08",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-44-q08-p01-6224d32725.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-mj-44-q08-p02-6a2b3c9c75.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "Fig. 8.1 is a transmission electron micrograph of striated muscle. 8 (a)\nFig. 8.1\nOn Fig. 8.1:\n• use the letter with a label line to show a region containing only actin P\n• use the letter with a label line to show a region containing only myosin Q\n• use the letter with a label line to show a region containing both actin and myosin. R\n[3]\nStriated muscle contraction is explained by the sliding filament model. (b)\nOutline the role of the proteins troponin and tropomyosin in the sliding filament model.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [4]\nStriated muscles can sometimes become less efficient at contracting if they have been active (c)\nfor a long time. This is called muscle fatigue.\nSuggest why muscles may become fatigued.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [2]\n[Total: 9]",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-mj-44-q08-d94f06a9e5.pdf"
    },
    {
      "id": "9700-2025-mj-44-q09",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-44-q09-p01-37128debce.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-mj-44-q09-p02-fe769eca73.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "Fig. 9.1 is a diagram outlining non-cyclic photophosphorylation. 9 (a)\n2e–\nreduced\nNADP\n2e– increasing\nenergy level\n2e–\nI photosystem\nH O 1 O 2 2 2\nlight\n2H+\nII photosystem\nlight\nFig. 9.1\nWith reference to Fig. 9.1, describe the process of non-cyclic photophosphorylation.\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n............................................................................................................................................. [7]\nFig. 9.2 shows the relationship between the rate of photosynthesis and light intensity. (b)\nrate of\nphotosynthesis\nlight intensity\nFig. 9.2\nDescribe explain the relationship shown in Fig. 9.2. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-mj-44-q09-e5f9bf1ea8.pdf"
    },
    {
      "id": "9700-2025-mj-44-q10",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 10,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_44.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_44.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-44-q10-p01-a4a832d9de.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-mj-44-q10-p02-480202f382.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-44-q10",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "The respiratory quotient (RQ) is used to indicate what type of substrate is being metabolised 10 (a)\nin respiration.\nDefine the term respiratory quotient. (i)\n...........................................................................................................................................\n...........................................................................................................................................\n...........................................................................................................................................\n..................................................................................................................................... [1]\nWhen the unsaturated fatty acid linoleic acid is respired aerobically the equation is: (ii)\nH O + ................ O 18CO + 16H O + energy C\n18 32 2 2 2 2\nCalculate how many molecules of oxygen are used when one molecule of linoleic acid is\nrespired aerobically.\nanswer ......................................................... [1]\nCalculate the RQ for linoleic acid. (iii)\nanswer ......................................................... [1]\nHummingbirds feed on nectar from flowers. Nectar is rich in sugars. (b)\nFig. 10.1 shows a hummingbird.\nFig. 10.1\nA study of aerobic respiration in captive hummingbirds was carried out. The hummingbirds\nwere allowed to feed freely and then made to fast for 4 hours. During the fasting period their\nRQ values were calculated every 40 minutes.\nFig. 10.2 shows the results from this study.\n1.0\n0.9\n0.8\nmean RQ 0.7\n0.6\n0.5\n0.4\n0 40 80 120 160 200 240\ntime after feeding / min\nFig. 10.2\nDescribe suggest explanations for the results shown in Fig. 10.2. and\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n...................................................................................................................................................\n.........................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-mj-44-q10-8145b28e35.pdf"
    },
    {
      "id": "9700-2025-mj-51-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 17,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p01-a208719201.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p02-a2ee111218.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p03-2a38d35351.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p04-a4bacb51ab.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p05-366b5d93f1.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p06-2de84be206.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q01-p07-1baf056acd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Pepsin is an enzyme that is present in gastric juice. Gastric juice is secreted into the stomach of\nSIHT\nhumans and many other animals.\nNI\nETIRW\nPepsin catalyses the hydrolysis of proteins into peptides, as shown in Fig. 1.1.\nTON\nOD\nNIGRAM\nSIHT\nprotein peptides\nNI\nETIRW\nFig. 1.1\nTON\nEgg albumen (egg white) contains a high proportion of protein.\nOD\n10% albumen solution is a cloudy‑white colour. This becomes colourless when pepsin is added to\nthe 10% albumen solution.\nNIGRAM\nA student used a colorimeter to follow the progress of the hydrolysis of protein by pepsin.\n(a) (i) The student used a blue filter in the colorimeter.\nSIHT\nNI Outline one other step the student should carry out to prepare the colorimeter so that\nETIRW\ncorrect measurements of absorbance can be obtained.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [1]\nNIGRAM\n(ii) After preparing the colorimeter, the student:\ncm3 • SIHT made a pH 2.0 albumen solution by mixing 10 of 10% albumen solution with\ncm3 of pH 2.0 buffer solution 10\nNI\ncm3 • added 3.0 of the pH 2.0 albumen solution to a colorimeter tube\nETIRW\n• added a small volume of pepsin solution to the colorimeter tube\n• immediately placed the colorimeter tube in the colorimeter\nTON\n• recorded the absorbance of the mixture every 20 seconds for 5 minutes\n• OD calculated the rate of change in absorbance as a measure of the rate of\nprotein hydrolysis.\nThe results are shown in Fig. 1.2.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n0.7\nSIHT\nNI\nETIRW\n0.6\nTON\nOD\n0.5\nNIGRAM\nSIHT 0.4\nNI absorbance\nETIRW\n/ au\n0.3 TON\nOD\n0.2\nNIGRAM\nSIHT\n0.1\nNI\nETIRW\nTON\n0.0\n0 40 80 120 160 200 240 280 320\nOD\n/ s time\nFig. 1.2\nNIGRAM\nUse Fig. 1.2 to calculate the rate of change in absorbance at pH 2.0 between\n40 s and 120 s.\nSIHT\nShow your working and include appropriate units. NI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nrate of change in absorbance at pH 2.0 = ......................................................... [3]\nETIRW\nTON\nNIGRAM\n(b) The student decided to investigate the effect of pH on the rate of protein hydrolysis by pepsin.\nSIHT\nNI From the internet the student found out that pepsin is inactive at pH 6.5 and above.\nETIRW\n(i) The student was provided with a colorimeter and standard laboratory apparatus.\nTON\nDescribe a method that the student could use to investigate the ef fect of pH on the rate\nOD\nof protein hydrolysis by pepsin and to determine the optimum pH of pepsin.\nYour method should be set out in a logical order and be detailed enough to allow another\nperson to follow it.\nNIGRAM\nDeta",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-51-q01-bcefd35d44.pdf"
    },
    {
      "id": "9700-2025-mj-51-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_51.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-51-q02-p01-f868e2dffc.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q02-p02-406df46185.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q02-p03-2326f36864.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q02-p04-497c50ca0c.png",
        "9700-experimental-skills/assets/9700-2025-mj-51-q02-p05-76f2585a1d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:10)(cid:44)\n2 Golden orb weaver spiders, Nephila pilipes, are found in East Asia, South‑east Asia and Australia.\nSIHT\nThey are active in the day and at night. Golden orb weaver spiders build webs in trees and shrubs\nNI to catch different species of insects for food.\nETIRW\nAll female golden orb weaver spiders are black with yellow spots on their legs and body. Biologists\nTON think that this adaptation evolved due to natural selection. Spiders with yellow spots m ay be able\nto attract more insects to their webs.\nOD\nFig. 2.1 shows a female golden orb weaver spider sitting on a web.\nNIGRAM\nSIHT\nNI\nETIRW\nbody length\nTON\n= 50 mm\nOD\nblack body with\nyellow spots\nNIGRAM\nSIHT\nNI\nETIRW\nblack legs with\nyellow spots\nTON\nOD\nFig. 2.1\nNIGRAM Some biologists visited a forest in East Asia in July 2008. The biologists found five webs that were\napproximately the same size. Each web belonged to a female golden orb weaver spider.\nSIHT\nThe biologists decided to investigate how the colour and pattern of spots on the spiders affect the\nnumber of insects attracted per hour to each web (insect attraction rate). NI\nETIRW\nThe biologists made two‑dimensional (2D) models of spiders, as shown in Table 2.1.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nTable 2.1\nSIHT\nNI\nmodel description diagram\nETIRW\n• A black body with yellow spots\n• TON black legs with yellow spots\n• same colour and pattern of spots as\nOD\ngolden orb weaver spiders\nNIGRAM\n• B black body with blue spots\n• black legs with blue spots\nSIHT\n• same pattern of spots as golden orb\nweaver spiders NI\nETIRW\nTON\n• C black body with large yellow spot OD\n• black legs\n• total area of yellow colour is the same\nas model A\nNIGRAM\nSIHT\n• D yellow body\n• NI yellow legs\nETIRW\nTON\nOD\n• E black body\n• black legs\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nAt the first web, the biologists:\nSIHT\n• NI removed the living golden orb weaver spider from the web\nETIRW • placed model A in the centre of the web\n• placed a video camera 1 m from the web\nTON • filmed the web and model for 6 hours\n• recorded an insect attraction event whenever an insect flew towards the model, touched the\nOD\nmodel, or touched the web\n• calculated the insect attraction rate of model A.\nThe procedure was repeated by placing models B, C, D and E on the four other webs.\nNIGRAM\nThe whole investigation was repeated 25 times, and a mean insect attraction rate for each model\nwas calculated. SIHT\nNI\n(a) (i) Identify the dependent variable in this investigation.\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n..................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-51-q02-a3f53ba181.pdf"
    },
    {
      "id": "9700-2025-mj-52-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p01-e256adff5f.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p02-4a17700f91.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p03-a5b9f0af95.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p04-c459b40f22.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p05-cd3484a728.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q01-p06-a24488a7be.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Phosphatases are enzymes that catalyse the removal of phosphate from other molecules,\nSIHT\n). A phosphatase can be extracted from mung bean seedlings, releasing inorganic phosphate (P\ni\nNI Vigna radiata.\nETIRW\nFig. 1.1 shows mung bean seedlings.\nTON\nmung bean seedlings\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 1.1\nTON\nA student found a published method to extract phosphatase from mung bean seedlings.\nOD\n• Use a pestle and mortar to grind mung bean seedlings, with a small volume of distilled water,\nto make a paste.\nNIGRAM\ncm3. • Add distilled water to the paste to make a mixture with a total volume of 50\n• Filter the mixture and put the filtrate into a clean centrifuge tube. SIHT\nNI\n• Centrifuge the filtrate until a solid pellet is formed, as shown in Fig. 1.2.\nETIRW\n• Pour the liquid extract containing phosphatase into a clean test-tube.\nTON\nlid of centrifuge tube OD\nNIGRAM\nplastic centrifuge tube\nSIHT\nNI\nliquid phosphatase extract ETIRW\nTON\nsolid pellet\nOD\nFig. 1.2\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) State two other variables that should be standardised in the published method so that extracts\nSIHT\nwith the same concentration of phosphatase can be produced.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [2]\nSIHT (b) The student read that the optimum pH for phosphatase extracted from mung bean seedlings\nis less than pH 7.0.\nNI\nETIRW\nThe student decided to investigate phosphatase activity , using the substrate\nphenolphthalein phosphate (PPP).\nTON\nOD In the first four steps of the method, the student:\ncm3 cm3 • mixed 5 of pH 6.0 buffer solution with 1 of a 1.0% solution of PPP in a test-tube\ncm3 • NIGRAM added 1 of the phosphatase extract to the test-tube and started a timer\n• incubated the test-tube for 10 minutes in a water-bath at 30 °C\nSIHT\ncm3 • stopped the enzyme reaction after 10 minutes, by adding 5 of 10.0% solution of NI\nETIRW sodium carbonate. Sodium carbonate solution is alkaline.\n(i) The buffered PPP solution and the phosphatase extract were mix ed and then placed in TON\nthe water-bath. The student identified this as a source of error in the method.\nOD\nExplain why this is a source of error and state how you would modify the method to\nrem",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-52-q01-edc6a9595e.pdf"
    },
    {
      "id": "9700-2025-mj-52-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_52.pdf?download=true",
      "source_pages": [
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-52-q02-p01-92a23cae03.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q02-p02-610407547c.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q02-p03-745938b48a.png",
        "9700-experimental-skills/assets/9700-2025-mj-52-q02-p04-486cc454cd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:10)(cid:44)\n2 Sponges are immobile, aquatic animals that live on rocks and sediment at the bottom of salt\nSIHT\nwater environments.\nNI\nETIRW\nFig. 2.1 shows an example of a sponge, growing on the seabed, in a marine habitat.\nTON\nOD\nscuba diver NIGRAM\nphotographing\nsponges\nSIHT\nNI\nETIRW\nTON\nsponge\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 2.1\nNIGRAM\nThere are many different species of sponge. Most species of sponge are sensitive to environmental\nstress. Human activity causes some environmental stress.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nScientists sampled the marine habitats along a length of coastline in Algeciras Bay in southern\nSIHT\nSpain to study the species diversity of sponges.\nNI\nETIRW\nThe scientists selected 12 sampling stations, A to L, as shown in Fig. 2.2. The scientists noted the\nland use or human activity along the coast next to each sampling station.\nTON\nKey to diagram:\nOD\nsampling type of land use\nstation or human activity\nA natural habitat NIGRAM\nB natural habitat\nH G\nSIHT I C tourist beach\nJ F\nNI\nD housing\nETIRW land\nE shipping port\nE TON K\nF tourist beach\nland\nOD\nG thermal power D\nL\nstation\nC\nH oil industry Algeciras Bay\nNIGRAM\nI ship building and B\nmarine habitat\nrepairs\nSIHT\nA\nJ tourist boats\nNI\nsampling stations\nETIRW K tourist boats\nL natural habitat\nTON\nOD\nFig. 2.2\nThe sampling stations were chosen to compare the ef fects of land use or human activity on the\nspecies diversity of sponges in the bay.\nNIGRAM\nAt each marine sampling station, the scientists:\nSIHT\n• placed permanent line transects, 50 m in length, on the seabed\nNI\nETIRW\n• photographed all sponges sighted at a distance of 1 m either side of the transect\nTON\n• sampled each transect for the same length of time\nOD\n• sampled each transect four times a year.\n(a) The scientists standardised some variables.\nNIGRAM\nState two other variables that the scientists should standardise in this investigation.\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n............................................................................................................................................. [2]\nTON\nNIGRAM\n(b) The scientists compared the species diversity of sponges between different sampling stations\nSIHT\nusing an index of diversity known as beta diversity.\nNI\nThe higher the beta diversity index, the higher the species diversity at a sampling station. ETIRW\nFig. 2.3 shows a graph of the beta diversity index for each sampling station.\nTON\n1.7 OD\n1.6\nNIGRAM 1.5\n1.4\nSIHT\nNI\n1.3\nETIRW\nbeta\ndiversity 1.2\nTON\nindex\nOD\n1.1\n1.0\nNIGRAM\n0.9\nSIHT 0.8\nNI\nETIRW 0.7\nTON 0.6\nA B C D E F G H I J K L\nOD\nsampling station\nFig.",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-52-q02-70e1465c59.pdf"
    },
    {
      "id": "9700-2025-mj-53-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_53.pdf?download=true",
      "source_pages": [
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-53-q01-p01-e18c1e227b.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q01-p02-9c49e82f4f.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q01-p03-73b5348cd2.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q01-p04-3f985d3f1e.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q01-p05-1165dd4169.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44)\n1 Drosophila melanogaster is a species of fruit fly.\nSIHT\nNI Fig. 1.1 shows a female fruit fly and a male fruit fly. A typical fruit fly is 3 mm in length.\nETIRW\nTON\nfemale\nOD\nNIGRAM\nrounded abdomen\npointed abdomen\nSIHT\nmale\nNI\nETIRW\nTON\nOD\nFig. 1.1\nNIGRAM Scientists have studied the inheritance patterns of many genetic traits in fruit flies.\nTo carry out genetic crosses with fruit flies:\nSIHT\n• a specimen tube is prepared with food for the fruit flies, as shown in Fig. 1.2 NI\nETIRW\n• adult male and female fruit flies are added to the specimen tube to allow mating to take place\nTON\n• the specimen tube is kept in warm conditions for several days\nOD\n• the eggs laid by female fruit flies develop into pupae\n• adults are removed from the specimen tube before pupae mature into adult fruit flies\nNIGRAM\n• offspring emerge as adult flies 10-15 days after eggs are laid.\nSIHT\ncotton\nNI wool\nETIRW\nspecimen\nTON tube\nOD\nagar jelly\ncontaining\nnutrients\nand water\nNIGRAM\nSIHT Fig. 1.2\nNI\nETIRW\nTON\nNIGRAM\nOne of the genetic traits studied in fruit flies is eye colour . The normal (wild type) eye colour of\nSIHT\nD. melanogaster is red.\nNI\nETIRW\nEye colour in D. melanogaster is controlled by several genes, including two genes that are located\non separate chromosomes, A/a and D/d.\nTON\n• Allele A is dominant to allele a.\nOD\n• Allele D is dominant to allele d.\nTable 1.1 summarises eye colour in D. melanogaster for these two genes.\nNIGRAM\nTable 1.1\nSIHT\neye colour phenotype genotypes NI\nETIRW\nred AADD, AaDD, AaDd, AADd\nbrown AAdd, Aadd TON\nscarlet (bright red) aaDD, aaDd OD\nwhite aadd\n(a) A student was provided with two populations of fruit fly:\nNIGRAM\n• brown-eyed fruit flies with genotype AAdd\nSIHT\n• scarlet-eyed fruit flies with genotype aaDD.\nNI\nETIRW\nIn each population, males and females were provided in separate specimen tubes.\nTON\nThe student decided to carry out two genetic crosses.\nOD\nThe first cross used fruit flies from the initial populations to produce of fspring that are\nheterozygous for each of the two genes (double heterozygotes).\nNIGRAM The second cross used the double heterozygotes produced from the first cross.\nT o carry out the genetic crosses, the student was provided with standard laboratory\nSIHT\nequipment and:\nNI\n• ETIRW specimen tubes containing food\n• a chemical to anaesthetise the flies – this chemical, when given at a particular dose, TON\nmakes the flies immobile for more than 30 minutes\nOD\n• small brushes for sorting immobile flies without harming them.\n(i) Identify a hazard in this investigation and state a risk associated with the hazard and\nNIGRAM\nstate one precaution that the student should take.\n........................................................................................................................................... SIHT\nNI\n......................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-53-q01-945a3d57d5.pdf"
    },
    {
      "id": "9700-2025-mj-53-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 17,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_53.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p01-50c9a5cce8.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p02-f000bc2f0f.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p03-cdaaaf16d6.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p04-98e9d2b825.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p05-bde67ce939.png",
        "9700-experimental-skills/assets/9700-2025-mj-53-q02-p06-81a71b32a4.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n2 Echidnas are mammals that live in Australia and New Guinea.\nSIHT\nNI Fig. 2.1 shows an echidna.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 2.1\nScientists analysed the milk produced by female echidnas and identified a protein that they named\nEchAMP. The scientists predicted that EchAMP may have antibacterial properties.\nNIGRAM\nThe scientists tested the effect of EchAMP on the bacterium Escherichia coli.\nSIHT\n1. 100 E. coli cells were added to each well on a cell culture plate with 96 wells.\nNI\n2. A treatment solution that contained EchAMP was added to each well on the plate.\nETIRW\n3. A chemical that causes living E. coli cells to fluoresce was added to each well.\n4. The plate was incubated at 37°C.\nTON\n5. Every hour for 7 hours, the fluorescence emitted by the E. coli on the plate was recorded as a\nOD measure of E. coli population growth.\n6. Steps 1–5 were repeated eight times.\nThe scientists also carried out two control experiments.\nNIGRAM\n• A negative control experiment repeated the procedure (steps 1–6), but the treatment solution\ndid not contain EchAMP.\nSIHT\n• A positive control experiment repeated the procedure (steps 1–6), but the treatment solution NI\nETIRW contained an antibiotic called bacitracin instead of EchAMP.\n(a) Identify the independent variable in this investigation. TON\nOD\n............................................................................................................................................. [1]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The scientists standardised the temperature and the initial number of E. coli cells.\nSIHT\nNI State two other variables that the scientists should standardise in this investigation.\nETIRW\n1 ................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n2 ................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n[2]\nSIHT (c) (i) Explain why the scientists included the negative control experiment in their investigation.\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\nOD ..................................................................................................................................... [1]\n(ii) Explain why the scientists included the positive ",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-53-q02-6f42ed064d.pdf"
    },
    {
      "id": "9700-2025-mj-54-q01",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "54",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 21,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_54.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-54-q01-p01-a81144057c.png",
        "9700-experimental-skills/assets/9700-2025-mj-54-q01-p02-ed0f14a5f8.png",
        "9700-experimental-skills/assets/9700-2025-mj-54-q01-p03-ed3aed9c41.png",
        "9700-experimental-skills/assets/9700-2025-mj-54-q01-p04-925c887772.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-54-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 A student investigated the effect of gibberellin (GA) and abscisic acid (ABA) on the germination of\nSIHT\nwheat grains.\nNI\nETIRW\nThe student found out the following information.\nTON • The molecular mass of GA is 346 g.\nOD\n• The molecular mass of ABA is 264 g.\n−6 to • The range of concentrations at which GA is active in plant tissues is from 1 × 10\n10−5 dm−3. mol 1 ×\nNIGRAM\n−6 • The range of concentrations at which ABA is active in plant tissues is from 1 × 10 to\n10−4 dm−3. mol 1 ×\nSIHT\n• Newly formed wheat grains contain a high concentration of ABA which decreases over time.\nNI\n• W hen the concentration of ABA falls low enough the GA is able to promote germination.\nETIRW\n• Wheat germinates best at temperatures from 10 °C to 25 °C and without light.\nTON\nOD\nFig. 1.1 shows the germination and early growth of wheat.\nfirst leaf\nNIGRAM\nshoot\nSIHT young shoot\nyoung shoot visible\nNI beneath grain wall\nETIRW\nroot hairs TON\nOD\nyoung root appears\nroots more roots appear\ngerminated grain seedling at 7 days\nNIGRAM\nFig. 1.1\nSIHT\nTo test the effects on germination of adding ABA, the student started by soaking wheat grains for\nNI\n10−6 dm−3 24 hours in a solution of 1 × mol GA. These were then split into several groups of 50 grains\nETIRW\nand a different concentration of ABA was added to each group.\nTON\n10−3 dm−3 mol which was then The student made a stock solution of ABA with a concentration of 1 ×\ndiluted to make the solutions needed for the investigation. OD\n(a) (i) Using the information about ABA above, describe how the student made a stock solution\n10−3 dm−3 mol ABA. of 1 ×\nNIGRAM\n......................................................................................................................................................\nSIHT\n......................................................................................................................................................\nNI\nETIRW ......................................................................................................................................................\n................................................................................................................................................ [2] TON\nNIGRAM\n10−3 dm−3 (ii) The student used the stock solution of 1 × mol ABA to prepare a range of\nSIHT\n10–4 dm−3 10–7 dm−3 dilutions from 1 × mol to 1 × mol to use in their investigation.\nNI\nETIRW cm3. Each dilution had a volume of 100\nDescribe how the student prepared these dilutions. TON\nOD ......................................................................................................................................................\n......................................................................................................................................................\n............................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-54-q01-35bdbe865a.pdf"
    },
    {
      "id": "9700-2025-mj-54-q02",
      "subject": "9700",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "54",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-may-june/9700_s25_qp_54.pdf?download=true",
      "source_pages": [
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2025-May-June/qp/9700_s25_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-mj-54-q02-p01-c4e6e3544a.png",
        "9700-experimental-skills/assets/9700-2025-mj-54-q02-p02-1b01973d5d.png",
        "9700-experimental-skills/assets/9700-2025-mj-54-q02-p03-0315671ab1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-mj-54-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n2 A student investigated the oxygen consumption of isolated mitochondria under different conditions.\nSIHT\nNI Mitochondria were isolated by crushing liver cells in ice cold buf fer and filtering to remove cell\nETIRW\ndebris. The filtrate was centrifuged to separate the mitochondria. The mitochondria were\nsuspended in ice cold buffer.\nTON\n(a) State one reason for keeping the mitochondria in cold buffer solution during isolation.\nOD\n...................................................................................................................................................\n............................................................................................................................................. [1]\nNIGRAM\nFig. 2.1 shows the apparatus used to measure the oxygen consumption of the isolated\nmitochondria. SIHT\nNI\ncomputer to record data\nETIRW\nfrom the oxygen meter\nairtight seal syringe for injecting isolated TON\nmitochondria and test reagents\nOD one after the other\nchamber containing\noxygenated buffer\nsolution NIGRAM\nmagnetic stirrer\n000.00\nSIHT\noxygen meter to measure NI\nETIRW oxygen sensor oxygen consumption of\nisolated mitochondria\nTON\nFig. 2.1\nOD\nThe apparatus was used as follows.\n• The apparatus was set up as shown in Fig. 2.1 and left for 1 minute to equilibrate.\nNIGRAM\n• A t 1 minute, a standard volume of mitochondria in buffer solution was injected through the seal.\nSIHT\n• A t 2 minutes, a solution of succinate, a Krebs cycle intermediate, was injected through the seal.\nNI\n• At 5 minutes, a solution of ADP was injected through the seal.\nETIRW\n• At 6 minutes, a solution of cyanide was injected through the seal.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nDuring this procedure the oxygen concentration in the chamber was measured continuously and\nSIHT\ndisplayed as a trace on the computer screen as shown in Fig. 2.2.\nNI\nETIRW succinate added\nADP added\nTON\nOD\noxygen mitochondria\ncyanide added\nconcentration added\n/ nmol cm–3\nNIGRAM\nSIHT\n0 1 2 3 4 5 6 7 8 9\ntime / min NI\nETIRW\nFig. 2.2\nTON\nUsing the figures from the trace, the computer calculated the rate of oxygen consumption\nOD following each addition. This procedure was repeated three more times using fresh samples of\nmitochondria.\nTable 2.1 shows the rates of oxygen consumption for each of the four trials.\nNIGRAM\nTable 2.1\nmin−1 rate of oxygen consumption / nmol SIHT\ntrial 1 trial 2 trial 3 trial 4 mean NI\nETIRW\nmitochondria alone 0.03 0.02 0.03 0.01 0.02\nTON\nmitochondria with succinate 50.22 49.10 48.53 50.15\nOD\nmitochondria with ADP and 139.23 170.10 142.67 138.10 147.53\nsuccinate\nmitochondria with cyanide, ADP 0.00 0.10 0.00 0.01 0.03\nand succinate NIGRAM\n(b) (i) Calculate the mean rate of oxygen consumption for mitochondria with succinate. SIHT\nNI\nWrite your answer in Table 2.1. [1]\nETIRW\n(ii) On Table 2.1 indicate, by placing a circle around each value, two result",
      "pdf_path": "9700-experimental-skills/print/9700-2025-mj-54-q02-a655f1ff9d.pdf"
    },
    {
      "id": "9700-2025-on-41-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-41-q01-p01-db9b6483c5.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-41-q01-p02-96f808b471.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q01",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Guinea pigs, Cavia porcellus, vary in the length and colour of their fur.\nSIHT\nNI Fig. 1.1 shows a guinea pig with short black fur.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 1.1\nTON\nTwo genes that determine the length and colour of the fur occur at the A/a locus and the B/b OD\nlocus. These two gene loci are on separate autosomal chromosomes.\n• The allele A results in short fur.\n• The allele a results in long fur. NIGRAM\n• A is dominant to a.\nSIHT • The allele B results in black fur.\n• The allele b results in chocolate fur.\nNI\n• B is dominant to b. ETIRW\n(a) (i) List all the possible genotypes of a guinea pig with short black fur.\nTON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\n(ii) A test cross could be used to determine the genotype of a female guinea pig with short\nSIHT\nblack fur.\nNI\nETIRW\nDescribe the phenotype of the male guinea pig that could be used to carry out this\ntest cross.\nTON\n..................................................................................................................................... [1]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) A black guinea pig with long fur that was homozygous at both loci was crossed with a\nSIHT\nchocolate guinea pig with short fur that was homozygous at both loci. The F1 offspring of this\nNI cross had short black fur. F1 offspring were mated together to produce the F2 offspring.\nETIRW\nComplete the Punnett square to:\nTON • show the cross between the F1 offspring\n• predict the F2 offspring genotypes.\nOD\nY ou should include the gametes in your answer.\nState the ratio of F2 of fspring phenotypes. You should include a key to link phenotypes to\ngenotypes.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nratio of F2 of fspring phenotypes:\nNIGRAM\nSIHT\n[4] NI\nETIRW\n(c) Some genes in guinea pigs are structural genes and some are regulatory genes.\nTON\nDescribe the dif ference between a structural gene and a regulatory gene.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n.......................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-41-q01-45da0fbcd0.pdf"
    },
    {
      "id": "9700-2025-on-41-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-41-q02-p01-5de3844654.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-41-q02-p02-3c4475f691.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 Exserohilum turcicum is a fungal pathogen. The growth of the mycelium of the fungus damages\nSIHT\nthe leaves of maize plants, Zea mays. Leaf damage reduces crop yield.\nNI\nETIRW\n(a) (i) Complete Table 2.1 to show one structural dif ference and one functional dif ference\nbetween E. turcicum and Z. mays.\nTON\nTable 2.1\nOD\nE. turcicum Z. mays\nNIGRAM\nstructural\nSIHT\ndifference\nNI\nETIRW\nTON\nOD\nfunctional\ndifference\nNIGRAM\n[2]\nSIHT\nNI (ii) Describe the principles by which organisms such as E. turcicum and Z. m ays are\nETIRW classified in the taxonomic hierarchy.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [4] OD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Two inbred varieties of maize, SKV50 and CML153, were crossed. The resulting F1 hybrids\nSIHT\nwere self‑crossed to produce F2 of fspring. The F2 plants were grown, and the percentage\nNI area of leaf damage caused by E. turcicum was measured.\nETIRW\nFig. 2.1 shows the results for the F2 generation. The arrows show the mean percentage area\nof leaf damage for the two parent varieties. TON\n50 OD\n40\nNIGRAM\n30\nnumber\nof plants\n20 SIHT\nNI\n10\nETIRW\n0\nTON\n20 30 40 50 60\nOD\npercentage area of\nleaf damage SKV50 CML153\nFig. 2.1\nNIGRAM\n(i) Explain how Fig. 2.1 can be used to determine which parent maize variety shows the\ngreatest resistance to infection by E. turcicum.\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n..............................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-41-q02-5da610ff26.pdf"
    },
    {
      "id": "9700-2025-on-41-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-on-41-q03-p01-c7da1d04ef.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-41-q03-p02-aee99fdc02.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-41-q03-p03-dcbd18fde5.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-41-q03-p04-6b1f5246d3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q03",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 Plants have several different photosynthetic pigments in their chloroplasts.\nSIHT\nNI (a) A student separated and identified the chloroplast pigments present in a leaf extract from a\nETIRW\nspinach plant using two slightly different methods.\nTON Method A\n• A type of chromatography known as thin layer chromatography (TLC) was used to\nOD\nseparate the pigments.\n• A mixture of ether and cyclohexane was used as a solvent in TLC.\nMethod B\nNIGRAM\n• The student repeated TLC but treated the spinach leaf extract with a chemical. The\nchemical causes a magnesium ion in a pigment to be replaced by two hydrogen ions.\n• The student used a leaf from the same spinach plant, and used the same solvent as in SIHT\nmethod A.\nNI\nETIRW\nvalues and compared these to reference values to identify The student calculated R\nf\nthe pigments.\nTON\nOD Fig. 3.1 shows the results for method A and method B.\nsolvent front\nNIGRAM\nβ-carotene\nSIHT\nsolvent front NI\nβ-carotene ETIRW\nTON\nOD\nphaeophytin a\nchlorophyll a\nNIGRAM\nphaeophytin b\nchlorophyll b\nSIHT\nchlorophyll b\nNI\nETIRW\nxanthophyll\nxanthophyll\nTON\nOD line of origin line of origin\nA B\nNIGRAM\nFig. 3.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Calculate the R value of β‑carotene in Fig. 3.1.\nSIHT f\nNI\nETIRW\nTON R = ...............................................................\nf\n[2]\nOD\n(ii) Suggest two explanations for the differences in appearance of chromatograms A and B\nin Fig. 3.1.\nNIGRAM\n1 ........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n2 ........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\n[2]\nNIGRAM\nvalues for the chloroplast pigments of spinach in a (iii) The student found some different R\nf\nscientific paper.\nSIHT\nThe values in the scientific paper were dif ferent from the reference values that the NI\nETIRW student originally used to identify the pigments on chromatograms A and B in Fig. 3.1.\nvalues. Suggest one reason, other than measurement error, for the different R TON\nf\nOD\n...........................................................................................................................................\n........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-on-41-q03-0ea68869c3.pdf"
    },
    {
      "id": "9700-2025-on-41-q04",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-41-q04-p01-989190d8b9.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-41-q04-p02-4266fd9144.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 The wolf, Canis lupus, lives in North America. Wolves may have a grey or a black coat colour. The\nSIHT\ncolour of an individual wolf depends on the DNA it inherits at the CPD103 gene locus.\nNI\nETIRW • Wolves inherit two copies of CPD103, one from each parent.\n• Wolves that inherit one copy of the black form of the CPD103 gene have a black coat.\nTON\n(a) State the term used to describe:\nOD\n• an organism that has two copies of each gene ..........................................................\n• a form of a gene ..........................................................\nNIGRAM\n• a form of a gene that gives a phenotypic\neffect in a heterozygote. .......................................................... SIHT\n[3]\nNI\nETIRW\nIn addition to producing black coat colour, the protein coded for by the CPD103 gene also defends\nagainst infectious lung disease.\nTON\nOD Canine distemper virus (CDV) causes serious lung disease in wolves. W olves that have been\npreviously infected by CDV have antibodies against CDV (anti‑CDV antibodies) in their blood.\n(b) CDV can be passed from domestic dogs to wolves.\nNIGRAM\n• Domestic dogs are more numerous in the southern part of the area occupied by wolves.\n• Domestic dogs are less numerous in the northern part of the area occupied by wolves.\nSIHT\n• The relative frequency of black wolves compared to grey wolves increases from the\nnorth to the south of the area they occupy. NI\nETIRW\nExplain how natural selection causes this trend in the distribution of black wolves.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Several different populations of wolves were compared.\nSIHT\nNI Fig. 4.1 shows the relationship between the percentage of ",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-on-41-q04-d2e6fbb7b0.pdf"
    },
    {
      "id": "9700-2025-on-41-q05",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-41-q05-p01-e315af8a4d.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-41-q05-p02-b551c30637.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 The endocrine system and the nervous system both coordinate responses in mammals.\nSIHT\nNI (a) Complete Table 5.1 to show the features of three cell‑signalling molecules of the endocrine\nETIRW\nsystem: antidiuretic hormone (ADH), glucagon and insulin.\nTON 7) if the molecule does not have Use a tick (3) if the molecule has the feature and a cross (\nthe feature.\nOD\n7) in every box. Put a tick (3) or a cross (\nTable 5.1\nNIGRAM\nfeature ADH glucagon insulin\nSIHT\nbinds to receptors\nNI\non cell surface\nETIRW\nmembranes\nresults in molecules TON\nmoving from cells\nOD\ninto the blood\nis secreted as a\nresult of detection by\nosmoreceptors NIGRAM\n[3]\nSIHT\n(b) The endocrine system has a slower transmission speed than the nervous system.\nNI\nETIRW\nDescribe other ways in which the endocrine system and the nervous system dif fer.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The endocrine system and the nervous system can affect muscle function.\nSIHT\nNI Fig. 5.1 shows a transmission electron micrograph of a longitudinal section of striated muscle\nETIRW\ntissue that is in a relaxed state.\nTON\nOD\nB\nNIGRAM\nC SIHT\nNI\nETIRW\nTON\nD\nE\nOD\nNIGRAM\nSIHT\nFig. 5.1\nNI\nETIRW\n(i) State the letter on Fig. 5.1 that indicates the length of a sarcomere.\nTON\n..................................................................................................................................... [1]\nOD\n(ii) State the letter on Fig. 5.1 that indicates a region where actin and myosin overlap.\n.................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-41-q05-9011c069a9.pdf"
    },
    {
      "id": "9700-2025-on-41-q06",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-41-q06-p01-0352f3c8a8.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-41-q06-p02-76e227614d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q06",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 The distribution of the large blue butterfly, Phengaris arion, extends across Europe and Asia. It is\nSIHT\nassessed by the International Union for Conservation of Nature (IUCN) on the Red List™ as ‘Near\nNI Threatened’ globally and ‘Endangered’ in Europe.\nETIRW\nIn Europe, P. arion became extinct in the Netherlands in 1964 and in the United Kingdom in 1979.\nTON\nFig. 6.1 lists the conservation status categories in the IUCN Red List™.\nOD\nconservation status\nExtinct (EX)\nNIGRAM\nExtinct in the Wild (EW)\nSIHT\nCritically Endangered (CR)\nNI\nETIRW\nEndangered (EN) increasing risk of extinction\nTON\nVulnerable (VU)\nOD\nNear Threatened (NT)\nLeast Concern (LC)\nNIGRAM\nFig. 6.1\nSIHT Fig. 6.2 shows P. arion.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 6.2\nTON\n(a) (i) Explain how IUCN Red List™ assessments help to conserve biodiversity.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n..................................................................................................................................... [3] TON\nNIGRAM\n(ii) With reference to Fig. 6.1 and the IUCN assessments for P . arion, suggest how the\nSIHT\nabundance of the butterfly differs across its distribution.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ..................................................................................................................................... [2]\nNI\n(b) P. arion has been successfully re‑introduced in the United Kingdom at 12 sites. These sites\nETIRW\nwere restored to flower‑rich grassland.\nTON\nThe conservation",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-on-41-q06-ab4ccdaad1.pdf"
    },
    {
      "id": "9700-2025-on-41-q07",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-on-41-q07-p01-261bab4857.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q07",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 In aerobic respiration, most ATP is produced by oxidative phosphorylation.\nSIHT\nNI (a) Outline the features of ATP that make it suitable as the universal energy currency.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................................................................................................................................. [3]\nTON\n(b) Rotenone is a compound that affects oxidative phosphorylation.\nOD\nRotenone disrupts the first carrier in the electron transport chain by stopping the transfer of\nelectrons from this carrier.\nNIGRAM Suggest and explain how rotenone reduces the production of ATP and water in\naerobic respiration.\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-on-41-q07-68424f2533.pdf"
    },
    {
      "id": "9700-2025-on-41-q08",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        17
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-on-41-q08-p01-d55ce7ffdb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:8)(cid:44)\n8 LibertyLink® soybean is a genetically modified crop. It was first grown in 1996 and used in food\nSIHT\nproducts from 1998. It has been grown in 6 countries and used in food products in 21 countries.\nNI\nETIRW\nTable 8.1 summarises the modifications made to the soybean plant to produce LibertyLink®\nsoybean.\nTON\nTable 8.1\nOD\nname of introduced\ngene donor organism gene product function of gene product\ngene\nNIGRAM\nStreptomyces phosphinothricin stops action of glufosinate,\npat\nviridochromogenes N‑acetyltransferase a herbicide\nSIHT\n(a) Explain how the modification made to produce LibertyLink® soybean may help to solve the\nNI\nglobal demand for food.\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [4]\nTON\n(b) (i) Name the type of enzyme that could be used to cut out the pat gene from\nOD\nS. viridochromogenes.\n..................................................................................................................................... [1]\nNIGRAM\n(ii) Name the enzyme that could be used to join the pat gene to a plasmid by forming\nphosphodiester bonds.\nSIHT\n..................................................................................................................................... [1]\nNI\nETIRW\n(c) Suggest reasons why LibertyLink® soybean is used in food products in 21 countries but only\ngrown in 6 countries.\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n...............................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-on-41-q08-55f5179861.pdf"
    },
    {
      "id": "9700-2025-on-41-q09",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        18
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-on-41-q09-p01-ffa4928c32.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n9 (a) Fig. 9.1 shows a longitudinal section of a human kidney.\nSIHT\nNI\nETIRW\nTON\nOD\nX\nNIGRAM\nY\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 9.1\nName the regions of the kidney labelled X and Y in Fig. 9.1.\nNIGRAM\nX ...............................................................................................................................................\nSIHT Y ...............................................................................................................................................\n[2]\nNI\nETIRW\n(b) A biosensor can be used to measure the concentration of glucose in urine.\nTON\nOutline how a biosensor measures the concentration of glucose in urine.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n[Total: 6]\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-on-41-q09-aee503e3d0.pdf"
    },
    {
      "id": "9700-2025-on-41-q10",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_41.pdf?download=true",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-41-q10-p01-cdec87d07b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-41-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:10)(cid:44)\n10 Explain the relationship between genes, proteins and phenotype, with reference to two examples\nSIHT\nof genetic diseases in humans.\nNI\nETIRW\n..........................................................................................................................................................\nTON ..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM\n..........................................................................................................................................................\nSIHT ..........................................................................................................................................................\nNI\n..........................................................................................................................................................\nETIRW\n..........................................................................................................................................................\nTON\nOD ..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM ..........................................................................................................................................................\n..........................................................................................................................................................\nSIHT\n.......................................................................................................................................................... NI\nETIRW\n..........................................................................................................................................................\nTON\n..........................................................................................................................................................\nOD\n.................................................................................................................................................... [6]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-41-q10-aec2565f0d.pdf"
    },
    {
      "id": "9700-2025-on-42-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q01-p01-d076c04794.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q01-p02-84eb86f555.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Rice, Oryza sativa, is a crop plant that has adapted to grow in flooded fields.\nSIHT\nNI Fig. 1.1 outlines the effects on rice plants of growing in flooded fields.\nETIRW\nflooded fields\nTON\nOD\nlower concentration of mineral ions and\nA in the soil available for absorption\nNIGRAM compared to non-flooded fields\nSIHT\nhigh concentration of\npresence of tissue C production of\nNI waste product B in\nin stems and roots ethylene (ethene)\nETIRW\nroot cells\nTON\nOD\nbreakdown of\nabscisic acid (ABA)\nNIGRAM\nFig. 1.1\n(a) With reference to Fig. 1.1, name: SIHT\nNI\nsubstance A ..........................................................................\nETIRW\nsubstance B ..........................................................................\nTON\ntissue C .......................................................................... OD\n[3]\n(b) Describe how root cells respond to a high concentration of substance B.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\n............................................................................................................................................. [1] ETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Gibberellin is involved in the growth of stems in rice plants.\nSIHT\nNI Fig. 1.2 shows the ef fect of gibberellin concentration on the length of stems in rice plants.\nETIRW\n60\nTON\n50\nOD\n40\nlength\nNIGRAM 30 of stem\nmm /\n20\nSIHT\nNI 10\nETIRW\n0\nTON\n2 3 4 5 1\nconcentration of gibberellin OD\n/ arbitrary units\nFig. 1.2\nNIGRAM\n(i) Describe the relationship shown in Fig. 1.2.\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\n(ii) ABA in rice plant cells inhibits the action of gibberellin.\nWith reference to Fig. 1.1 and Fig. 1.2, explain the importance of ethylene production for SIHT\nthe growth of rice plants in flooded fields.\nNI\nETIRW\n...........................................................................................................................................\nTON\n........................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-42-q01-bcae634dd0.pdf"
    },
    {
      "id": "9700-2025-on-42-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-42-q02-p01-c32c73b0f1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 (a) Explain the terms gene, genotype and phenotype.\nSIHT\nNI gene ..........................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\ngenotype ...................................................................................................................................\nOD\n...................................................................................................................................................\nphenotype .................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n[3] SIHT\nNI\n(b) A monohybrid genetic cross can be carried out to produce an F1 and an F2 generation.\nETIRW\n• Outline how a monohybrid genetic cross is carried out.\nTON\n• State the expected percentage of each of the dif ferent of fspring genotypes in the\nOD F2 generation.\nAssume that the inheritance pattern is for autosomal dominant and recessive alleles.\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................................................................................................................................. [5]\nTON\n[Total: 8]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-42-q02-594e49ec3b.pdf"
    },
    {
      "id": "9700-2025-on-42-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-42-q03-p01-b0a705c048.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-42-q03-p02-94184eabe2.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-42-q03-p03-8deeb8bb86.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-42-q03-p04-e0b6a1e069.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q03",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 The lac operon is present in the genome of the bacterium Escherichia coli.\nSIHT\nNI When glucose is not available, the presence of lactose in the extracellular environment leads\nETIRW\nto the expression of the genes of the lac operon. This leads to an increase in the uptake and\nmetabolism of lactose.\nTON\nWhen glucose and lactose are available in the extracellular medium, lactose is prevented from\nOD\nentering the bacterial cell.\nFig. 3.1 shows what happens when glucose enters a bacterial cell.\nNIGRAM\n• Glucose enters the bacterial cell using a transport protein, A.\n• B is an enzyme found in the cytoplasm of the bacterial cell.\n• B catalyses the phosphorylation of glucose to produce glucose 6‑phosphate. SIHT\nNI\nglucose extracellular\nlactose ETIRW\nlactose environment\npermease\nTON\nOD\ncell\nA surface\nmembrane\nNIGRAM\nP\nB\nSIHT\ncytoplasm\nNI\nETIRW\nTON\nB\nOD\nglucose 6-phosphate\nNIGRAM\nrepressor\nSIHT\nlacI lacZ lacY lacA P O\nNI\nETIRW\nFig. 3.1\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) (i) With reference to Fig. 3.1, suggest and explain how the presence of glucose in the\nSIHT\nextracellular environment prevents lactose entering the bacterial cell.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ...........................................................................................................................................\nNI\n...........................................................................................................................................\nETIRW\n..................................................................................................................................... [3]\nTON\nOD (ii) Suggest the advantages for the bacterial cell in preventing the entry of lactose when\nglucose is present in the extracellular environment.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n.........................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-42-q03-49aed95117.pdf"
    },
    {
      "id": "9700-2025-on-42-q04",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        10
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-42-q04-p01-9593d6d291.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 The evolution of antibiotic resistance in bacteria has occurred as a result of natural selection.\nSIHT\nNI (a) Name two ways in which a bacterium can become resistant to an antibiotic.\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [2]\nOD\n(b) The World Health Organisation regularly analyses bacterial DNA sequence data.\nSuggest one way in which this contributes to solving the problem of antibiotic resistance\nNIGRAM\nin bacteria.\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n............................................................................................................................................. [1]\nTON\nOD (c) Some infectious bacterial diseases are treated with the antibiotic streptomycin.\nIf a person does not finish the prescribed course of streptomycin, bacteria are more likely to\nbecome resistant to the antibiotic.\nNIGRAM\nExplain why a streptomycin‑resistant strain of bacteria is more likely to develop as a result of\nnatural selection when a person does not complete the prescribed course of antibiotics.\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\n[Total: 6]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-on-42-q04-438f267fdf.pdf"
    },
    {
      "id": "9700-2025-on-42-q05",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 14,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        12,
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-on-42-q05-p01-dbf3075b4d.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-on-42-q05-p02-765c1a77ac.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-on-42-q05-p03-441f370759.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-on-42-q05-p04-416e549ddb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q05",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 Metachromatic leukodystrophy (MLD) is a genetic disease that affects the nervous system.\nSIHT\nNI MLD is caused by mutations in the ARSA gene located on chromosome 22. The ARSA gene,\nETIRW\nwhich is 3150 base pairs (bp) in length, includes 8 exons and is shown in Fig. 5.1.\nexon TON\nOD\n3.15 kb\nNIGRAM\nFig. 5.1\nA genetic test using DNA sequencing is available to identify mutations associated with MLD in the\nSIHT\nARSA gene. The sequencing method can only work if a DNA fragment size is less than 1000 bp.\nNI\nETIRW\nThe test involves a number of different stages:\nTON • Genomic DNA is extracted from a blood sample.\n• Polymerase chain reaction (PCR) using 5 pairs of primers selects 5 dif ferent DNA fragments\nOD\nof the ARSA gene.\n• Gel electrophoresis is carried out on the DNA fragments.\n• The DNA fragments are sequenced and analysed for mutations.\nNIGRAM\nTable 5.1 shows the length of the DNA fragments and the exons present in each DNA fragment.\nSIHT Table 5.1\nNI\nETIRW fragment DNA fragment exons in DNA\nnumber length / bp fragment\nTON\n1 405 1\nOD\n2 737 1‑2\n3 706 2‑4\n4 860 5‑7\nNIGRAM\n5 916 7‑8\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) Genomic DNA and primers are added to the PCR machine.\nSIHT\nNI Name two other substances that are added to the PCR machine, and explain why each\nETIRW\nsubstance is added.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Gel electrophoresis is carried out on the 5 dif ferent DNA fragments produced as a result of\nSIHT\nPCR. ",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-on-42-q05-e4720b543b.pdf"
    },
    {
      "id": "9700-2025-on-42-q06",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-on-42-q06-p01-5a5fa33713.png",
        "9700-topic-14-homeostasis/assets/9700-2025-on-42-q06-p02-e2d657b620.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n6 (a) Phosphoinositide 3‑kinase (PI3K) and protein kinase B (PKB) are enzymes involved in the\nSIHT\nregulation of blood glucose concentration.\nNI\nETIRW\nFig. 6.1 shows a cell‑signalling pathway involving insulin, PI3K and PKB in a muscle cell.\nTON insulin glucose\nOD cell receptor\nsurface\nmembrane\nNIGRAM\nPI3K\nactivation\nPKB\nSIHT cytoplasm\nNI\nETIRW\nGLUT4 transport protein\nTON\nOD\nvesicle\nFig. 6.1\nNIGRAM\nT ype 2 diabetes mellitus is a common disease. Some people with type 2 diabetes have a\nlow concentration of PI3K in their muscle cells and cannot maintain their blood glucose SIHT\nconcentration within normal limits.\nNI\nETIRW\nWith reference to Fig. 6.1, suggest why a low concentration of PI3K can lead to type 2 diabetes.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Describe how a biosensor can be used to measure blood glucose concentration.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n......................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-on-42-q06-4f8d8e1528.pdf"
    },
    {
      "id": "9700-2025-on-42-q07",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q07-p01-4494ec3ea7.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q07-p02-b5df1f77c0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q07",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n7 (a) The Venus fly trap, Dionaea muscipula, is a plant that can capture insects. Each leaf of a\nSIHT\nVenus fly trap is modified to form 2 lobes.\nNI\nETIRW\nFig. 7.1 shows how the leaf of a V enus fly trap appears folded (closed) when an insect has\nbeen captured.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 7.1\n(i) Suggest a reason why the Venus fly trap needs to capture insects, even though it carries\nNIGRAM\nout photosynthesis.\nSIHT ...........................................................................................................................................\nNI\n..................................................................................................................................... [1]\nETIRW\n(ii) The leaves are specialised to form 2 lobes. The lobes are red on the upper surface.\nTON\nOD Suggest why the lobes are red.\n...........................................................................................................................................\nNIGRAM ..................................................................................................................................... [1]\n(iii) The lobes have sensory hairs. Touching 1 sensory hair will not cause a response by the\nSIHT\nleaf. At least 2 sensory hairs need to be touched within 20 seconds to cause a response\nin the leaves. NI\nETIRW\nState the advantage to the plant of producing a response only when 2 sensory hairs are\ntouched within 20 seconds. TON\nOD\n...........................................................................................................................................\n..................................................................................................................................... [1]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(iv) When an insect has been trapped, the leaves have to remain closed for a number\nSIHT\nof days.\nNI\nETIRW\nSuggest why this needs to happen.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n..................................................................................................................................... [1]\nNIGRAM\n(b) Action potentials are produced by the Venus fly trap during the closure of a leaf. Each action\npotential is also associated with a refractory period. This is similar to the action potential and\nSIHT refractory period observed in a mammalian neurone during nerve impulse transmission.\nNI\nExplain the role played by the refractory period in the transmission of an impulse in a\nETIRW\nmammalian neurone.\nTON\n.....................................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-42-q07-28d7a1b2b7.pdf"
    },
    {
      "id": "9700-2025-on-42-q08",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-42-q08-p01-f23f5536cf.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-42-q08-p02-5bddcd1965.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-42-q08-p03-70618c0862.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q08",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n8 (a) Palm oil is the most widely used vegetable oil. It is obtained from the African oil palm, Elaeis\nSIHT\nguineensis. Oil palms are the highest yielding vegetable oil crop, needing only 10% of the\nNI land area required by other crops to produce the same quantity of oil.\nETIRW\nOn the island of Borneo, tropical rainforests are cut down and cleared to create oil palm\nTON plantations. This has a large effect on biodiversity.\nOD\nThe Bornean orangutan, Pongo pygmaeus, is only found on the island of Borneo and is\nclassified as critically endangered on the International Union for Conservation of Nature\n(IUCN) Red List of Threatened Species™.\nNIGRAM\nFig. 8.1 shows Bornean orangutans.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 8.1\nSuggest ways in which the Bornean orangutan may be conserved.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n............................................................................................................................................. [4]\nETIRW\nTON\nNIGRAM\n(b) Biodiversity can be assessed at different levels.\nSIHT\nNI Outline the main levels at which biodiversity can be assessed.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...........................................................................................................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-on-42-q08-a184f51fdd.pdf"
    },
    {
      "id": "9700-2025-on-42-q09",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        23,
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-on-42-q09-p01-a4112f9b62.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-42-q09-p02-86805227df.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-42-q09-p03-4c8a24204e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q09",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:4)(cid:44)\n9 (a) Fig. 9.1 shows an absorption spectrum for chlorophyll a and the corresponding action\nSIHT\nspectrum for a species of plant.\nNI\nETIRW\nTON\nabsorption action\nOD spectrum spectrum\nNIGRAM\nSIHT\nNI\n400 500 600 700\nETIRW\nwavelength of light / nm\nTON\nFig. 9.1\nOD\n(i) Explain what is meant by an absorption spectrum and an action spectrum.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [2]\nOD\n(ii) Suggest and explain why the curve shown in Fig. 9.1 for the absorption spectrum is\ndifferent from the curve for the action spectrum for wavelengths of light between 450 nm\nand 550 nm.\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\n........................................................................................................................................... ETIRW\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ..................................................................................................................................... [3]\nETIRW\nTON\nNIGRAM\n(b) Scientists investigated the effect of temperature on the rate of photosynthesis and the rate of\nSIHT\nrespiration of the trailing azalea, Kalmia procumbens, at a high light intensity.\nNI\nETIRW\nFig. 9.2 shows the results of this investigation.\n4.0 TON 2.0\nOD\npphhoottoossyynntthheessiiss\n3.0 1.5\nNIGRAM\nrate of rate",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-on-42-q09-f08d76a2ca.pdf"
    },
    {
      "id": "9700-2025-on-42-q10",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_42.pdf?download=true",
      "source_pages": [
        26,
        27
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q10-p01-dae5ccef7a.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-42-q10-p02-ca93431289.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-42-q10",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:7)(cid:44)\n10 (a) Nerve impulses can be transmitted along a myelinated motor neurone to a neuromuscular\nSIHT\ns–1. junction at fast speeds of up to 100 m\nNI\nETIRW s–1 is achieved by the neurone. Outline how a transmission speed of 100 m\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n............................................................................................................................................. [3]\nTON\nOD (b) When a nerve impulse reaches the neuromuscular junction, acetylcholine is released and\ndiffuses to the sarcolemma.\nDescribe how the release of acetylcholine can result in the binding of calcium ions to troponin\nNIGRAM in the sarcomere.\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n.............................................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-42-q10-ef90169146.pdf"
    },
    {
      "id": "9700-2025-on-43-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-43-q01-p01-db9b6483c5.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-43-q01-p02-96f808b471.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q01",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Guinea pigs, Cavia porcellus, vary in the length and colour of their fur.\nSIHT\nNI Fig. 1.1 shows a guinea pig with short black fur.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 1.1\nTON\nTwo genes that determine the length and colour of the fur occur at the A/a locus and the B/b OD\nlocus. These two gene loci are on separate autosomal chromosomes.\n• The allele A results in short fur.\n• The allele a results in long fur. NIGRAM\n• A is dominant to a.\nSIHT • The allele B results in black fur.\n• The allele b results in chocolate fur.\nNI\n• B is dominant to b. ETIRW\n(a) (i) List all the possible genotypes of a guinea pig with short black fur.\nTON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\n(ii) A test cross could be used to determine the genotype of a female guinea pig with short\nSIHT\nblack fur.\nNI\nETIRW\nDescribe the phenotype of the male guinea pig that could be used to carry out this\ntest cross.\nTON\n..................................................................................................................................... [1]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) A black guinea pig with long fur that was homozygous at both loci was crossed with a\nSIHT\nchocolate guinea pig with short fur that was homozygous at both loci. The F1 offspring of this\nNI cross had short black fur. F1 offspring were mated together to produce the F2 offspring.\nETIRW\nComplete the Punnett square to:\nTON • show the cross between the F1 offspring\n• predict the F2 offspring genotypes.\nOD\nY ou should include the gametes in your answer.\nState the ratio of F2 of fspring phenotypes. You should include a key to link phenotypes to\ngenotypes.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nratio of F2 of fspring phenotypes:\nNIGRAM\nSIHT\n[4] NI\nETIRW\n(c) Some genes in guinea pigs are structural genes and some are regulatory genes.\nTON\nDescribe the dif ference between a structural gene and a regulatory gene.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n.......................................................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-43-q01-cf065349e4.pdf"
    },
    {
      "id": "9700-2025-on-43-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-43-q02-p01-5de3844654.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-43-q02-p02-3c4475f691.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 Exserohilum turcicum is a fungal pathogen. The growth of the mycelium of the fungus damages\nSIHT\nthe leaves of maize plants, Zea mays. Leaf damage reduces crop yield.\nNI\nETIRW\n(a) (i) Complete Table 2.1 to show one structural dif ference and one functional dif ference\nbetween E. turcicum and Z. mays.\nTON\nTable 2.1\nOD\nE. turcicum Z. mays\nNIGRAM\nstructural\nSIHT\ndifference\nNI\nETIRW\nTON\nOD\nfunctional\ndifference\nNIGRAM\n[2]\nSIHT\nNI (ii) Describe the principles by which organisms such as E. turcicum and Z. m ays are\nETIRW classified in the taxonomic hierarchy.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [4] OD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Two inbred varieties of maize, SKV50 and CML153, were crossed. The resulting F1 hybrids\nSIHT\nwere self‑crossed to produce F2 of fspring. The F2 plants were grown, and the percentage\nNI area of leaf damage caused by E. turcicum was measured.\nETIRW\nFig. 2.1 shows the results for the F2 generation. The arrows show the mean percentage area\nof leaf damage for the two parent varieties. TON\n50 OD\n40\nNIGRAM\n30\nnumber\nof plants\n20 SIHT\nNI\n10\nETIRW\n0\nTON\n20 30 40 50 60\nOD\npercentage area of\nleaf damage SKV50 CML153\nFig. 2.1\nNIGRAM\n(i) Explain how Fig. 2.1 can be used to determine which parent maize variety shows the\ngreatest resistance to infection by E. turcicum.\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n..............................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-43-q02-905e35e1f2.pdf"
    },
    {
      "id": "9700-2025-on-43-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-on-43-q03-p01-c7da1d04ef.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-43-q03-p02-aee99fdc02.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-43-q03-p03-dcbd18fde5.png",
        "9700-topic-13-photosynthesis/assets/9700-2025-on-43-q03-p04-6b1f5246d3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q03",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 Plants have several different photosynthetic pigments in their chloroplasts.\nSIHT\nNI (a) A student separated and identified the chloroplast pigments present in a leaf extract from a\nETIRW\nspinach plant using two slightly different methods.\nTON Method A\n• A type of chromatography known as thin layer chromatography (TLC) was used to\nOD\nseparate the pigments.\n• A mixture of ether and cyclohexane was used as a solvent in TLC.\nMethod B\nNIGRAM\n• The student repeated TLC but treated the spinach leaf extract with a chemical. The\nchemical causes a magnesium ion in a pigment to be replaced by two hydrogen ions.\n• The student used a leaf from the same spinach plant, and used the same solvent as in SIHT\nmethod A.\nNI\nETIRW\nvalues and compared these to reference values to identify The student calculated R\nf\nthe pigments.\nTON\nOD Fig. 3.1 shows the results for method A and method B.\nsolvent front\nNIGRAM\nβ-carotene\nSIHT\nsolvent front NI\nβ-carotene ETIRW\nTON\nOD\nphaeophytin a\nchlorophyll a\nNIGRAM\nphaeophytin b\nchlorophyll b\nSIHT\nchlorophyll b\nNI\nETIRW\nxanthophyll\nxanthophyll\nTON\nOD line of origin line of origin\nA B\nNIGRAM\nFig. 3.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Calculate the R value of β‑carotene in Fig. 3.1.\nSIHT f\nNI\nETIRW\nTON R = ...............................................................\nf\n[2]\nOD\n(ii) Suggest two explanations for the differences in appearance of chromatograms A and B\nin Fig. 3.1.\nNIGRAM\n1 ........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n2 ........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\n[2]\nNIGRAM\nvalues for the chloroplast pigments of spinach in a (iii) The student found some different R\nf\nscientific paper.\nSIHT\nThe values in the scientific paper were dif ferent from the reference values that the NI\nETIRW student originally used to identify the pigments on chromatograms A and B in Fig. 3.1.\nvalues. Suggest one reason, other than measurement error, for the different R TON\nf\nOD\n...........................................................................................................................................\n........................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-on-43-q03-bd493600e8.pdf"
    },
    {
      "id": "9700-2025-on-43-q04",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-43-q04-p01-989190d8b9.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-43-q04-p02-4266fd9144.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 The wolf, Canis lupus, lives in North America. Wolves may have a grey or a black coat colour. The\nSIHT\ncolour of an individual wolf depends on the DNA it inherits at the CPD103 gene locus.\nNI\nETIRW • Wolves inherit two copies of CPD103, one from each parent.\n• Wolves that inherit one copy of the black form of the CPD103 gene have a black coat.\nTON\n(a) State the term used to describe:\nOD\n• an organism that has two copies of each gene ..........................................................\n• a form of a gene ..........................................................\nNIGRAM\n• a form of a gene that gives a phenotypic\neffect in a heterozygote. .......................................................... SIHT\n[3]\nNI\nETIRW\nIn addition to producing black coat colour, the protein coded for by the CPD103 gene also defends\nagainst infectious lung disease.\nTON\nOD Canine distemper virus (CDV) causes serious lung disease in wolves. W olves that have been\npreviously infected by CDV have antibodies against CDV (anti‑CDV antibodies) in their blood.\n(b) CDV can be passed from domestic dogs to wolves.\nNIGRAM\n• Domestic dogs are more numerous in the southern part of the area occupied by wolves.\n• Domestic dogs are less numerous in the northern part of the area occupied by wolves.\nSIHT\n• The relative frequency of black wolves compared to grey wolves increases from the\nnorth to the south of the area they occupy. NI\nETIRW\nExplain how natural selection causes this trend in the distribution of black wolves.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Several different populations of wolves were compared.\nSIHT\nNI Fig. 4.1 shows the relationship between the percentage of ",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-on-43-q04-ace89e7f4d.pdf"
    },
    {
      "id": "9700-2025-on-43-q05",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-43-q05-p01-e315af8a4d.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-43-q05-p02-b551c30637.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 The endocrine system and the nervous system both coordinate responses in mammals.\nSIHT\nNI (a) Complete Table 5.1 to show the features of three cell‑signalling molecules of the endocrine\nETIRW\nsystem: antidiuretic hormone (ADH), glucagon and insulin.\nTON 7) if the molecule does not have Use a tick (3) if the molecule has the feature and a cross (\nthe feature.\nOD\n7) in every box. Put a tick (3) or a cross (\nTable 5.1\nNIGRAM\nfeature ADH glucagon insulin\nSIHT\nbinds to receptors\nNI\non cell surface\nETIRW\nmembranes\nresults in molecules TON\nmoving from cells\nOD\ninto the blood\nis secreted as a\nresult of detection by\nosmoreceptors NIGRAM\n[3]\nSIHT\n(b) The endocrine system has a slower transmission speed than the nervous system.\nNI\nETIRW\nDescribe other ways in which the endocrine system and the nervous system dif fer.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The endocrine system and the nervous system can affect muscle function.\nSIHT\nNI Fig. 5.1 shows a transmission electron micrograph of a longitudinal section of striated muscle\nETIRW\ntissue that is in a relaxed state.\nTON\nOD\nB\nNIGRAM\nC SIHT\nNI\nETIRW\nTON\nD\nE\nOD\nNIGRAM\nSIHT\nFig. 5.1\nNI\nETIRW\n(i) State the letter on Fig. 5.1 that indicates the length of a sarcomere.\nTON\n..................................................................................................................................... [1]\nOD\n(ii) State the letter on Fig. 5.1 that indicates a region where actin and myosin overlap.\n.................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-43-q05-040111f98d.pdf"
    },
    {
      "id": "9700-2025-on-43-q06",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-43-q06-p01-0352f3c8a8.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-43-q06-p02-76e227614d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q06",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 The distribution of the large blue butterfly, Phengaris arion, extends across Europe and Asia. It is\nSIHT\nassessed by the International Union for Conservation of Nature (IUCN) on the Red List™ as ‘Near\nNI Threatened’ globally and ‘Endangered’ in Europe.\nETIRW\nIn Europe, P. arion became extinct in the Netherlands in 1964 and in the United Kingdom in 1979.\nTON\nFig. 6.1 lists the conservation status categories in the IUCN Red List™.\nOD\nconservation status\nExtinct (EX)\nNIGRAM\nExtinct in the Wild (EW)\nSIHT\nCritically Endangered (CR)\nNI\nETIRW\nEndangered (EN) increasing risk of extinction\nTON\nVulnerable (VU)\nOD\nNear Threatened (NT)\nLeast Concern (LC)\nNIGRAM\nFig. 6.1\nSIHT Fig. 6.2 shows P. arion.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 6.2\nTON\n(a) (i) Explain how IUCN Red List™ assessments help to conserve biodiversity.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n..................................................................................................................................... [3] TON\nNIGRAM\n(ii) With reference to Fig. 6.1 and the IUCN assessments for P . arion, suggest how the\nSIHT\nabundance of the butterfly differs across its distribution.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ..................................................................................................................................... [2]\nNI\n(b) P. arion has been successfully re‑introduced in the United Kingdom at 12 sites. These sites\nETIRW\nwere restored to flower‑rich grassland.\nTON\nThe conservation",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-on-43-q06-c8d35596cf.pdf"
    },
    {
      "id": "9700-2025-on-43-q07",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        16
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-on-43-q07-p01-261bab4857.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q07",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 In aerobic respiration, most ATP is produced by oxidative phosphorylation.\nSIHT\nNI (a) Outline the features of ATP that make it suitable as the universal energy currency.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................................................................................................................................. [3]\nTON\n(b) Rotenone is a compound that affects oxidative phosphorylation.\nOD\nRotenone disrupts the first carrier in the electron transport chain by stopping the transfer of\nelectrons from this carrier.\nNIGRAM Suggest and explain how rotenone reduces the production of ATP and water in\naerobic respiration.\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-on-43-q07-2413d9907c.pdf"
    },
    {
      "id": "9700-2025-on-43-q08",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        17
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-on-43-q08-p01-d55ce7ffdb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:8)(cid:44)\n8 LibertyLink® soybean is a genetically modified crop. It was first grown in 1996 and used in food\nSIHT\nproducts from 1998. It has been grown in 6 countries and used in food products in 21 countries.\nNI\nETIRW\nTable 8.1 summarises the modifications made to the soybean plant to produce LibertyLink®\nsoybean.\nTON\nTable 8.1\nOD\nname of introduced\ngene donor organism gene product function of gene product\ngene\nNIGRAM\nStreptomyces phosphinothricin stops action of glufosinate,\npat\nviridochromogenes N‑acetyltransferase a herbicide\nSIHT\n(a) Explain how the modification made to produce LibertyLink® soybean may help to solve the\nNI\nglobal demand for food.\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [4]\nTON\n(b) (i) Name the type of enzyme that could be used to cut out the pat gene from\nOD\nS. viridochromogenes.\n..................................................................................................................................... [1]\nNIGRAM\n(ii) Name the enzyme that could be used to join the pat gene to a plasmid by forming\nphosphodiester bonds.\nSIHT\n..................................................................................................................................... [1]\nNI\nETIRW\n(c) Suggest reasons why LibertyLink® soybean is used in food products in 21 countries but only\ngrown in 6 countries.\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n...............................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-on-43-q08-8ec68be986.pdf"
    },
    {
      "id": "9700-2025-on-43-q09",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        18
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-on-43-q09-p01-ffa4928c32.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q09",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n9 (a) Fig. 9.1 shows a longitudinal section of a human kidney.\nSIHT\nNI\nETIRW\nTON\nOD\nX\nNIGRAM\nY\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 9.1\nName the regions of the kidney labelled X and Y in Fig. 9.1.\nNIGRAM\nX ...............................................................................................................................................\nSIHT Y ...............................................................................................................................................\n[2]\nNI\nETIRW\n(b) A biosensor can be used to measure the concentration of glucose in urine.\nTON\nOutline how a biosensor measures the concentration of glucose in urine.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n[Total: 6]\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-on-43-q09-edd17fde24.pdf"
    },
    {
      "id": "9700-2025-on-43-q10",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 6,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_43.pdf?download=true",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-43-q10-p01-cdec87d07b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-43-q10",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:10)(cid:44)\n10 Explain the relationship between genes, proteins and phenotype, with reference to two examples\nSIHT\nof genetic diseases in humans.\nNI\nETIRW\n..........................................................................................................................................................\nTON ..........................................................................................................................................................\nOD\n..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM\n..........................................................................................................................................................\nSIHT ..........................................................................................................................................................\nNI\n..........................................................................................................................................................\nETIRW\n..........................................................................................................................................................\nTON\nOD ..........................................................................................................................................................\n..........................................................................................................................................................\nNIGRAM ..........................................................................................................................................................\n..........................................................................................................................................................\nSIHT\n.......................................................................................................................................................... NI\nETIRW\n..........................................................................................................................................................\nTON\n..........................................................................................................................................................\nOD\n.................................................................................................................................................... [6]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-43-q10-f84c9e15d5.pdf"
    },
    {
      "id": "9700-2025-on-44-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-44-q01-p01-25f58330f8.png",
        "9700-topic-15-control-and-coordination/assets/9700-2025-on-44-q01-p02-0988d12942.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) Fig. 1.1 is a diagram of a sensory neurone.\nSIHT\nA D NI F\nETIRW\nB\nTON\nOD\nC E\nNIGRAM\nFig. 1.1\nSIHT\nUse the letters A–F in Fig. 1.1 to identify:\nNI\nETIRW a receptor cell ................................................\nan area where dendron membrane depolarisation occurs ................................................ TON\nOD\na structure that forms a synapse with an intermediate neurone ................................................\na structure that allows rapid transmission of impulses ................................................\n[4]\nNIGRAM\n(b) Opioid drugs can bind to opioid receptors in the presynaptic membrane of a cholinergic\nsynapse. SIHT\nNI\nFig. 1.2 is a diagram of a presynaptic membrane with an opioid receptor.\nETIRW\nsynaptic cleft\nTON voltage-gated opioid\nopioid\nCa2+ channel drug OD receptor\nprotein\npresynaptic\nmembrane\nNIGRAM\nactivated causes\nSIHT blockage\nG protein\nNI\ncytoplasm ETIRW\nFig. 1.2 TON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nOpioid drugs have an effect on the normal events that occur at a cholinergic synapse.\nSIHT\nNI Suggest and explain the effect that an opioid drug will have on the normal events that occur\nETIRW\nat a cholinergic synapse.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM [Total: 8]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2025-on-44-q01-12a2fc5b45.pdf"
    },
    {
      "id": "9700-2025-on-44-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-44-q02-p01-d9826e3ad0.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-44-q02-p02-3564855b48.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 (a) Meiosis and cytokinesis occur in the male reproductive organs (anthers) of plants to make\nSIHT\npollen grains. Cells which carry out meiosis are known as pollen mother cells.\nNI\nETIRW\nFig. 2.1 shows five stages of meiosis in a pollen mother cell.\nTON\nOD\nNIGRAM\nJ K\nSIHT\nNI\nETIRW\nTON\nOD\nL M N\nNIGRAM\nFig. 2.1\nUse the letters J–N in Fig. 2.1 to state all the stages that show:\nSIHT\nNI anaphase ................................................\nETIRW\ncells containing pairs of homologous chromosomes ................................................\nTON\ncrossing over ................................................\nOD\nhaploid cells ................................................\n[4]\nNIGRAM\n(b) Maize plants have male and female reproductive organs on the same plant. The male anthers\nare located on structures known as tassels. When selective breeding is carried out to create\nSIHT an F1 hybrid, the tassels are removed.\nNI\nSuggest why the tassels are removed when selective breeding is carried out to create an F1\nETIRW\nhybrid.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [2]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Outline how selective breeding is used to produce vigorous, uniform varieties of maize.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n.....................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-44-q02-ad7f6c0ce7.pdf"
    },
    {
      "id": "9700-2025-on-44-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2025-on-44-q03-p01-a72022bfeb.png",
        "9700-topic-19-genetic-technology/assets/9700-2025-on-44-q03-p02-b60a050584.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 (a) Genetic engineering is a modern method for producing crop plants with improved\nSIHT\ncharacteristics.\nNI\nETIRW\nOne example of a crop plant with improved characteristics is soybean, Glycine max, which\nhas been genetically modified to make it resistant to a herbicide. This genetically modified\nTON soybean is called GM soybean.\nOD\nTo create GM soybean, a bacterial gene and a section of regulatory DNA were introduced\ninto soybean cells.\nOutline the roles of enzymes and a section of regulatory DNA in the creation of genetically\nNIGRAM\nmodified organisms such as GM soybean.\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [5]\nNIGRAM\n(b) In 2018, an area of 123.5 million hectares was used to grow soybean crops worldwide.\nGM soybean was grown in 73% of this area.\nSIHT\nNI In 2018, soybeans accounted for 50% of the total GM crop area worldwide.\nETIRW\nCalculate the total area used to grow GM crops worldwide in 2018.\nTON\nShow your working. OD\nNIGRAM\nSIHT\n........................................... hectares [2]\nNI\nETIRW\nTON\nNIGRAM\n(c) Discuss the social implications of growing GM crops.\nSIHT\nNI ...........",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2025-on-44-q03-3e994e69a2.pdf"
    },
    {
      "id": "9700-2025-on-44-q04",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        8,
        9,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-44-q04-p01-fb9c0fbc5b.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-44-q04-p02-063d99dd41.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-44-q04-p03-12dd75739d.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2025-on-44-q04-p04-25e3b60ce8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 Alleles are alternative forms of a gene. For example, there may be a dominant allele, T, and a\nSIHT\nrecessive allele, t, at the same gene locus.\nNI\nETIRW\nThe relative frequency of each allele of a gene in a population can change over time due to factors\nsuch as selection, genetic drift and the bottleneck effect.\nTON\n(a) Fig. 4.1 shows the relative frequency of the T allele in a population of 50 individuals over 20\nOD\ngenerations.\nAt generation 0, the number of T alleles and the number of t alleles in the population was\nequal, so the relative frequency of each allele was 0.5. The relative frequencies of the two\nNIGRAM\nalleles of the gene add up to 1.\n1.0 SIHT\nNI\n0.9\nETIRW\n0.8\nTON\nOD\n0.7\n0.6\nrelative NIGRAM\nfrequency 0.5\nof T allele\nSIHT\n0.4\nNI\nETIRW 0.3\nTON\n0.2\nOD\n0.1\n0\n0 2 4 6 8 10 12 14 16 18 20 NIGRAM\ngeneration number\nSIHT\nFig. 4.1\nNI\nETIRW\n(i) With reference to Fig. 4.1, state the relative frequency of the t allele after 20 generations.\nTON\n..................................................................................................................................... [1]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Suggest possible explanations for the change in the relative frequency of the T allele\nSIHT\nbetween generation 0 and generation 13.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ...........................................................................................................................................\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n..................................................................................................................................... [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The relative frequency of the T allele in a population of only 10 individuals wa",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2025-on-44-q04-ce00643e65.pdf"
    },
    {
      "id": "9700-2025-on-44-q05",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        12,
        13,
        14
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2025-on-44-q05-p01-992b637d52.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-44-q05-p02-164f8a277f.png",
        "9700-topic-16-inheritance/assets/9700-2025-on-44-q05-p03-d303f6833f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits.\nSIHT\nNI A fruit fly with normal features is described as wild type. It has a grey body and its wings are\nETIRW\nlonger than its abdomen. The genes for body colour and wing length are located on different\nchromosomes.\nTON\nA fruit fly with mutations in these two genes has a black body and short wings.\nOD\nFig. 5.1 shows a wild type fruit fly and a mutant fruit fly.\nwild type mutant\nNIGRAM\ngrey body black body\nSIHT\nNI\n3mm\nETIRW\nlong wing short wing\nTON\nOD\nFig. 5.1\n(a) Fruit flies were first used for genetic crosses by Thomas Morgan in 1908. They are one of the\nmost studied animals in current genetic research. NIGRAM\n• Male fruit flies are easily distinguished from female fruit flies.\nSIHT • Fruit flies have a short life cycle and a female can lay hundreds of eggs in a few days.\n• Some genes for development and cell signalling in fruit flies are similar to those of\nNI\nhumans. ETIRW\nSuggest why fruit flies are still used in genetic crosses.\nTON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [3]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) When a wild type fruit fly was crossed with a mutant fruit fly with a black body and short\nSIHT\nwings, all the F1 offspring had grey bodies and long wings.\nNI\nETIRW\nUsing appropriate symbols, complete Fig. 5.2 to show the expected results of a cross between\ntwo of these F1 fruit flies.\nTON\nOD symbols:\nNIGRAM\nF1 phenotypes: grey body × grey body\nSIHT\nlong wing long wing\nNI\nETIRW F1 genotypes:\nTON\nPunnett\nOD\nsquare\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM offspring phenotypes:\nSIHT\nNI\nETIRW\nTON\nOD\nphenotypic ratio ............................................................................\nNIGRAM\nFig. 5.2\n[6]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Describe how you would determine the genotype of an F2 fruit fly with a grey body and long\nSIHT\nwings.\nNI\nETIRW",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2025-on-44-q05-ce129588bb.pdf"
    },
    {
      "id": "9700-2025-on-44-q06",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-on-44-q06-p01-2235d13252.png",
        "9700-topic-14-homeostasis/assets/9700-2025-on-44-q06-p02-0010591753.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n6 Some people can develop a condition called type 2 diabetes. In people with type 2 diabetes,\nSIHT\nglucose uptake from the blood is decreased. In some cases, the pancreas cannot make enough\nNI insulin to keep blood glucose concentration within a healthy range.\nETIRW\n(a) A person with type 2 diabetes and a person without type 2 diabetes were given a glucose\nTON drink. The blood glucose concentration of each person was measured at regular intervals for\n120 minutes.\nOD\nThe results are shown in Fig. 6.1.\n14\nNIGRAM\nperson with\ntype 2 diabetes\n12\nSIHT\n10\nNI\nETIRW\nblood\n8\nglucose\nTON\nconcentration\n6\nOD dm–3 / mmol\nperson without\ntype 2 diabetes 4\nNIGRAM 2\n0\nSIHT\n90 120 0 30 60\n/ min time NI\nETIRW\nglucose\ndrink\nTON\nOD\nFig. 6.1\n(i) Describe the curve for the person without type 2 diabetes.\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ...........................................................................................................................................\nETIRW\n..................................................................................................................................... [2]\nTON\n(ii) Calculate the percentage increase in blood glucose concentration, between 0 and\nOD\n60 minutes, for the person with type 2 diabetes.\nNIGRAM\nSIHT\nanswer ......................................................... [2]\nNI\nETIRW\nTON\nNIGRAM\n(iii) Suggest ways in which people with type 2 diabetes can help to control their condition.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n..................................................................................................................................... [2]\n(b) Glucagon has a role in the control of blood glucose concentration.\nNIGRAM\nDescribe the cell signalling pathway involving glucagon, and describe how this pathway leads\nSIHT to a change in the blood glucose concentration.\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-on-44-q06-62c4e58e78.pdf"
    },
    {
      "id": "9700-2025-on-44-q07",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 7,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        18,
        19,
        20
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-44-q07-p01-8f722b0c6e.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-44-q07-p02-5c2b5343fa.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2025-on-44-q07-p03-6ad4cc4309.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q07",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n7 The kakapo, Strigops habroptila, is a species of large, nocturnal (active at night) parrot and is\nSIHT\nfound only in New Zealand. The bird does not fly and lives on the ground.\nNI\nETIRW\nThe kakapo is classified as critically endangered on the International Union for Conservation of\nNature (IUCN) Red List of Threatened Species™.\nTON\nFig. 7.1 shows a kakapo.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 7.1\nSIHT\n(a) Complete Table 7.1 to show the classification of the kakapo.\nNI\nETIRW Table 7.1\nTON\ntaxonomic group name\nOD\ndomain Eukarya\nkingdom Animalia\nNIGRAM\nChordata\n..........................\nSIHT\nclass Aves NI\nETIRW\nPsittaciformes\n.......................... TON\nOD\nfamily Strigopidae\ngenus\n..........................\nNIGRAM\n[3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The kakapo was widely distributed before humans arrived in New Zealand. The Kakapo\nSIHT\nRecovery Programme started in 1995. Birds were moved to protected areas. In 2024 they\nNI were only found in these protected areas.\nETIRW\nThis is shown in Fig. 7.2.\nTON\nkey\nOD\nkakapo distribution before humans\narrived in New Zealand\nprotected areas 2024\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFig. 7.2\nNI\nETIRW Suggest the ideal features of a protected area for the kakapo.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n............................................................................................................................................. [3]\nTON\nNIGRAM\n(c) DNA has shown that the kakapo population may have gone through one or more genetic\nSIHT\nbottlenecks after the arrival of humans in New Zealand.\nNI\nETIRW\nSuggest the consequences of a genetic bottleneck to the kakapo population in New Zealand.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n......",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2025-on-44-q07-55d0090c09.pdf"
    },
    {
      "id": "9700-2025-on-44-q08",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 8,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2025-on-44-q08-p01-06563df4b9.png",
        "9700-topic-14-homeostasis/assets/9700-2025-on-44-q08-p02-1938e23094.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q08",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n8 (a) When the water supply to a plant is reduced, the concentration of abscisic acid (ABA) in the\nSIHT\nleaves increases.\nNI\nETIRW\nABA binds to receptors in the cell surface membranes of guard cells. This triggers a series of\nevents in the cells, which results in stomatal closure.\nTON\nFig. 8.1 is a diagram of part of a guard cell showing some of the events that occur when ABA\nOD\nbinds to its receptor.\nNIGRAM proton pump\nCa2+ cell wall X\nABA\nSIHT\nNI cell surface\nETIRW\nmembrane\nTON\ninhibition OD\nCa2+ X\nreceptor cytoplasm\nstimulation\nNIGRAM\nFig. 8.1\nFig. 8.1 shows that calcium ions are involved in the events within a guard cell that result in\nSIHT\nstomatal closure.\nNI\nETIRW\nOutline the role of calcium ions in the response of the guard cell to a reduced water supply\nand explain how this response results in stomatal closure.\nTON\nIdentify X in your answer.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [4]\nNI\nETIRW\nTON\nNIGRAM\n(b) Explain, using examples, why the homeostatic control involving the opening and closing of\nSIHT\nstomata is important for the efficient functioning of plants.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n........",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2025-on-44-q08-0724ca6fdd.pdf"
    },
    {
      "id": "9700-2025-on-44-q09",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        24,
        25,
        26
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2025-on-44-q09-p01-655946d985.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-on-44-q09-p02-70df255fa5.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2025-on-44-q09-p03-19c89d2723.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:5)(cid:44)\n9 (a) In aerobic conditions, pyruvate that has formed during glycolysis enters the mitochondrion\nSIHT\nand takes part in the link reaction.\nNI\nETIRW\nName the two coenzymes involved in the link reaction.\nTON ......................................................................................\nOD\n......................................................................................\n[2]\n(b) Enzymes play an important role in the functioning of the Krebs cycle.\nNIGRAM\nThe enzymes in the Krebs cycle can be affected by the presence of hydrogen peroxide.\nHydrogen peroxide is a product of some of the reactions that occur within the mitochondrion. SIHT\nNI\nAn investigation was carried out to measure the effect of two different concentrations of\nETIRW\nhydrogen peroxide on an enzyme of the Krebs cycle. The activity of the enzyme was measured\nover 10 minutes when exposed to the two different concentrations of hydrogen peroxide.\nTON\nOD The results are shown in Fig. 9.1.\n100\nNIGRAM\n80\nSIHT\nhydrogen peroxide NI\nETIRW concentration\ndm–3 10–3 mol 0.1 ×\n60\nTON\npercentage\nOD\nenzyme\nactivity\nhydrogen peroxide\n40\nconcentration\nNIGRAM 10–3 dm–3 mol 0.5 ×\nSIHT 20\nNI\nETIRW\nTON 0\n0 2 4 6 8 10\nOD\ntime / minutes\nFig. 9.1\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nDescribe the effect shown in Fig. 9.1 of hydrogen peroxide concentration on the activity of the\nSIHT\nKrebs cycle enzyme.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ............................................................................................................................................. [3]\nNI\n(c) Reduced NAD and reduced FAD are produced during the Krebs cycle. They carry hydrogen\nETIRW\nto the inner mitochondrial membrane where oxidative phosphorylation occurs.\nTON\nThe first two steps in oxidative phosphorylation are:\nOD\n1. Hydrogen atoms split into protons and electrons.\n2. Electrons move along the electron transport chain, releasing energy.\nNIGRAM Outline the steps that occur to complete oxidative phosphorylation.\n...................................................................................................................................................\nSIHT\n...................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2025-on-44-q09-415d85fac9.pdf"
    },
    {
      "id": "9700-2025-on-44-q10",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 10,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_44.pdf?download=true",
      "source_pages": [
        27
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_44.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2025-on-44-q10-p01-34ef780a9b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-44-q10",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:8)(cid:44)\n10 (a) Cyclic photophosphorylation and non‑cyclic photophosphorylation occur during the\nSIHT\nlight‑dependent stage of photosynthesis.\nNI\nETIRW\nOutline the differences between cyclic photophosphorylation and non‑cyclic\nphotophosphorylation.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [3]\n(b) Complete the passage about the Calvin cycle using the most appropriate word or words.\nNIGRAM\nA molecule of .................................................... combines with a five‑carbon molecule,\nSIHT\nribulose bisphosphate (RuBP), catalysed by the enzyme ................................................... .\nNI\nETIRW This reaction produces a six‑carbon compound that splits into two molecules of a three‑carbon\ncompound, glycerate 3‑phosphate (GP). TON\nOD\nATP and .................................................... are used to convert GP molecules into molecules\nof a three‑carbon sugar, triose phosphate (TP).\nNIGRAM\nSome TP molecules are used to make ...................................................., while others are\nrecycled to regenerate RuBP using ATP.\nSIHT\n[4]\nNI\nETIRW\n[Total: 7]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2025-on-44-q10-6934aac181.pdf"
    },
    {
      "id": "9700-2025-on-51-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 18,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_51.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p01-b29a6784a8.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p02-80ea879af1.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p03-e6d11c0399.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p04-b55ae1e3b3.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p05-e768cd6c38.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q01-p06-d4b8af5153.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Juglone is a chemical produced by plants in the genus Juglans, which includes dif ferent species\nSIHT\nof walnut tree.\nNI\nETIRW\nFig. 1.1 shows a black walnut tree.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 1.1\nNIGRAM\nJuglone produced by a black walnut tree can diffuse into the soil surrounding the tree.\nGroups of scientists have observed that juglone affects different species of plant growing close to SIHT\nblack walnut trees.\nNI\nETIRW\nJuglone can reduce the percentage of seeds that germinate, delay the start of germination or\nreduce growth. In some studies, juglone has no effect on germination or growth.\nTON\nA student planned to investigate the effect of juglone on the cucumber plant, Cucumis sativus. OD\nIn the investigation, the student:\n• placed some damp soil containing juglone in a tray NIGRAM\n• added some cucumber seeds to the soil\n• placed the container in a suitable environment for germination and growth.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nFig. 1.2 shows cucumber seedlings (young plants) a few days after germination.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 1.2\nTON\nThe student planned to compare the effect of exposing cucumber seeds to:\nOD\n10–3 dm–3, • juglone solution with a concentration of 1.0 × mol which is a concentration that has\nbeen measured in soils surrounding black walnut trees\n• a control treatment.\nNIGRAM\nThe student planned to determine:\nSIHT • the percentage of cucumber seeds that germinated in soil\n• the growth of each cucumber seedling 10 days after the appearance of the seedling above\nNI\nthe soil. ETIRW\n(a) Identify the independent variable in this investigation. TON\nOD\n............................................................................................................................................. [1]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Juglone solutions are prepared using a mixture of solvents.\nSIHT\nNI The student was provided with:\nETIRW\ndm–3 • a juglone solution with a concentration of 0.1 mol\nTON • the mixture of solvents.\nOD\ncm3 Using the solution provided, the student prepared a juglone solution with a volume of 500\n10–3 dm–3. mol and a concentration of 1.0 ×\n–3 dm–3 mol solution using measuring The student carried out 1 dilution to prepare the 1.0 × 10\nNIGRAM\ncylinders.\ncm3 of (i) State the volumes that the student used to carry out the dilution to produce 500 SIHT\n10–3 dm–3 mol juglone solution. 1.0 ×\nNI\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [2]\nOD\n(ii) Outline one improvement the student could make to reduce the percentage error while\ncarrying out the dilution.\nNIGRAM .............................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-51-q01-5445b6bfd5.pdf"
    },
    {
      "id": "9700-2025-on-51-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_51.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-51-q02-p01-a96d3639bb.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q02-p02-1ed631719d.png",
        "9700-experimental-skills/assets/9700-2025-on-51-q02-p03-b833efb9f2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n–) 2 Compounds containing nitrite ions (NO are present in many foods eaten by humans.\nSIHT 2\nNI Scientists tested the ef fect of different concentrations of nitrite ions on the population growth of\nETIRW\n5 species of bacterium. All 5 species are pathogenic and can infect the human digestive system.\nTON For each species of bacterium, the scientists placed suspensions of the bacteria in a microwell\nplate. A microwell plate is a plastic plate containing 96 wells, which are similar to small test-tubes.\nOD\nIn each well, the scientists added:\n• a suspension of the bacterial species\nNIGRAM\n• a solution containing nitrite ions\n• nutrient broth.\nSIHT\nThe scientists incubated the microwell plate at 37 °C for 24 hours. After 24 hours, the scientists\nNI\nestimated the number of bacteria in each well by measuring optical density , using a microwell\nETIRW\nplate reader.\nTON\nFor each species of bacterium, the scientists repeated the experiment with several dif ferent\nOD concentrations of solution containing nitrite ions.\nThe scientists determined the lowest nitrite concentration at which no bacterial population growth\nhad occurred after 24 hours.\nNIGRAM\n(a) The scientists standardised temperature, pH (pH 4.8) and time in the investigation.\nSIHT\nState two other variables that the scientists should standardise in this investigation.\nNI\nETIRW 1 ................................................................................................................................................\n2 ................................................................................................................................................ TON\n[2]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The results of the investigation are shown in Fig. 2.1.\nSIHT\n6.0 NI\nETIRW\n5.0\nTON\n4.0 OD lowest nitrite concentration\nat which no bacterial\n3.0 population growth occurred\nafter 24 hours\nNIGRAM cm–3 μmol / 2.0\n1.0\nSIHT\nNI\n0.0\nETIRW Yersinia Salmonella Salmonella Shigella Escherichia\nenterocolitica enteritidis typhimurium sonnei coli\nTON\nspecies of bacterium\nOD\nFig. 2.1\n(i) State the type of data represented by the species of bacterium and the lowest nitrite\nNIGRAM\nconcentration in Fig. 2.1.\nspecies of bacterium .........................................................................................................\nSIHT\nNI lowest nitrite concentration ................................................................................................\nETIRW\n[2]\nTON (ii) The 2 species of Salmonella shown in Fig. 2.1 are closely related.\nOD\nCalculate the percentage dif ference between the results shown in Fig. 2.1 for the\n2 species of Salmonella.\nShow your working.\nNIGRAM\nSIHT\nNI\nETIRW\n.............................................................%\nTON\n[3]\nOD\n(iii) Use Fig. 2.1 to state:\n• the species of bacterium that is most affected by nitrite ions\n• NIGRAM t",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-51-q02-0b7aeb2133.pdf"
    },
    {
      "id": "9700-2025-on-52-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 18,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_52.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p01-0eb15c303a.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p02-41c28bd300.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p03-b271bfb1ac.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p04-4990d689ef.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p05-24a216e330.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q01-p06-9c9677d1c8.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 The plantain lily, Hosta plantaginea, as shown in Fig. 1.1, is a flowering plant found in Asia.\nSIHT\nNI\nETIRW\nTON\nOD\nleaf with\nstomata on\nthe lower\nNIGRAM\nepidermis\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 1.1 SIHT\nNI The hydrostatic pressure increases inside guard cells when water enters the guard cells down a\nETIRW\nwater potential gradient. This affects the width of the stomata.\nTON\nFig. 1.2 shows 1 open stoma from the plantain lily surrounded by 2 guard cells.\nOD\nstoma\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nguard cells\nFig. 1.2\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nA student investigated the effect of different sucrose solutions on the width of stomata in the leaves\nSIHT\nof plantain lily.\nNI\nETIRW\nThe student:\nTON • prepared 5 microscope slides as shown in Table 1.1\nOD\nTable 1.1\nmicroscope slide number liquid added to microscope slide\nNIGRAM\n1 large drop of distilled water\n2 large drop of 2.5% sucrose solution\nSIHT\n3 large drop of 5.0% sucrose solution\nNI\nETIRW 4 large drop of 10.0% sucrose solution\n5 large drop of 20.0% sucrose solution\nTON\nOD • removed the lower epidermis from a leaf of plantain lily\n• immersed 1 small piece of lower epidermis in the drop on microscope slide 1\n• placed a cover slip over the piece of lower epidermis\n• observed the stomata using a light microscope fitted with an eyepiece graticule\nNIGRAM • repeated the steps for microscope slides 2, 3, 4 and 5.\n(a) (i) State the independent and dependent variables in the investigation.\nSIHT\nNI independent .......................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\ndependent ..........................................................................................................................\nOD\n...........................................................................................................................................\n[2]\nNIGRAM\ncm3 of 2.5%, 5.0% and 10.0% sucrose solutions from a 20.0% (ii) The student prepared 20\nw/v stock solution.\nSIHT\nComplete the description of the student’s dilution method for the 2.5% and 5.0% sucrose\nNI\nsolutions by writing the correct volumes in the following sentences.\nETIRW\nm3 m3 of 20.0% sucrose stock solution with ..................c. The student mixed ...................c\nTON\nof distilled water to produce a 2.5% sucrose solution.\nOD\nm3 m3 of 20.0% sucrose stock solution with ..................c. The student mixed ...................c\nof distilled water to produce a 5.0% sucrose solution.\n[2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(iii) The student observed stomata on the microscope slides using the high‑power objective\nSIHT\nlens.\nNI\nETIRW\nOutline a method the student could use to investigate the effect of the sucrose solutions\nin Table 1.1",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-52-q01-4f044e454f.pdf"
    },
    {
      "id": "9700-2025-on-52-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_52.pdf?download=true",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-52-q02-p01-3498833a12.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q02-p02-a826755602.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n2 Scientists investigated the ef fect of abscisic acid (ABA) on reducing water loss in plants. The\nSIHT\nscientists predicted that ABA, as well as stimulating stomatal closure, also reduces water loss in\nplants in other ways. NI\nETIRW\nThe scientists used a mutant variety of thale cress, Arabidopsis thaliana, that has stomata that\ndo not respond to ABA. When ABA is present, the stomata of the mutant variety remain open.\nTON\nThe scientists sprayed dif ferent concentrations of ABA on the leaves of the mutant variety and OD\nmeasured the transpiration rates of the plants.\nThe results are shown in Fig. 2.1.\nNIGRAM\n2.5\nError bars show ± 1 standard error (SE).\ndm−3 0 μmol ABA = distilled water\nSIHT\nNI\n2.0\nETIRW\nTON\nOD\n1.5\nmean\ntranspiration\nrate\nm−2 s−1 / mmol NIGRAM\n1.0\nSIHT\nNI\nETIRW 0.5\nTON\nOD\n0.0\n0 10 50 100 400\nABA concentration\nNIGRAM dm−3 / μmol\nFig. 2.1\nSIHT\ndm–3 of ABA for (a) Calculate the percentage decrease in transpiration rate from 0 to 400 µmol NI\nthe mutant variety. ETIRW\nShow your working.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\npercentage decrease = ...............................................................\n[2] TON\nNIGRAM\n(b) With reference to Fig. 2.1, suggest conclusions that can be made about the effect of different\nSIHT\nconcentrations of ABA on the transpiration rates of the mutant variety.\nNI\nETIRW\nState the evidence that supports your conclusions.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n............................................................................................................................................. [2]\nETIRW\n(c) The scientists used a mutant variety of thale cress that has stomata that do not respond\nTON\nto ABA.\nOD\nThe scientists concluded that ABA reduces water loss in thale cress by ways other than\nstomatal closure.\nNIGRAM Suggest additional information that is required to increase the confidence in this conclusion.\n...................................................................................................................................................\nSIHT\n..............................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-52-q02-a7c4b3b0f8.pdf"
    },
    {
      "id": "9700-2025-on-52-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "52",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 5,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_52.pdf?download=true",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-52-q03-p01-56933dd9ae.png",
        "9700-experimental-skills/assets/9700-2025-on-52-q03-p02-ff93111cd1.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-52-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n3 Seeds of the common stork’s‑bill, Erodium cicutarium, are shown in Fig. 3.1.\nSIHT\nNI The awn is an extension on the E. cicutarium seed. The seed uses the awn to push the seed head\nETIRW\ninto the soil to allow the seed to germinate. The awn changes shape when the humidity changes.\nTON Humidity is the concentration of water vapour in the air . The higher the humidity , the higher the\nconcentration of water vapour in the air.\nOD\nawn in low humidity\nNIGRAM\nSIHT\nNI\nawn in high humidity\nETIRW\nTON\nseed heads\nOD\nFig. 3.1\nNIGRAM\nA student investigated the effect of humidity on the appearance of the awn.\nSIHT\nThe student set up the apparatus as shown in Fig. 3.2.\nNI\nWater can be added to the filter paper to change the humidity in the beaker surrounding the seed. ETIRW\nIncreasing the volume of water added to the filter paper increases the humidity.\nTON\ncm3 beaker 250\nOD\nNIGRAM\nawn seed head\nSIHT\nNI\nmaterial to hold ETIRW\nseed in place\nfilter paper\nTON\nPetri dish\nOD\nFig. 3.2\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) State two variables the student would need to standardise for the investigation using the\nSIHT\napparatus shown in Fig. 3.2.\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [2]\nSIHT (b) Outline a method the student could use to measure the effect of different levels of humidity on\nthe appearance of the awn, using the apparatus in Fig. 3.2.\nNI\nETIRW\nDo not include standardised variables from (a) or a risk assessment.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n..................................................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-52-q03-cedfdabffb.pdf"
    },
    {
      "id": "9700-2025-on-53-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 18,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_53.pdf?download=true",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p01-b29a6784a8.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p02-80ea879af1.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p03-e6d11c0399.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p04-b55ae1e3b3.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p05-e768cd6c38.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q01-p06-d4b8af5153.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Juglone is a chemical produced by plants in the genus Juglans, which includes dif ferent species\nSIHT\nof walnut tree.\nNI\nETIRW\nFig. 1.1 shows a black walnut tree.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFig. 1.1\nNIGRAM\nJuglone produced by a black walnut tree can diffuse into the soil surrounding the tree.\nGroups of scientists have observed that juglone affects different species of plant growing close to SIHT\nblack walnut trees.\nNI\nETIRW\nJuglone can reduce the percentage of seeds that germinate, delay the start of germination or\nreduce growth. In some studies, juglone has no effect on germination or growth.\nTON\nA student planned to investigate the effect of juglone on the cucumber plant, Cucumis sativus. OD\nIn the investigation, the student:\n• placed some damp soil containing juglone in a tray NIGRAM\n• added some cucumber seeds to the soil\n• placed the container in a suitable environment for germination and growth.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nFig. 1.2 shows cucumber seedlings (young plants) a few days after germination.\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 1.2\nTON\nThe student planned to compare the effect of exposing cucumber seeds to:\nOD\n10–3 dm–3, • juglone solution with a concentration of 1.0 × mol which is a concentration that has\nbeen measured in soils surrounding black walnut trees\n• a control treatment.\nNIGRAM\nThe student planned to determine:\nSIHT • the percentage of cucumber seeds that germinated in soil\n• the growth of each cucumber seedling 10 days after the appearance of the seedling above\nNI\nthe soil. ETIRW\n(a) Identify the independent variable in this investigation. TON\nOD\n............................................................................................................................................. [1]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Juglone solutions are prepared using a mixture of solvents.\nSIHT\nNI The student was provided with:\nETIRW\ndm–3 • a juglone solution with a concentration of 0.1 mol\nTON • the mixture of solvents.\nOD\ncm3 Using the solution provided, the student prepared a juglone solution with a volume of 500\n10–3 dm–3. mol and a concentration of 1.0 ×\n–3 dm–3 mol solution using measuring The student carried out 1 dilution to prepare the 1.0 × 10\nNIGRAM\ncylinders.\ncm3 of (i) State the volumes that the student used to carry out the dilution to produce 500 SIHT\n10–3 dm–3 mol juglone solution. 1.0 ×\nNI\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [2]\nOD\n(ii) Outline one improvement the student could make to reduce the percentage error while\ncarrying out the dilution.\nNIGRAM .............................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-53-q01-5ad4032bda.pdf"
    },
    {
      "id": "9700-2025-on-53-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_53.pdf?download=true",
      "source_pages": [
        8,
        9,
        10
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-53-q02-p01-a96d3639bb.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q02-p02-1ed631719d.png",
        "9700-experimental-skills/assets/9700-2025-on-53-q02-p03-b833efb9f2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n–) 2 Compounds containing nitrite ions (NO are present in many foods eaten by humans.\nSIHT 2\nNI Scientists tested the ef fect of different concentrations of nitrite ions on the population growth of\nETIRW\n5 species of bacterium. All 5 species are pathogenic and can infect the human digestive system.\nTON For each species of bacterium, the scientists placed suspensions of the bacteria in a microwell\nplate. A microwell plate is a plastic plate containing 96 wells, which are similar to small test-tubes.\nOD\nIn each well, the scientists added:\n• a suspension of the bacterial species\nNIGRAM\n• a solution containing nitrite ions\n• nutrient broth.\nSIHT\nThe scientists incubated the microwell plate at 37 °C for 24 hours. After 24 hours, the scientists\nNI\nestimated the number of bacteria in each well by measuring optical density , using a microwell\nETIRW\nplate reader.\nTON\nFor each species of bacterium, the scientists repeated the experiment with several dif ferent\nOD concentrations of solution containing nitrite ions.\nThe scientists determined the lowest nitrite concentration at which no bacterial population growth\nhad occurred after 24 hours.\nNIGRAM\n(a) The scientists standardised temperature, pH (pH 4.8) and time in the investigation.\nSIHT\nState two other variables that the scientists should standardise in this investigation.\nNI\nETIRW 1 ................................................................................................................................................\n2 ................................................................................................................................................ TON\n[2]\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The results of the investigation are shown in Fig. 2.1.\nSIHT\n6.0 NI\nETIRW\n5.0\nTON\n4.0 OD lowest nitrite concentration\nat which no bacterial\n3.0 population growth occurred\nafter 24 hours\nNIGRAM cm–3 μmol / 2.0\n1.0\nSIHT\nNI\n0.0\nETIRW Yersinia Salmonella Salmonella Shigella Escherichia\nenterocolitica enteritidis typhimurium sonnei coli\nTON\nspecies of bacterium\nOD\nFig. 2.1\n(i) State the type of data represented by the species of bacterium and the lowest nitrite\nNIGRAM\nconcentration in Fig. 2.1.\nspecies of bacterium .........................................................................................................\nSIHT\nNI lowest nitrite concentration ................................................................................................\nETIRW\n[2]\nTON (ii) The 2 species of Salmonella shown in Fig. 2.1 are closely related.\nOD\nCalculate the percentage dif ference between the results shown in Fig. 2.1 for the\n2 species of Salmonella.\nShow your working.\nNIGRAM\nSIHT\nNI\nETIRW\n.............................................................%\nTON\n[3]\nOD\n(iii) Use Fig. 2.1 to state:\n• the species of bacterium that is most affected by nitrite ions\n• NIGRAM t",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-53-q02-9b5dbee3c8.pdf"
    },
    {
      "id": "9700-2025-on-54-q01",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "54",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_54.pdf?download=true",
      "source_pages": [
        2,
        3,
        4
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-54-q01-p01-c122474446.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q01-p02-cad41c78d9.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q01-p03-72f29259b6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-54-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Beta vulgaris vulgaris, as shown in Fig. 1.1, is a food crop with a swollen, edible root known as\nSIHT\nbeetroot.\nNI\nETIRW\nTON\nOD\nbeetroot\nNIGRAM with outer skin\nleaves SIHT\nNI\nETIRW\nmain taproot\nTON\nOD\nNIGRAM\nFig. 1.1\nA student determined the concentration of a sucrose solution that has the same water potential as\nSIHT\nthe cells of the beetroot.\nNI\nETIRW dm–3 sucrose to prepare a range of The student used distilled water and a stock solution of 2.0 mol\nsucrose solutions.\nTON\ncm3. Each sucrose solution was prepared using proportional dilution and had a final volume of 30\nOD\n(a) Complete Table 1.1 to show how the concentrations were made.\nTable 1.1\nNIGRAM\ndm–3 final concentration of volume of volume of 2.0 mol\nsucrose solution distilled water stock sucrose solution\nSIHT\ndm–3 cm3 cm3 / mol / /\nNI\nETIRW 2.0 0.0 30.0\nTON\n........................ ........................ ........................\nOD\n........................ ........................ ........................\nNIGRAM\n........................ ........................ ........................\nSIHT\n........................ ........................ ........................\nNI\nETIRW 0.0 30.0 0.0\n[2]\nTON\nNIGRAM\n(b) Describe a method the student could use to collect the data needed to determine the\nSIHT\nconcentration of a sucrose solution that has the same water potential as the cells of the\nNI beetroot.\nETIRW\nThe student was supplied with the sucrose solutions prepared in 1(a) and standard laboratory\nTON equipment.\nOD\nDo not include details of how the student:\n• prepared the sucrose solutions in 1(a)\n• would use the data collected to determine the concentration of a sucrose solution that\nNIGRAM\nhas the same water potential as the cells of the beetroot.\nYour method should be set out in a logical order and be detailed enough to allow another SIHT\nperson to follow it.\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...............................................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-54-q01-200700c3b2.pdf"
    },
    {
      "id": "9700-2025-on-54-q02",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "54",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_54.pdf?download=true",
      "source_pages": [
        6,
        7,
        8,
        9,
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p01-c69a9d8927.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p02-7ba0f5d668.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p03-89116c527e.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p04-cadd656678.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p05-bcb30cb516.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p06-e8a8e1d7ea.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p07-8eac9ab98d.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q02-p08-f8b58fc96f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-54-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n2 The buff‑tailed bumblebee, Bombus terrestris, is an insect that uses its tongue to feed on the\nSIHT\nnectar and pollen of flowers.\nNI\nETIRW\nBuff‑tailed bumblebees can feed on the nectar and pollen of flowers by either gripping (holding)\nonto flower petals or by hovering (flying) in front of the flowers.\nTON\nFig. 2.1 shows the buff‑tailed bumblebee.\nOD\n10 mm\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nclaw\nNIGRAM\nFig. 2.1\nSIHT\nThe buff‑tailed bumblebee has adaptations, such as claws, to help it to grip on to flower petals to\nNI\nobtain nectar and pollen.\nETIRW\nFig. 2.2 shows the buff‑tailed bumblebee using its claws to grip on to the flower petals as the bee\nTON\nfeeds.\nOD\nNIGRAM\nflower\npetal\nSIHT\nNI\nETIRW\nTON\nOD\nclaw used to\ngrip the flower\npetal\nNIGRAM\nSIHT\nNI\nETIRW\nFig. 2.2\nTON\nNIGRAM\nFig. 2.3 is a magnified image of the claw of a bee.\nSIHT\nNI\nETIRW\nTON\nOD\nclaws NIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT Fig. 2.3\nNI\n(a) Suggest a method that can be used to measure the image length of the longest part of the\nETIRW\ncurved claw in Fig. 2.3.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [1]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The surfaces of flower petals of different plant species have different textures.\nSIHT\nNI Fig. 2.4 shows a scanning electron micrograph of a rough surface of a flower petal.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFig. 2.4\nFig. 2.5 shows a scanning electron micrograph of a smoother surface of a flower petal. SIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFig. 2.5\nNI\nETIRW\nTON\nNIGRAM\nA scientist made an artificial flower apparatus, as shown in Fig. 2.6.\nSIHT\nNI\nETIRW\n60 mm diameter disc to\nrepresent a flower petal\nTON\nOD\nsmall drop of concentrated\nsucrose solution to\nrepresent nectar\nNIGRAM\nsupport\nSIHT\nNI\nETIRW\nFig. 2.6\nTON\nTo model flower petals with different levels of roughness, the scientist used different discs.\nOD\nEach disc was made using particles of different diameters attached to the surface of the disc,\nas shown in Table 2.1.\nTable 2.1 NIGRAM\nlevel of mean particle diameter\nSIHT\nµµm roughness on the disc /\nNI\n5 smooth ETIRW\n9\nTON\n12\nOD\n16\n30\nNIGRAM rough 53\n(i) The diameter of each disc was standardised as 60 mm.\nSIHT\nNI Identify one other variable the scientist should standardise when making the discs.\nETIRW\n...........................................................................................................................................\nTON\n.................................",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-54-q02-15837e106c.pdf"
    },
    {
      "id": "9700-2025-on-54-q03",
      "subject": "9700",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 5,
      "variant": "54",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2025-oct-nov/9700_w25_qp_54.pdf?download=true",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2025-Oct-Nov/qp/9700_w25_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2025-on-54-q03-p01-f291c1666b.png",
        "9700-experimental-skills/assets/9700-2025-on-54-q03-p02-ec2052558a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2025-on-54-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n3 (a) Pectin is a polysaccharide found in plant cell walls. Pectinase is an enzyme that hydrolyses\nSIHT\npectin.\nNI\nETIRW\nA student investigated the use of pectinase to extract more juice from different types of fruit.\nTON The student:\nOD\n• removed the outer skin and cut each type of fruit into small pieces\ncm3 • placed each type of fruit into a separate beaker with 5 of 1% pectinase solution\n• incubated the mixture at 60 °C for 10 minutes\n• passed the contents of the beaker through a filter funnel into a measuring cylinder\nNIGRAM\n• measured the volume of juice extracted from each type of fruit.\nTable 3.1 shows the results. SIHT\nNI\nTable 3.1\nETIRW\ncm3 volume of juice extracted / TON\ntype of fruit\nmean of OD\nsample 1 sample 2 sample 3\nsamples\napple 16 17 17\n........................\nNIGRAM\norange 30 37 33\n........................\nSIHT\nNI pineapple 41 44 45\n........................ ETIRW\ngrapes 17 15 11\nTON ........................\nOD\n(i) Complete Table 3.1 by calculating the means. [1]\n(ii) The independent variable is the type of fruit and the dependent variable is the volume of\nNIGRAM juice extracted.\nComplete Table 3.2 to show the type of variable and the type of data.\nSIHT\nTable 3.2 NI\nETIRW\ntype of variable type of data\nTON\ntype of fruit\nOD\nvolume of juice extracted\n[2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(iii) State two changes the student should make to their method to improve the quality of\nSIHT\ntheir results.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\nSIHT (b) The student wanted to determine the quantity of pectin in the cell walls of different types of\nfruit.\nNI\nETIRW\nSuggest why the student could not use the experiment in 3(a) to determine the quantity of\npectin in the cell walls of different types of fruit.\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n........",
      "pdf_path": "9700-experimental-skills/print/9700-2025-on-54-q03-b46d95cb20.pdf"
    },
    {
      "id": "9700-2026-m-42-q01",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-m-42-q01-p01-281789bd62.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-m-42-q01-p02-5b415d4b13.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q01",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Batrachochytrium dendrobatidis is a fungus that lives in a warm and damp environment, with\nSIHT\noptimal temperatures for growth of between 17 °C and 25 °C.\nNI\nETIRW\nB. dendrobatidis causes a disease in the skin of frogs, which can be fatal in up to 100% of\ninfected frogs. The disease is known as chytridiomycosis.\nTON\nThe mountain chicken frog, Leptodactylus fallax, is a large species of frog and is critically\nOD\nendangered. The frog is native to the Caribbean islands of Dominica and Montserrat.\nPopulations of L. fallax were believed to be in some decline in the years before 1995, but\nbetween 1995 and 2011 the total number of frogs decreased by more than 88%.\nNIGRAM\n• In 1995, ash and acid rain from a volcanic eruption killed a large number of frogs\non Montserrat. SIHT\n• In 2002, chytridiomycosis arrived on Dominica where it caused a severe decrease in the\nNI\nnumber of frogs within 18 months.\nETIRW\n• In 2009, chytridiomycosis arrived on Montserrat where it caused a severe decrease in the\nnumber of frogs within 12 months.\nTON\nOD Figure 1.1 shows a mountain chicken frog.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 1.1\n(a) Suggest reasons that may have contributed to L. fallax becoming critically endangered,\nother than the volcanic eruption and the effects of chytridiomycosis.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Captive breeding was used to increase the number of L. fallax individuals. Releasing\nSIHT\nthese frogs into the wild was not successful because B. dendrobatidis was still present on\nNI Dominica and Montserrat.\nETIRW\nConservationists have trialled the re‑introduction of L. fallax into safer, semi‑wild areas on\nTON Dominica and Montserrat. Here, conditions have been created that protect L. fallax from\ninfection by B. dendrobatidis but still allow the frog to survive and reproduce.\nOD\nSuggest measur",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2026-m-42-q01-3e9eac3ae3.pdf"
    },
    {
      "id": "9700-2026-m-42-q02",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-m-42-q02-p01-77ac99e43c.png",
        "9700-topic-16-inheritance/assets/9700-2026-m-42-q02-p02-a5efe367b5.png",
        "9700-topic-16-inheritance/assets/9700-2026-m-42-q02-p03-75f2b8ea00.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 The Labrador retriever is a breed of dog that, as an adult, shows continuous variation in\nSIHT\nbody mass.\nNI\nETIRW\nFigure 2.1 shows a Labrador retriever.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD Figure 2.1\n(a) (i) Explain what is meant by continuous variation.\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [1] NI\nETIRW\n(ii) Describe the causes of continuous variation in the body mass of adult mammals, such\nas Labrador retriever dogs. TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n..................................................................................................................................... [3] OD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Scientists have discovered that the gene POMC helps to regulate the body mass of\nSIHT\nLabrador retrievers.\nNI\nETIRW\nApproximately 25% of Labrador retrievers have at least one mutant POMC allele.\nTON Figure 2.2 shows the mean body mass of Labrador retrievers with each of the three possible\ngenotypes for this gene:\nOD\n• two normal POMC alleles (homozygous normal)\n• one normal POMC allele and one mutant POMC allele (heterozygous)\n• two mutant POMC alleles (homozygous mutant).\nNIGRAM\nEach error bar shows the mean ±1 standard deviation.\nSIHT\n34\nNI\nETIRW\n33\nTON\n32\nOD\nmean\nbody mass 31\nkg /\n30\nNIGRAM\n29\nSIHT\n28 NI\nhomozygous heterozygous homozygous\nETIRW\nnormal mutant\ngenotype\nTON\nOD\nFigure 2.2\n(i) With reference to Figure 2.2, describe the effect of the mutant POMC allele on the body\nmass of Labrador retrievers.\nNIGRAM\n...........................................................................................................................................\nSIHT\n.....................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-m-42-q02-943b8bbbfc.pdf"
    },
    {
      "id": "9700-2026-m-42-q03",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-m-42-q03-p01-ae045dad71.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-m-42-q03-p02-128435b963.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-m-42-q03-p03-516554045e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44)\n3 Huntington’s disease is caused by a mutation in a gene coding for a protein that has a number of\nSIHT\ncellular functions.\nNI\nETIRW\n(a) Identify the gene and state the name of the protein that are affected by the mutation that\ncauses Huntington’s disease.\nTON\ngene ...............................................................\nOD\nprotein ...............................................................\n[2]\nNIGRAM\n(b) In many countries, a genetic screening test for Huntington’s disease is available.\nSIHT\nAdults over the age of 18 years who are considered to be at risk of developing the disease\nNI\nare offered the genetic screening test.\nETIRW\n(i) Suggest which adults are considered to be at risk of developing Huntington’s disease.\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ..................................................................................................................................... [1]\n(ii) Suggest and explain why the genetic screening test for Huntington’s disease is only\nSIHT\noffered to adults who are at least 18 years old.\nNI\nETIRW ...........................................................................................................................................\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(c) Part of the gene associated with Huntington’s disease has a repeated DNA triplet nucleotide\nSIHT\nsequence of CAG. There are 35 or fewer CAG repeats in the normal allele. Alleles with 40 or\nNI more CAG repeats are expected to result in the development of Huntington’s disease.\nETIRW\nThe genetic screening test for Huntington’s disease includes a polymerase chain reaction\nTON (PCR). This results in amplification of the gene that is associated with Huntington’s disease.\nOD\nThe DNA fragments produced by PCR are then observed using agarose gel electrophoresis.\nThe genetic screening test results of four 18‑year‑old adults, A, B, C and D, who had no\nsymptoms of Huntington’s disease are shown in Figure 3.1. Control DNA containing the\nNIGRAM\nfragments from amplification of alleles with 35 and 40 CA",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-m-42-q03-a13aa46871.pdf"
    },
    {
      "id": "9700-2026-m-42-q04",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2026-m-42-q04-p01-76915f289a.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-m-42-q04-p02-a1dc70410b.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-m-42-q04-p03-53d9cbca02.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 (a) Outline the theory of evolution.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [3]\nNI\nETIRW\n(b) Approximately 95% of bird species living today belong to an evolutionary group known\nTON\nas Neoaves.\nOD\nBird species in the Neoaves group are believed to have evolved from a common ancestor\nthat existed around 67 million years ago. The Neoaves that exist today can be divided into\nfour groups:\nNIGRAM\n• Mirandornithes, referred to as group 1\n• Columbimorphae, referred to as group 2\nSIHT • Otidimorphae, referred to as group 3\n• all other Neoaves, referred to as group 4.\nNI\nETIRW\nIn 2014, scientists used DNA sequence data from 48 bird species to produce an evolutionary\n(phylogenetic) tree for the Neoaves. This is shown in Figure 4.1. The evolutionary tree\nTON\nshows possible evolutionary relationships between the four groups of species within\nOD the Neoaves.\n67 million years ago Neoaves\nNIGRAM\nnot to scale\nSIHT\ngroup 1 group 2 group 3 group 4 present day\nNI\nETIRW Figure 4.1\nIn 2024, scientists used DNA sequence data from 363 bird species to produce a new TON\nevolutionary tree for the four groups of species within the Neoaves. This is shown in\nOD\nFigure 4.2.\n67 million years ago Neoaves\nNIGRAM\nnot to scale\nSIHT\ngroup 1 group 2 group 3 group 4 present day\nNI\nETIRW\nFigure 4.2\nTON\nNIGRAM\n(i) Describe the similarities and differences between the evolutionary relationships shown\nSIHT\nin Figure 4.1 and Figure 4.2.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n......",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2026-m-42-q04-1819941baf.pdf"
    },
    {
      "id": "9700-2026-m-42-q05",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        13
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-m-42-q05-p01-dac1d22a7b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q05",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:4)(cid:44)\n5 (a) The sarcomere is composed of many structures and is the functional unit in the contraction\nSIHT\nof striated muscle.\nNI\nETIRW\nComplete Table 5.1 to match each description with the correct region of the sarcomere.\nTON Table 5.1\nOD\nregion of\ndescription\nsarcomere\nNIGRAM only myosin present\n.................................\nSIHT actin and myosin present\n.................................\nNI\nETIRW\nonly actin present\n.................................\nTON\nattachment for myosin\n................................. OD\nattachment for actin\n.................................\n[5]\nNIGRAM\nSIHT (b) Hypocalcaemia is a condition where the blood calcium ion concentration is below the\nnormal range.\nNI\nETIRW\nExplain the effect of a low blood calcium ion concentration on the functioning of the\nsarcomeres of striated muscle fibres.\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n[Total: 9]\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-m-42-q05-02936a4ff3.pdf"
    },
    {
      "id": "9700-2026-m-42-q06",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-m-42-q06-p01-82a305370a.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-m-42-q06-p02-d8d6f2ed3d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q06",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 (a) Protein production in a prokaryote can be genetically controlled.\nSIHT\nNI One example of the control of protein production in a prokaryote is the lac operon. The lac\nETIRW\noperon is functional when glucose is absent and lactose is present.\nTON Complete Table 6.1, which shows information about the four genes associated with the\nlac operon.\nOD\nTable 6.1\nNIGRAM name of protein function of protein name of\ntype of gene\nproduced produced gene\nSIHT\nbinds to operator to\nNI\nprevent RNA polymerase ETIRW\nrepressor protein lacI\nfrom binding to .................................\nthe promoter\nTON\nOD\n........................................\nlacZ structural β‑galactosidase\nNIGRAM ........................................\n........................................\nSIHT\nNI\nETIRW\nmembrane transport\nstructural lactose permease protein to increase\n.................. TON\ncellular uptake of lactose\nOD\nacetylates galactoside NIGRAM\nlacA structural transacetylase sugar residues to\ndetoxify the cell\nSIHT\nNI\nETIRW\n[3]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Insulin can be produced as a recombinant human protein using genetically engineered\nSIHT\nEscherichia coli bacteria. Recombinant insulin can be used to treat type 1 diabetes.\nNI\nETIRW\nDescribe the advantages of using recombinant insulin, instead of insulin from animal\nsources, to treat type 1 diabetes.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-m-42-q06-1db3d93aa6.pdf"
    },
    {
      "id": "9700-2026-m-42-q07",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2026-m-42-q07-p01-3f65230b66.png",
        "9700-topic-14-homeostasis/assets/9700-2026-m-42-q07-p02-84aff2e778.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q07",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 (a) The process of ultrafiltration occurs in the Bowman’s capsule region of a kidney nephron.\nSIHT\nNI Figure 7.1 is a diagram representing part of a Bowman’s capsule.\nETIRW\nTON lumen of\nA\nBowman’s\nOD\ncapsule\nB\nNIGRAM\nSIHT\nlumen of\nNI capillary in\nETIRW\nglomerulus\nC\nTON\nFigure 7.1\nOD\nState the name of cell A, structure B and cell C.\nA ...............................................................\nNIGRAM\nB ...............................................................\nSIHT\nC ...............................................................\nNI\n[3]\nETIRW\nTON\n(b) The kidney function of a person can be assessed by measuring the glomerular filtration rate\nOD (GFR). The GFR is the rate at which fluid filters from the blood in the glomerulus into the\nBowman’s capsule.\nSuggest factors that are likely to decrease the GFR.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The kidney has an important role in homeostasis.\nSIHT\nNI Explain what is meant by the term homeostasis and describe the principles of homeostasis\nETIRW\nin mammals.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n....................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2026-m-42-q07-0320241fd5.pdf"
    },
    {
      "id": "9700-2026-m-42-q08",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-m-42-q08-p01-10f0fc898d.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-m-42-q08-p02-e347abff15.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q08",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 (a) Figure 8.1 shows three processes that require energy in living organisms.\nSIHT\nNI\nmovement of ETIRW\nsubstance X A\nTON\nhigh concentration low concentration\nOD of substance X of substance X\nmembrane\nB\nNIGRAM\nSIHT\nmonomers\nNI\nETIRW\nC TON\nlight electrons chlorophyll\nOD\nFigure 8.1\nNIGRAM\nState the name of each type of process shown in A, B and C.\nSIHT\nA ...............................................................\nNI\nETIRW\nB ...............................................................\nTON\nC ...............................................................\n[3] OD\n(b) ATP is produced in mitochondria by a process called chemiosmosis.\nNIGRAM\nOutline the process of chemiosmosis in mitochondria.\nSIHT\n...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Different respiratory substrates have different energy values.\nSIHT\nNI Table 8.1 shows the approximate energy values of carbohydrate and lipid.\nETIRW\nTable 8.1\nTON\nOD respiratory approximate energy value\ng–1 substrate / kJ\ncarbohydrate 17.0\nNIGRAM lipid 37.8\nCalculate the mass of carbohydrate needed to release the same quantity of energy as 100 g SIHT\nof lipid.\nNI\nETIRW\nGive your answer to one decimal place.\nTON\nOD\nmass = ....................................................... g [2]\n[Total: 7]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-m-42-q08-844030a2ae.pdf"
    },
    {
      "id": "9700-2026-m-42-q09",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        20,
        21,
        22
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-m-42-q09-p01-3b474cef8a.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-m-42-q09-p02-a8670c3f2d.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-m-42-q09-p03-1fad2df524.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 (a) Nitric oxide is a cell‑signalling molecule that has a role in a number of physiological\nSIHT\nprocesses in the body. It is also used as a therapeutic drug. Nitric oxide is known to have\nNI harmful effects in body cells if conditions are not controlled.\nETIRW\nScientists investigated the effect of nitric oxide on the use of oxygen by kidney cells grown in\nTON a culture medium.\nOD\nThe scientists recorded the oxygen concentration in the culture medium over a period of\n24 minutes. During this 24‑minute period, a small volume of nitric oxide was added to the\nculture medium on three occasions.\nNIGRAM\nThe results are shown in Figure 9.1.\n16\nSIHT\nnitric oxide\nadded NI\nETIRW\n12\nTON\nnitric oxide\nadded OD\noxygen concentration\nin culture medium 8\n/ arbitrary units nitric oxide\nNIGRAM\nadded\nSIHT 4\nNI\nETIRW\n0 TON\n0 4 8 12 16 24 20\nOD\ntime / minutes\nFigure 9.1\nNIGRAM\nDescribe the results of the investigation and suggest an explanation for these results.\nSIHT ...................................................................................................................................................\nNI\n................................................................................................................................................... ETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [4]\nTON\nNIGRAM\n(b) Some organisms are able to carry out respiration in anaerobic conditions.\nSIHT\nNI Describe the differences in respiration in anaerobic conditions between mammalian muscle\nETIRW\ncells and yeast cells.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n......................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-m-42-q09-d06a715f6a.pdf"
    },
    {
      "id": "9700-2026-m-42-q10",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        23,
        24,
        25
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-m-42-q10-p01-f76e01d58c.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-m-42-q10-p02-3da951b734.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-m-42-q10-p03-7e0efb2618.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-42-q10",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:4)(cid:44)\n10 (a) Stomata have an important role in photosynthesis and transpiration.\nSIHT\nNI Figure 10.1 shows the formula for calculating the stomatal index for an area of leaf.\nETIRW\nnumber of stomata\nstomatal index = × 100\nnumber of stomata + number of epidermal cells TON\nFigure 10.1 OD\n(i) Table 10.1 shows the stomatal index of the lower epidermis of the leaves of some\ncrop plants.\nNIGRAM\nComplete Table 10.1 for cotton and tomato using the formula shown in Figure 10.1.\nSIHT\nTable 10.1\nNI\nETIRW\nnumber of number of\ncrop stomata per epidermal cells stomatal index\nTON unit area per unit area\nOD\ncotton 6 15.8\n..........................\ngroundnut 9 36 20.0\nNIGRAM\nmung bean 5 12 29.4\nSIHT\nsorghum 8 20 28.6 NI\nETIRW\ntomato 7 25\n.......................... TON\nOD [2]\n(ii) The Russian wormwood, Artemisia gmelinii, has a stomatal index of 13.8 for its lower\nleaf surface and 1.7 for its upper leaf surface.\nNIGRAM\nExplain the advantages to the plant for this difference in stomatal index values.\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) Figure 10.2 is a graph showing changes over a 24‑hour period in the carbon dioxide\nSIHT\nconcentration of the air in a glasshouse containing plants. The glasshouse did not have\nNI artificial lighting so the only source of light for the plants was sunlight.\nETIRW\nThe carbon dioxide concentration in the external atmosphere over the 24‑hour period is\nTON also shown.\nOD\nThe air in the glasshouse was not completely sealed off from the external atmosphere so\nany difference in the carbon dioxide concentration between the glasshouse and external\natmosphere will gradually decrease unless there are processes to maintain the difference.\nNIGRAM\n0.08\nSIHT\nglasshouse\nNI\n0.06 ETIRW\nTON\nexternal atmosphere percentage\nOD\ncarbon dioxide 0.04\nconcentration\nNIGRAM\n0.02\nSIHT\nNI\nETIRW\n0.00\n06:00 10:00 14:00 18:00 22:00 02:00 06:00\nTON\ntime of day\nOD\nFigure 10.2\nSuggest how different limiting factors affect the rate of photosynthesis of the plants in the\nNIGRAM\nglasshouse over the 24‑hour period shown in Figure 10.2.\n.......................................................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-m-42-q10-9ded23715d.pdf"
    },
    {
      "id": "9700-2026-m-52-q01",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-52.pdf?download=1",
      "source_pages": [
        2,
        3,
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-m-52-q01-p01-c22ea083ce.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q01-p02-91f17408a0.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q01-p03-382df7276e.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q01-p04-e1ffb39592.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q01-p05-299831207e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 Blood pressure in the arteries changes during the cardiac cycle. The highest blood pressure\nSIHT\noccurs during ventricular systole when the ventricles contract and push blood out of the heart.\nNI The lowest blood pressure occurs during diastole when the atria and the ventricles are relaxed.\nETIRW\nBlood pressure is measured as two values:\nTON\n• systolic blood pressure\nOD\n• diastolic blood pressure.\nA blood pressure monitor, as shown in Figure 1.1, is used to measure the blood pressure in the\nbrachial artery in the arm.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nblood pressure cuff\ndigital display\nNIGRAM\nair tube\nSIHT\nNI\nETIRW\nTON\nFigure 1.1\nOD\nSome scientists investigated the effect of a 12-week exercise programme on the blood pressure\nof a group of 18 people. The people were aged from 40 years to 64 years, did not exercise\nregularly and had jobs that mainly involved sitting.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nThe exercise programme consisted of a sequence of three activities, repeated three times a\nSIHT\nweek for 12 weeks.\nNI\nETIRW\nThe three activities in each sequence were:\nTON • walking for 5 minutes\n• running slowly for 30 minutes\nOD\n• walking for 5 minutes.\n(a) The time spent walking and running, the number of repeats and the 12-week period were\nall standardised.\nNIGRAM\nIdentify two other variables in this exercise programme that the scientists\nshould standardise. SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ............................................................................................................................................. [2]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) On the morning before the start of the 12-week exercise programme, blood pressure\nSIHT\nmeasurements were taken for the group of 18 people, as shown in Figure 1.2. Every blood\nNI pressure measurement included a systolic and a diastolic blood pressure reading.\nETIRW\n1 Resting blood pressure was measured while each person was sitting.\nTON\n2 Each person then carried out a walking test on a treadmill for 15 minutes. The OD\ngradient of the treadmill was increased every minute.\nNIGRAM\ntreadmill\n(walking machine)\nSIHT\nNI\nETIRW\n3 After the walking test, the mean blood pressure of each person was determined from TON\nblood pressure measure",
      "pdf_path": "9700-experimental-skills/print/9700-2026-m-52-q01-dcf1dac79c.pdf"
    },
    {
      "id": "9700-2026-m-52-q02",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-52.pdf?download=1",
      "source_pages": [
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-m-52-q02-p01-1836b04e83.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q02-p02-974b0a7b9d.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q02-p03-81b8feb364.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44)\n2 Daphnia magna is a small invertebrate that is found in many fresh water ecosystems.\nSIHT\nNI Figure 2.1 shows one D. magna, as seen with a light microscope. The body of D. magna is\nETIRW\ntransparent, and the beating of the heart can be easily observed.\nTON\nhead heart\nOD\neggs NIGRAM\nantennae\nSIHT\nNI body\nETIRW\nTON\nOD\n4 mm\nFigure 2.1\nNIGRAM\nA student obtained a sample of pond water containing several D. magna.\nSIHT\nTo measure the time taken for one cardiac cycle in D. magna, the student:\nNI\nETIRW • used a pipette to place one D. magna in the centre of a Petri dish\n• added a small amount of cotton wool to reduce the movement of the D. magna\n• added two drops of pond water to the D. magna TON\n• observed the heart of the D. magna using a light microscope\nOD\n• counted the number of cardiac cycles (heart beats) in one minute.\n(a) In a trial experiment at 20 °C, the student counted 232 cardiac cycles in one minute.\nNIGRAM\nCalculate the time taken for one cardiac cycle.\nGive your answer in seconds and to two significant figures. SIHT\nNI\nETIRW\nTON\nOD\ntime for one cardiac cycle = ...................................................... s [1] NIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) Caffeine is a drug found in tea and coffee.\nSIHT\nNI The student decided to investigate the effect of caffeine concentration on the time taken for\nETIRW\none cardiac cycle in D. magna.\nTON (i) The student was given a 0.5% stock solution of caffeine from which to prepare a range\nof caffeine solutions of known concentrations to use in this investigation.\nOD\ncm3 of each solution. The student made 20\nDescribe a method the student could use to make a proportional dilution of the 0.5%\nNIGRAM\nstock solution of caffeine to obtain a 0.3% solution of caffeine.\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n..................................................................................................................................... [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Describe a method that the student could use to investigate the effect of caffeine\nSIHT\nconcentration on the time taken for one cardiac cycle in D. magna.\nNI\nETIRW\nYour method should be set out in a logical order and be detailed enough to allow\nanother person to follow it.\nTON\nYou can assume that a detailed risk assessment has been prep",
      "pdf_path": "9700-experimental-skills/print/9700-2026-m-52-q02-95bfed1db9.pdf"
    },
    {
      "id": "9700-2026-m-52-q03",
      "subject": "9700",
      "year": 2026,
      "session": "March",
      "session_code": "m",
      "paper": 5,
      "variant": "52",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-march/9700-biology-march-2026-question-paper-52.pdf?download=1",
      "source_pages": [
        10,
        11,
        12,
        13,
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2026-March/qp/9700_m26_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p01-f253b17c8d.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p02-2701696afa.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p03-29c6b0a65e.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p04-9034803820.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p05-2cfc1dff4f.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p06-36a622fe89.png",
        "9700-experimental-skills/assets/9700-2026-m-52-q03-p07-9007643a44.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-m-52-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n3 The Armand Bayou is a slow-flowing river that forms an aquatic ecosystem in Texas,\nSIHT\nNorth America.\nNI\nETIRW\nSome biologists decided to study two invasive alien species of aquatic plants that are commonly\nfound in the Armand Bayou: water hyacinth, Eichhornia crassipes, and malancha, Alternanthera\nTON philoxeroides. Water hyacinth and malancha reduce biodiversity in the Armand Bayou.\nOD\nWater hyacinth is a floating plant that grows on the surface of the river. Water currents and the\nwind can move water hyacinth plants from one place to another.\nMalancha has roots that attach the plant to the soil at the edge of the river. Malancha grows\nNIGRAM\nand reproduces to form dense mats on the river surface, which can trap floating water\nhyacinth plants.\nSIHT\nFigure 3.1 shows water hyacinth and Figure 3.2 shows malancha.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 3.1\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFigure 3.2\nTON\nNIGRAM\nThe biologists carried out a survey at 69 locations in the Armand Bayou. At each location, both\nSIHT\ninvasive alien species were present.\nNI\nETIRW\n(a) The biologists estimated the percentage cover of water hyacinth and the percentage cover\nof malancha at each of the 69 locations. For these two variables, the biologists calculated\nTON Pearson’s linear correlation coefficient, r, as +0.389, which was significant at p < 0.001.\nOD\nComplete the sketch graph in Figure 3.3 to show the relationship between these two\nvariables. Include axis labels in your answer.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT Figure 3.3\n[2]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) The biologists carried out an investigation at 12 of the 69 locations in the Armand Bayou\nSIHT\nwhere both invasive alien species were present. The biologists wanted to test the\nNI hypothesis that:\nETIRW\nThe removal of malancha from the Armand Bayou will reduce the abundance of\nTON water hyacinth.\nOD\nFigure 3.4 shows a map of the Armand Bayou and the surrounding area.\nkey\nN\nNIGRAM\nriver\nSIHT\nwoodland\nNI\nETIRW\nroad\nTON\nOD\nArmand Bayou\nNIGRAM\nSIHT\nNI\nETIRW direction of water\nflow\nTON\n1 km\nOD\nFigure 3.4\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nAt each of the 12 locations, a 3 m × 5 m study area was marked out as shown in Figure 3.5.\nSIHT\nNI\nETIRW\nriver\nm 5\nTON land\nm 1 OD\ncentral strip\nused for\n3 m measurement\nNIGRAM\nSIHT\nNI study area\nETIRW\nedge of\nriver\nTON\nOD\nFigure 3.5\n(i) Describe a method that the biologists could use to select 12 locations at random from\nthe 69 locations used in 3(a).\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-m-52-q03-4c7efbb061.pdf"
    },
    {
      "id": "9700-2026-mj-41-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        3
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-41-q01-p01-05cecf4f3c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q01",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44)\n1 Aerobic respiration in cells proceeds in four stages.\nSIHT\nNI (a) Table 1.1 shows the first three stages of aerobic respiration and the location in the cell where\nETIRW\nglycolysis occurs.\nTON Complete Table 1.1 to show the locations in the cell and the products of the first three stages\nof aerobic respiration.\nOD\nTable 1.1\nNIGRAM stage location in the cell products of this stage\nglycolysis cytoplasm\nSIHT\nNI\nlink reaction ETIRW\nTON\nKrebs cycle\nOD\n[5]\n(b) The fourth stage of aerobic respiration, oxidative phosphorylation, involves the process of NIGRAM\nchemiosmosis. Chemiosmosis also occurs in photosynthesis.\nSIHT\nDuring chemiosmosis, ions move across a membrane down their electrochemical gradient.\nNI\nIdentify: ETIRW\nthe ions that move in chemiosmosis in aerobic respiration and in photosynthesis\nTON\nOD\n...................................................................................................................................................\nthe membrane across which these ions move in photosynthesis\nNIGRAM\n...................................................................................................................................................\nthe process by which these ions move in chemiosmosis.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n[3]\nTON [Total: 8]\nOD\nNIGRAM\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-mj-41-q01-406a887b65.pdf"
    },
    {
      "id": "9700-2026-mj-41-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2026-mj-41-q02-p01-7fbb072b05.png",
        "9700-topic-14-homeostasis/assets/9700-2026-mj-41-q02-p02-9fb9dd666b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q02",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 (a) Homeostasis allows mammals to maintain relatively constant internal conditions despite\nSIHT\nchanges in the external environment. One condition that is controlled is the water potential of\nNI the blood.\nETIRW\n(i) Complete the list to show the receptors, coordination systems, effectors and response\nTON that achieve the homeostatic control of the water potential of the blood. Two of the\nanswers have been started for you.\nOD\nstimulus: decrease in blood water potential\ninitial receptors: .................................................................................................................\nNIGRAM\nnervous system and\ncoordination systems: .......................................................................................................\nSIHT\ndistal convoluted tubule and\nfinal effectors: ....................................................................................................................\nNI\nETIRW\nresponse: ..........................................................................................................................\n[4]\nTON\nOD (ii) Explain why homeostasis of the water potential of the blood is necessary.\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n...........................................................................................................................................\nSIHT\n........................................................................................................................................... NI\nETIRW\n..................................................................................................................................... [2]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\ndm3 day–1. (b) The glomerular filtration rate of the kidneys is 200\nSIHT\ndm–3. NI The mean concentration of glucose in the blood is 5 mmol\nETIRW\n1 mmol of glucose weighs 0.18 g.\nTON\n(i) Calculate the mass of glucose that is filtered from the blood by the kidneys in one day.\nOD\nNIGRAM\nSIHT\n........................................................g [1]\nNI\nETIRW (ii) A healthy person does not lose the mass of glucose calculated in 2(b)(i) in urine each\nday due to the processes occurring in the kidneys.\nTON\nDescribe the processes that occur in the kidneys to prevent the loss of this glucose\nOD\nin urine.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n..............................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2026-mj-41-q02-99bd6cc587.pdf"
    },
    {
      "id": "9700-2026-mj-41-q03",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-41-q03-p01-10916aba53.png",
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-41-q03-p02-9eb9f5def9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q03",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 Stomata have daily rhythms of opening and closing. Stomatal aperture width and the concentration\nSIHT\nof potassium ions in guard cells were measured in the leaves of basil plants, Ocimum basilicum. The\nNI basil plants were grown in conditions of natural light, with optimum temperature and soil moisture.\nETIRW\nFigure 3.1 shows the results obtained between the times of 06:00 and 17:00. Error bars have\nbeen included. TON\nOD\nkey\nmean stomatal aperture width\nconcentration of potassium ions in guard cells\nNIGRAM\n8 5\nSIHT\nNI\n4\nETIRW\n6\nTON\n3 mean concentration of\nOD\nstomatal potassium ions\n4\naperture in guard cells\n/ μm width arbitrary units / 2\nNIGRAM\n2\n1\nSIHT\nNI\nETIRW\n0 0\nlight dark\nTON 04:00 06:00 08:00 10:00 12:00 14:00 16:00 18:00 20:00\nOD time of day\nFigure 3.1\n(a) (i) With reference to Figure 3.1, describe the changes that occur in mean stomatal NIGRAM\naperture width and suggest reasons why these changes benefit basil plants.\nSIHT\n...........................................................................................................................................\nNI\n........................................................................................................................................... ETIRW\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ..................................................................................................................................... [4]\nETIRW\nTON\nNIGRAM\n(ii) Figure 3.1 shows there is a correlation between mean stomatal aperture width and the\nSIHT\nconcentration of potassium ions in guard cells.\nNI\nETIRW\nExplain how a change in one of these factors may cause a change in the other factor.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n......................................................................................................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-41-q03-2eb2ec0d8e.pdf"
    },
    {
      "id": "9700-2026-mj-41-q04",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 7,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-41-q04-p01-94918b8993.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-41-q04-p02-ff7a8527b3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q04",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 A student separated the pigments in a plant leaf using paper chromatography. The student\nSIHT\nanalysed the positions of the separated pigments to identify the pigments.\nNI\nETIRW\nFigure 4.1 shows the chromatogram results and analysis.\nTON\nR value pigment\nf\nOD\n0.95 carotene\nNIGRAM\n0.71 xanthophyll\nSIHT\n0.65 chlorophyll a\nNI\nETIRW 0.45 chlorophyll b\npencil line\nTON\nOD\nFigure 4.1\n(a) The student was provided with a prepared extract containing the leaf pigments.\nNIGRAM\nOutline how the student obtained the results shown in Figure 4.1 and analysed them to\ndeduce the names of the pigments present.\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n............................................................................................................................................. [4]\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Describe the role of these pigments in a leaf.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\n[Total: 7]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nO",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-mj-41-q04-313effeab5.pdf"
    },
    {
      "id": "9700-2026-mj-41-q05",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 15,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-41-q05-p01-39c9fd8968.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-41-q05-p02-82ecff5204.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-41-q05-p03-8b49a98ea3.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-41-q05-p04-4a1017675d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 Shell colour in the brown-lipped snail, Cepaea nemoralis, involves more than one gene.\nSIHT\nCB, CP CY. NI and These alleles form a dominance hierarchy. Gene 1 has three alleles:\nETIRW\nCB, • The dominant allele, results in a brown shell.\nTON CP • The allele results in a pink shell.\nCY, • The recessive allele, results in a yellow shell.\nOD\nCB CP CP CY. • The allele is dominant to and the allele is dominant to\nGene 2 has two alleles, N and n.\nNIGRAM\n• The dominant allele, N, results in a shell without stripes.\n• The recessive allele, n, results in a shell with stripes.\nSIHT\nFigure 5.1 shows a snail with a pink shell without stripes.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nFigure 5.1\nETIRW\n(a) List all the possible genotypes of the snail shown in Figure 5.1.\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\n(b) A student predicted the shell colour of the F1 offspring that are produced when a cross\nwas carried out between a snail with a brown shell with stripes and a snail with a yellow\nSIHT\nshell without stripes. Each parent snail was known to be homozygous for gene 1 and\nhomozygous for gene 2. NI\nETIRW\nThe student wrote down the cross between the two parent snails:\nTON\nparental phenotypes brown with stripes × yellow without stripes\nOD\nCBCB CYCY nn × NN parental genotypes\n(i) State the genotype and phenotype of the F1 offspring of the cross between these two\nNIGRAM\nparent snails.\ngenotype ........................................................................................................................... SIHT\nNI phenotype .........................................................................................................................\nETIRW\n[2]\nTON\nNIGRAM\nCYCY (ii) In a second breeding experiment, snails with the genotype nn were used in a test\nSIHT\ncross carried out on the F1 offspring.\nNI\nETIRW\nThe student assumed that each F1 snail would produce four different gametes in\nCYCY nn equal numbers. The student predicted the results of the test cross between\nTON and the F1 offspring. They predicted that the offspring would show a 1 : 1 : 1 : 1 ratio\nof phenotypes.\nOD\nExplain how an understanding of meiosis caused the student to assume that there\nwould be four different gametes in equal numbers.\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...................................................................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-41-q05-f0af7b2189.pdf"
    },
    {
      "id": "9700-2026-mj-41-q06",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        14,
        15,
        16
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-41-q06-p01-db3c5bd833.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-41-q06-p02-117f09853f.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-41-q06-p03-019e9489a0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q06",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 (a) The red panda, Ailurus fulgens, is an endangered animal found in parts of Asia.\nSIHT\nNI Figure 6.1 shows a red panda.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 6.1\nNIGRAM\n(i) Table 6.1 shows part of the classification of the red panda.\nSIHT\nTable 6.1\nNI\nETIRW\ndomain\n................................... TON\nOD\nkingdom Animalia\nphylum Chordata\nNIGRAM\nclass Mammalia\nSIHT\nNI order Carnivora\nETIRW\nfamily Ailuridae TON\nOD\ngenus\n...................................\nspecies\nNIGRAM ...................................\nComplete Table 6.1 to show the red panda’s domain, genus and species. [2]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Most animals in the order Carnivora eat other animals. Red pandas have adapted to\nSIHT\nfeed on a specialised diet of the leaves and shoots of bamboo plants.\nNI\nETIRW\nSuggest why red pandas are classified in the order Carnivora.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\nSIHT\n(b) Many zoos around the world have been breeding red pandas for over 80 years. The total\nNI\nnumber of red pandas in these zoos is currently estimated as 1000 individuals.\nETIRW\nIn the last 80 years the total number of red pandas living in the wild has halved to\nTON\napproximately 10 000 individuals. This is mostly due to loss of suitable habitat.\nOD\nIn 2011, for the first time, red pandas that had been born in a zoo were released into the wild.\nEvaluate the success of efforts to conserve the red panda.\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNI",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2026-mj-41-q06-b6e3b591f9.pdf"
    },
    {
      "id": "9700-2026-mj-41-q07",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        17,
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-41-q07-p01-33df40839b.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-41-q07-p02-e340937929.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-41-q07-p03-3cfbfe79fa.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q07",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:8)(cid:44)\n7 Algae are aquatic photosynthetic protoctists. The alga Alexandrium is red in colour. Alexandrium\nSIHT\ncauses harmful algal blooms (HABs) that turn seawater red when Alexandrium numbers rise to\nNI high levels. The abundance of Alexandrium depends mainly on abiotic factors.\nETIRW\nAlexandrium releases a toxin that accumulates in shellfish such as oysters and mussels. Humans\nTON who eat shellfish contaminated with Alexandrium toxin show signs of poisoning, which in some\ncases may cause death.\nOD\nFigure 7.1 shows the global number of cases of poisoning linked to Alexandrium HABs\nbetween 1986 and 2018. The line shows the long-term trend in cases of poisoning due to\nAlexandrium HABs.\nNIGRAM\n120\nSIHT\nNI\nETIRW\n100\nTON\nOD\n80\nNIGRAM\nnumber of cases\nof poisoning due to 60\nSIHT\nAlexandrium HABs\nNI\nETIRW\n40\nTON\nOD\n20\nNIGRAM\n0 SIHT\n6 0 4 8 2 4 6 8 0 2 4 8 2 6 0 6 8\n8 8 9 9 9 9 9 0 0 0 0 0 1 1 1 1 1\n9 9 9 9 9 9 9 0 0 0 0 0 0 0 0 0 0\n1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 NI\nETIRW\nyear\nTON Figure 7.1\nOD\n(a) (i) Use the long-term trend line shown in Figure 7.1 to calculate the percentage change\nin cases of poisoning due to Alexandrium HABs between the start of 1986 and the end\nof 2018.\nNIGRAM\nShow your working and give your answer to three significant figures.\nSIHT\nNI\nETIRW\nchange ......................................................% [2]\nNIGRAM\n(ii) The number of cases of poisoning due to Alexandrium HABs in Figure 7.1 is an indirect\nSIHT\nmeasure of the abundance of Alexandrium. Changes in the abundance of Alexandrium\nNI can be related to changes in abiotic factors that affect the metabolism, growth and\nETIRW\nreproduction of the alga.\nTON Identify, with reasons, two abiotic factors that will affect the metabolism of the\nphotosynthetic alga Alexandrium.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n..............................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-mj-41-q07-7990952d30.pdf"
    },
    {
      "id": "9700-2026-mj-41-q08",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-41-q08-p01-56e79f1e4d.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-41-q08-p02-13c977f893.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n8 (a) Petunia axillaris is a plant with flowers that are pollinated by night-flying moths. To\nSIHT\nattract moths, P. axillaris synthesises compounds at night that are released as scents\nNI (attractive smells).\nETIRW\nFigure 8.1 shows a flower of P. axillaris.\nTON\nFigure 8.2 shows a moth with large sensory structures known as antennae.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 8.1 Figure 8.2\n(i) Moths detect scent chemicals using sensory receptor cells on their antennae.\nNIGRAM\nOutline the role of sensory receptor cells.\nSIHT ...........................................................................................................................................\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ..................................................................................................................................... [3]\n(ii) The compounds that are released as scents by P. axillaris are lipids.\nSIHT\nOutline how P. axillaris could control the production of scent chemicals by regulating NI\nETIRW gene expression, so that scent chemicals are synthesised only at night.\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI ..................................................................................................................................... [3]\nETIRW\nTON\nNIGRAM\n(b) Microarrays could be used to investigate whether gene expression in P. axillaris is different\nSIHT\nduring the day compared to the night.\nNI\nETIRW\nFigure 8.3 shows some steps that would be performed in this microarray analysis.\nday sample night sample TON\nOD remove samples of P. axillaris flower cells\nextract chemical A\nNIGRAM\nSIHT synthesise chemical B from A\nNI\nETIR",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-mj-41-q08-a5f1f53ed5.pdf"
    },
    {
      "id": "9700-2026-mj-41-q09",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-41-q09-p01-5b31db1928.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-41-q09-p02-e184171a65.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n9 Adenosine triphosphate (ATP) is used by organisms as an immediate source of energy.\nSIHT\nNI (a) List two anabolic reactions in cells that need ATP.\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [2]\nOD\n(b) Figure 9.1 shows three stages in the life cycle of a marine invertebrate, the jellyfish\nAurelia aurita.\nNIGRAM\nadult jellyfish\nSIHT\nNI\nETIRW planula larva drifts\nin ocean current\nTON\nOD\nNIGRAM\npolyp attached\nnot to scale\nto seabed\nSIHT\nNI Figure 9.1\nETIRW\n(i) Suggest why the adult jellyfish needs more ATP than the other life cycle stages.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [2]\nSIHT\n(ii) State two ways in which ATP can be synthesised in cells, such as those in\nNI\nAurelia aurita.\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ..................................................................................................................................... [2]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) The respiratory quotient (RQ) can identify the respiratory substrate that is being used to\nSIHT\nsynthesise ATP in an organism. A respirometer can be used to calculate the RQ.\nNI\nETIRW\nTable 9.1 shows the volumes of oxygen used and carbon dioxide produced when three\ndifferent types of substrate are respired.\nTON\nTable 9.1\nOD\nvolume of carbon dioxide respiratory volume of oxygen\ncm3 cm3 produced / substrate used /\nNIGRAM\nA 830 830\nB 2020 1430 SIHT\nC 970 780 NI\nETIRW\n(i) Use Table 9.1 to calculate the RQ for respiratory substrate B.\nTON\nShow your working and give your answer to two significant figures. OD\nNIGRAM\nSIHT\nNI\nETIRW\nRQ = ......................................................... [2]\nTON\n(ii) With reference to Table 9.1, state the letter of the respiratory substrate that is a\ncarbohydr",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-mj-41-q09-b867034d79.pdf"
    },
    {
      "id": "9700-2026-mj-41-q10",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 10,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-41.pdf?download=1",
      "source_pages": [
        24
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_41.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-41-q10-p01-ce11d32372.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-41-q10",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:5)(cid:44)\n10 Genetic engineering is a technology that can be applied to medicine.\nSIHT\nNI (a) Explain what is meant by genetic engineering.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\n(b) Outline how severe combined immunodeficiency (SCID) can be treated by gene therapy.\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [6]\nNI\nETIRW\n[Total: 9]\nTON\nOD\nThe boundaries and names shown, the designation",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-mj-41-q10-bad6cf32c7.pdf"
    },
    {
      "id": "9700-2026-mj-42-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        2
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-42-q01-p01-1029019f4a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) Figure 1.1 is a diagram of a mammalian motor neurone.\nSIHT\nF NI\nA\nETIRW\nTON\nOD\nNIGRAM\nD\nC E B\nSIHT\nFigure 1.1\nNI\nETIRW\nUse the letters A–F in Figure 1.1 to identify:\nTON\n• a structure that forms a synapse with an intermediate neurone ..........................................\nOD\n• a structure that forms part of a neuromuscular junction ......................................................\n• a structure that provides insulation ......................................................................................\nNIGRAM\n• two structures that allow saltatory conduction. ....................................................................\n[4]\nSIHT\nNI\nETIRW\n(b) Outline the events that occur during an action potential.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ............................................................................................................................................. [4]\n[Total: 8]\nSIHT\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-42-q01-b72d1a98ed.pdf"
    },
    {
      "id": "9700-2026-mj-42-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-42-q02-p01-90cf9c1b74.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-42-q02-p02-12e89bdd76.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-42-q02-p03-49ee0013d2.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44)\n2 When a barley seed germinates, gibberellin is released from structure A. Gibberellin diffuses to\nSIHT\nits target cells located in tissue B. This causes the cells at B to synthesise the enzyme amylase.\nNI\nETIRW\n(a) Figure 2.1 is a diagram of a longitudinal section through a barley seed.\nTON On Figure 2.1, use label lines and the letters A and B to identify:\nOD\nA – the structure where gibberellin is released\nB – the location of the target cells for gibberellin that synthesise amylase.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFigure 2.1\n[2]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) A gene in the cells of B is switched on so that the cells can synthesise amylase.\nSIHT\nNI Figure 2.2 shows the sequence of molecular events that occurs in the cells of B during\nETIRW\nseed germination.\nTON\ngibberellin binds to protein C\nOD\ngibberellin-protein C complex\nNIGRAM binds to protein D\nSIHT\nNI protein D breaks down and\nETIRW\nreleases protein E\nTON\nOD\nprotein E binds to DNA to allow\nthe binding of RNA polymerase\nNIGRAM\nFigure 2.2\n(i) Identify proteins C, D and E. SIHT\nNI\nC ...............................................................\nETIRW\nD ...............................................................\nTON\nE ............................................................... OD\n[3]\n(ii) After protein E binds to DNA, RNA polymerase can bind to DNA.\nNIGRAM\nIdentify:\nSIHT\n• the region of the DNA that RNA polymerase binds to\n• the gene product.\nNI\nETIRW\nregion of DNA ....................................................................................................................\nTON\ngene product .....................................................................................................................\nOD\n[2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) A variety of pea plant has a homozygous recessive genotype that prevents it from making\nSIHT\nactive gibberellin.\nNI\nETIRW\nSuggest what effect this homozygous recessive genotype will have on the height of these\npea plants.\nTON\n...................................................................................................................................................\nOD\n............................................................................................................................................. [1]\n[Total: 8]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-42-q02-4496eecfc1.pdf"
    },
    {
      "id": "9700-2026-mj-42-q03",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-42-q03-p01-3deebb3f83.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-42-q03-p02-312a4430b6.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-42-q03-p03-f9e2171a37.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q03",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 (a) The green-eyed tree-frog, Litoria genimaculata, lives in tropical rainforests in Australia.\nSIHT\nAround 5 million years ago, the tree-frog population split into northern (N) and southern (S)\nNI populations that were separated by a dry area with no rainforest.\nETIRW\n6500 years ago, the N and S tree-frog populations came into contact again as a warmer\nTON climate allowed rainforest to grow again in the area between the N and S populations.\nOD\nFigure 3.1 is a map showing an area, labelled the contact zone, where N and S populations\nof tree-frog currently come into contact.\nNIGRAM\nN\nSIHT\nNI\nETIRW\nTON\nOD Barron River\nN\ncontact\nNIGRAM\nzone\nN\nSIHT\nNI S\nETIRW\n0 4 km\nTON\nFigure 3.1\nOD\n(i) N tree-frogs and S tree-frogs can be distinguished by genetic techniques.\nOutline the genetic techniques that could be used to distinguish between N and S\nNIGRAM\ntree-frog DNA.\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n........................................................................................................................................... OD\n...........................................................................................................................................\n........................................................................................................................................... NIGRAM\n..................................................................................................................................... [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Explain how a period of geographical separation, such as the one that occurred\nSIHT\nbetween the N and S tree-frog populations, can eventually lead to speciation.\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT .......................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2026-mj-42-q03-106179776d.pdf"
    },
    {
      "id": "9700-2026-mj-42-q04",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-42-q04-p01-43a8fdc858.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-42-q04-p02-802ac71278.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q04",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 Genetically modified (GM) crops are grown in many countries around the world. These crops are\nSIHT\nproduced by the process of genetic engineering.\nNI\nETIRW\n(a) Explain what is meant by the genetic engineering of crops.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n............................................................................................................................................. [3]\nETIRW\nTON\n(b) Figure 4.1 shows how the global area of GM crop production changed in the 17 years since\nOD GM crops were first grown.\n175\nNIGRAM 150\n125\nSIHT\narea of\nNI 100\nGM crop\nETIRW\nproduction\n75 / million\nTON\nhectares\nOD 50\n25\nNIGRAM\n0\n6991 7991 8991 9991 0002 1002 2002 3002 4002 5002 6002 7002 8002 9002 0102 2102 1102\nSIHT\nyear\nNI\nETIRW\nFigure 4.1\nTON\nCalculate the percentage change in global area of GM crop production between 2000\nOD and 2012.\nNIGRAM\nSIHT\nNI\npercentage change = ...................................................... % [2] ETIRW\nTON\nNIGRAM\n(c) With reference to examples, explain how GM crops can improve our food supply.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n............................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-mj-42-q04-82da5afa1a.pdf"
    },
    {
      "id": "9700-2026-mj-42-q05",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-42-q05-p01-1c6478e09b.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-42-q05-p02-b8efc60d9c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 (a) Pigeons are birds that belong to the Columbidae family. They are distributed throughout\nSIHT\nthe world.\nNI\nETIRW • The two sex chromosomes in female pigeons are known as the Z chromosome and the\nW chromosome.\nTON • Male pigeons have two Z chromosomes.\n• One of the genes controlling feather colour in pigeons is located on the Z chromosome.\nOD\nThe W chromosome does not have this gene.\n• This gene has three alleles, each coding for a different feather colour.\n• The allele for ash-red feathers is dominant to the allele for blue feathers, which is\ndominant to the allele for brown feathers.\nNIGRAM\nUsing the symbols, complete Figure 5.1 to show how a cross between an ash-red male\npigeon and a blue female pigeon can produce a brown offspring. SIHT\nNI\nETIRW ZA = Z chromosome with allele coding for ash-red colour symbols:\nZBl = Z chromosome with allele coding for blue colour\nZBr = Z chromosome with allele coding for brown colour TON\nW = W chromosome with no gene controlling feather colour\nOD\nparental phenotypes: ash-red male blue female\nparental genotypes: .................... ....................\nNIGRAM\noffspring\nSIHT\nparental\nNI\ngametes ETIRW\nTON\nOD\ngenotype genotype\nNIGRAM\nphenotype phenotype\nSIHT\nNI\ngenotype genotype ETIRW\nTON\nOD\nphenotype phenotype\nNIGRAM\nFigure 5.1\n[4] SIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) State the type of inheritance shown in this cross.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n............................................................................................................................................. [2]\nTON\nOD\n(c) Describe how you would determine the genotype of an ash-red male pigeon.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM [Total: 9]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-42-q05-287bcaef53.pdf"
    },
    {
      "id": "9700-2026-mj-42-q06",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-42-q06-p01-9584d8ab42.png",
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-42-q06-p02-6fcc8d6eeb.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q06",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 (a) Neurones in the brain metabolise glucose to produce ATP. Glucose enters brain neurones\nSIHT\nthrough transport proteins in the cell surface membrane. These proteins are similar to the\nNI transport proteins for glucose found in liver and muscle cells.\nETIRW\nState the process by which glucose passes through transport proteins into brain neurones.\nTON\n............................................................................................................................................. [1]\nOD\n(b) Brain neurones use ATP very quickly.\nNIGRAM\n109 molecules of ATP per second. One brain neurone uses 4.7 ×\n106 molecules of ATP per second. One striated muscle cell uses 9.8 × SIHT\nNI\nCalculate how many muscle cells would use the same number of ATP molecules per second\nETIRW\nas one brain neurone.\nTON\nGive your answer to the nearest whole number.\nOD\nNIGRAM\nSIHT\nnumber of muscle cells = .......................................................... [2]\nNI\nETIRW\n(c) Suggest reasons why neurones, such as those in the brain, use large numbers of ATP TON\nmolecules per second.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [4]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(d) When ATP is completely broken down, the compound adenosine is released which then\nSIHT\nleaves the cell.\nNI\nETIRW • During the day, the extracellular concentration of adenosine increases.\n• In brain synapses, adenosine binds to adenosine receptors in the cell surface\nTON membrane of postsynaptic brain neurones and inhibits postsynaptic neurone activity.\n• This action promotes sleep.\nOD\n• During sleep, ade",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-42-q06-08003dcbc6.pdf"
    },
    {
      "id": "9700-2026-mj-42-q07",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        16,
        17,
        18
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-42-q07-p01-a0e29069b5.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-42-q07-p02-97e21b9494.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-42-q07-p03-b089eebd3b.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q07",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 The southern damselfly, Coenagrion mercuriale, is a flying insect found in most of Europe.\nSIHT\nNI Figure 7.1 shows a southern damselfly.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFigure 7.1\nSIHT\n(a) Complete Table 7.1 to show the classification of the southern damselfly.\nNI\nETIRW Table 7.1\nTON\ndomain Eukarya\nOD\nkingdom Animalia\nNIGRAM\nphylum Arthropoda\nSIHT\nInsecta\n......................................\nNI\nETIRW\norder Odonata\nTON\nCoenagrionidae\nOD ......................................\ngenus\n......................................\nNIGRAM\nspecies\n......................................\n[2] SIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The domain Eukarya includes the kingdoms Animalia and Plantae.\nSIHT\nNI Describe the differences between organisms in Animalia and Plantae.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [4]\nNIGRAM\n(c) Southern damselflies require areas of open vegetation with slow-flowing water in which to\nSIHT\nlay their eggs.\nNI\n• ETIRW Damselfly eggs hatch into larvae within five weeks.\n• Damselfly larvae spend two years underwater before emerging as an adult.\n• Damselfly larvae eat other insect larvae and freshwater shrimps. TON\n• Adult damselflies live for only a few weeks.\nOD\nSouthern damselfly numbers are decreasing and this species is listed as endangered in\nthe UK.\nNIGRAM\n(i) Suggest possible reasons why the southern damselfly has become endangered in\nthe UK.\nSIHT\n...................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2026-mj-42-q07-315dd133e7.pdf"
    },
    {
      "id": "9700-2026-mj-42-q08",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 8,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2026-mj-42-q08-p01-c4111d5e6c.png",
        "9700-topic-14-homeostasis/assets/9700-2026-mj-42-q08-p02-81af7a9934.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q08",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n8 (a) In the kidney, glomerular filtrate forms in the Bowman’s capsule. Some substances are then\nSIHT\nreabsorbed in the proximal convoluted tubule.\nNI\nETIRW\nFigure 8.1 is a diagram showing the relationship between the proximal convoluted tubule\nand a blood capillary.\nTON\nP\nOD\nS\nblood\nlumen of proximal NIGRAM plasma\nconvoluted tubule\nSIHT\nR\nNI\nETIRW\nTON\nOD\nQ\nproximal convoluted\ntubule cell\nNIGRAM\nFigure 8.1\nSIHT\nComplete Table 8.1 using the letters P, Q, R and S. Each letter may be used once, more\nNI\nthan once or not at all.\nETIRW\nTable 8.1\nTON\nOD\nNa+ K+ location of / pumps\n..................\nlocation of cotransport proteins\n.................. NIGRAM\nstructures through which\nglucose molecules pass .................. SIHT\nNI\nprevents leakage of fluid\nETIRW ..................\n[4]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Osmoregulation is the control of blood water potential in mammals.\nSIHT\nNI Describe the role of ADH in osmoregulation when the water potential of blood decreases\nETIRW\nbelow the set point and is detected by osmoreceptors in the hypothalamus.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n.........................................................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2026-mj-42-q08-afe563224f.pdf"
    },
    {
      "id": "9700-2026-mj-42-q09",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-42-q09-p01-1443c5a9e7.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-42-q09-p02-232324499d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n9 (a) Mature red blood cells do not contain mitochondria. The only respiratory pathway carried\nSIHT\nout in red blood cells is glycolysis. The initial reaction in glycolysis is the phosphorylation of\nNI glucose using ATP. This reaction is catalysed by the enzyme hexokinase.\nETIRW\nAn autosomal recessive gene mutation results in less hexokinase being produced, which\nTON can lead to hexokinase deficiency disease.\nOD\nSuggest the effect of the production of less hexokinase on red blood cells and suggest a\npossible symptom for a person with hexokinase deficiency disease.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [4]\nNI\nETIRW\n(b) Muscle cells contain mitochondria. This means that the Krebs cycle can take place in\nmuscle cells. TON\nOD\nOutline the main events in the Krebs cycle.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-mj-42-q09-66d50e9ba1.pdf"
    },
    {
      "id": "9700-2026-mj-42-q10",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 10,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-42.pdf?download=1",
      "source_pages": [
        24,
        25,
        26
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_42.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-42-q10-p01-af67770b0d.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-42-q10-p02-cd24796ceb.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-42-q10-p03-feb6b16d03.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-42-q10",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:5)(cid:44)\n10 (a) The structure of a chloroplast is related to its function in photosynthesis.\nSIHT\nNI Figure 10.1 is a diagram of a chloroplast.\nETIRW\nstroma\nTON\nthylakoid\nOD\nspace\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nthylakoid\nDNA membrane\nNIGRAM Figure 10.1\nWith reference to Figure 10.1, outline the relationship between the structure and function of\nSIHT\na chloroplast in photosynthesis.\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n............................................................................................................................................. [4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Chloroplast pigments can be separated and identified by a technique called chromatography.\nSIHT\nNI A student carried out an investigation using paper chromatography to separate the pigments\nETIRW\nin a solution containing two chloroplast pigments.\nTON The results of the investigation are shown in Figure 10.2.\nOD\nNIGRAM\n15 cm\ndistance travelled\nby solvent\nSIHT\ndistance travelled NI\n10 cm ETIRW by pigment A\nTON\ndistance travelled\nOD by pigment B\n5 cm\nbaseline\nNIGRAM\nchromatography 0 cm\npaper\nSIHT\nNI\nETIRW\nFigure 10.2 TON\nOD\nvalue of pigment A is 0.96. (i) The R\nf\nvalue of pigment B. Using Figure 10.2, calculate the R\nf\nNIGRAM\nR value of pigment B ......................................................... [1]\nf\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(ii) Table 10.1 shows the R values of some photosynthetic pigments.\nSIHT f\nNI Table 10.1\nETIRW\npigment R value\nf TON\nchlorophyll a 0.59 – 0.68\nOD\nchlorophyll b 0.40 – 0.54\ncarotene 0.94 – 0",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-mj-42-q10-03931fe1e6.pdf"
    },
    {
      "id": "9700-2026-mj-43-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-43-q01-p01-fb9c050f59.png",
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-43-q01-p02-e1f1b2c0f9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q01",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) Figure 1.1 is a diagram of a neuromuscular junction.\nSIHT\nNI\nETIRW\nTON\nA\nF OD\nE\nB\nNIGRAM\nC\nSIHT\nD\nNI\nETIRW\nTON\nOD\nFigure 1.1\nUse the letters A–F in Figure 1.1 to identify a structure:\nNIGRAM\n• that contains a neurotransmitter ........................................................................................\nSIHT\n• where the Krebs cycle occurs ...........................................................................................\nNI\nETIRW\n• that depolarises when a neurotransmitter binds to its receptors .......................................\nTON • that contains troponin. .......................................................................................................\n[4] OD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) In mammals, control and coordination is provided by the nervous system and the\nSIHT\nendocrine system.\nNI\nETIRW\nOne role of the endocrine system is to allow the body to exercise for an extended time, such\nas running continuously for an hour.\nTON\nExplain the importance of the endocrine system when a mammal runs continuously for\nOD\nan hour.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [5] NI\nETIRW\n[Total: 9]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-43-q01-b8f4f67c41.pdf"
    },
    {
      "id": "9700-2026-mj-43-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-43-q02-p01-e71e77a1fb.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-43-q02-p02-69c973a53a.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q02",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 Meiosis is a process that includes reduction division.\nSIHT\nNI (a) The cells that carry out meiosis in the male reproductive organ of plants are known as pollen\nETIRW\nmother cells.\nTON Figure 2.1 shows photomicrographs of three different stages of meiosis in pollen mother\ncells from Lilium, a lily plant.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nA B C\nOD\nFigure 2.1\nName the stages of meiosis shown in cells A, B and C in Figure 2.1.\nNIGRAM\nA ...............................................................................................................................................\nSIHT\nB ...............................................................................................................................................\nNI\nETIRW\nC ...............................................................................................................................................\n[3]\nTON\nOD\n(b) Homologous chromosomes pair up and then separate into different daughter cells\nduring meiosis.\nExplain what is meant by a homologous pair of chromosomes. NIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) Explain the need for reduction division to occur during meiosis.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n.........................................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-43-q02-9e28ba1cba.pdf"
    },
    {
      "id": "9700-2026-mj-43-q03",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-43-q03-p01-3f0cdb2979.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-43-q03-p02-f013ff50ab.png",
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-43-q03-p03-46d08d4dd9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q03",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n3 Sickle cell anaemia is a disease caused by a mutation in the HBB gene.\nSIHT\nNI The presence of the mutant HBB allele in the genome of a person can be detected using a\nETIRW\nvariety of techniques.\nTON (a) DNA can be extracted and purified from cell samples taken from a person. The polymerase\nchain reaction (PCR) can be used to amplify the region of DNA containing the mutant HBB\nOD\nallele, if it is present in the genome of the person.\nThis PCR process amplifies only the mutant HBB allele and not other DNA sequences.\nNIGRAM\nOutline the steps involved in PCR and explain how PCR can be carried out so that only the\nmutant HBB allele is amplified.\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n............................................................................................................................................. [4] TON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Restriction enzymes and electrophoresis can be used to determine the alleles of the\nSIHT\nHBB gene that make up the genotype of a person.\nNI\nETIRW\nThe restriction enzymes:\nTON • produce one fragment when only the mutant HBB allele is present\n• produce two smaller fragments when only the normal HBB allele is present.\nOD\nFigure 3.1 shows the results of electrophoresis of DNA samples from four people, 1 to 4.\nOnly DNA fragments from the HBB gene locus are shown.\nNIGRAM\n1 2 3 4\nloading wells\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 3.1\nNIGRAM\nState which people, 1– 4, show:\nSIHT\na genotype that is homozygous for the normal HBB allele ..................\nNI\nETIRW a genotype that is homozygous for the mutant HBB allele ..................\nTON a genotype that is heterozygous for the HBB gene ...........",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-mj-43-q03-51f2aa2bab.pdf"
    },
    {
      "id": "9700-2026-mj-43-q04",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-43-q04-p01-64c542a45c.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-43-q04-p02-2eecd112d5.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n4 The leopard, Panthera pardus, is a large mammal found in a number of countries in Africa\nSIHT\nand Asia.\nNI\nETIRW\n(a) Most leopards have yellow‑brown fur with black spots, but 11% of leopards have black fur.\nTON Figure 4.1 shows the two types of leopard.\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFigure 4.1 TON\nOD\nIn the gene that controls fur colour in leopards, the allele that results in black fur is recessive.\nAssuming the Hardy‑Weinberg principle applies to leopard populations, calculate the\npercentage of leopards that are heterozygous for the gene that controls fur colour.\nNIGRAM\nUse the equations:\nSIHT\np + q = 1\nNI\nETIRW\np2 q2 + 2pq + = 1\nTON key\nOD\np = frequency of the dominant allele\nq = frequency of the recessive allele\np2 = frequency of homozygous dominant genotype\n2pq = frequency of heterozygous genotype\nNIGRAM\nq2 = frequency of homozygous recessive genotype\nSIHT Show your working and give your answer to the nearest whole number.\nNI\nETIRW\nTON\nOD\nNIGRAM\npercentage of heterozygous leopards ......................................................... [3]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Fur colour in leopards shows discontinuous variation. The mass of leopards shows\nSIHT\ncontinuous variation.\nNI\nETIRW\nCompare the genetic basis of discontinuous variation and continuous variation.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM (c) Scientists think that the leopard populations in Asia were established by the migration of\nleopards from Africa approximately 500 000 years ago.\nSIHT\nAfrican and Asian leopards are considered to be the same species, but there ",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2026-mj-43-q04-b011dd470c.pdf"
    },
    {
      "id": "9700-2026-mj-43-q05",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-43-q05-p01-01dbd7ee5b.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-43-q05-p02-f4541d689e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q05",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n5 There are several different human blood group systems.\nSIHT\nNI Each blood group system is based on the presence or absence of certain antigens attached to\nETIRW\nthe cell surface membranes of red blood cells.\nTON In the ABO blood group system, the ABO gene, known as gene I, has three alleles. The gene is\nlocated on an autosome.\nOD\nIA • Allele codes for an enzyme that adds antigen A to the cell surface membrane.\nIB • Allele codes for an enzyme that adds antigen B to the cell surface membrane.\nIO • Allele codes for a non‑functional enzyme, so no antigen is added.\nNIGRAM\nIO IA IB. • is recessive to and\nIA IB • and are codominant.\nSIHT\n(a) Explain what is meant by the term recessive and the term codominant.\nNI\nETIRW\nrecessive ..................................................................................................................................\nTON\n...................................................................................................................................................\nOD\ncodominant ...............................................................................................................................\n...................................................................................................................................................\nNIGRAM [2]\nSIHT\n(b) Complete Table 5.1 to show the possible genotypes of the blood groups in the ABO system.\nNI\nETIRW Table 5.1\nTON\nblood group genotypes\nOD\nA\nB\nNIGRAM\nAB\nSIHT\nO\nNI\nETIRW\n[3]\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(c) In the Rhesus blood group system, a gene codes for a protein known as the Rhesus factor.\nSIHT\nThis protein is an antigen found on the surface of red blood cells.\nNI\nETIRW • The dominant allele D codes for the Rhesus factor.\n• The recessive allele d results in no Rhesus factor.\nTON • A person who has the Rhesus factor is described as Rhesus positive (Rh+).\n• A person who lacks the Rhesus factor is described as Rhesus negative (Rh–).\nOD\n• The gene coding for the Rhesus factor is on a different autosome from the ABO gene.\nComplete the Punnett square in Figure 5.1 to show the possible blood groups of children\nfrom these parents.\nNIGRAM\n• Parent 1 is heterozygous for blood group B and heterozygous for the Rhesus factor.\n• Parent 2 has blood group O and is heterozygous for the Rhesus factor. SIHT\nNI\nETIRW\nparental phenotypes: B Rh+ O Rh+\nTON\nparental genotypes:\nOD\nparent 1 gametes\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nparent 2\ngametes\nNIGRAM\nSIHT\nNI\nratio of offspring phenotypes:\nETIRW\nTON\nOD\nFigure 5.1\n[5]\nNIGRAM\n[Total: 10]\nSIHT\nNI\nETIRW",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-43-q05-a7fc82ea26.pdf"
    },
    {
      "id": "9700-2026-mj-43-q06",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        14
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2026-mj-43-q06-p01-bc69674cef.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q06",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:5)(cid:44)\n6 (a) In the kidney, ultrafiltration occurs when blood enters the glomerulus through the\nSIHT\nafferent arteriole.\nNI\nETIRW\nDescribe the process of ultrafiltration.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [6]\nTON\nOD\n(b) Caffeine is a natural compound present in tea and coffee. Caffeine is also added to some\ndrinks. Some people produce more urine after having a drink containing caffeine than after\ndrinking an equivalent volume of water.\nNIGRAM\nSuggest ways in which caffeine may cause an increase in urine production in some people.\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n..........................................................................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2026-mj-43-q06-2cfc90e185.pdf"
    },
    {
      "id": "9700-2026-mj-43-q07",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-43-q07-p01-b83ddcd4a1.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-43-q07-p02-00f60e3a06.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-43-q07-p03-d154e2f996.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q07",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:6)(cid:44)\n7 The ring‑tailed lemur, Lemur catta, is a primate that is found only on the island of Madagascar.\nSIHT\nNI Figure 7.1 shows a ring‑tailed lemur.\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nFigure 7.1 SIHT\nNI\n(a) Ring‑tailed lemurs are classified as endangered on the International Union for Conservation\nETIRW\nof Nature (IUCN) Red List of Threatened Species™. There are approximately 2000\nring‑tailed lemurs in the wild in Madagascar.\nTON\nSuggest reasons why the ring‑tailed lemur has become endangered. OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [3]\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) Zoos have a role in protecting endangered species.\nSIHT\nNI Outline the methods of assisted reproduction used by zoos in the conservation of\nETIRW\nendangered mammals such as the ring‑tailed lemur.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n....................",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2026-mj-43-q07-f77aa5bc84.pdf"
    },
    {
      "id": "9700-2026-mj-43-q08",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 8,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        18,
        19
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-43-q08-p01-239c54436c.png",
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-43-q08-p02-1242578e3f.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q08",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:9)(cid:44)\n8 (a) Figure 8.1 is a diagram of a section through a barley seed.\nSIHT\nNI\nETIRW\naleurone layer seed coat\nTON\nOD\nendosperm\nembryo\nNIGRAM\nSIHT\nFigure 8.1\nNI\nETIRW\nComplete Table 8.1 by naming the part of the seed from Figure 8.1 that matches the\nfunction stated.\nTON\nTable 8.1 OD\nfunction part of seed\nNIGRAM\ngibberellin synthesis\n...............................................................................\nSIHT starch store\n...............................................................................\nNI\nETIRW\namylase synthesis\n...............................................................................\nTON\nmaltose production\n............................................................................... OD\n[4]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The plant hormone auxin has an important role in the growth of plants. When auxin\nSIHT\nmolecules bind to receptors in the cell surface membrane of young cells, a series of events\nNI is triggered that results in cell elongation.\nETIRW\nFigure 8.2 outlines this series of events.\nTON\nevent 1 event 3 event 2\nOD\nA B H O\n2 cell wall auxin\nNIGRAM\ncell surface\nmembrane\nSIHT\nNI\nstimulation ETIRW\nA B H O\n2 receptor\nTON cytoplasm\nOD\nFigure 8.2\n(i) Describe event 1 in Figure 8.2 and explain how this event is important for the process\nof cell elongation.\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..................................................................................................................................... [3]\nSIHT (ii) Explain how event 2 leads to event 3 and results in cell elongation.\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n........................................................................................................................................... OD\n......................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-43-q08-9f2f8dd468.pdf"
    },
    {
      "id": "9700-2026-mj-43-q09",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 9,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-43-q09-p01-95e0b8f4fd.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-43-q09-p02-cac5a8027c.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q09",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 (a) The structure of a mitochondrion is related to its function in respiration.\nSIHT\nNI Figure 9.1 is a diagram of a mitochondrion.\nETIRW\nintermembrane\nDNA\nTON space\nOD matrix\nNIGRAM\nSIHT\nNI\nETIRW\nouter\nTON\nmembrane\nOD\ninner\nmembrane\nNIGRAM\nFigure 9.1\nSIHT\nWith reference to Figure 9.1, outline the relationship between the structure and function of a\nNI mitochondrion in respiration.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n............................................................................................................................................. [4]\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) Table 9.1 shows the mean percentage of cell volume occupied by mitochondria in three\nSIHT\ndifferent cell types.\nNI\nETIRW\nTable 9.1\nTON\nmean percentage of cell volume\ncell type\nOD occupied by mitochondria\nheart muscle cell 48\nliver cell 17\nNIGRAM\nred blood cell 0\nSIHT\nSuggest why the three cell types shown in Table 9.1 have different mean percentages of cell\nvolume occupied by mitochondria. NI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n..........................................................................................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-mj-43-q09-f7fac98a3e.pdf"
    },
    {
      "id": "9700-2026-mj-43-q10",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 10,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-43.pdf?download=1",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_43.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-43-q10-p01-b561c7a6bb.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-43-q10-p02-6ac00532dd.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-43-q10",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 (a) The photoactivation of chlorophyll is an essential process in the conversion of light energy to\nSIHT\nchemical energy during the light‑dependent stage of photosynthesis.\nNI\nETIRW\nPhotoactivation of chlorophyll leads to the synthesis of ATP.\nTON Outline how the photoactivation of chlorophyll leads to the production of ATP.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nOD ...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM ...................................................................................................................................................\n............................................................................................................................................. [4]\nSIHT\nNI\nETIRW (b) (i) ATP produced in the light‑dependent stage passes to the Calvin cycle of the\nlight‑independent stage.\nTON\nName two compounds in the Calvin cycle that are formed using this ATP. Do not give\nOD\nabbreviations for these compounds.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n[2]\nSIHT\n(ii) Reduced NADP is also produced in the light‑dependent stage.\nNI\nETIRW\nOutline the role of reduced NADP in the light‑independent stage.\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n.....................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-mj-43-q10-85de5f5717.pdf"
    },
    {
      "id": "9700-2026-mj-44-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        2,
        3
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-44-q01-p01-f627186462.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-44-q01-p02-3cef382b25.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q01",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 The bacterium Escherichia coli uses glucose as the main respiratory substrate, but other\nSIHT\nsubstances, such as lactose, can be used when necessary.\nNI\nETIRW\nThe enzyme β-galactosidase hydrolyses lactose to form glucose and galactose, which can then\nbe used by E. coli.\nTON\nProduction of β-galactosidase in E. coli is regulated and involves a length of DNA known as the\nOD\nlac operon.\n(a) Figure 1.1 is a simple diagram of the lac operon.\nNIGRAM\nP O lacZ lacY lacA\nSIHT\nFigure 1.1 NI\nETIRW\nWhen lactose is absent, a repressor molecule is attached to part of the lac operon.\nTON\nDraw a cross (X) on Figure 1.1 to show the location where the repressor attaches to the lac\nOD\noperon when lactose is absent. [1]\n(b) Structural genes lacZ, lacY and lacA are parts of the lac operon.\nNIGRAM\nComplete Table 1.1 to:\nSIHT\n• name the gene product of lacY and of lacA\nNI • state the role of the gene product of lacY.\nETIRW\nTable 1.1\nTON\nOD structural gene gene product role of gene product\nhydrolyses lactose to form glucose\nlacZ β-galactosidase\nand galactose\nNIGRAM\n.......................................................................\nlacY ...................................\nSIHT\n.......................................................................\nNI\ntransfers an acetyl group from one molecule\nETIRW\nlacA\nto another ...................................\n[2] TON\nOD\n(c) The enzyme β-galactosidase is an inducible enzyme.\nExplain what is meant by an inducible enzyme.\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [1]\nNI\nETIRW\nTON\nNIGRAM\n(d) A scientist investigated population growth in E. coli.\nSIHT\n• NI A sample of E. coli was removed from a liquid nutrient medium containing glucose.\nETIRW • The sample was added to a liquid nutrient medium containing limited quantities of\nglucose and lactose.\nTON • Estimates of the total number of cells in the sample (population size) were made at\nregular intervals.\nOD\nFigure 1.2 shows the results of this investigation.\nphases\nNIGRAM\nA B C\nSIHT\nNI\nETIRW\nE. coli\nTON population\nsize\nOD\nNIGRAM\ntime\nSIHT\nFigure 1.2 NI\nETIRW\nWith reference to Figure 1.2, suggest explanations for phase B and phase C.\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n..............................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-44-q01-27824cba74.pdf"
    },
    {
      "id": "9700-2026-mj-44-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        4
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-44-q02-p01-d53204c1f9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q02",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)\n2 (a) The movement of ions can result in different actions in plant cells and animal cells.\nSIHT\nNI Table 2.1 lists several ions, the direction of movement of each ion and the cell type\nETIRW\nconcerned, and the action that results from the movement of each ion.\nTON Complete Table 2.1.\nOD\nTable 2.1\ndirection of movement of ion action that results from\nion\nNIGRAM and type of cell movement of ion\nK+ opening of stomata SIHT\n....................................................\nNI\ninto synaptic knob of ETIRW\nNa+\npresynaptic neurone\n....................................................\nTON\ninto synaptic knob of\nCa2+\nOD\npresynaptic neurone\n....................................................\nfrom sarcoplasmic reticulum\nto sarcoplasm of cell of binding to troponin\nstriated muscle ................ NIGRAM\n[4]\nSIHT\n(b) In the nephron of the kidney, selective reabsorption occurs in the cells of the proximal NI\nETIRW convoluted tubule. Sodium ions are actively transported out of these cells into the\nsurrounding tissue fluid and pass into the blood.\nTON\nOutline how the active transport of sodium ions out of the proximal convoluted tubule cells\nOD\nresults in the absence of glucose in urine.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n................................................................................................................................................... SIHT\nNI\n...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n................................................................................................................................................... OD\n...................................................................................................................................................\n................................................................................................................................................... NIGRAM\n............................................................................................................................................. [4]\nSIHT\n[Total: 8]\nNI\nETIRW\nTON",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-44-q02-2283c557bc.pdf"
    },
    {
      "id": "9700-2026-mj-44-q03",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic_number": 18,
      "topic": "Classification, biodiversity and conservation",
      "topic_slug": "9700-topic-18-classification-biodiversity-and-conservation",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-44-q03-p01-a6d3416fbc.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-44-q03-p02-70438183cd.png",
        "9700-topic-18-classification-biodiversity-and-conservation/assets/9700-2026-mj-44-q03-p03-12980e5ed3.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q03",
        "html": "9700-topic-18-classification-biodiversity-and-conservation/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:6)(cid:44)\n3 The tuatara, Sphenodon punctatus, is a reptile that is native to New Zealand and is found in the\nSIHT\nwild only in that country. Fossil evidence shows that there has been little morphological change in\nNI the tuatara over the last 200 million years.\nETIRW\nFigure 3.1 shows a tuatara.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFigure 3.1\nNI\nETIRW (a) The tuatara inhabited the main islands and the smaller islands of New Zealand for most of\nthe last 200 million years. Around 800 years ago, humans arrived in New Zealand. In the last\nTON hundred years, a rapid decrease in the number of tuatara has occurred and now they are\nfound in the wild on only 32 small islands.\nOD\nSuggest why the number of tuatara has decreased in most parts of New Zealand.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n............................................................................................................................................. [4]\nNI\nETIRW\nNIGRAM\n(b) Moutoki Island is one of the small islands where there is a wild population of tuatara. The\nSIHT\nsize of the population of tuatara was estimated using the mark–release–recapture method:\nNI\nETIRW • A sample of 23 tuatara was captured using a humane (harmless) trap.\n• The tuatara were marked in a way that did not harm them.\nTON • The marked tuatara were released into the area where they were captured.\n• A period of time was allowed for the marked tuatara to mix into the population.\nOD\n• A second sample of 38 tuatara was captured, in which 6 individuals were marked.\nUse the Lincoln Index to calculate the size of the tuatara population on Moutoki Island.\nNIGRAM\nGive your answer to the nearest whole number.\nLincoln Index SIHT\nn × n\nNI",
      "pdf_path": "9700-topic-18-classification-biodiversity-and-conservation/print/9700-2026-mj-44-q03-c7ab0117e1.pdf"
    },
    {
      "id": "9700-2026-mj-44-q04",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic_number": 17,
      "topic": "Selection and evolution",
      "topic_slug": "9700-topic-17-selection-and-evolution",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-44-q04-p01-04da83e742.png",
        "9700-topic-17-selection-and-evolution/assets/9700-2026-mj-44-q04-p02-85d8e827da.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q04",
        "html": "9700-topic-17-selection-and-evolution/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n4 (a) Twenty million years ago there was an ocean between the continents of North America and\nSIHT\nSouth America. The snapping shrimp, Alpheus sp., lived in this ocean.\nNI\nETIRW\nFigure 4.1 shows a snapping shrimp.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 4.1\nAbout 3 million years ago, volcanic activity and sedimentation formed a narrow strip of land,\nNIGRAM\nPanama, joining North America and South America.\nSIHT Figure 4.2 shows the area 20 million years ago and now.\nNI\n20 million years ago now\nETIRW\nNorth America North America\nAtlantic Ocean Atlantic Ocean\nTON\nOD\nCaribbean Sea Caribbean Sea\nNIGRAM\nPacific Ocean Pacific Ocean\nSouth America South America\nSIHT\nNI\nPanama\nETIRW\nFigure 4.2\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nSnapping shrimp living in the Caribbean Sea near Panama look similar to snapping shrimp\nSIHT\nliving in the Pacific Ocean near Panama. However, if they are paired together, they do\nNI not mate.\nETIRW\nExplain how, over time, the snapping shrimp living in the Caribbean Sea and the\nTON snapping shrimp living in the Pacific Ocean developed so that they now do not mate with\neach other.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n..........................................................................................",
      "pdf_path": "9700-topic-17-selection-and-evolution/print/9700-2026-mj-44-q04-d91595e099.pdf"
    },
    {
      "id": "9700-2026-mj-44-q05",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic_number": 14,
      "topic": "Homeostasis",
      "topic_slug": "9700-topic-14-homeostasis",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-14-homeostasis/assets/9700-2026-mj-44-q05-p01-7e0b825136.png",
        "9700-topic-14-homeostasis/assets/9700-2026-mj-44-q05-p02-5fb5f92253.png",
        "9700-topic-14-homeostasis/assets/9700-2026-mj-44-q05-p03-595f4cc0f6.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q05",
        "html": "9700-topic-14-homeostasis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n5 (a) Insulin has an important role in the maintenance of blood glucose concentration.\nSIHT\nNI An investigation measured how blood glucose concentration and blood insulin concentration\nETIRW\nchanged after a person had eaten a glucose-rich meal.\nTON The results are shown in Figure 5.1.\nOD\n7.5 300\n7.0 280\nNIGRAM\n6.5 260\nSIHT\n6.0 240\nNI\nETIRW\n5.5 220\nTON\n5.0 200\nOD\n4.5 180\nblood insulin 4.0 160 blood glucose\nNIGRAM\nconcentration concentration\ndm–3 dm–3 mmol pmol / / 3.5 140\nSIHT\n3.0 120\nNI\nETIRW\n2.5 100\nTON\n2.0 80\nOD\n1.5 60\n1.0 40\nNIGRAM\n0.5 20\nSIHT\n0.0 0 NI\n07:00 08:00 09:00 10:00 11:00 12:00 ETIRW\ntime of day\nTON key\nglucose-rich meal eaten\nOD blood glucose\nconcentration\nblood insulin\nNIGRAM concentration\nFigure 5.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(i) Use Figure 5.1 to calculate the percentage increase in the blood insulin concentration\nSIHT\nbetween 07:00 and 08:00.\nNI\nETIRW\nGive your answer to four significant figures.\nTON\nOD\n...................................................... % [2]\n(ii) With reference to Figure 5.1, explain the change that occurs in the blood glucose NIGRAM\nconcentration between 08:00 and 09:00.\nSIHT\n...........................................................................................................................................\nNI\nETIRW ...........................................................................................................................................\n........................................................................................................................................... TON\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n...........................................................................................................................................\nSIHT\nNI ...........................................................................................................................................\nETIRW\n..................................................................................................................................... [4]\nTON\nOD\n(b) Type 1 diabetes in humans is a condition in which the pancreas does not produce enough\ninsulin to control blood glucose concentration. People with type 1 diabetes measure their\nblood glucose concentration at regular intervals using either a biosensor or test strips.\nNIGRAM\n(i) Name the two enzymes that can be used in biosensors and test strips to measure\nblood glucose concentration.\nSIHT\n........................................................",
      "pdf_path": "9700-topic-14-homeostasis/print/9700-2026-mj-44-q05-7304365f5c.pdf"
    },
    {
      "id": "9700-2026-mj-44-q06",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic_number": 13,
      "topic": "Photosynthesis",
      "topic_slug": "9700-topic-13-photosynthesis",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        13,
        14,
        15
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-44-q06-p01-8a6f26f1b1.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-44-q06-p02-48460ab87b.png",
        "9700-topic-13-photosynthesis/assets/9700-2026-mj-44-q06-p03-6eeebcb59d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q06",
        "html": "9700-topic-13-photosynthesis/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:4)(cid:44)\n6 (a) Figure 6.1 shows a transmission electron micrograph of part of a chloroplast of Timothy\nSIHT\ngrass, Phleum pratense.\nNI\nETIRW\nTON P\nOD\nS\nNIGRAM\nQ\nSIHT\nR\nNI\nETIRW\nTON\nOD\nFigure 6.1\nState the letter P, Q, R or S from Figure 6.1 that identifies:\nNIGRAM\n• a structure containing photosynthetic pigments ...................\nSIHT\n• an area where the light-independent stage of photosynthesis is carried out ...................\nNI [2]\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) A student carried out an experiment to investigate the effect of light intensity on the\nSIHT\nrate of photosynthesis in the brown alga Fucus serratus. Brown algae are aquatic\nNI multicellular protoctists.\nETIRW\n• Discs were cut from the photosynthetic tissue of F. serratus.\nTON cm3 • Ten discs were added to each of four beakers containing 50 of sea water.\n• The beakers were illuminated with identical LED lamps placed at different distances (d),\nOD\nin centimetres, from the beakers.\n2. • The relative light intensity was determined by calculating 1 / d\n• When the discs, which were denser than sea water, had sunk to the bottom of the\nbeakers, the student started a timer.\nNIGRAM\n• The student observed the discs rising to the surface of the water in the beakers and\nrecorded the time taken (t), in minutes, for the fifth disc from each beaker to reach\nthe surface. SIHT\n• The rate of photosynthesis was determined by calculating 1000 / t.\nNI\nETIRW\nThe results are shown in Table 6.1.\nTON\nTable 6.1\nOD\nrate time (t ) for fifth disc distance (d ) of\nrelative\nof photosynthesis to reach the surface beaker from lamp\nlight intensity\nmin–1 / / min / cm NIGRAM\n5 0.0400 43.5\n.................... SIHT\nNI 10 0.0100 36 27.8\nETIRW\n15 52 19.2\n....................\nTON\nOD 20 0.0025 88 11.4\n(i) Complete Table 6.1 to show:\nNIGRAM\n• the time for the fifth disc to reach the surface of the water in the beaker when the\ndistance (d) between the beaker and lamp was 5 cm\n2) • the relative light intensity (1 / d when the distance between the beaker and lamp SIHT\nwas 15 cm. [2]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(ii) Use the data in Table 6.1 to:\nSIHT\n• NI describe the relationship between light intensity and the rate of photosynthesis\nETIRW • explain why the discs rose to the surface of the water in the beakers.\nTON ...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n..............................................................................................................................",
      "pdf_path": "9700-topic-13-photosynthesis/print/9700-2026-mj-44-q06-aa2e0c5160.pdf"
    },
    {
      "id": "9700-2026-mj-44-q07",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 7,
      "topic_number": 16,
      "topic": "Inheritance",
      "topic_slug": "9700-topic-16-inheritance",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-16-inheritance/assets/9700-2026-mj-44-q07-p01-c86a4e8189.png",
        "9700-topic-16-inheritance/assets/9700-2026-mj-44-q07-p02-a22ce09eea.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q07",
        "html": "9700-topic-16-inheritance/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:7)(cid:44)\n7 Neurofibromatosis type 1 (NF1) is a genetically inherited autosomal condition in which tumours\nSIHT\ngrow in the nervous tissue.\nNI\nETIRW\n(a) If both parents of a child do not have NF1, then the chance that the child will have inherited\nNF1 is 0%.\nTON\nIn a family where one parent has NF1 and the other parent is unaffected, each child has a\nOD\n50% chance of having NF1.\nUsing appropriate symbols, complete Figure 7.1 to show how there is a 50% chance of each\nchild in this family having the condition.\nNIGRAM\nsymbols:\nSIHT\nNI\nparental phenotypes: parent with NF1 unaffected parent ETIRW\nTON\nparental genotypes:\nOD\ngametes:\nNIGRAM\noffspring genotypes:\nSIHT\nNI offspring phenotypes:\nETIRW\nTON\nOD\nFigure 7.1\n[4]\nNIGRAM\n(b) Genetic screening during pregnancy can help identify the risk of a fetus having NF1.\nScreening tests can include blood tests, ultrasounds and other procedures.\nSIHT\nDiscuss the social and ethical considerations of using genetic screening for NF1.\nNI\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ............................................................................................................................................. [3]\nTON\nNIGRAM\n(c) One symptom of NF1, which can develop around the age of 20 years, is hearing loss\nSIHT\ncaused by tumours of the auditory nerve. The auditory nerve contains sensory neurones that\nNI transmit impulses to the brain.\nETIRW\nSuggest how a tumour on the auditory nerve could prevent the transmission of nerve\nTON impulses to the brain.\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ..........................................",
      "pdf_path": "9700-topic-16-inheritance/print/9700-2026-mj-44-q07-e4bf38efbe.pdf"
    },
    {
      "id": "9700-2026-mj-44-q08",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 8,
      "topic_number": 19,
      "topic": "Genetic technology",
      "topic_slug": "9700-topic-19-genetic-technology",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        19
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-19-genetic-technology/assets/9700-2026-mj-44-q08-p01-2bb66b12cc.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q08",
        "html": "9700-topic-19-genetic-technology/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:10)(cid:44)\n8 The polymerase chain reaction (PCR) is an automated technique used in many applications of\nSIHT\ngenetic technology. The main function of PCR is to amplify selected regions of DNA.\nNI\nETIRW\n(a) Name the substances that are added to the PCR machine at the beginning of the process.\nTON ...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT ...................................................................................................................................................\nNI\n...................................................................................................................................................\nETIRW\n............................................................................................................................................. [3]\nTON\nOD\n(b) Describe and explain the steps involved in the PCR technique.\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ............................................................................................",
      "pdf_path": "9700-topic-19-genetic-technology/print/9700-2026-mj-44-q08-c1d482b725.pdf"
    },
    {
      "id": "9700-2026-mj-44-q09",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 9,
      "topic_number": 15,
      "topic": "Control and coordination",
      "topic_slug": "9700-topic-15-control-and-coordination",
      "marks": 9,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        20,
        21
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-44-q09-p01-1b08cc1df0.png",
        "9700-topic-15-control-and-coordination/assets/9700-2026-mj-44-q09-p02-736a29fccc.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q09",
        "html": "9700-topic-15-control-and-coordination/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:1)(cid:44)\n9 (a) The Venus fly trap, Dionaea muscipula, is a carnivorous plant that captures small insects in\nSIHT\na trap made from a specialised leaf.\nNI\nETIRW\n(i) The first step in capturing an insect occurs when the insect touches at least two sensory\nhairs on the leaf of the plant within 35 seconds. The second step occurs when an action\nTON potential is produced and spreads over the leaf.\nOD\nOutline how the stimulation of sensory hair cells of a Venus fly trap plant leads to an\ninsect being trapped.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW ..................................................................................................................................... [4]\n(ii) A Venus fly trap can carry out photosynthesis. TON\nOD\nSuggest why the Venus fly trap also captures and feeds on insects.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n..................................................................................................................................... [1]\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(b) The nervous and endocrine systems of a mammal coordinate responses to changes in the\nSIHT\ninternal and external environment.\nNI\nETIRW\nTable 9.1 compares some features of the nervous and endocrine systems.\nTON Complete Table 9.1.\nOD\nTable 9.1\nfeature nervous system endocrine system\nNIGRAM\nsignal impulse\n............................",
      "pdf_path": "9700-topic-15-control-and-coordination/print/9700-2026-mj-44-q09-22698c6994.pdf"
    },
    {
      "id": "9700-2026-mj-44-q10",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 10,
      "topic_number": 12,
      "topic": "Energy and respiration",
      "topic_slug": "9700-topic-12-energy-and-respiration",
      "marks": 12,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-44.pdf?download=1",
      "source_pages": [
        22,
        23
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_44.pdf",
      "image_paths": [
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-44-q10-p01-3dbba96e44.png",
        "9700-topic-12-energy-and-respiration/assets/9700-2026-mj-44-q10-p02-a8ccd29440.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-44-q10",
        "html": "9700-topic-12-energy-and-respiration/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:3)(cid:44)\n10 (a) Glycolysis is a stage of respiration that occurs in the cytoplasm of eukaryotic cells.\nSIHT\nNI Outline the glycolysis stage of respiration.\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\nSIHT\n...................................................................................................................................................\nNI\nETIRW\n...................................................................................................................................................\nTON\n............................................................................................................................................. [4]\nOD\n(b) A group of drugs called statins are taken by many people to help lower the concentration of\ncholesterol in the blood. Sometimes statins can inhibit the production of coenzyme A.\nNIGRAM\nSuggest the effects of a low concentration of coenzyme A on aerobic respiration.\nSIHT\n...................................................................................................................................................\nNI\nETIRW ...................................................................................................................................................\n................................................................................................................................................... TON\nOD\n...................................................................................................................................................\n...................................................................................................................................................\nNIGRAM\n...................................................................................................................................................\n...................................................................................................................................................\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................",
      "pdf_path": "9700-topic-12-energy-and-respiration/print/9700-2026-mj-44-q10-5ce5bc162b.pdf"
    },
    {
      "id": "9700-2026-mj-51-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 17,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-51.pdf?download=1",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p01-1e5ea5c217.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p02-d84ee8487b.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p03-28490a4002.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p04-2dea7bd530.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p05-89b35ae2e4.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q01-p06-7abdf42e11.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-51-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 (a) The common woodlouse, Oniscus asellus, is a small invertebrate animal that is found in\nSIHT\nmany areas of western and northern Europe.\nNI\nETIRW\nFigure 1.1 shows some O. asellus.\nTON\nOD\nNIGRAM\nSIHT\nNI 10 mm\nETIRW\nFigure 1.1 TON\nOD\nA student wanted to investigate the effect of temperature on the rate of respiration\nin O. asellus.\nThe student set up a respirometer as shown in Figure 1.2. A stand and clamp were used to\nNIGRAM\nhold the respirometer in position.\nSIHT\nNI\nsyringe\nETIRW\nstart point\nruler\nTON\nthree-way tap OD\nbung\nNIGRAM\nglass tube drop of coloured liquid\nlarge test-tube\nSIHT\nO. asellus NI\nETIRW\ngauze\nTON\nsodium hydroxide pellets\nOD\nFigure 1.2\nTo determine the rate of respiration in O. asellus, the student:\nNIGRAM\n• opened the three‑way tap and used the syringe to move the drop of coloured liquid to\nSIHT the start point\n• closed the three‑way tap\nNI\n• measured the distance moved by the drop of coloured liquid in a known time period.\nETIRW\nTON\nNIGRAM\n(i) Predict the direction of movement of the drop of coloured liquid during the investigation.\nSIHT\nExplain your answer.\nNI\nETIRW\ndirection .............................................................................................................................\nTON explanation ........................................................................................................................\nOD\n...........................................................................................................................................\n[1]\n(ii) Suggest a suitable control for this investigation.\nNIGRAM\n...........................................................................................................................................\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n..................................................................................................................................... [1]\nTON (iii) Describe a method, using the respirometer shown in Figure 1.2, that the student could\nuse to investigate the effect of temperature on the rate of respiration in O. asellus.\nOD\nYou should use temperatures from 5 °C to 35 °C, so that the O. asellus are not harmed\nin the investigation.\nNIGRAM\nYour method should be set out in a logical order and be detailed enough to allow\nanother person to follow it.\nSIHT\n...........................................................................................................................................\nNI\nETIRW\n...........................................................................................................................................\nTON ...........................................................................................................................................\nOD\n...........",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-51-q01-a4ec7377d1.pdf"
    },
    {
      "id": "9700-2026-mj-51-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "51",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 13,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-51.pdf?download=1",
      "source_pages": [
        8,
        9,
        10,
        11,
        12
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_51.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-51-q02-p01-ceaf77fb2d.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q02-p02-32b9ea069e.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q02-p03-48d0cbf5cc.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q02-p04-96385366b0.png",
        "9700-experimental-skills/assets/9700-2026-mj-51-q02-p05-6d310e49c9.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-51-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)\n2 The Galápagos Islands are a group of islands in the Pacific Ocean that are approximately 900 km\nSIHT\nfrom Ecuador in South America. Charles Darwin developed his theory of evolution by natural\nNI selection after visiting the Galápagos Islands in the 1830s.\nETIRW\nFinches are small birds that are commonly found on the Galápagos Islands.\nTON\nSome biologists studied one species of finch, Geospiza conirostris, living on Genovesa Island in\nOD\nthe Galápagos. Genovesa Island is a small island with no mountains.\nFigure 2.1 shows G. conirostris.\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nFigure 2.1\nNI\nETIRW\nThe biologists studied several breeding pairs of G. conirostris. Adult male G. conirostris sing a\nsong to attract females for mating. Female G. conirostris do not sing.\nTON\nThe biologists noticed that adult male G. conirostris could be separated into two groups based on\nOD\ntheir song type. Adult male G. conirostris sing either song type A or song type B.\nAt the start of the investigation, the biologists caught the birds in a mist net.\nNIGRAM Figure 2.2 shows a bird caught in a mist net. The mist net does not harm the bird.\nSIHT\nbill of bird\nNI\nETIRW\nTON\nOD\nmist net\nNIGRAM\nSIHT\nNI\nETIRW\nFigure 2.2\nTON\nNIGRAM\nFor each male G. conirostris caught in the mist net, the biologists:\nSIHT\n• NI attached a coloured ring to the leg of the bird so that it could be easily identified in the future\nETIRW • measured the mass of the bird\n• measured the depth, width and length of the bill of the bird\nTON • released the bird back into the ecosystem\n• observed the song type and other behaviours of the bird over a period of four months.\nOD\n(a) Table 2.1 shows the data from the investigation.\nTable 2.1\nNIGRAM\nsong type of male number of mean mean bill mean bill mean bill\nSIHT\nG. conirostris birds mass / g depth / mm width / mm length / mm\nNI\nsong type A 11 25.85 10.79 9.99 15.47\nETIRW\nsong type B 13 24.96 11.02 10.23 14.55\nTON\nStatistical analysis showed that there was no significant difference between the mean mass, OD\nthe mean bill depth and mean bill width of male G. conirostris with song type A and male\nG. conirostris with song type B.\nThe biologists then carried out a t‑test to compare the mean bill lengths of male G. conirostris NIGRAM\nwith song type A and male G. conirostris with song type B.\nSIHT (i) State two conditions of the data that allowed the biologists to carry out a t‑test.\nNI\n........................................................................................................................................... ETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...............................................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-51-q02-477108a49d.pdf"
    },
    {
      "id": "9700-2026-mj-52-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 20,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-52.pdf?download=1",
      "source_pages": [
        2,
        3,
        4,
        5,
        6,
        7,
        8
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p01-3b3a451364.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p02-50bb9c33d6.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p03-054d5606fb.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p04-f879d16ac3.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p05-fc7c778a82.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p06-f91e301785.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q01-p07-e5745dfc8d.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-52-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 The avocado tree, Persea americana, is a cultivated food crop. The fruits of the tree are\nSIHT\ncalled avocados.\nNI\nETIRW\nFigure 1.1 shows a whole avocado and an avocado that has been cut to show the inside.\nTON\nwhole avocado\nOD\nmesocarp\nseed\nNIGRAM\nSIHT\ncut avocado\n(skin) exocarp\nNI\nETIRW\nFigure 1.1\nTON\nAvocados are removed from the tree (harvested) when they are mature (maximum size).\nOD\nAvocados ripen only after harvesting. As avocados ripen, the colour of the skin changes from\nbright green to dark purple and the mesocarp changes from hard to soft.\nMesocarp cells in the avocado are a source of the enzyme catalase. The quantity of catalase in\nNIGRAM\nthe mesocarp changes as the avocado ripens.\nSIHT Catalase catalyses the breakdown of hydrogen peroxide, as shown in Figure 1.2.\nNI\n2H O + O 2H O\nETIRW 2 2 2 2\nFigure 1.2 TON\nOD\n(a) A student decided to see the effect of catalase activity in avocados.\nThe student removed the skin from an avocado and cut up some pieces of the mesocarp.\nThe student placed some of the pieces of mesocarp into a beaker containing hydrogen\nNIGRAM\nperoxide solution. Tiny bubbles of oxygen were immediately released from the surface of the\nmesocarp, as shown in Figure 1.3.\nSIHT\ncm3 250 beaker\nNI\nETIRW\nhydrogen peroxide solution\nbubble of oxygen gas TON\nOD pieces of mesocarp\nNIGRAM\nFigure 1.3\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nSuggest one hazard of this method and the risk associated with the hazard and the\nSIHT\nprecaution that should be taken.\nNI\nETIRW\nhazard .......................................................................................................................................\nTON ...................................................................................................................................................\nOD\nrisk ............................................................................................................................................\n...................................................................................................................................................\nNIGRAM\nprecaution .................................................................................................................................\nSIHT ...................................................................................................................................................\n[1]\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) The student planned to investigate the changes in catalase activity in avocados as they ripen.\nSIHT\nNI (i) Identify the independent variable in the investigation.\nETIRW\n..................................................................................................................................... [1]\nTON\n(ii) Describe a method the student could use to measure changes in catalase activity",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-52-q01-93e29697f9.pdf"
    },
    {
      "id": "9700-2026-mj-52-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "52",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 10,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-52.pdf?download=1",
      "source_pages": [
        10,
        11,
        12,
        13
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_52.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-52-q02-p01-31b0dddb60.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q02-p02-7f91c8a90b.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q02-p03-31775b3da4.png",
        "9700-experimental-skills/assets/9700-2026-mj-52-q02-p04-5786926b6e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-52-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)\n2 A butterfly is a flying insect. One species of butterfly, the cabbage white butterfly, Pieris brassicae,\nSIHT\nis shown in Figure 2.1.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nFigure 2.1 TON\nOD\nSome scientists are carrying out a long-term project to estimate the population size of all butterfly\nspecies in the UK. The scientists have asked people living in the UK to help them to collect\nthe data.\nNIGRAM\nA modified belt transect is one of the methods used to collect the data.\nThe recommended method for the modified belt transect is to:\nSIHT\nNI • walk at a slow pace along a transect line\nETIRW\n• identify all the butterfly species within 2.5 m on each side of the transect line\n• count the number of individuals of each species within 2.5 m on each side of the\nTON\ntransect line.\nOD\n(a) (i) Suggest how each species of butterfly could be identified.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT ..................................................................................................................................... [1]\nNI\n(ii) The scientists standardised the width of the belt transect. ETIRW\nState two other variables that should be standardised when carrying out this\nTON\nbelt transect.\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\nSIHT\n..................................................................................................................................... [2]\nNI\nETIRW\nTON\nNIGRAM\n(iii) Discuss the limitations of the method used to collect data.\nSIHT\nNI ...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\n...........................................................................................................................................\nOD\n...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM\n...............................................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-52-q02-f4051f1c28.pdf"
    },
    {
      "id": "9700-2026-mj-53-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 19,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-53.pdf?download=1",
      "source_pages": [
        2,
        3,
        4,
        5,
        6
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-53-q01-p01-741f4b6b0e.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q01-p02-72458b9f2e.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q01-p03-a33a3f2ebc.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q01-p04-fa1848450a.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q01-p05-2427e37130.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-53-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 A student planned to investigate the rate of respiration in germinating seeds and then determine\nSIHT\nthe respiratory quotient (RQ) of the germinating seeds.\nNI\nETIRW\nThe student used the respirometer shown in Figure 1.1 for the investigation.\nTON\nOD\nNIGRAM\nSIHT\ncm3 syringe 1\nNI\nETIRW tap\nTON\nscale\nOD\ngerminating\nseeds NIGRAM\nSIHT\ngauze\nNI\nETIRW\nsodium hydroxide solution TON\ncapillary tubing containing\ncoloured liquid OD\nFigure 1.1\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) The student investigated the effect of temperature on the rate of respiration in the\nSIHT\ngerminating seeds.\nNI\nETIRW\nThe student planned to use a temperature range of 10–50 °C in the investigation.\nTON The student read that in some species:\nOD\n• at lower temperatures, the rate of respiration approximately doubles for each 10 °C\nincrease in temperature\n• the optimum temperature for respiration is approximately 40 °C.\nNIGRAM\n(i) Sketch a curve on Figure 1.2 to show the results that you predict the student will\nproduce from this investigation.\nSIHT\nInclude labelled axes. Do not include units or values on the axes.\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 1.2\n[3]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(ii) The student was supplied with the respirometer in Figure 1.1, standard laboratory\nSIHT\nequipment and standard laboratory chemicals.\nNI\nETIRW\nDescribe a method the student could use to take the measurements needed for\ncalculating the rates of respiration at different temperatures.\nTON\nThe description of your method should be set out in a logical way and be detailed\nOD\nenough for another person to follow.\n...........................................................................................................................................\nNIGRAM\n...........................................................................................................................................\n........................................................................................................................................... SIHT\nNI\n...........................................................................................................................................\nETIRW\n...........................................................................................................................................\nTON\nOD ...........................................................................................................................................\n...........................................................................................................................................\nNIGRAM ...........................................................................................................................................\n.................................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-53-q01-9681fc9ccf.pdf"
    },
    {
      "id": "9700-2026-mj-53-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "53",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-53.pdf?download=1",
      "source_pages": [
        7,
        8,
        9
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_53.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-53-q02-p01-37c558448c.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q02-p02-bcf33bd6c4.png",
        "9700-experimental-skills/assets/9700-2026-mj-53-q02-p03-b7dd7479ef.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-53-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44)\n2 Approximately 50% of cases of coronary heart disease are thought to be a result of high blood\nSIHT\npressure. Approximately 25% of adults worldwide have high blood pressure.\nNI\nETIRW Blood pressure changes during the cardiac cycle, so the measurement of blood pressure uses\ntwo values: systolic blood pressure and diastolic blood pressure.\nTON\nA person who is described as having high blood pressure will have a resting systolic blood\nOD\npressure of 140 mm Hg or higher and a resting diastolic blood pressure of 90 mm Hg or higher.\nThere is some evidence that eating garlic can have positive effects on human health, including\nbenefits to the cardiovascular system.\nNIGRAM\nThe storage organ (bulb) of the garlic plant, Allium sativum, is eaten worldwide. Figure 2.1 shows\na garlic bulb.\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 2.1\nGarlic extract can be obtained from garlic bulbs and placed into capsules that can be swallowed.\nNIGRAM\nScientists investigated the effect of eating garlic capsules on the blood pressure of people that\nhad been diagnosed with high blood pressure.\nSIHT\nThe investigation studied 84 men and women of various ages. Each person was randomly placed\nNI\nin one of four groups and asked to follow a specific treatment for 12 weeks. ETIRW\n• 21 people ate one garlic capsule each day. TON\n• 21 people ate two garlic capsules each day.\nOD\n• 21 people ate four garlic capsules each day.\n• 21 people were placed in a control group.\nAll the people continued taking any medication for high blood pressure that they had previously\nNIGRAM\nbeen prescribed.\nThe resting systolic blood pressure of each person was measured at the start of the investigation\nSIHT\nand at intervals during the 12 weeks.\nNI\nETIRW (a) Identify the independent variable and the dependent variable in this investigation.\nindependent variable ................................................................................................................ TON\nOD\ndependent variable ...................................................................................................................\n[2]\nNIGRAM\n(b) Suggest a suitable treatment for the people in the control group.\n...................................................................................................................................................\nSIHT\n................................................................................................................................................... NI\nETIRW\n............................................................................................................................................. [1]\nNIGRAM\n(c) The results for the blood pressure measurements are shown in Table 2.1.\nSIHT\nNI The standard errors (SE) for each value are shown in brackets.\nETIRW\nTable 2.1\nTON\nOD mean resting systolic blood pressure / mm Hg (± SE)\nnumber of\nweeks after\n1 garlic 2 garlic 4 garl",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-53-q02-b3ef501cd9.pdf"
    },
    {
      "id": "9700-2026-mj-54-q01",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "54",
      "question_number": 1,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-54.pdf?download=1",
      "source_pages": [
        2,
        3,
        4,
        5
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-54-q01-p01-de27d91142.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q01-p02-3a866e1d63.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q01-p03-0e1722a28c.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q01-p04-d311f43746.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-54-q01",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)\n1 The stinging nettle, Urtica dioica, is a plant species that is commonly found in many areas of the\nSIHT\nworld. Stinging nettles grow in woodland, in gardens and in farmland.\nNI\nETIRW\nFigure 1.1 shows a stinging nettle.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nFigure 1.1 NIGRAM\nA student decided to investigate photosynthesis in stinging nettles.\nSIHT\n(a) The student made a leaf extract containing chloroplast pigments by grinding some leaves\nNI\nfrom stinging nettles. The student used chromatography to identify the chloroplast pigments ETIRW\npresent in the leaves.\nTON\nFigure 1.2 shows the chromatogram.\nOD\nsolvent front\npigment 1\nNIGRAM\npigment 2\nSIHT\nNI\npigment 3\nETIRW\npigment 4 TON origin\nOD\nNIGRAM\nFigure 1.2\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nTable 1.1 shows the reference R values of some chloroplast pigments from a\nSIHT f\npublished source.\nNI\nETIRW\nTable 1.1\nTON\nchloroplast pigment R value\nf\nOD\ncarotene 0.95\nchlorophyll a 0.65\nchlorophyll b 0.45 NIGRAM\nphaeophytin 0.83\nSIHT\nxanthophyll 0.71\nNI\nETIRW\nUse Figure 1.2 and Table 1.1 to identify pigment 1 and pigment 2.\nTON pigment 1 ..................................................................................................................................\nOD\npigment 2 ..................................................................................................................................\n[2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) (i) The student read that different light intensities result in different combinations of\nSIHT\nchloroplast pigments present in the leaves of stinging nettles.\nNI\nETIRW\nThe student decided to investigate the different combinations of chloroplast pigments\npresent in the leaves of stinging nettles growing in:\nTON\n• direct sunlight, with high light intensity\nOD\n• the shade, with low light intensity.\nDescribe a method that the student could use to:\nNIGRAM\n• collect leaves from stinging nettles growing in direct sunlight and in the shade\n• use chromatography to separate the chloroplast pigments present in the leaves\n• collect the data needed to identify each chloroplast pigment. SIHT\nNI\nThe student had access to standard laboratory equipment and a suitable solvent for\nETIRW\nchromatography. Your method should be set out in a logical order and be detailed\nenough to allow another person to follow it.\nTON\nOD You should not include a risk assessment.\n...........................................................................................................................................\n........................................................................................................................................... NIGRAM\n...........................................................................................................................................\nSIHT\n...................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-54-q01-e531ae2689.pdf"
    },
    {
      "id": "9700-2026-mj-54-q02",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "54",
      "question_number": 2,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 8,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-54.pdf?download=1",
      "source_pages": [
        6,
        7,
        8,
        10,
        11
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-54-q02-p01-9d400ce6c9.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q02-p02-093754f74a.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q02-p03-b90430b8ae.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q02-p04-608efbae73.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q02-p05-59de6a531e.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-54-q02",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)\n2 Anabaena doliolum is a species of photosynthetic prokaryote that is found in many freshwater\nSIHT\necosystems. The individual cells of A. doliolum form long chains.\nNI\nETIRW\nFigure 2.1 shows a photomicrograph of several chains of A. doliolum.\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\n×600\nOD\nFigure 2.1\nSome scientists investigated the effect of sodium chloride concentration on the rate of\nNIGRAM\nphotosynthesis of A. doliolum by carrying out two different experiments.\nSIHT In the first experiment, the scientists estimated the rate of photosynthesis of A. doliolum by\ndetermining the rate of oxygen production.\nNI\nETIRW\ndm−3 • solution of A sample of A. doliolum cells was placed in a set volume of 0.25 mol\nsodium chloride.\nTON\n• An oxygen electrode was used to determine the rate of oxygen production.\nOD dm−3 • The procedure was repeated with 0.20, 0.15, 0.10, 0.05 and 0.00 mol solutions of\nsodium chloride.\nIn the second experiment, the scientists estimated the rate of photosynthesis of A. doliolum by\nNIGRAM determining the rate of DCPIP reduction. DCPIP changes colour from blue to colourless when it\nis reduced.\nSIHT\n• An enzyme was added to a sample of A. doliolum cells. The enzyme partially hydrolysed the\nbacterial cell walls of A. doliolum. NI\nETIRW dm−3 • The sample of A. doliolum cells was placed in a test-tube with a set volume of 0.25 mol\nsolution of sodium chloride. This mixture was green in colour.\n• A set volume of DCPIP was added to the test-tube to give a blue-green colour. TON\n• The time taken for the mixture in the test-tube to change colour from blue-green to green\nOD\nwas measured and the rate of DCPIP reduction was determined. The rate of reduction\nof DCPIP is a measure of the rate of NADP reduction during the light-dependent stage\nof photosynthesis.\ndm−3 • The procedure was repeated with 0.20, 0.15, 0.10, 0.05 and 0.00 mol solutions of\nNIGRAM\nsodium chloride.\nSIHT\nNI\nETIRW\nTON\nNIGRAM\n(a) Identify the two dependent variables in this investigation.\nSIHT\nNI first experiment .........................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\nsecond experiment ...................................................................................................................\nOD\n...................................................................................................................................................\n[2]\nNIGRAM\n(b) The scientists standardised the light intensity, temperature and volumes of solutions in the\nfirst and second experiments. SIHT\nNI\nIdentify one other variable that should be standardised in the second experiment.\nETIRW\n.............................................................................",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-54-q02-894e9e4951.pdf"
    },
    {
      "id": "9700-2026-mj-54-q03",
      "subject": "9700",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 5,
      "variant": "54",
      "question_number": 3,
      "topic_number": null,
      "topic": "Planning, analysis and evaluation",
      "topic_slug": "9700-experimental-skills",
      "marks": 11,
      "source_pdf": "https://pastpapers.co/api/file/caie/a-level/biology-9700/2026-may-june/9700-biology-june-2026-question-paper-54.pdf?download=1",
      "source_pages": [
        12,
        13,
        14,
        15,
        16,
        17
      ],
      "local_pdf": "_source-pdfs/2026-May-June/qp/9700_s26_qp_54.pdf",
      "image_paths": [
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p01-38987631b0.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p02-f88d4ae6e3.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p03-df29d4e73f.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p04-f04ae8991c.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p05-3f2c8921d1.png",
        "9700-experimental-skills/assets/9700-2026-mj-54-q03-p06-60ee9fb9f0.png"
      ],
      "answer": {
        "status": "pending",
        "id": "9700-2026-mj-54-q03",
        "html": "9700-experimental-skills/answers.html"
      },
      "text_excerpt": "NIGRAM\n(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44)\n3 Female Anopheles mosquitoes are the vectors that transmit malaria. Use of insecticides\nSIHT\nto control the transmission of malaria has resulted in many Anopheles mosquitoes\nNI becoming insecticide-resistant.\nETIRW\nThe fungus Metarhizium pingshaense infects and kills mosquitoes. Some biologists investigated\nTON how this fungus could be used to prevent the transmission of malaria by Anopheles mosquitoes.\nThe biologists also genetically modified M. pingshaense to produce a toxin that is usually made\nOD\nby spiders.\n(a) The biologists carried out a laboratory procedure to investigate the effect of M. pingshaense\non the reproduction of Anopheles coluzzii mosquitoes. The percentage of female A. coluzzii\nNIGRAM\nmosquitoes that produced eggs was determined for:\n• a control group of uninfected A. coluzzii mosquitoes SIHT\n• A. coluzzii mosquitoes infected with M. pingshaense that had not been genetically\nNI\nmodified (non-GM)\nETIRW\n• A. coluzzii mosquitoes infected with M. pingshaense that had been genetically\nmodified (GM).\nTON\nOD Figure 3.1 shows the results of the investigation. Error bars show 95% confidence intervals\n(95% CI).\n100\nNIGRAM\nSIHT\n80\nNI\nETIRW\nTON\n60\nOD mean percentage of female\nA. coluzzii mosquitoes that\nproduced eggs\n40\nNIGRAM\nSIHT\n20\nNI\nETIRW\nTON\n0\nOD control non-GM GM\nM. pingshaense M. pingshaense\nmosquito treatment\nNIGRAM\nFigure 3.1\nSIHT\nNI\nETIRW\nTON\nNIGRAM\nUse Figure 3.1 to outline what can be concluded from the data.\nSIHT\nNI ...................................................................................................................................................\nETIRW\n...................................................................................................................................................\nTON\n...................................................................................................................................................\nOD\n...................................................................................................................................................\n............................................................................................................................................. [2]\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nTON\nOD\nNIGRAM\nSIHT\nNI\nETIRW\nNIGRAM\n(b) Malaria is a common infectious disease in Burkina Faso, a country in west Africa.\nSIHT\nNI The biologists investigated the effect of GM M. pingshaense on the reproduction of\nETIRW\nA. coluzzii mosquitoes in a natural environment in Burkina Faso.\nTON The biologists set up two 9.55 m × 9.10 m experimental areas in Burkina Faso in 2017. With\nthe help of people from Burkina Faso, the biologists:\nOD\n• built a traditional hut inside each experimental area\n• planted fruit trees as a source of sugar for the mosquitoes to feed on\n• made small ponds as a source",
      "pdf_path": "9700-experimental-skills/print/9700-2026-mj-54-q03-f82827f69b.pdf"
    }
  ]
}