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AQA · GCSE · Biology

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  • 1

    세포 생물학

    1.1

    Cell biology: the unit of life

    Every living thing is built from cells, and the differences between them decide what each can do. This reference covers AQA GCSE Biology 8461, topic 4.1 Cell biology.

    How the exam treats this topic:

    • Paper 1 (4.1–4.4) carries Cell biology. Required practicals: RP1 (light microscope), RP2 (antiseptics/antibiotics on agar, biology only) and RP3 (osmosis in plant tissue).
    • Culturing microorganisms is biology only; standard-form answers for bacterial numbers are HT only.
    • Calculations the exam demands: magnification (image ÷ real), real size from image and magnification, area by πr², percentage change in mass, surface-area-to-volume ratio, bacterial division counts.
    1.1

    Cell structure: eukaryotes and prokaryotes

    Syllabus

    Cell structure, specialisation, microscopy and culturing (AQA 8461 statements 4.1.1.1-4.1.1.6, RP1, RP2).

    1. Describe plant, animal and bacterial cells, naming structures and their functions.
    2. Relate specialised cells' structures to their functions.
    3. Explain the importance of cell differentiation.
    4. Calculate magnification, real size and image size, with standard form and prefixes.
    5. Describe aseptic technique with reasons, bacterial division counts and zone areas (biology only).

    출처: Cambridge International syllabus

    An animal cell, a plant cell and a bacterial cell with their key structures labelled.
    Learn the labels — the exam asks you to name and function them.
    Structure Function
    nucleus genetic material (DNA); controls the cell
    cytoplasm jelly where reactions happen
    cell membrane controls what enters and leaves
    mitochondria aerobic respiration releases energy
    ribosomes protein synthesis
    chloroplast (plant) photosynthesis
    permanent vacuole (plant) cell sap, keeps the cell firm
    cell wall (plant, cellulose) strength and support
    • Eukaryotic 真核 cells (plant and animal): genetic material enclosed in a nucleus.
    • Prokaryotic 原核 cells (bacteria) are much smaller; cytoplasm and cell membrane surrounded by a cell wall; genetic material is not in a nucleus — a single DNA loop plus one or more plasmids (small rings of DNA).

    Scale: use the prefixes centi (10⁻²), milli (10⁻³), micro (10⁻⁶), nano (10⁻⁹) and order-of-magnitude comparisons in standard form.

    English 한국어
    eukaryotic/ˌjuːkərɪˈɒtɪk/ 진핵
    prokaryotic/ˌprɒkərɪˈɒtɪk/ 원핵
    1.1

    Cell specialisation, differentiation and microscopy

    Syllabus

    Cell structure, specialisation, microscopy and culturing (AQA 8461 statements 4.1.1.1-4.1.1.6, RP1, RP2).

    1. Describe plant, animal and bacterial cells, naming structures and their functions.
    2. Relate specialised cells' structures to their functions.
    3. Explain the importance of cell differentiation.
    4. Calculate magnification, real size and image size, with standard form and prefixes.
    5. Describe aseptic technique with reasons, bacterial division counts and zone areas (biology only).

    출처: Cambridge International syllabus

    Specialised cells relate structure to function (learn one feature ↔ one advantage each):

    Cell Adaptation Function
    sperm cell tail for swimming, many mitochondria, acrosome with enzymes fertilisation
    nerve cell long axon, insulating sheath carrying impulses
    muscle cell many proteins fibres, many mitochondria contraction
    root hair cell large surface area, thin wall absorbing water and minerals
    xylem cell hollow, lignified walls, no end walls transporting water
    phloem cell sieve plates, few organelles transporting sugars

    Differentiation 分化: cells acquire different sub-cellular structures to become specialised. Animal cells differentiate mostly at an early stage; plant cells throughout life. In mature animals, cell division is mainly for repair and replacement.

    Microscopy: a light microscope uses lenses; an electron microscope has much higher magnification and resolving power 分辨能力, so far more sub-cellular structures could be seen and understood.

    $$\text{magnification} = \frac{\text{size of image}}{\text{size of real object}}$$

    Worked example. A cell image is 4.8 cm wide at magnification ×1200.

    • Convert first: $4.8\ \text{cm} = 4.8\times10^{-2}$ m $= 48\,000\ \mu\text{m}$.
      $$\text{real size} = \frac{\text{image size}}{\text{magnification}} = \frac{48\,000\ \mu\text{m}}{1200} = 40\ \mu\text{m}$$

    RP1: observe, draw and label plant and animal cells with a magnification scale included (e.g. a scale bar or the printed magnification).

    English 한국어
    Differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/ 분화
    resolving power/rɪˈzɒlvɪŋ ˈpaʊə/ 해상능
    1.1

    Culturing microorganisms (biology only)

    Syllabus

    Cell structure, specialisation, microscopy and culturing (AQA 8461 statements 4.1.1.1-4.1.1.6, RP1, RP2).

    1. Describe plant, animal and bacterial cells, naming structures and their functions.
    2. Relate specialised cells' structures to their functions.
    3. Explain the importance of cell differentiation.
    4. Calculate magnification, real size and image size, with standard form and prefixes.
    5. Describe aseptic technique with reasons, bacterial division counts and zone areas (biology only).

    출처: Cambridge International syllabus

    Bacteria multiply by binary fission 二分裂 as often as once every 20 minutes with enough nutrients and a suitable temperature. They grow in nutrient broth or as colonies on agar gel.

    Aseptic technique 无菌操作 (each point earns a mark — know the why):

    • sterilise Petri dishes and culture media before use — kills unwanted microorganisms;
    • flame the inoculating loop before transferring bacteria — kills contaminants on the loop;
    • secure the lid with adhesive tape and store upside down — stops airborne contamination and condensation dripping;
    • in schools, incubate at a maximum of 25 °C — prevents growth of pathogens harmful to humans.

    Worked example (division count). One bacterium divides every 20 minutes for 3 hours.

    • 3 hours = 180 min = 9 division times; number $= 2^9 = 512$.

    Worked example (zone area). A clear zone of radius 4 mm around an antibiotic disc.

    $$A = \pi r^2 = \pi \times 4^2 = 50\ \text{mm}^2\ \text{(2 s.f.)}$$

    RP2: effect of antiseptics/antibiotics on bacterial growth — measure zones of inhibition with a ruler (several diameters, mean, ÷2), controls with no substance.

    English 한국어
    binary fission/ˈbaɪnəri ˈfɪʃn/ 이분열
    Aseptic technique/æˈseptɪk tekˈniːk/ 무균 조작법
    1.2

    Cell division: chromosomes, mitosis and the cell cycle

    Syllabus

    세포 분열 및 줄기세포 (AQA 8461 문항 4.1.2.1-4.1.2.3).

    1. 염색체와 체세포에서의 쌍 형성 과정을 설명하시오.
    2. 세포 주기를 설명하되, 총 3단계와 미토oses을 포함하시오.
    3. 배아, 성인 동물 및 식물 생장점에서의 줄기세포 기능을 설명하시오.
    4. 의학적 용도와 식물 복제에서의 줄기세포 사용을 평가하되, 위험성과 윤리적 반대 의견을 포함하시오.

    출처: Cambridge International syllabus

    • The nucleus contains chromosomes 染色体 made of DNA; each chromosome carries many genes; body cells have chromosomes in pairs (humans: 23 pairs).
    • The cell cycle: DNA replicates → growth (more ribosomes and mitochondria) → mitosis 有丝分裂 (one set of chromosomes pulled to each end; nucleus divides) → cytoplasm and membranes divide → two identical cells.

    Mitosis matters for growth and development of multicellular organisms, and repair. Recognise contexts where mitosis is occurring (a growing root tip, healing skin, a tumour).

    English 한국어
    chromosomes/ˈkrəʊməsəʊmz/ 염색체
    mitosis/maɪˈtəʊsɪs/ 유사분열
    1.2

    Stem cells

    Syllabus

    세포 분열 및 줄기세포 (AQA 8461 문항 4.1.2.1-4.1.2.3).

    1. 염색체와 체세포에서의 쌍 형성 과정을 설명하시오.
    2. 세포 주기를 설명하되, 총 3단계와 미토oses을 포함하시오.
    3. 배아, 성인 동물 및 식물 생장점에서의 줄기세포 기능을 설명하시오.
    4. 의학적 용도와 식물 복제에서의 줄기세포 사용을 평가하되, 위험성과 윤리적 반대 의견을 포함하시오.

    출처: Cambridge International syllabus

    A stem cell 干细胞 is an undifferentiated cell that can give rise to many more of the same type and, by differentiation, to other cell types.

    Source Can become
    embryo most types of human cell
    adult bone marrow many cells, including blood cells
    plant meristems any type of plant cell, throughout life

    Uses and evaluation (the credited pairs — benefit and risk/objection):

    • Medicine: potential treatment for diabetes and paralysis; therapeutic cloning makes an embryo with the patient's own genes, so cells are not rejected.
    • Risks/ethics: transfer of viral infection; ethical and religious objections to using embryos.
    • Plants: meristem clones — rare species preserved from extinction; disease-resistant crops cloned quickly and economically.
    English 한국어
    stem cell/stem sel/ 줄기세포
    1.3

    Transport in cells

    Syllabus

    세포 내 수송 (AQA 8461 문장 4.1.3, RP3).

    1. 확산, 삼투 및 능동수송을 정의하고 예시와 에너지 차이를 설명하시오.
    2. 확산 속도에 영향을 미치는 요인을 설명하시오.
    3. 표면적 대 부피 비율을 계산하고 비교하며, 교환 표면의 적응 현상을 설명하시오.
    4. 용질 농도가 식물 조직 질량에 미치는Effect를 조사하고 백분율 변화율을 계산하시오 (RP3).

    출처: Cambridge International syllabus

    Process Direction Energy Examples
    diffusion 扩散 high → low concentration (net) no O₂/CO₂ in gas exchange; urea from cells to blood plasma
    osmosis 渗透 water from dilute → concentrated solution through a partially permeable membrane no water into plant cells
    active transport 主动运输 low → high (against the gradient) yes — from respiration mineral ions into root hairs; glucose from gut to blood

    Rate of diffusion increases with: concentration gradient, temperature, surface area of the membrane.

    Diffusion, osmosis and active transport compared in three panels.
    Direction and energy separate the three processes.
    Two cubes showing how doubling the side halves the surface-area-to-volume ratio.
    Bigger organisms need exchange surfaces.

    Surface area to volume ratio: single-celled organisms have a large SA:V — enough exchange across the surface alone. Multicellular organisms need exchange surfaces and a transport system. Exchange surfaces are effective by: large surface area, thin membrane (short diffusion path), (animals) good blood supply, (animals, gas exchange) ventilation.

    Adapted exchange surfaces to know: small intestine and lungs in mammals, gills in fish, roots and leaves in plants.

    Worked example (SA:V). A cube of side 2 mm: SA $= 6\times2^2 = 24$ mm², V $= 8$ mm³, SA:V $= 3:1$. A cube of side 4 mm: SA $= 96$, V $= 64$, SA:V $= 1.5:1$ — doubling the side halves the ratio.

    RP3 (osmosis): potato pieces (same surface area, blotted dry, exact masses) in a range of sugar or salt solutions; measure percentage change in mass:

    $$\%\ \text{change} = \frac{\text{change in mass}}{\text{initial mass}} \times 100$$

    Interpretation: gain in dilute solutions (water enters); no change where the solution matches the cell concentration; loss in concentrated solutions (water leaves). Control variables: volume of solution, temperature, time, surface area.

    English 한국어
    diffusion/dɪˈfjuːʒn/ 확산
    osmosis/ɒzˈməʊsɪs/ 삼투
    active transport/ˈæktɪv ˈtrænspɔːt/ 능동 수송
    1.3

    Checklist before you call this topic done

    • Label animal, plant and bacterial cells; match every structure to its function.
    • Specialised cells: one structural feature linked to one function each.
    • Magnification, real size and image size — convert units before dividing.
    • Aseptic technique whys; binary-fission counts; πr² zones (RP1, RP2).
    • Cell cycle stages and where mitosis occurs; stem cell sources with benefits and risks.
    • Diffusion, osmosis, active transport: direction, energy, examples; the three rate factors.
    • SA:V calculations; four exchange-surface features; percentage change in mass (RP3).
  • 2

    조직

    2.1

    Organisation: from cells to systems

    Cells form tissues, tissues form organs, organs form systems — and the digestive system, blood and plants each show the pattern. This reference covers AQA GCSE Biology 8461, topic 4.2 Organisation.

    How the exam treats this topic:

    • Paper 1 carries Organisation. Required practicals: RP4 (food tests) and RP5 (pH and amylase rate).
    • You must evaluate heart-disease treatments and stem-cell/lifestyle risk data, and interpret graphs and tables of risk factors.
    • Recall exactly: the enzymes table, the heart's vessels, blood components, plant tissues, transpiration factors.
    2.1

    Principles of organisation

    Syllabus

    Principles of organisation and the digestive system (AQA 8461 statements 4.2.1-4.2.2.1, RP4, RP5).

    1. Order cells, tissues, organs, organ systems and organisms.
    2. Describe enzyme nature and lock-and-key action; relate activity to temperature and pH.
    3. Recall sites and actions of amylase, proteases and lipases, with word equations.
    4. Explain bile's two functions; test foods (RP4) and investigate pH on amylase (RP5).

    출처: Cambridge International syllabus

    Cells → tissues → organs → organ systems → organism.

    • A tissue 组织 is a group of cells with a similar structure and function.
    • An organ 器官 aggregates tissues performing specific functions.
    • Organ systems 器官系统 work together to form the whole organism.

    The digestive system is the named example of an organ system.

    English 한국어
    tissue/ˈtɪʃuː/ 조직
    organ/ˈɔːɡən/ 장기
    organ systems/ˈɔːɡən ˈsɪstəmz/ 장기계
    2.1

    The human digestive system (4.2.2.1)

    Syllabus

    Principles of organisation and the digestive system (AQA 8461 statements 4.2.1-4.2.2.1, RP4, RP5).

    1. Order cells, tissues, organs, organ systems and organisms.
    2. Describe enzyme nature and lock-and-key action; relate activity to temperature and pH.
    3. Recall sites and actions of amylase, proteases and lipases, with word equations.
    4. Explain bile's two functions; test foods (RP4) and investigate pH on amylase (RP5).

    출처: Cambridge International syllabus

    Enzymes 酶 are biological catalysts: proteins with an active site whose shape fits one substrate — the lock and key model. Temperature and pH change the rate; far from the optimum the enzyme denatures (active site changes shape).

    Enzyme Made in Acts on Products
    amylase (a carbohydrase) salivary glands, pancreas, small intestine starch simple sugars (glucose)
    protease stomach, pancreas, small intestine protein amino acids
    lipase pancreas, small intestine lipids glycerol + 3 fatty acids

    Digestive enzymes turn food into small soluble molecules that can be absorbed into the bloodstream. The products build new carbohydrates, lipids and proteins; some glucose is used in respiration.

    Bile 胆汁: made in the liver, stored in the gall bladder. It is alkaline, neutralising hydrochloric acid from the stomach, and emulsifies fat into small droplets, increasing the surface area — both raise the rate of fat breakdown by lipase.

    RP4 food tests (learn reagent, colour change, substance):

    Test Reagent Positive result
    sugars (reducing) Benedict's, heat blue → brick-red
    starch iodine orange/brown → blue-black
    protein Biuret blue → purple
    lipids Sudan III red-stained layer on top

    RP5 (pH and amylase): buffer solutions at pH values, amylase + starch, continuous sampling onto a spotting tile with iodine, record the time for the starch to disappear — shortest time = fastest rate at the optimum pH.

    English 한국어
    enzyme/ˈenzaɪm/ 효소
    Bile/baɪl/ 담즙
    2.2

    The heart, blood vessels and blood (4.2.2.2–4.2.2.3)

    Syllabus

    심장, 혈관, 혈액 및 관상혈관질환 (AQA 8461 문장 4.2.2.2-4.2.2.4).

    1. 심장과 이중 순환계를 서술하며, 다섯 개의 명명된 혈관과_journal maker_ 세포를 기술하시오.
    2. 동맥, 정맥 및 모세혈관의 구조가 기능과 어떻게 연결되는지 서술하시오.
    3. 혈장, 적혈구, 백혈구 및 혈소판의 특징과 적응 현상을 알고 있되 서술하시오.
    4. CHD를 설명하고, 스텐트, 스타틴, 판막, 이식 수술 및 인공 심장의 평가를 논하시오.

    출처: Cambridge International syllabus

    The human heart with its four chambers and the five named vessels.
    Right side to the lungs; left side to the body.
    • The heart pumps blood in a double circulatory system: the right ventricle pumps blood to the lungs (gas exchange); the left ventricle pumps it around the rest of the body.
    • Vessels to know: aorta, vena cava, pulmonary artery, pulmonary vein, coronary arteries.
    • The resting heart rate is set by cells in the right atrium — the natural pacemaker; artificial pacemakers correct irregular rates.
    • Lungs: trachea → bronchi → alveoli 肺泡, surrounded by capillaries; alveoli are adapted by a large surface area, thin walls, good blood supply and ventilation.
    Vessel Structure Function
    artery thick, elastic, muscular walls; small lumen carries blood away at high pressure
    vein thin walls, large lumen, valves returns blood at low pressure
    capillary one cell thick exchange with tissues
    Red cells, white cells, plasma and platelets.

    Blood is a tissue: plasma 血浆 (carries dissolved sugars, amino acids, CO₂, urea, hormones, and heat), red blood cells 红细胞 (no nucleus, packed with haemoglobin, biconcave — large surface area for oxygen transport), white blood cells 白细胞 (engulf pathogens or make antibodies), platelets 血小板 (clotting).

    English 한국어
    alveoli/ˈælvɪɒli/ 포포
    plasma/ˈplæzmə/ 혈장
    red blood cells/red blʌd selz/ 적혈구
    white blood cells/waɪt blʌd selz/ 백혈구
    platelets/ˈpleɪtlɪts/ 혈song
    2.2

    Coronary heart disease (4.2.2.4) and treatments

    Syllabus

    심장, 혈관, 혈액 및 관상혈관질환 (AQA 8461 문장 4.2.2.2-4.2.2.4).

    1. 심장과 이중 순환계를 서술하며, 다섯 개의 명명된 혈관과_journal maker_ 세포를 기술하시오.
    2. 동맥, 정맥 및 모세혈관의 구조가 기능과 어떻게 연결되는지 서술하시오.
    3. 혈장, 적혈구, 백혈구 및 혈소판의 특징과 적응 현상을 알고 있되 서술하시오.
    4. CHD를 설명하고, 스텐트, 스타틴, 판막, 이식 수술 및 인공 심장의 평가를 논하시오.

    출처: Cambridge International syllabus

    In CHD layers of fatty material build up inside the coronary arteries, narrowing them — less blood, so less oxygen for the heart muscle. Evaluate each treatment with benefits and risks:

    Treatment Benefit Risk/limitation
    stents 支架 keeps the artery open, restores flow surgery risk; doesn't treat the cause
    statins 他汀 reduce blood cholesterol, slowing fatty deposits long-term medication with side effects (liver)
    valve replacement (biological or mechanical) restores one-way flow major surgery; mechanical need lifelong anti-clotting drugs
    transplant (heart / heart and lungs) cures failure donor shortage; rejection — immunosuppressants
    artificial heart keeps the patient alive while waiting, or lets the heart rest infection, clotting, bulky
    English 한국어
    stents/stents/ 스텐트
    statins/ˈstætɪnz/ 스타틴
    2.3

    Health, lifestyle and cancer (4.2.2.5–4.2.2.7)

    Syllabus

    건강, 생활 양식 및 암 (AQA 8461 문장 4.2.2.5-4.2.2.7).

    1. 건강을 정의하고, 질환과 기타 요인 간의 상호작용을 서술하시오.
    2. 비전염성 질환에 대한 인과 기제가 입증된 위험 요인을 나열하시오.
    3. 양성 종양과 악성 종양의 차이를 구별하고, 이차 종양을 서술하시오.
    4. 위험 요인 데이터를 해석하며, 상관관계와 인과관계를 구분하시오.

    출처: Cambridge International syllabus

    Health is the state of physical and mental well-being — not just the absence of disease. Diet, stress and life situations affect both; different diseases interact (immune defects → more infections; viruses → some cancers; immune reactions trigger asthma; severe physical illness → depression).

    Risk factors 危险因素 are linked to an increased rate of a disease — aspects of lifestyle or substances in the body/environment. A causal mechanism is proven for some, not others: diet/smoking/exercise → cardiovascular disease; obesity → Type 2 diabetes; alcohol → liver damage and brain function; smoking → lung disease and lung cancer; smoking and alcohol → unborn babies; carcinogens including ionising radiation → cancer. Interpret risk-factor data as correlation; proven mechanism makes it causation.

    Cancer results from changes in cells producing uncontrolled growth and division:

    • Benign tumours: contained in one area, usually within a membrane; do not invade.
    • Malignant tumours (cancers) invade neighbouring tissues and spread in the blood to form secondary tumours.

    Risk factors include lifestyle and, for some cancers, genes.

    English 한국어
    Risk factors/rɪsk ˈfæktəz/ 위험 요인
    2.4

    Plant tissues and transport (4.2.3)

    Syllabus

    식물 조직, 기관 및 수송 (AQA 8461 지문 4.2.3).

    1. 식물 조직(표피, 상피판, 공극판, 관목, 체관, 분생조직, 기공수호세포)과 기능을 연결한다.
    2. 뿌리모, 관목 및 체관의 적응과 증발 및 이송 과정을 설명한다.
    3. 온도, 습도, 공기 흐름 및 빛의 강도가 증발율에 미치는 영향을 설명한다.

    출처: Cambridge International syllabus

    A leaf in cross-section with its tissues labelled.
    Each tissue has one main job.
    • Epidermal tissue: protective covering; waxy cuticle on top.
    • Palisade mesophyll 栅栏组织: packed chloroplasts — most photosynthesis.
    • Spongy mesophyll: air spaces for gas exchange.
    • Xylem 木质部: hollow tubes strengthened by lignin — carries water and mineral ions from roots to leaves in the transpiration stream.
    • Phloem 韧皮部: sieve tubes — carries dissolved sugars from leaves to the rest of the plant (translocation 运输).
    • Meristem: at the growing tips of shoots and roots (stem cells).
    • Guard cells 保卫细胞 surround stomata 气孔, controlling gas exchange and water loss.
    • Root hair cells: large surface area; water enters by osmosis, mineral ions by active transport.

    Transpiration 蒸腾作用 is the loss of water vapour from the leaves (mostly through stomata); it pulls the transpiration stream. The rate increases with: temperature ↑, air movement ↑, light intensity ↑ (stomata open); decreases with humidity ↑. Measure rate by water uptake with a potometer.

    English 한국어
    palisade mesophyll/ˈpælɪseɪd ˈmesəfɪl/ 판자상 피층
    Xylem/ˈzaɪləm/ 관부
    Phloem/ˈfləʊɪm/ 선부
    Guard cells/ɡɑːd selz/ 경계 세포
    stomata/ˈstəʊmətə/ 기공
    Transpiration/trænspəˈreɪʃn/ 증산 작용
    translocation/trænsləʊˈkeɪʃn/ 이송
    2.4

    Checklist before you call this topic done

    • Order cells → tissues → organs → systems with one example each.
    • Enzyme table: name, site, substrate, products; lock and key; denaturing.
    • Bile: two functions with reasons; RP4 tests and colours; RP5 method.
    • Label the heart with all five named vessels; double circulation; pacemaker.
    • Artery/vein/capillary structure ↔ function; four blood components with adaptations.
    • CHD: cause + each treatment's benefit and risk.
    • Risk factors with causal mechanisms; benign vs malignant tumours.
    • Plant tissues and their jobs; xylem vs phloem; four transpiration factors.
  • 3

    감염 및 반응

    3.1

    감염과 반응: 몸속의 전쟁

    병원체가 침입하고, 방어 체계가 대응하며, 약물이 싸움을 준비합니다. 이 참고 자료는 AQA GCSE 생물학 8461, 주제 4.3 감염과 반응을 다룹니다.

    시험에서 이 주제를 다루는 방식:

    • 시험지 1에 이 주제가 포함되며, RP2(agar 배지에서의 항생제/항세균제 험)은 주제 1와 연결됩니다.
    • 단클론 항체(4.3.2)와 식물 질병(4.3.3)은 생물학 전과목이며, 단클론 항체 및 식물 질병 검출은 고급(High Tier) 과정만 해당합니다.
    • 정확히 암기하십시오: 명명된 각 질환 → 병원체 유형, 증상 및 전파 경로; 방어 체계; 약물 테스트 단계.
    3.1

    전염병과 병원체 (4.3.1.1)

    Syllabus

    Communicable diseases, pathogen types and named diseases (AQA 8461 statements 4.3.1.1-4.3.1.5).

    1. Explain how viruses, bacteria, protists and fungi spread in animals and plants, and how spread is reduced.
    2. Recall measles, HIV, TMV, Salmonella, gonorrhoea, rose black spot and malaria with pathogen type, spread, symptoms and control.

    출처: Cambridge International syllabus

    네 가지 병원체 유형과 주요 특징.

    병원체는 전염병을 일으키는 미생물로, 바이러스, 세균, 원생생물, 균류를 의미합니다. 식물이나 동물에 감염되어 직접 접촉, 물 또는 공기를 통해 확산됩니다. 세균은 조직을 손상시키는 독소를 생성할 수 있으며, 바이러스는 세포 내에서 생존하고 복제됩니다.

    확산 방지: 위생(손 씻기), 상처 소독, 깨끗한 물, 하수 처리, 콘돔 사용, 매개체 통제(모기장, 물 고임 제거).

    English 한국어
    Pathogens/ˈpæθədʒnz/ 병원체
    toxins/ˈtɒksɪnz/ 독소
    phagocytosis/ˌfæɡəsɪˈtəʊsɪs/ 식세포작용
    antibody production/ˈæntɪbɒdi prəˈdʌkʃn/ 항체 생성
    antitoxin production/ˌæntɪˈtɒksɪn prəˈdʌkʃn/ 항독소 생성
    Vaccination/ˌvæksɪˈneɪʃn/ 예방접종
    Antibiotics/ˌæntɪbaɪˈɒtɪks/ 항생제
    Monoclonal antibodies/ˈmɒnəʊklɒnl ˈæntɪbɒdiz/ 단일 클론 항체
    hybridoma/ˌhaɪbrɪˈdəʊmə/ 하이브리드ومة
    aphids/ˈeɪfɪdz/ 진드기
    chlorosis/ˈklɔːrəʊsiz/ 황화증
    3.1

    명명된 질환 (4.3.1.2–4.3.1.5)

    Syllabus

    Communicable diseases, pathogen types and named diseases (AQA 8461 statements 4.3.1.1-4.3.1.5).

    1. Explain how viruses, bacteria, protists and fungi spread in animals and plants, and how spread is reduced.
    2. Recall measles, HIV, TMV, Salmonella, gonorrhoea, rose black spot and malaria with pathogen type, spread, symptoms and control.

    출처: Cambridge International syllabus

    질환 병원체 전파 증상/징후 통제
    麻疹 바이러스 비말 흡입(재채기, 기침) 발열, 붉은 피부 발진; 치명적일 수 있음 유아 예방접종
    HIV 바이러스 성접촉, 체액 교환(공유 바늘) 감기 유사 증상; 면역 세포 공격 → 후기 AIDS 항역病毒제; 안전한 성관계; 청결한 바늘
    TMV(담배 모자이크 바이러스) 바이러스(식물) 식물 간 접촉 잎의 모자이크 변색; 광합성 및 성장 저하 내병성 품종; 위생 관리
    살모넬라 세균 비위생적 환경下的 섭취 발열, 경련, 구토, 설사(세균 + 독소) 가금류 예방접종(영국); 위생적인 식품 처리
    임질 세균 성접촉 두꺼운 노란색/녹색 분비물, 배뇨 통증 항생제(내성 증가 중); 콘돔
    장미 흑점病 균류(식물) 물/공기 속 포자 잎에 검은 점; 잎 황변 및 낙엽 잎 제거/소각; 방진제
    말라리아 원생생물 모기 매개체 재발성 발열; 치명적일 수 있음 모기장; 모기 번식 억제
    3.2

    인간 방어 시스템 및 면역 (4.3.1.6–4.3.1.7)

    Syllabus

    인간 방어 시스템, 예방접종 및 약물 (AQA 8461 지문 4.3.1.6-4.3.1.9).

    1. 비특이적 방어 시스템을 기술하고 세 가지 백혈구 작용을 설명한다.
    2. 개인 및 집단 수준에서의 예방접종을 설명한다.
    3. 항생제와 진통제의 차이를 구별하고, 항생제가 바이러스 치료에 사용할 수 없는 이유를 설명하며 내성을 기술한다.
    4. 약물 개발의 출처와 이중-blind 시험을 포함한 임상/비임상 테스트 단계를 설명한다.

    출처: Cambridge International syllabus

    방어 계층 및 약물 테스트 과정.

    비특이적 방어: 피부(장벽, 지은), 코(모, 점액), 기도 및 기관지(점액, 섬모), 위(염산).

    면역계는 침입한 병원체를 파괴한다. 백혈구는 다음과 같이 방어한다:

    1. 포식작용 — 병원체 포식;
    2. 항체 생성 — 특정 항원에 결합하는 단백질;
    3. 해독 물질 생성 — 독소 중화.

    접종: 사멸하거나 비활성화된 병원체의 소량이 백혈구를 자극하여 항체를 생성하게 합니다. 재감염 시 기억 세포가 정확한 항체를 빠르게 반응하여 감염을 예방합니다. 집단 면역: 대다수의 접종으로 전파를 줄여 모든 사람에게 보호 효과를 줍니다. 평가: 이점(매사병과 같은 역병 통제)과 드문 부작용 간 균형.

    3.2

    항생제, 진통제 및 약물 개발 (4.3.1.8–4.3.1.9)

    Syllabus

    인간 방어 시스템, 예방접종 및 약물 (AQA 8461 지문 4.3.1.6-4.3.1.9).

    1. 비특이적 방어 시스템을 기술하고 세 가지 백혈구 작용을 설명한다.
    2. 개인 및 집단 수준에서의 예방접종을 설명한다.
    3. 항생제와 진통제의 차이를 구별하고, 항생제가 바이러스 치료에 사용할 수 없는 이유를 설명하며 내성을 기술한다.
    4. 약물 개발의 출처와 이중-blind 시험을 포함한 임상/비임상 테스트 단계를 설명한다.

    출처: Cambridge International syllabus

    • 항생제(페니실린 등)는 신체 내부의 박테리아를 죽여 세균 질환을 치료합니다. 특정 항생제는 특정 박테리아에 특이적입니다. 바이러스를 죽일 수 없습니다(바이러스는 세포 내에 존재하며, 이를 제거할 경우 신체 조직이 손상됩니다). 내성은 주요 문제이며, 신규 항생제 개발 경쟁이 지속되고 있습니다.
    • 진통제는 증상을 완화하지만 병원체를 죽이지는 않습니다.

    약물 개발: 전통적으로 식물/미생물에서 추출되었음—심장 약용 디ج리탈리스(포기초), 아스피린(버드나무), 페니실린(Penicillium 곰팡이, 플레밍). 대부분의 신약은 화학 합성되지만, 초기 원료로는 여전히 식물 유래 화합물이 사용될 수 있음.

    테스트 단계: 임상 전(세포, 조직, 생체 동물) → 건강자 대상 임상시험(저용량, 안전성 확인) 및 환자 대상 임상시험(최적 용량, 효능 검증) → 위약(cp) 포함 이중 맹검试验; 결과 동료 심사.

    3.3

    단클론 항체(생물학 전공, HT)

    Syllabus

    단일 클론 항체, 생물학 전용 및 HT (AQA 8461 지문 4.3.2).

    1. 하이브리드마에서 단일 클론 항체를 생성하는 과정을 설명한다.
    2. 진단, 측정, 연구 및 암 치료에서의 용도를 기술한다.

    출처: Cambridge International syllabus

    단클론 항체는 단일 클론의 세포에서 생성되며, 하나의 단백질 항원에 있는 하나의 결합 부위에 특이적입니다. 생산 방법: 마우스 림프구를 자극하여 항체 생성 유도 → 암세포와 융합시켜 하이브리드마 생성 — 하이브리드마는 분열하고 동시에 항체를 생산함 → 하이브리드마 클oning → 다량의 특정 항체를 수집 및 정제.

    용도: 임신 검사; 혈액 내 호르몬/화학 물질 수량 측정; 병원체 탐지; 연구—형광 항체가 분자 위치 파악; 암 치료—항체에 방사성/독성 물질을 결합하여 암세포에만 전달.

    3.4

    식물 병해(생물학 전공)

    Syllabus

    식물 질병, 생물학 전용 (AQA 8463 지문 4.3.3).

    1. 징후에서 식물 병질을 감지하고 수동, 실험실 또는 단일 클론 검사로 식별한다.
    2. 질산 및 마그네슘 이온 결핍 현상을 설명한다.
    3. 물리적, 화학적 및 기계적 식물 방어 반응을 기술한다.

    출처: Cambridge International syllabus

    감지(HT): 성장 저해, 잎 반점, 썩음 부위, 종양, 변형된 줄기/잎, 색소 침착, 해충 존재.

    특정 identification(HT): 원예 가이드/웹사이트; 실험실 분석; 단클론 항체를 활용한 검출 키트 사용.

    명시된 식물 문제: TMV(바이러스성), 장미 흑반병(균류), 진딧벌(곤충).

    이온 결핍: 질산염 부족 → 생장 저해 (아미노산/단백질 합성에 필요); 마그네슘 결핍 → 황화현상 (잎 노랑 — 클로로필 합성에 마그네슘 필요).

    방어 반응: 물리적 — 셀룰로스 세포벽, 왁질 각피, 껍질/사층; 화학적 — 항균 물질, 독소; 기계적 — 가시, 털, 잎의 처짐/말림, 모조.

    3.4

    이 주제를 완료한 후 체크리스트

    • 네 가지 병원체 유형; 세 가지 전파 경로; 각 명명된 질병의 전파를 줄이는 두 가지 방법.
    • 질병 표를 암기: 병원체, 전파, 증상, 통제.
    • 네 가지 비특이적 방어 기전; 세 가지 백혈구 작용.
    • 예방접종 기전 + 집단 효과; 항생제와 진통제의 차이; 항생제가 바이러스에 실패하는 이유.
    • 약물 원료 (디지탈리스, 버드나무, 페니실리움) 및 전임상/임상 단계와 위약, 이중맹검법.
    • (고등수업, 생물학) 하이브리도마 생성 순서와 단클론형 항체의 네 가지 용도.
    • (생물학) 식물 병증 징후, 식별 방법, 질산염/마그네슘 결핍, 세 가지 방어 유형.
  • 4

    생물에너지학

    4.1

    Bioenergetics: photosynthesis and respiration

    Plants capture the Sun's energy; every cell releases it again. This reference covers AQA GCSE Biology 8461, topic 4.4 Bioenergetics.

    How the exam treats this topic:

    • Paper 1 carries Bioenergetics. RP6: light intensity and the rate of photosynthesis using pondweed.
    • Limiting-factor graphs with two or three factors, the inverse square law and greenhouse economics are HT only.
    • Equations are word equations — know the symbol forms too (CO₂, H₂O, O₂, C₆H₁₂O₆).
    4.1

    Photosynthesis (4.4.1.1)

    Syllabus

    Photosynthesis (AQA 8461 statements 4.4.1.1-4.4.1.2, RP6).

    1. Write the photosynthesis word and symbol equations, and describe it as endothermic.
    2. Explain how light intensity, CO2 concentration and temperature limit the rate of photosynthesis, including interpreting limiting-factor graphs.
    3. (HT) Use the inverse-square relation between light intensity and distance, and reasons for greenhouse economics.
    4. Describe RP6: light intensity and the rate of photosynthesis with pondweed, its controls and the rate calculation.

    출처: Cambridge International syllabus

    $$\text{carbon dioxide} + \text{water} \xrightarrow{\ \text{light}\ } \text{glucose} + \text{oxygen}$$
    $$6\,\text{CO}_2 + 6\,\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2$$

    Photosynthesis is an endothermic 吸热 reaction: energy is transferred from the environment to the chloroplasts by light.

    English 한국어
    endothermic/ˌendəʊˈθɜːmɪk/ 吸热(吸热反应)
    4.1

    Rate of photosynthesis and limiting factors (4.4.1.2)

    Syllabus

    Photosynthesis (AQA 8461 statements 4.4.1.1-4.4.1.2, RP6).

    1. Write the photosynthesis word and symbol equations, and describe it as endothermic.
    2. Explain how light intensity, CO2 concentration and temperature limit the rate of photosynthesis, including interpreting limiting-factor graphs.
    3. (HT) Use the inverse-square relation between light intensity and distance, and reasons for greenhouse economics.
    4. Describe RP6: light intensity and the rate of photosynthesis with pondweed, its controls and the rate calculation.

    출처: Cambridge International syllabus

    Rate of photosynthesis against one limiting factor at a time.
    Each curve rises then flattens where another factor limits.

    Rate increases with: light intensity, CO₂ concentration, temperature (to the enzyme optimum), and the amount of chlorophyll.

    Reading single-factor graphs: the flat top means something else has become the limiting factor 限制因素. (HT) On two- or three-factor graphs, decide which factor is limiting at each point.

    (HT) Inverse square law: doubling the distance from the lamp quarters the light intensity:

    $$\text{light intensity} \propto \frac{1}{\text{distance}^2}$$

    (HT) Greenhouse economics: adding heat, light or CO₂ raises the rate — but each costs money; the grower adds the cheapest factor that is limiting, up to the point where extra yield no longer pays.

    RP6 set-up: pondweed under an inverted funnel in a beaker, with a lamp at a measured distance.

    RP6: pondweed in water at different lamp distances; count the oxygen bubbles per minute (or collect the gas); control temperature and CO₂ (sodium hydrogencarbonate); repeat and mean; rate = bubbles ÷ time.

    Worked example. Bubbles: 30 per minute at 10 cm; at 20 cm the light intensity is a quarter, so expect about 8 per minute (if light is still limiting).

    English 한국어
    limiting factor/ˈlɪmɪtɪŋ ˈfæktə/ 제한 요인
    4.1

    Uses of glucose (4.4.1.3)

    Syllabus

    Photosynthesis (AQA 8461 statements 4.4.1.1-4.4.1.2, RP6).

    1. Write the photosynthesis word and symbol equations, and describe it as endothermic.
    2. Explain how light intensity, CO2 concentration and temperature limit the rate of photosynthesis, including interpreting limiting-factor graphs.
    3. (HT) Use the inverse-square relation between light intensity and distance, and reasons for greenhouse economics.
    4. Describe RP6: light intensity and the rate of photosynthesis with pondweed, its controls and the rate calculation.

    출처: Cambridge International syllabus

    Glucose from photosynthesis is used for:

    • respiration;
    • converted to insoluble starch for storage;
    • fat or oil for storage;
    • cellulose for cell walls;
    • amino acids for protein — this also needs nitrate ions from the soil.
    4.2

    Respiration (4.4.2.1)

    Syllabus

    Respiration and metabolism (AQA 8461 statements 4.4.1.3, 4.4.2.1-4.4.2.3).

    1. Name five uses of glucose from photosynthesis.
    2. Compare aerobic and anaerobic respiration in muscle and yeast, with word equations, and fermentation's economic importance.
    3. Explain the body's response to exercise and (HT) oxygen debt and the liver's role.
    4. Define metabolism and list the reactions it includes.

    출처: Cambridge International syllabus

    Respiration is an exothermic 放热 reaction occurring continuously in living cells, transferring the energy for: building larger molecules, movement, and keeping warm.

    Aerobic Anaerobic (muscle) Anaerobic (plant/yeast)
    oxygen needed not needed not needed
    equation glucose + oxygen → CO₂ + water glucose → lactic acid glucose → ethanol + CO₂
    energy much more much less much less

    Anaerobic respiration in yeast is fermentation 发酵, economically important in bread (CO₂ makes it rise) and alcoholic drinks.

    English 한국어
    exothermic/eɡzəˈðɜːmɪk/ 발열 반응
    fermentation/fɜːmənˈteɪʃn/ 발효
    4.2

    Response to exercise (4.4.2.2)

    Syllabus

    Respiration and metabolism (AQA 8461 statements 4.4.1.3, 4.4.2.1-4.4.2.3).

    1. Name five uses of glucose from photosynthesis.
    2. Compare aerobic and anaerobic respiration in muscle and yeast, with word equations, and fermentation's economic importance.
    3. Explain the body's response to exercise and (HT) oxygen debt and the liver's role.
    4. Define metabolism and list the reactions it includes.

    출처: Cambridge International syllabus

    During exercise the heart rate, breathing rate and breath volume increase to deliver more oxygenated blood to muscles. If oxygen supply is insufficient, muscles switch to anaerobic respiration: incomplete glucose oxidation builds up lactic acid 乳酸 → fatigue; an oxygen debt 氧债 builds.

    (HT) Blood carries lactic acid to the liver, which converts it back to glucose. The oxygen debt is the extra oxygen needed after exercise to react with and remove the accumulated lactic acid.

    English 한국어
    oxygen debt/ˈɒksɪdʒn det/ 산소 차액
    lactic acid/ˈlæktɪk ˈæsɪd/ 젖산
    4.2

    Metabolism (4.4.2.3)

    Syllabus

    Respiration and metabolism (AQA 8461 statements 4.4.1.3, 4.4.2.1-4.4.2.3).

    1. Name five uses of glucose from photosynthesis.
    2. Compare aerobic and anaerobic respiration in muscle and yeast, with word equations, and fermentation's economic importance.
    3. Explain the body's response to exercise and (HT) oxygen debt and the liver's role.
    4. Define metabolism and list the reactions it includes.

    출처: Cambridge International syllabus

    Metabolism 代谢 is the sum of all reactions in a cell or body; respiration fuels them. It includes:

    • glucose → starch, glycogen, cellulose;
    • glycerol + 3 fatty acids → lipids;
    • glucose + nitrate ions → amino acids → proteins;
    • respiration;
    • breakdown of excess protein → urea for excretion.
    English 한국어
    Metabolism/məˈtæbəlɪzəm/ 대사
    4.2

    Checklist before you call this topic done

    • Write the photosynthesis word and symbol equations; say endothermic + light→chloroplasts.
    • Explain the four rate factors; read flattening single-factor graphs; (HT) inverse square law and greenhouse economics.
    • Describe RP6 with its controls and the rate calculation.
    • List five uses of glucose (with nitrate for protein).
    • Compare aerobic with both anaerobic equations; name fermentation's two products and uses.
    • Exercise: three increases; lactic acid and oxygen debt; (HT) the liver's role.
    • Define metabolism with its five strands.
  • 5

    항상성 및 반응

    5.1

    홈오스타시스 및 반응: 상태 유지

    • 홈오스타시스는 내부 및 외부 변화에 대응하여 효소 작용에 최적의 조건과 모든 세포 기능을 유지하기 위해 세포나 생물의 내부 조건을 조절하는 것입니다.
    • 몸은 혈당 농도, 체온 및 수분 수준을 조절합니다.
    • 모든 자동 제어 시스템에는: 수용기(자극 감지), 조절 중심(뇌, 척수, 췌장 — 정보 수신 및 처리), 작용기(근육 또는 선장 — 최적 수치를 회복하는 반응을 유도)가 있습니다.
    • 신경 반응은 빠르고 짧지만, 호르몬 반응은 느리지만 더 오래 지속됩니다.
    5.1

    Homeostasis와 신경계 (4.5.1–4.5.2.1, RP7)

    Syllabus

    Homeostasis and the nervous system (AQA 8461 statements 4.5.1-4.5.2.1, RP7).

    1. Define homeostasis and name the three conditions the body controls.
    2. Describe receptors, coordination centres and effectors, and the stimulus-receptor-coordinator-effector-response pathway.
    3. Explain the reflex arc, including sensory neurone, synapse, relay neurone and motor neurone, and why reflexes are automatic and rapid.
    4. Describe RP7: the effect of a factor on human reaction time, with controls and data handling.

    출처: Cambridge International syllabus

    경로: 자극 → 수용체 → 조율자(CNS: 뇌 + 척수) → 효과기(근육 수축 / 분비선 분비) → 반응.

    정보는 **신경세포(neurones)**를 따라 전기 신호로 전달됩니다. CNS가 효과기의 반응을 조율합니다.

    반사 아치: 자극에서 수용체, 세포체를 가진 감각 신경세포, CNS 내 중계 신경세포, 효과기로 가는 운동 신경세포, 그리고 신경세포 간의 시냅스.

    반사 아치 — 의식적 뇌의 관여 없이 자동적이고 빠른 보호 기전:

    자극 → 수용체 → 감각 신경세포 → 시냅스 → 중계 신경세포(CNS 내) → 시냅스 → 운동 신경세포 → 효과기 → 반응.

    시냅스에서 전기 신호가 신경세포 말단에 도달하면 화학 물질이 방출되고, 이 물질은 간극을 확산하여 다음 신경세포에 전기 신호를 유발합니다.

    RP7 — 반응 시간: 한 변수(예: 카페인)가 인간 반응 시간에 미치는 영향을 계획하고 조사하십시오. 파트너의 손가락과 엄지 사이에서 자를 떨어뜨리고 잡는 거리를 측정하며 시간을 변환하거나 컴퓨터 테스트를 사용하십시오. 변수를 통제(같은 손, 같은 높이)하고 반복하여 평균을 구하십시오. 그래프와 표에서 반응 시간 데이터를 해석하십시오.

    English 한국어
    Homeostasis/ˌhəʊmiːəˈstɑːsiz/ 항상성
    neurone/ˈnjuːrəʊn/ 신경세포
    synapse/ˈsɪnæps/ 시냅스
    retina/ˈretɪnə/ 망막
    5.2

    뇌, 눈 및 온도 — 생물학 전담 (4.5.2.2–4.5.2.4)

    Syllabus

    The brain, the eye and body temperature — biology only (AQA 8461 statements 4.5.2.2-4.5.2.4).

    1. Identify cerebral cortex, cerebellum and medulla with their functions, and (HT) difficulties of investigating and treating the brain.
    2. Relate eye structures to functions and explain accommodation for near and distant objects.
    3. Explain myopia and hyperopia and their correction with lenses and new technologies.
    4. Explain how vasodilation, sweating, vasoconstriction and shivering control body temperature (HT in context).

    출처: Cambridge International syllabus

    뇌 (HT 추가)

    도면에서 식별: 대뇌피질(의식, 기억, 언어), 소뇌(균형, 운동 조율), 연수(심박수, 호흡).

    (HT) 뇌 연구 및 치료의 어려움: 복잡하고 예민하며 쉽게 손상되고 접근하기 어렵습니다. 매핑 방법: 뇌 손상이 있는 환자 연구, 영역의 전기 자극, 및 MRI 스캔. 뇌 수술의 장단점을 평가하십시오.

    눈

    눈의 수직 단면:巩膜, 각막, 동공 주변 근육(홍채), 수晶体, 비색근, 고랑인대, 망막 및 시각신경.

    식별 및 설명: 망막(빛의 강도와 색상에 민감한 수용 세포), 시각신경(망막에서 뇌로 신호 전달), 巩膜(강한 외막), 각막(빛 굴절), 동공 주변 근육(홍채)(동공 크기 조절 색소 근육 — 어두운/밝은 빛에 대한 반사 적응), 비색근 및 고랑인대(수晶体 모양 조절).

    초점 조절(Accommodation) — 초점을 맞추기 위해 수晶体 모양 변경:

    근거리 물체 원거리 물체
    비색근 수축 이완
    고랑인대 풀림 당김
    렌즈 두껍고, 빛을 강하게 굴절함 얇고, 빛을 약하게 굴절함

    시각 결함: 근시 (원거리 물체가 망막 앞의 초점에 형성됨; 오목 렌즈로 교정), 원시 (근거리 물체가 망막 뒤의 초점에 형성됨; 볼록 렌즈). 치료법: 안경 렌즈, 콘택트 렌즈, 각막 형상을 바꾸는 레이저 수술, 인공수체 교체.

    체온 조절

    뇌에 있는 체온 조절 중추는 혈액 온도에 민감한 수용체를 가지고 있으며, 피부 수용체는 이 중추로 신호를 보낸다.

    • 너무 더울 때: 혈관이 확장되어 (혈관 확장) 표면에 가까운 혈액이 증가하여 열이 환경으로 전달되고, 땀이 분비된다. 증발하는 땀이 열을体外로 방출한다.
    • 너무 추울 때: 혈관이 수축되어 (혈관 수축); 땀 분비가 멈추고, 골격근이 수축하여 (진전 — 호흡이 열을 생성하여 혈액을 따뜻하게 함). 진전은 호흡이 열을 생성하여 혈액을 따뜻하게 한다.

    (HT) 각 메커니즘이 어떻게 온도 optimum으로 되돌리는지 설명하시오.

    English 한국어
    accommodation/əˌkɒməˈdeɪʃn/ 초점 조절
    myopia/maɪˈəʊpɪə/ 근시
    hyperopia/ˌhaɪpəˈrəʊpɪə/ 원시
    vasodilation/ˌvæsədɪˈleɪʃn/ 혈관 확장
    vasoconstriction/ˌvæsəkənˈstrɪkʃn/ 혈관 수축
    hormone/ˈhɔːməʊn/ 호르몬
    insulin/ˈɪnsjuːlɪn/ 인슐린
    glycogen/ˈɡlaɪkədʒn/ 당원
    glucagon/ˈɡluːkæɡən/ 글루카곤
    negative feedback/ˈneɡətɪv ˈfiːdbæk/ 부정 피드백
    urea/juːˈrɪə/ 요소
    5.3

    인간의 호르몬 조절 (4.5.3)

    Syllabus

    Hormonal coordination in humans (AQA 8461 statements 4.5.3).

    1. Describe the endocrine system: glands, hormones, target organs, and the six named glands.
    2. Explain insulin control of blood glucose, (HT) glucagon and negative feedback, and compare Type 1 and Type 2 diabetes.
    3. (Bio) Describe water and nitrogen balance: kidneys, filtration and selective reabsorption, (HT) ADH, and dialysis vs transplant.
    4. (HT) Explain FSH, LH, oestrogen and progesterone interactions in the menstrual cycle, contraception methods and IVF.
    5. (HT) Explain adrenaline and thyroxine, with negative feedback.

    출처: Cambridge International syllabus

    내분비계는 호르몬을 직접 혈액으로 분비하는 선으로, 혈액은 이를 표적 장기로 운반한다. 신경계보다 느리지만 지속 시간이 길다. 신체 도식에서 다음을 식별할 것: 하안체 (뇌의 '주령선'), 췌장, 갑상선, 부신, 난소, 고환.

    혈당은 췌장에 의해 감시되고 조절된다:

    • 너무 높음 → 췌장이 인슐린을 분비 → 포도당이 혈액에서 세포로 이동; 간 및 근육 세포에서 여분의 포도당은 저장용 글리코겐으로 전환된다.
    • (HT) 너무 낮음 → 췌장이 글루카곤을 분비 → 글리코겐이 다시 포도상으로 전환되어 혈액으로 방출된다. 인슐린-글루카곤은 음성 피드백 사이클이다.

    당뇨병: Type 1 — 췌장이 충분한 인슐린을 생산하지 못함; 통제되지 않은 고혈당; 인슐린 주사로 치료함. Type 2 — 몸 세포가 더 이상 인슐린에 반응하지 않음; 탄수화물 조절 식단과 운동 계획으로 치료함; 비만은 위험 요인이다. 당뇨병 유무에 따른 혈당 그래프를 해석하시오.

    수분 및 질소 균형 — 생물학 관련 내용

    수분은 폐(호흡)와 피부(땀, 이온 및 요소 포함)를 통해 배출되는데, 이는 통제되지 않는다. 여분의 수분, 이온 및 요소는 신장에 의해 소변으로 제거된다. 삼투압에 의해 수분을 너무 많이 잃거나 얻은 세포는 효율적으로 기능하지 못한다. (HT) 식이에서 과잉 아미노산은 간에서 탈아민화됩니다: 암모니아(독성) → 요소로 변환되어 안전하게 배설됨.

    신장은 혈액을 여과하고 포도당, 일부 이온 및 물을 선택적 재흡수한다. (HT) 혈액이 너무 농축되면 하안체가 ADH를 분비하여 신장 관의 투과성을 높이고, 더 많은 물이 재흡수된다 — 다시 한번 음성 피드백이다.

    신장 실패: 이식 또는 혈액 투석으로 치료함 — 투석液中에 적절한 수분/이온 농도의 반투과성 막 사이를 혈액이 흐르므로 노폐물이扩散出去while glucose and useful substances are retained; mechanical device vs transplant evaluation.

    인간 생식에서의 호르몬

    에스트로겐(난소) — 주요 여성 호르몬; 사춘기에는 난포가 성숙하고 약 28일마다 하나가 방출됨 (배란). 테스토스테론(고환) — 정자 생성을 촉진함.

    (HT) 월경 주기:

    호르몬 산지 역할
    FSH 전하상비 난소(포낭) 내 알란의 성숙을 유도함
    LH 전하상비 알란 방출(배란)을 자극함
    에스트로겐 난소(포낭) 자궁내막 유지 및 증식; LH 급증 자극
    프로게스테론 난소(황체) 자궁내막 유지; FSH 및 LH 억제
    28일 월경주기에 따른 호르몬 수준: 초기 FSH, 에스트로겐 상승, 14일째 LH 급증 및 배란, 이후 프로게스테론.

    (HT) 주기에 따른 호르몬 수준 그래프 해석: FSH가 포낭 성장을 시작 → 에스트로겐 상승 → LH 급증이 배란 유발(약 14일째) → 프로게스테론이 자궁내막 두께를 유지하다가 감소하며 주기 재시작.

    임신 방지법 — 호르몬 및 비호르몬 방법 평가: 경구 피임약(FSH 억제하여 알란 성숙 차단), 서방형 프로게스테론 주사/植入/패치, 장벽 방식(콘돔, 디아프람), 자궁내장기, 정자살균제, 알란이 난관 내에 있을 때 절주, 영구 불임술.

    불임증 (HT): '임신 촉진제'(FSH + LH)가 알란 성숙을 자극함. IVF: FSH와 LH가 여러 알란을 자극 → 알란 채취 후 실험실에서 정자와 수정 → 배아가 미세 세포 덩어리로 성장 → 하나 또는 두 개를 자궁으로 이식. 평가: 정서적/신체적 스트레스, 낮은 성공률, 다태아 임신 위험.

    음성 피드백 (HT): 아드레날린(부신, 공포/스트레스)은 심박수 증가 및 뇌와 근육으로의 산소 및 포도당 공급을 높여 '투쟁 또는 도주' 반응을 유도함. 티록신(갑상선)은 기초대사율을 자극하고 성장 및 발달에 중요하며, 그 수준은 음성 피드백에 의해 조절됨. 간단한 음성 피드백 도해 해석.

    English 한국어
    ADH/ˌeɪ diː ˈeɪtʃ/ 항이뇨호르몬(ADH)
    oestrogen/ˈiːstrədʒn/ 에스트로겐
    ovulation/ˌɒvjʊˈleɪʃn/ 배란
    testosterone/teˈstɒstərəʊn/ 테스토스테론
    auxin/ˈɔːksɪn/ 옥신
    phototropism/ˌfəʊtəʊˈtrəʊpɪzəm/ 광반사성
    gravitropism/ˈɡrævɪtrəʊpɪzəm/ 지성(지향성)
    sweat/swet/ 땀
    5.4

    식물 호르몬 — 생물학만 (4.5.4)

    Syllabus

    식물 호르몬 — 생물학 전담 (AQA 8461 지문 4.5.4, RP8).

    1. 뿌리와 줄기에서의 불균일한 옥신 분포를 통해 광유성 및 중력유성을 설명하시오.
    2. RP8 설명: 빛 또는 중력이 싹 트는 새싹에 미치는 영향, 길이 측정 및 표시된 도식.
    3. (고수) 옥신, 자피릴린 및 에틸렌의 농업, 원예 및 식품 산업에서의 활용 설명.

    출처: Cambridge International syllabus

    식물은 호르몬을 생성하여 성장과 빛(광감성) 및 중력(중력감성/지연감성)에 대한 반응의 조율 및 통제를 담당함. 옥신의 불균형 분포는 뿌리와 줄기의 불균형 성장률을 유발 → 줄기는 빛 방향으로 굽고, 뿌리는 아래로 자라남.

    RP8: 싹트刚한 묘목에 빛이나 중력의 영향을 조사 — 길이 측정치 및 신중한 라벨이 붙은 생물학적 그림으로 기록.

    (HT) 지베렐린은 종자 발아를 유도하고, 에틸렌은 세포 분열 및 과실 숙성을 조절함.

    (HT) 용도: 옥신을 제초제, 발근제 및 ** tissue culture**에서의 생장 촉진제로 사용; 에틸렌은 보관 및 운송 중 숙성 조절에 사용; 지베렐린은 종자 휴면 종료, 개화 촉진 및 과실 크기 증가에 사용. 제초제가 생태계 다양성에 미치는 영향을 고려.

    5.4

    이 주제를 완료한 후 체크리스트

    • 가정정의를 정의; 수용체-조절중심-효과자의 순서로 설명; 시냅스가 포함된 반사 아크 그림 그리기.
    • 눈 구조 및 초점 조절 표 암기; 근시 및 원시의 교정법 비교.
    • 온도: 혈관 확장/땀 배출 vs 혈관 수축/진동, 맥락 내에서 (HT).
    • 인슐린 및 (HT) 글루카곤과 음성 피드백; Type 1 vs Type 2 치료법.
    • (Bio) 신장 여과 + 선택적 재흡수; (HT) ADH; 투석 및 이식 비교.
    • (HT) FSH–에스트로겐–LH–프로게스테론 상호작용 및 그래프; 피임법 목록; IVF 단계; 아드레날린과 티록신.
    • (Bio) 호르몬에 의한 유성운동; RP8; (HT) 세 가지 호르몬의 용도.
  • 6

    유전, 변이 및 진화

    6.1

    Inheritance, variation and evolution: passing it on

    • Sexual reproduction mixes genetic information; asexual reproduction clones it. Meiosis halves chromosomes for gametes; fertilisation restores them.
    • DNA → genes → alleles → phenotype: the Punnett square turns this into probabilities.
    • Natural selection changes inherited characteristics across generations. Selective breeding, genetic engineering and cloning (biology only) use different mechanisms. Fossils and resistant bacteria provide evolutionary evidence; classification uses evidence of relationships.
    6.1

    Reproduction, DNA and genomes (4.6.1.1–4.6.1.5)

    Syllabus

    생식, DNA 및 게놈 (AQA 8461 지문 4.6.1.1-4.6.1.5).

    1. 유성 생식과 무성 생식을 비교하되, 두 방법을 모두 사용하는 특정 생물 예시 포함.
    2. 감수분열이 염색체 수를 반으로 줄이고 수정이 이를 회복하는 과정 설명하며, 배 발달을 추적하시오.
    3. DNA, 유전자 및 게놈 설명 및 인간 게놈 연구의 중요성 논평.
    4. (생물) DNA의 뉴클레오티드 구조 및 염기 서열 설명; (고수) 단백질 합성 및 돌연변이의 효과 설명.

    출처: Cambridge International syllabus

    Sexual reproduction — fusion of male and female gametes (sperm + egg in animals; pollen + egg in flowering plants). Meiosis forms gametes; mixing of genetic information gives variety in offspring. Asexual reproduction — one parent, no fusion, mitosis only: genetically identical offspring (clones 克隆).

    Meiosis: one parent cell's chromosomes are copied, then two divisions give four genetically different gametes, each with a single set of chromosomes.

    Meiosis 减数分裂: in reproductive organs, copies of the genetic information are made first, then the cell divides twice to form four gametes, each with a single set of chromosomes — all genetically different. Fertilisation restores the normal number; the new cell divides by mitosis and cells differentiate as the embryo develops.

    (Bio) Sexual vs asexual: sexual gives variation — a survival advantage if the environment changes, and the variation selective breeding uses; asexual needs no mate (time and energy efficient), is faster, and produces many identical offspring when conditions are favourable. Malarial parasites: asexual in the human host, sexual in the mosquito; many fungi: asexual spores + sexual; strawberry plants: sexual seeds + asexual runners; daffodils: bulb division.

    DNA and the genome: DNA is a polymer of two strands forming a double helix, held in the nucleus as chromosomes. A gene 基因 is a small section of DNA that codes for a sequence of amino acids → a specific protein. The genome 基因组 is the entire genetic material of the organism. Importance of the human genome: searching for genes linked to disease; understanding and treating inherited disorders; tracing human migration patterns from the past.

    (Bio) DNA structure: a polymer of four different nucleotides 核苷酸 — common sugar + phosphate group + one of four bases — A 腺嘌呤, C 胞嘧啶, G 鸟嘌呤, T 胸腺嘧啶; alternating sugar–phosphate backbone; a sequence of three bases codes for one amino acid, and the base order controls the amino-acid order of the protein.

    (HT Bio) Protein synthesis: on ribosomes, from a template; carrier molecules bring specific amino acids in the correct order; the chain folds into a unique shape that lets the protein work as an enzyme, hormone or structural protein (e.g. collagen). In complementary strands C pairs with G, T with A. Mutations change the base sequence: a mutation may leave the protein unchanged, alter it slightly, or change its shape so an enzyme no longer fits its substrate or a structural protein loses strength. Variants in non-coding DNA can switch genes on and off, changing how genes are expressed.

    Apply the reproduction and DNA knowledge

    Try first — strawberries. Seeds form after fertilisation; runners form without gamete fusion. Identify the two routes and explain which may help a population survive a new disease.

    Worked reasoning. Seeds result from sexual reproduction: meiosis makes gametes and fertilisation mixes genetic information. Variation means some offspring may resist the disease. Runners form by mitosis from one parent: clones, apart from mutation. Asexual reproduction needs no mate and is fast in favourable conditions; neither route guarantees survival.

    Try first — chromosome number. A human cell has 46 chromosomes before meiosis. Predict the number of gametes, their chromosome number, the number after fertilisation and the division growing the embryo.

    Worked reasoning. DNA is copied before two divisions form four gametes, each with 23 chromosomes (one set). DNA copying does not change the count to 92 chromosomes at that point. Fertilisation combines 23 + 23 = 46; mitosis increases embryo cell number and cells differentiate. The diagram models chromosome numbers, not the stages of meiosis.

    Try first — scale and code. Distinguish a gene, chromosome and genome. (HT Bio) Write the bases complementary to A C G T and explain whether a mutation must stop an enzyme working.

    Worked reasoning. A gene is a DNA section coding an amino-acid sequence for a protein; a chromosome contains a long DNA molecule and many genes; a genome is all the organism’s genetic material. Complementary bases are T G C A. At ribosomes, a template and carrier molecules determine the amino-acid order; the chain folds. A mutation may leave the protein unchanged or change its shape and function. Non-coding variants can change whether genes are expressed.

    English 한국어
    meiosis/meɪˈəʊsɪs/ 감수분열
    clone/kləʊn/ 복제
    gene/dʒiːn/ 유전자
    genome/ˈdʒiːnəʊm/ 게놈
    nucleotide/ˈnjuːklɪɒtaɪd/ 염기
    6.2

    Genetic inheritance, disorders and sex determination (4.6.1.6–4.6.1.8)

    Syllabus

    유전 양식, 장애 및 성 결정 (AQA 8461 지문 4.6.1.6-4.6.1.8).

    1. 배우자, 염색체, 유전자, 대립유전자, 우성, 열성, 동형접합, 이형접합, 유전자형 및 표현형 정의.
    2. 판네트 사각형 및 가계도 완성 및 해석, 비율 및 확률 계산.
    3. 다지증 및 낭포성 섬유증의 유전 양식 설명 및 배 스크리닝 문제 판단.
    4. XX/XY 교배를 통한 성 결정 과정 및 비율 산출.

    출처: Cambridge International syllabus

    Learn the terms: gamete 配子 (sex cell), chromosome 染色体 (DNA structure), gene (section coding a protein), allele 等位基因 (a form of a gene), dominant 显性 (expressed with one copy), recessive 隐性 (expressed only with two), homozygous 纯合 (two same alleles), heterozygous 杂合 (two different), genotype 基因型 (alleles present), phenotype 表现型 (characteristics expressed).

    Single-gene examples: fur colour in mice; red-green colour blindness. Most characteristics result from multiple genes interacting.

    Punnett squares and crosses: complete a Punnett square, extract ratios (e.g. 3:1, 1:1) and probabilities from genetic crosses and family trees. (HT) Construct a cross from parent genotypes and predict outcomes with probability.

    Inherited disorders: polydactyly 多指症 (extra fingers or toes) — dominant allele; cystic fibrosis 囊性纤维化 (a disorder of cell membranes) — recessive allele. Make informed judgements about the economic, social and ethical issues of embryo screening.

    Left: the XX × XY sex cross giving a 1:1 ratio. Right: two cystic-fibrosis carriers (Cc × Cc), with a 1/4 probability of cc.

    Sex determination: body cells have 23 pairs of chromosomes; 22 pairs control characteristics, one pair carries the sex genes — female XX, male XY. A sex cross (X×X, X×Y) gives a 1:1 ratio, 50 % chance each.

    From terms to family inference and probabilities

    Try first — terminology. D causes dominant polydactyly; d does not. Classify Dd and predict its phenotype and gamete alleles.

    Worked reasoning. Dd is heterozygous (two different alleles); DD and dd are homozygous. Genotype means the alleles present; phenotype means the characteristics expressed. Dd is affected because one D is expressed. “Dominant” does not mean stronger or more common. Gametes contain D or d, not Dd.

    Guided family inference — adapted from AQA June 2024 Paper 2H Q06.2. An affected father and unaffected mother have an unaffected son. Deduce the father’s genotype before reading the answer.

    Worked reasoning. The mother and son are dd. The father passed d to his son, but must also carry D because he is affected. Therefore he is Dd. The child’s genotype supplies evidence that the father’s phenotype alone cannot provide.

    Construct a cross (HT). Two unaffected CF carriers are Cc and Cc. Each produces gametes C or c. Combining them gives CC, Cc, Cc, cc: probability of CF = 1/4; probability of being a carrier = 1/2. Carriers do not have CF in this recessive model. The separate XX × XY cross gives XX, XX, XY, XY: probability 1/2 each in the GCSE sex-determination model.

    Exam transfer — adapted from AQA June 2024 Paper 2H Q06.3. An affected mother Dd and unaffected father dd have three unaffected sons. Construct a cross and predict whether the fourth child will be affected.

    Worked reasoning. Egg alleles D/d combine with sperm alleles d/d to give Dd, dd, Dd, dd. Two of four equally likely combinations are affected: probability = 2/4 = 1/2. Earlier births do not change this probability. The four boxes describe possible combinations, not four promised children.

    Evaluate — teacher-written screening scenario. Screening can identify embryos with the CF genotype, but treatment is costly and some embryos may not be used. A biological benefit is identifying cc embryos to inform the couple’s decisions. Cost can limit access and creates funding choices. Some families object to selecting or not using embryos; others prioritise reducing inherited disease. An informed judgement uses the stated evidence and recognises different values.

    English 한국어
    gamete/ˈɡæmiːt/ 배세포
    chromosome/ˈkrəʊməsəʊm/ 염색체
    allele/əˈliːl/ 대립유전자
    dominant/ˈdɒmɪnənt/ 우성
    recessive/rɪˈsesɪv/ 열성
    homozygous/ˌhɒməˈzɪɡəs/ 순합자
    heterozygous/ˌhetrəˈzɪɡəs/ 이형접합
    genotype/ˈdʒenətaɪp/ 형질
    phenotype/ˈfenətaɪp/ 표현형
    polydactyly/ˌpɒlɪˈdæktili/ 다지증
    cystic fibrosis/ˈsɪstɪk fɪˈbrəʊsɪs/ 낭성 섬유증
    6.3

    Variation, selective breeding, genetic engineering and cloning (4.6.2)

    Syllabus

    변이, 인공선택육종, 유전자 공학 및 복제 (AQA 8461 지문 4.6.2).

    1. 유전체-환경 상호작용과 변이의 세 가지 원인, 그리고 돌연변이의 역할을 설명하시오.
    2. 자연선택에 의한 진화와 신종 형성을 설명하시오.
    3. 인공선배, 그 용도 및 근친교배의 위험성을 설명하시오.
    4. (HT) 유전자 공학의 주요 단계, 장점, 위험성 및 반대를 설명하시오.
    5. (Bio) 조직 배양, 절취, 배아 이식 및 성체 세포 복제를 설명하시오.

    출처: Cambridge International syllabus

    Variation 变异 — differences between individuals in a population — comes from genes inherited (genetic), conditions of development (environmental), or a combination. There is usually extensive genetic variation within a population; all genetic variants arise from mutations — most have no effect on the phenotype, some influence it, very few determine it. A rare mutation giving a new phenotype suited to an environmental change can change the species rapidly.

    Evolution: a change in the inherited characteristics of a population over time, through natural selection 自然选择, which may result in a new species. All species evolved from simple life forms that first developed over three billion years ago. Natural selection: variation → the phenotype best suited to the environment survives and breeds → those characteristics are passed on. If two populations become so different they cannot interbreed to produce fertile offspring, two new species have formed.

    Selective breeding (artificial selection): choose parents with the desired characteristic from a mixed population → breed → choose the best offspring → repeat over many generations. Uses: disease resistance in food crops; animals with more meat or milk; dogs with gentle natures; large or unusual flowers. Risk: inbreeding 近交 — breeds prone to disease or inherited defects.

    Genetic engineering: modifying the genome by introducing a gene from another organism to give a desired characteristic — GM crops resistant to insect attack or herbicides (increased yields); bacteria engineered to make human insulin. (HT) Steps: enzymes isolate the required gene → inserted into a vector (bacterial plasmid or virus) → vector inserts the gene into the required cells → transferred early in development so the organism develops with the desired characteristic. Weigh benefits (medicine, agriculture) against risks (wild-flower and insect populations, unexplored health effects) and ethical objections.

    (Bio) Cloning: tissue culture — small groups of plant cells grown into identical new plants (preserving rare species; nurseries); cuttings — simple, older gardeners' method; embryo transplants — splitting unspecialised cells of a developing animal embryo into identical embryos placed in host mothers; adult cell cloning — nucleus removed from an unfertilised egg cell → nucleus from an adult body cell inserted → electric shock makes it divide into an embryo → ball of cells placed in the womb.

    Apply variation, selection and biotechnology

    Start with variation — teacher-written seedling case. Seedlings differ in inherited disease resistance. Genetically identical plants grow to different heights under different light conditions. Identify genetic and environmental variation, then explain what may happen during a disease outbreak.

    Worked reasoning. Resistance alleles provide genetic variation; light conditions can cause environmental differences in height. Phenotype often depends on both. Resistant plants may survive the disease and reproduce more successfully, passing favourable alleles to offspring. Over generations, resistance alleles can become more common. Mutations change DNA; they do not occur because an organism needs a particular change.

    Exam transfer — adapted from AQA June 2024 Paper 2H Q01.2. Describe natural selection using insects that already vary in inherited insecticide resistance.

    Worked reasoning. The insecticide is a selection pressure: more susceptible insects die. Resistant survivors reproduce and pass resistance alleles to offspring. The population’s inherited characteristics change across generations; an individual does not evolve just by being sprayed. Speciation requires populations to become unable to interbreed to produce fertile offspring.

    Compare mechanisms — teacher-written crop case. A grower wants inherited disease resistance without transferring a gene. Select resistant parents from a mixed population, breed them, select resistant offspring and repeat over many generations. This is selective breeding, not natural selection or genetic engineering. Repeatedly breeding close relatives risks inherited defects and reduced genetic variation.

    Linked exam explanation — adapted from AQA June 2024 Paper 2H Q09.1. GM soya plants resist glyphosate. Explain how spraying the field can increase yield.

    Worked reasoning. Glyphosate kills weeds but the resistant crop survives. Less competition gives the crop more light, water and mineral ions. More light and water can support photosynthesis, producing glucose for respiration and building biomass. Nitrate ions support amino-acid and protein synthesis. Link resources to growth and harvested yield; “GM means higher yield” omits the mechanism. These are authored explanation points checked against the scheme, not an official model or a guaranteed score.

    HT — follow a gene. Enzymes isolate a required gene (for example, human insulin). Insert it into a vector such as a bacterial plasmid; use the vector to introduce it into the required cells. The modified cells express the gene and produce the substance. For a developing animal or plant, introduce the gene early so the organism develops with the desired characteristic.

    Evaluate — teacher-written GM scenario. An insect-resistant crop suffers less target-pest damage, but a study reports fewer nearby non-target insects without establishing why. Reduced damage may improve harvest. The insect decline raises a food-web concern; compare modified and unmodified fields while accounting for habitat and pesticide use. Association alone does not prove the inserted gene caused the decline. Judge the particular modification and evidence.

    Biology — choose and order cloning methods. A nursery can use tissue culture to grow plants from small groups of cells, or cuttings from a parent plant. Embryo transplants split cells before they specialise and place identical embryos into host mothers. Adult cell cloning removes an unfertilised egg’s nucleus, inserts an adult body-cell nucleus, stimulates division with an electric shock, and transfers the ball of cells to a womb. The adult body-cell donor supplies the nuclear genetic information. Environmental differences can still affect a clone’s phenotype.

    English 한국어
    variation/ˌveərɪˈeɪʃn/ 변이
    natural selection/ˈnætʃərəl sɪˈlekʃn/ 자연선택
    inbreeding/ˈɪnbriːdɪŋ/ 근친교배
    adenine/ˈædəniːn/ 아데닌
    cytosine/ˈsaɪtəsaɪn/ 시토신
    guanine/ˈɡwɑːnaɪn/ 구안신
    thymine/ˈθaɪmaɪn/ 티민
    6.4

    Evolution, genetics and evidence (4.6.3, biology only)

    Syllabus

    진화, 유전 및 증거 (AQA 8461 서술문 4.6.3, 생물학 전담).

    1. 다윈의 이론을 설명하고, 그것이なぜ 점진적으로 받아들여졌는지, 그리고 라마르크의 이론을 설명하시오.
    2. 월리스와 멘델의 기여와 유전학의 발전을 설명하시오.
    3. 화석과 그 형성 과정, 화석 기록이 불완전한 이유, 멸종 원인 및 진화樹를 설명하시오.
    4. 항생제 내성 세균과 내성 감속 방법을 설명하시오.

    출처: Cambridge International syllabus

    (Bio) Darwin's theory: wide variation within a species; individuals best suited to the environment survive to breed; the useful characteristics are passed on. Published in On the Origin of Species (1859). Accepted only gradually: it challenged the idea that God made all living things; insufficient evidence at publication; the mechanism of inheritance was unknown for another 50 years. Lamarck's rival theory — that changes acquired during an organism's lifetime are inherited — is now known to be wrong in almost all cases.

    (Bio) Wallace: independently proposed evolution by natural selection; joint 1858 publications with Darwin prompted Darwin to publish; best known for warning colouration and pioneering work on speciation — the steps by which new species arise.

    (Bio) Mendel and the growth of genetics: mid-19th-century breeding experiments showed each characteristic is inherited through 'units' passed unchanged to descendants; late 19th century — chromosomes observed in cell division; early 20th century — the units behave like chromosomes → genes on chromosomes; mid-20th century — DNA structure determined and gene function worked out. Mendel's work was unrecognised in his lifetime.

    (Bio) Evidence for evolution: genes show characteristics pass to offspring; fossils; antibiotic resistance in bacteria.

    (Bio) Fossils form when parts do not decay (a decay condition is absent), when parts are replaced by minerals as they decay, or as preserved traces (footprints, burrows, rootlet traces). Early soft-bodied life left few traces, mostly destroyed by geological activity — so science cannot be certain how life began. Extinction means no individuals of a species remain — causes include new diseases, new predators, competition, environmental change, catastrophic events. Evolutionary trees use current classification data and fossil data.

    (Bio) Resistant bacteria: mutations produce genetic variants; some are resistant to a particular antibiotic. Susceptible bacteria are killed, while resistant survivors reproduce quickly and pass on resistance genes. The resistant strain becomes more common and can spread (for example, MRSA). This population change is evidence for evolution. Resistance can make infection harder to treat; it does not mean every antibiotic is ineffective or that people have no immune defence. Reducing inappropriate prescribing and unnecessary agricultural antibiotic use reduces avoidable selection pressure. Developing new antibiotics takes time and resources.

    Try first — resistance explanation. A population contains susceptible and resistant variants before antibiotic exposure. Explain why resistant bacteria become more common, and correct “the antibiotic makes every bacterium mutate because it needs resistance”.

    Worked reasoning. Genetic variation already exists; mutations are not directed by need. Antibiotic exposure selects survivors. Resistant bacteria reproduce and pass resistance genes to offspring, increasing their share of the population. Selection and inheritance explain the change without claiming that every individual becomes resistant.

    Use historical and fossil evidence

    Explain the history (Biology). Darwin and Wallace independently proposed natural selection; their joint writings appeared in 1858 and Darwin published his book in 1859. Acceptance was gradual: the theory challenged prevailing creation beliefs, some scientists found the evidence insufficient, and inheritance mechanisms were unknown. Mendel’s plant breeding suggested inherited units. His work’s importance was recognised after his death; later chromosome behaviour, genes on chromosomes and DNA structure connected these units to physical mechanisms.

    Apply speciation — teacher-written case. A barrier separates a population into different habitats. Different selection pressures favour different variants; survivors reproduce and pass favourable alleles to offspring. Mutations supply genetic variants. Over many generations the populations may diverge. Separation alone does not prove a new species: inability to interbreed to produce fertile offspring is the stated species criterion.

    Interpret evidence — teacher-written fossil case. A footprint preserved in rock is a trace fossil; a shell replaced by minerals illustrates mineral replacement. Conditions that prevent decay can also preserve remains. Soft-bodied organisms may leave no trace, and geological activity destroys fossils. An incomplete record still supplies evidence of past organisms and change. A species missing from one site is not necessarily extinct: extinction means no living individuals remain anywhere.

    Read the diagram. Trace A and B back to their first shared fork: this is their most recent common ancestor. A and C share the older ancestor of all four species, so A and B are more closely related. This is a schematic without a time scale; do not infer dates from branch lengths or relationships from vertical spacing. Trees use classification and fossil evidence.

    6.5

    Classification of living organisms (4.6.4)

    Syllabus

    생물 분류 (AQA 8461 서술문 4.6.4).

    1. 리네우스 체계와 이명법(이중명명법)을 설명하시오.
    2. 우즈(Woese)의 삼도 체계와 이를 주도한 요인을 설명하시오.
    3. 분류 및 화석 데이터를 사용하여 진화樹를 해석하시오.

    출처: Cambridge International syllabus

    Linnaeus: classification by structure and characteristics into kingdom, phylum, class, order, family, genus, species; organisms named by the binomial system (genus + species).

    Three-domain system (Carl Woese), from chemical-analysis evidence: archaea (a distinct domain of prokaryotes, including many organisms from extreme environments), bacteria (true bacteria), eukaryota (protists, fungi, plants, animals). Improvements in microscopes and biochemistry drove the new models.

    An evolutionary tree: branches from a common ancestor; identify the first shared fork when tracing two tips back to their most recent common ancestor; drawn lengths here are not a time scale.

    Evolutionary trees show how scientists believe organisms are related — interpret them using classification and fossil data.

    Apply classification. In Panthera leo and Panthera tigris, Panthera identifies the shared genus; the second word distinguishes the species within it. The ranks are kingdom → phylum → class → order → family → genus → species. New microscopy and chemical-analysis evidence can reveal relationships that external appearance does not show, leading to revised classification. Woese’s three domains are archaea, bacteria and eukaryota.

    Final retrieval. Without the answers, explain (1) how 23-chromosome gametes restore 46 at fertilisation; (2) the next affected-child probability for Dd × dd after three unaffected births; (3) how antibiotic exposure changes a bacterial population. Check: two gamete sets combine; the cross gives a 1/2 probability independently at each birth; resistant survivors reproduce and pass on resistance genes. Then use the earlier worked examples to retrieve gene transfer, cloning order and common ancestry.

    6.5

    Checklist before you call this topic done

    • Meiosis (four gametes, halved number) vs mitosis; sexual vs asexual advantages with named organisms.
    • All ten genetics terms; Punnett squares with ratios; polydactyly vs cystic fibrosis; XX/XY cross 1:1.
    • (Bio) DNA nucleotides and bases; (HT) protein synthesis and mutations.
    • Variation three sources; natural selection sequence; selective breeding steps and inbreeding risk; (HT) genetic engineering steps; (Bio) four cloning methods.
    • (Bio) Darwin/Wallace/Mendel history, fossil formation and why the record is incomplete, MRSA story, Linnaeus ranks and the three domains.
  • 7

    생태학

    7.1

    Ecology: communities, cycles and human impacts

    • The Sun's energy passes through ecosystems; carbon and water cycle between the living and non-living world.
    • Communities compete and depend on each other; adaptations fit organisms to their conditions.
    • Human activity — waste, land use, deforestation, global warming — threatens biodiversity, and trophic-level biology (bio only) explains why food chains are short.
    7.1

    Adaptations, interdependence and competition (4.7.1)

    Syllabus

    적응, 상호의존 및 경쟁 (AQA 8461 서술문 4.7.1).

    1. 생태계의 조직 수준을 설명하고, 생태계, 군집, 생물적 요인 및 비생물적 요인의 정의를 하시오.
    2. 식물과 동물이 경쟁하는 요인을 제시하고, 상호의존성을 설명하시오.
    3. 주어진 비생물적 요인과 생물적 요인의 변화가 군집에 미치는 영향을 설명하시오.
    4. 구조적, 행동적, 기능적 적응과 극한생물(extrêmophile)을 설명하시오.

    출처: Cambridge International syllabus

    An ecosystem 生态系统 is the interaction of a community of living organisms (biotic 生物因子) with the non-living (abiotic factor 非生物因子) parts of their environment. Levels of organisation: individual → population → community → ecosystem.

    Competition 竞争: plants compete for light, space, water and mineral ions; animals compete for food, mates and territory. Interdependence 相互依赖: each species depends on others for food, shelter, pollination, seed dispersal — remove one species and the whole community can be affected.

    Abiotic factors affecting a community: light intensity, temperature, moisture levels, soil pH and mineral content, wind intensity and direction, carbon dioxide levels (plants), oxygen levels (aquatic animals). Biotic factors : availability of food, new predators arriving, new pathogens, one species outcompeting another until numbers fall too low to breed. Explain the effect of a change in any of these from given data.

    Adaptations 适应性 — features that enable survival in normal conditions — are structural (shape/body), behavioural (actions) or functional (processes, e.g. camouflage chemistry, venom). Extremophiles 极端微生物 live in extreme environments — high temperature, pressure or salt — e.g. bacteria in deep-sea vents.

    English 한국어
    ecosystem/ˈiːkəʊsɪstəm/ 생태계
    biotic factor/baɪˈɒtɪk ˈfæktə/ 생물학적 요인
    abiotic factor/ˌæbɪˈɒtɪk ˈfæktə/ 비생물적 요인
    competition/ˌkɒmpəˈtɪʃn/ 경쟁
    interdependence/ˌɪntədɪˈpendəns/ 상호 의존성
    adaptation/ˌædæpˈteɪʃn/ 적응
    extremophile/ekˈstreməfaɪl/ 극한생물
    7.2

    Organisation of an ecosystem and material cycles (4.7.2, RP9/RP10)

    Syllabus

    Organisation of an ecosystem and material cycles (AQA 8461 statement 4.7.2, RP9/RP10).

    1. Use food-chain vocabulary from producer to tertiary consumer and interpret predator-prey cycles.
    2. Describe RP9 sampling with transects and quadrats, with mean, mode and median.
    3. Explain the carbon and water cycles and the role of microorganisms.
    4. (Bio) Explain how temperature, water and oxygen affect decay rate, with RP10 and biogas.

    출처: Cambridge International syllabus

    Photosynthetic organisms are the producers 生产者 of biomass for life on Earth. Food chains: producer (green plant/alga making glucose by photosynthesis) → primary consumers 初级消费者 → secondary consumers 次级消费者 → tertiary consumers 三级消费者. A predator-prey cycle: prey numbers rise first, the predator population follows with a delay, and both oscillate.

    Predators kill and eat prey; in a stable community their numbers rise and fall in cycles — interpret the classic predator–prey graph (prey rises first, predator follows).

    Sampling (RP9): transects and quadrats measure the distribution and abundance of species; calculate mean, mode, median of abundance; plot graphs with suitable scales.

    The carbon cycle: photosynthesis takes CO2 into plants, feeding passes carbon to animals, decay by decomposers returns it, with respiration and combustion releasing CO2.

    Carbon cycle: returns carbon from organisms to the atmosphere as carbon dioxide for photosynthesis — respiration (plants, animals, decomposers), decay, combustion. Water cycle: evaporation and precipitation provide fresh water before it drains to the sea. Microorganisms recycle materials — returning carbon to the atmosphere as CO₂ and mineral ions to the soil. The nitrogen cycle is NOT required.

    (Bio) Decomposition (RP10): temperature, water and oxygen affect the rate of decay. Gardeners and farmers provide optimum conditions for rapid decay 腐解 of waste into compost — a natural fertiliser. Anaerobic decay produces methane — biogas generators. RP10: effect of temperature on the rate of decay of fresh milk by measuring pH change.

    (HT Bio) Environmental change: temperature, water availability, atmospheric-gas composition — changes may be seasonal, geographic or human-caused — shift the distribution of species.

    English 한국어
    producer/prəˈdjuːsə/ 생산자
    primary consumer/ˈpraɪməri kənˈsuːmə/ 1차 소비자
    secondary consumer/ˈsekəndəri kənˈsuːmə/ 2차 소비자
    tertiary consumer/ˈtɜːʃjəri kənˈsuːmə/ 3차 소비자
    decomposition/ˌdiːkɒmpəˈzɪʃn/ 부패 분해
    7.3

    Biodiversity and human interactions (4.7.3)

    Syllabus

    생물 다양성과 인간 상호작용 (AQA 8461 문항 4.7.3).

    1. 생물 다양성을 정의하고 그것이 생태계를 안정화시키는 이유를 설명한다.
    2. 폐기물, 토지 이용, 산림 벌채 및 글로벌 warming이 생물 다양성을 줄이는 방법을 설명한다.
    3. 생물 다양성을 유지하는 프로그램을 설명한다.

    출처: Cambridge International syllabus

    Biodiversity 生物多样性 is the variety of all different species on Earth or within an ecosystem; it stabilises ecosystems by reducing the dependence of one species on another, and our own future relies on maintaining it.

    • Waste management: population growth and living standards increase resource use and waste → pollution 污染 in water (sewage, fertiliser, toxic chemicals), air (smoke, acidic gases) and on land (landfill, toxic chemicals) — killing plants and animals.
    • Land use: building, quarrying, farming and dumping waste reduce land for other species; destroying peat bogs for garden compost destroys habitat and biodiversity, and decaying or burnt peat releases CO₂.
    • Deforestation in tropical areas: land for cattle and rice fields, crops for biofuels.
    • Global warming: rising atmospheric CO₂ and methane contribute to global warming; describe biological consequences (distribution shifts, migration changes, biodiversity loss).

    Maintaining biodiversity — programmes reducing negative human effects: breeding programmes for endangered species; protection and regeneration of rare habitats; field margins and hedgerows reintroduced in single-crop areas; reduced deforestation and CO₂ emissions; recycling rather than landfill.

    English 한국어
    biodiversity/ˌbaɪəʊdaɪˈvɜːsɪti/ 생물 다양성
    pollution/pəˈluːʃn/ 오염
    7.4

    Trophic levels and food security — biology only (4.7.4–4.7.5)

    Syllabus

    영양 단계 및 식량 안보, 생물학 전용 (AQA 8461 문항 4.7.4-4.7.5).

    1. 영양 단계를 명시하고 분해자를 설명하며 생체량 피라미드를 구성한다.
    2. 영양 단계 간 생체량 손실을 설명하고 전달 효율을 계산한다.
    3. 식량 안보를 위협하는 생물학적 요인을 설명한다.
    4. 농업 기법, 지속 가능한 어업 및 생명공학적 역할에 대해 평가한다.

    출처: Cambridge International syllabus

    Trophic levels: 1 producers (plants, algae); 2 primary consumers (herbivores); 3 secondary consumers (carnivores eating herbivores); 4 tertiary consumers (carnivores eating carnivores). Apex predators have no predators. Decomposers secrete enzymes onto dead material and absorb the small soluble food molecules.

    A pyramid of biomass with producers at the bottom and about 10 percent of biomass passing to each level above.

    Pyramids of biomass: trophic level 1 at the bottom; construct from data. Transfer of biomass: producers transfer ~1 % of incident light energy into biomass; only ~10 % of biomass passes from each level to the next — losses from egested faeces (not all ingested material is absorbed) and waste (CO₂ and water from respiration, water and urea in urine); glucose used in respiration. Calculate efficiency by percentage or fraction of mass.

    Food security 粮食安全 — having enough food for the population — is threatened by: rising birth rate; changing diets in developed countries; new pests and pathogens; environmental changes (famine when rains fail); the cost of agricultural inputs; conflicts affecting water or food availability.

    Farming techniques improve efficiency by restricting energy transfer from food animals to the environment — limiting movement and controlling temperature; high-protein feeds increase growth. Weigh these against ethical objections to intensive farming.

    Sustainable fisheries: maintain fish stocks where breeding continues — control net size and fishing quotas .

    Role of biotechnology: culturing microorganisms for food — Fusarium fungus grown on glucose syrup in aerobic conditions produces mycoprotein 菌蛋白 (protein-rich, vegetarian); GM bacteria produce insulin; GM crops such as golden rice add food or nutritional value.

    English 한국어
    food security/fuːd sɪˈkjʊərɪti/ 식량 안보
    mycoprotein/ˈmaɪkəprəʊtiːn/ 균단백질
    7.4

    Checklist before you call this topic done

    • Definitions chain: ecosystem, community, abiotic/biotic factors with examples.
    • Plant vs animal competition; interdependence consequences; three adaptation types + extremophiles.
    • Food-chain vocabulary; predator–prey cycle read off a graph; RP9 sampling with mean/mode/median.
    • Carbon and water cycles with decomposers' role; (bio) decay factors + biogas + RP10.
    • Four human threats + five biodiversity programmes named.
    • (Bio) Trophic levels, pyramid shape, 10 % rule with efficiency calculation.
    • (Bio) Six food-security threats; efficiency farming, net size and quotas, mycoprotein, golden rice.

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IGCSE, A-Level & AP