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Pearson Edexcel · International A-Level · 生物

  • 1

    分子、饮食、运输与健康

    1.1

    From molecules to a working circulation

    A red blood cell leaving your lung carries oxygen on haemoglobin, rides an artery, squeezes through a capillary narrower than itself, releases the oxygen to a respiring cell, and returns through a vein. Every structure in that journey is built from the molecules this unit starts with, and every exam question about it asks you to connect structure to function.

    WBI11 Molecules, Diet, Transport and Health is the first IAS unit: a written paper of 1 hour 30 minutes, 80 marks, all compulsory. It assesses Topic 1 (Molecules, Transport and Health) and Topic 2 (Membranes, Proteins, DNA and Gene Expression). Expect data questions on health studies, calculations with standard form, and extended answers that name structures and explain mechanisms.

    1.1

    水、碳水化合物和食物检测(陈述1.1–1.4,核心实践1)

    教学大纲

    培生爱德思国际高中生物课程大纲第2版(2021年2月),主题1陈述1.1-1.4及核心实践1(大纲页15-16)。这些是官方陈述的地方教学细分内容。

    作为运输溶剂的水:偶极性质、分子间氢键以及极性物质和离子化合物溶解的原因。碳水化合物:单糖/二糖/多糖(葡萄糖、果糖、半乳糖;麦芽糖、蔗糖、乳糖;糖原、直链淀粉、支链淀粉);缩合形成糖苷键及水解断裂它们;结构与能量供应及储存的关系(α-葡萄糖螺旋 vs 支链淀粉/糖原分支;此处不包括纤维素)。核心实践1:使用比色板进行半定量的本尼迪特试剂还原糖浓度估算及碘液淀粉估算。

    来源:Cambridge International 教学大纲

    Water is the transport medium because it is a polar solvent 极性溶剂. In a water molecule, oxygen pulls the shared electrons more strongly than hydrogen: each O–H bond has a slight negative end at O and a slight positive end at H. This is the dipole 偶极 nature of water. Opposite charges between neighbouring molecules form hydrogen bonds 氢键, and polar or charged particles (glucose, sodium ions, amino acids) dissolve because water clusters around them.

    Carbohydrates are joined and split by two reactions:

    • Condensation 缩合 removes a water molecule as a glycosidic bond 糖苷键 forms between two sugars.
    • Hydrolysis 水解 adds water back and breaks that bond.

    Two glucose molecules condense to maltose; glucose plus fructose give sucrose; glucose plus galactose give lactose. Long chains make polysaccharides: amylose 直链淀粉 (1,4-linked, coils into a helix, compact), amylopectin 支链淀粉 and glycogen 糖原 (both 1,4-chains with 1,6-branches; glycogen is the animal store in liver and muscle). Branching means many chain ends, so glucose can be released quickly; the helix means a lot of energy fits in a small space.

    Core Practical 1: estimating sugar and starch concentration

    Reducing sugars reduce blue Benedict's reagent on heating, giving a brick-red precipitate. To estimate an unknown concentration, compare the colour against standards of known concentration, or read the solution in a colourimeter.

    Calibration of Benedict's reagent: absorbance of colour standards against glucose concentration, with an unknown read off the straight part of the line.

    Worked check. Standards of 0, 0.5, 1, 2 and 4 % glucose give absorbances 0.02, 0.09, 0.18, 0.35, 0.64. An unknown gives 0.58. On the straight section, each 1 % adds about 0.16 absorbance, so the unknown is near $0.58/0.16 \approx 3.6\%$. Reading beyond the standards (extrapolating past 4 %) is less reliable than measuring within them — a point the examiners like you to make. Iodine turns blue-black with starch; intensity of the blue-black gives the same kind of estimate for starch.

    1.2

    脂类和酯键(陈述1.5)

    教学大纲

    陈述1.5(大纲页16):甘油三酯的合成(甘油与三个脂肪酸通过缩合形成酯键);饱和与不饱和脂肪酸(无双键 vs 一个或多个C=C双键、直链紧密堆积 vs 弯折链松散排列、熔点差异)。联系后续心血管病(CVD)风险(1.5子主题)的饮食脂质化学及乙醇乳化试验。

    来源:Cambridge International 教学大纲

    A triglyceride 甘油三酯 is one glycerol plus three fatty acids, joined by three condensation reactions that form ester bonds 酯键. Saturated 饱和 fatty acids have no C=C bonds: straight chains pack closely, so animal fats are solid at room temperature. Unsaturated 不饱和 acids have one or more double bonds that kink the chain, packing is looser, and oils are liquid. Lipids store more energy per gram than carbohydrates because they are more reduced.

    1.3

    心脏、血管和心动周期(陈述1.6–1.8)

    教学大纲

    陈述 1.6-1.8(见第16-17页):为什么多细胞动物需要体循环运输系统(扩散限制);毛细血管、动脉和静脉的结构与功能(管壁层次、管腔、瓣膜、弹性及平滑肌组织);哺乳动物心脏的心腔、瓣膜和主要血管;心动周期(心房收缩期、心室收缩期、舒张期)中的压力与容积变化及瓣膜关闭。心肌性刺激细节在 IAS 阶段不需要(参见说明)。

    来源:Cambridge International 教学大纲

    Diffusion supplies cells only a few tenths of a millimetre thick. A big organism needs mass transport 大规模运输: a pump and pipes that carry every cell's supply and waste. The three vessel types match three jobs:

    Vessel Wall Why
    Artery thick muscle and elastic fibres, small lumen carries blood at high, pulsing pressure; elastin smooths the pulse
    Capillary one cell thick endothelium short diffusion path, huge total surface area; tissue fluid leaks out at the start
    Vein thin wall, wide lumen, valves low pressure return; valves and the surrounding muscles stop backflow

    The mammalian heart is two pumps in one organ. The right side sends blood to the lungs; the left side, with the thicker ventricle wall, sends blood to the body. Atria receive; ventricles deliver. The coronary arteries feed the heart muscle itself — blockage there is a heart attack.

    The cardiac cycle

    One beat has three phases. Atrial systole 心房收缩期 squeezes the last blood into the ventricles. Ventricular systole 心室收缩期 raises ventricular pressure above arterial pressure, the semilunar valves open, and blood is driven out. Diastole 舒张期 relaxes the muscle, pressure falls, the semilunar valves slam shut (the second heart sound), and the chambers refill from the veins.

    One cardiac cycle: ventricular and aortic pressure with ventricular volume. Valves change state whenever the two pressure curves cross.

    Read the figure the way an examiner wants: valve state is decided by pressure difference, never by an instruction. Atrio-ventricular valves open when atrial pressure exceeds ventricular pressure; semilunar valves open when ventricular pressure exceeds aortic pressure.

    1.4

    血红蛋白和气体运输(陈述1.9)

    教学大纲

    陈述 1.9(见第17页):血红蛋白的四级结构(四条多肽链,每条含一个血红素基团),由S形解离曲线所示的协同结合,氧气和二氧化碳的运输,以及波尔效应——氢离子负荷使血红蛋白对氧亲和力降低,二氧化碳分压升高导致曲线右移。对比肌蛋白的双曲线,其为单链储存蛋白。

    来源:Cambridge International 教学大纲

    Haemoglobin is a protein of four polypeptide chains, each folded around a haem group carrying one Fe²⁺ ion that binds one O₂ molecule — four in total. Binding the first oxygen changes the shape of the whole molecule and makes the next bindings easier: cooperative binding 协同结合 gives the oxygen dissociation curve its shallow-then-steep S shape.

    Oxygen dissociation curves: the sigmoid curve for normal blood and the rightward Bohr shift when carbon dioxide rises.

    In the lungs (high pO₂) the curve flattens near 100 % saturation. In resting tissue near 1–2 kPa the curve is steep, so a small fall in pO₂ unloads a large amount of oxygen — exactly where it is needed. When tissue respires hard it releases more carbon dioxide; hydrogen ions attach to haemoglobin and shift the whole curve to the right (the Bohr effect 波尔效应), forcing even more oxygen to unload in the busiest tissue.

    Worked check. At pO₂ 2 kPa the normal curve reads about 30 % saturation and the shifted curve about 15 % — haemoglobin holds only about a third as much oxygen at that partial pressure, so most of its four sites are released to the tissue sooner.

    1.5

    动脉粥样硬化、血液凝固和心血管疾病风险(陈述1.10–1.13,核心实践2)

    教学大纲

    陈述 1.10-1.13 结合核心实验 2(见第17-18页):动脉粥样硬化的发展过程(内皮损伤、炎症反应、低密度脂蛋白沉积、斑块生长、动脉瘤及血栓形成风险);凝血级联反应(凝血活酶释放、凝血酶原转化为凝血酶、纤维蛋白原转化为纤维蛋白);血压(收缩压/舒张压,高血压作为风险因素)及核心实验 2 心率调查(水蚤或类似生物)。风险因素是相乘而非相加关系。

    来源:Cambridge International 教学大纲

    Atherosclerosis 动脉粥样硬化 begins with damage to the endothelium 内皮 lining an artery. The inflammatory response grows, low-density lipoproteins invade, smooth muscle and connective tissue build a plaque 斑块, the artery narrows and hardens, and a clot can finally block it or break loose.

    Clotting itself is a cascade. Damage releases thromboplastin 凝血酶原激酶, which (with calcium ions) converts the soluble plasma protein prothrombin 凝血酶原 into the enzyme thrombin 凝血酶; thrombin converts soluble fibrinogen 纤维蛋白原 into the insoluble mesh of fibrin 纤维蛋白 that traps platelets and cells.

    Blood pressure 血压 is reported as systolic over diastolic. Hypertension, smoking, high blood LDL-cholesterol, diet, inactivity, age, gender and genetics are risk factors 危险因素 — and they multiply, not add: two moderate risks can matter more than one severe risk.

    Reading health-risk data

    Questions 8 and 9 of every paper live here. Four habits cover almost every mark:

    • Correlation is not causation. 相关不等于因果 Two variables moving together may share a hidden cause, or the link may be chance.
    • Judge the study: How large was the sample? How was it selected? What was controlled? Was it blind? Are there confounders 混杂变量 such as age or smoking?
    • Do the arithmetic carefully: percentage change, ratios in standard form, per-capita rates.
    • Ask who or what is missing: non-responders, unpublished studies, extrapolated lines.

    Worked check. A study finds gum disease correlates with endothelial damage. Before claiming one causes the other you would want a mechanism, a dose–response pattern, and confounders (diet, smoking, age) excluded — and the question "were the groups matched for blood pressure?" is exactly a confounder question.

    1.7

    饮食、胆固醇和心血管疾病治疗(陈述1.19–1.20)

    教学大纲

    陈述 1.19-1.20(见第18页):关于饮食的科学知识(能量平衡、用身体质量指数和腰臀比衡量肥胖、盐、饱和与不饱和脂肪)如何指导选择;心血管病治疗的好处与风险——抗高血压药(β受体阻滞剂、血管紧张素转换酶抑制剂、利尿剂)、他汀类药物(上调LDL受体、肌肉和肝脏副作用)、抗凝剂(华法林、阿司匹林)及血小板抑制药物,并结合生活方式干预。

    来源:Cambridge International 教学大纲

    Energy balance decides body mass. Overweight and obesity are screened by BMI (mass in kg divided by height in metres squared; 18.5–24.9 is the healthy band) and waist-to-hip ratio (WHR). Dietary saturated fat raises blood LDL-cholesterol; unsaturated fat and soluble fibre lower it.

    Worked check. A woman of mass 74 kg and height 1.66 m: BMI $= 74 / 1.66^2 = 74/2.76 = 26.8$ — just into the overweight band. Waist 84 cm and hips 98 cm give WHR $= 84/98 = 0.86$.

    Treatments all carry risks alongside benefits:

    • Antihypertensives 抗高血压药 (beta blockers, ACE inhibitors, diuretics) lower blood pressure.
    • Statins 他汀类 (for example simvastatin) block a liver enzyme in cholesterol synthesis, so cells take up LDL from blood.
    • Anticoagulants 抗凝剂 (warfarin) and antiplatelet drugs (aspirin) reduce clotting, so bleeding risk rises.

    A question asking you to "discuss" wants both sides plus a judgement anchored in the data given.

    1.8 1.9

    气体交换表面和膜结构(陈述2.1–2.2,核心实践3)

    教学大纲

    主题 2 陈述 2.1-2.3 结合核心实验 3(见第19-20页):交换表面的特性(大表面积、薄、维持浓度梯度);基于流动镶嵌模型的膜结构——磷脂双分子层、内在和外在蛋白、胆固醇、糖脂和糖蛋白;膜模型历史(从欧弗顿到辛格-尼科尔森)及核心实验 3 膜通透性调查(在不同温度或溶剂浓度下的甜菜根)。

    陈述 2.4-2.5(见第20页):渗透作用作为自由水分子通过半透膜从高水势向低水势的移动;被动运输(扩散、载体蛋白和通道蛋白介导的易化扩散)与主动运输(需ATP,逆浓度梯度);胞吞作用和胞吐作用用于批量物质运输;菲克定律应用于表面积、浓度梯度和扩散距离(速率正比于 SA x 梯度 / 厚度)。

    来源:Cambridge International 教学大纲

    Every exchange surface is a membrane doing controlled business. The fluid mosaic model 流动镶嵌模型 is the accepted picture: a phospholipid bilayer (fluid because the tails can slide), studded with intrinsic proteins that span the bilayer (channels, carriers), extrinsic proteins, cholesterol that controls fluidity, and glycoproteins and glycolipids for recognition.

    Four transport routes:

    • Diffusion 扩散 — net movement down a concentration gradient, no ATP.
    • Facilitated diffusion 易化扩散 — down the gradient through a channel or carrier protein.
    • Active transport 主动运输 — against the gradient, powered by ATP, through carrier proteins that change shape.
    • Osmosis 渗透 — water moving from higher to lower water potential 水势 across a partially permeable membrane. Endocytosis and exocytosis move bulk material in vesicles.

    For diffusion, Fick's law: rate is proportional to (surface area × concentration difference) ÷ diffusion distance. Gas exchange surfaces answer all three: huge area, maintained gradient, and a wall one or two cells thick. This is why alveoli, capillaries and fish gills all look the way they do.

    1.10

    氨基酸、蛋白质和酶(陈述2.6–2.8,核心实践4)

    教学大纲

    陈述2.6-2.8及核心实践4(大纲页20):氨基酸结构(氨基、羧基、R基团、通过缩合形成的肽键);蛋白质结构层次(一级序列至由氢键/离子键/二硫键连接形成的三级和四级结构);酶的作用机制——活性位点特异性、诱导契合、温度/pH/底物浓度对反应速率的影响、竞争性与非竞争性抑制;核心实践4探究酶活性(温度或pH或[S],例如淀粉酶-淀粉体系)。固定化酶及偶联酶的应用。

    来源:Cambridge International 教学大纲

    An amino acid carries an amino group, a carboxyl group, a hydrogen and an R group on one central carbon; two join by a peptide bond 肽键 in a condensation reaction. The primary structure 一级结构 is the sequence; hydrogen bonds coil it into the secondary 二级结构; further hydrogen, ionic and disulfide bonds fold the tertiary 三级结构; separate chains assemble into the quaternary 四级结构 — haemoglobin is the classic example.

    An enzyme 酶 is a protein whose active site 活性部位 fits one substrate shape. Binding is not rigid: the induced-fit 诱导契合 model has both molecule and site flexing until complementary. This is why temperature and pH change rate — they act on the tertiary structure — and why the active site is specific.

    Enzyme rate curves: temperature rises then collapses as the protein denatures above its optimum; pH shows a sharp optimum; substrate concentration saturates, and a competitive inhibitor is out-competed at high concentration.

    Competitive inhibitors 竞争性抑制剂 bind the active site and are outnumbered as substrate concentration rises; non-competitive inhibitors 非竞争性抑制剂 bind elsewhere and distort the site, so rate falls whatever the concentration.

    Worked check. Q10 is quoted at 2.4: between 10 °C and 20 °C the rate is 2.4 times faster; between 45 °C and 55 °C the same factor means the denatured enzyme is collapsing at an increasing rate. Q10 only describes; it does not explain.

    1.11

    DNA、复制和遗传密码(陈述2.9–2.12)

    教学大纲

    陈述2.9-2.12(大纲页21):单核苷酸结构(戊糖、磷酸、含氮碱基;脱氧核糖与核糖的区别)、DNA聚合体通过磷酸二酯键形成、双螺旋结构及其互补碱基配对与反向平行链;半保留复制(解旋酶解开双链、DNA聚合酶、游离核苷酸、梅塞尔森-斯塔尔实验证据);基因作为编码氨基酸序列的碱基序列;三联体密码子——非重叠性、简并性、通用性,以及起始密码子和终止密码子。

    来源:Cambridge International 教学大纲

    A mononucleotide 单核苷酸 is a pentose (deoxyribose or ribose), a phosphate and a nitrogenous base. Condensation between phosphate and sugar of neighbours makes a strand held by phosphodiester bonds 磷酸二酯键. DNA is two antiparallel strands: A pairs with T (two hydrogen bonds), G with C (three), twisted into a double helix.

    Replication is semi-conservative 半保留: helicase unzips the two strands, each is a template, and DNA polymerase DNA聚合酶 joins free nucleotides in the 5' to 3' direction. Meselson and Stahl grew bacteria first on heavy ¹⁵N then light ¹⁴N nitrogen; after one round of replication all the DNA was one intermediate density, after two rounds half was light — exactly what semi-conservative predicts and the alternatives forbid.

    A gene is a sequence of bases coding for a sequence of amino acids. The code is a triplet 三联体 code: three bases per amino acid. It is non-overlapping 非重叠 (read in successive groups), degenerate 简并 (most amino acids have several codons, so some mutations are silent), and carries start and stop codons.

    Worked check. A gene of 300 base pairs coding a protein of 99 amino acids plus a stop codon: $99 \times 3 + 3 = 300$. ✓

    1.12

    蛋白质合成、突变和筛查(陈述2.13–2.18)

    教学大纲

    陈述2.13-2.18(大纲页21-22):转录(RNA聚合酶、模板链、前体mRNA剪接)与翻译(mRNA密码子、携带氨基酸的tRNA反密码子、核糖体);源自DNA复制错误的突变——替换、插入、缺失及其移码后果;等位基因、基因型、表现型、显隐性关系及伴性遗传直至WBI11要求范围(以囊性纤维化为例说明突变基因对CFTR氯离子转运的影响、黏液积聚及感染风险);基因筛查(携带者鉴定、胚胎植入前遗传学诊断、羊膜穿刺术及绒毛取样)及筛查项目的伦理与社会问题。

    来源:Cambridge International 教学大纲

    Transcription: RNA polymerase binds the gene, uses one strand as template and builds mRNA 信使RNA. Translation: in the cytoplasm, a ribosome 核糖体 reads the mRNA codons, and tRNA 转运RNA molecules whose anticodons 反密码子 pair with those codons deliver amino acids in order; peptide bonds join them.

    Mutations 突变 arise from replication errors. A substitution 替换 swaps one base: one amino acid changes, or nothing changes (degeneracy). An insertion 插入 or deletion 缺失 adds or removes a base and shifts every reading frame after it — a frameshift 移码 — usually destroying the protein.

    Cystic fibrosis is the course's model: deletion of three bases removes one phenylalanine from the CFTR protein, so chloride transport fails, mucus stays thick, and airways clog. The allele is recessive: carriers are healthy. Genetic screening 基因筛查 can identify carriers, test embryos (pre-implantation genetic diagnosis) or foetuses (amniocentesis, chorionic villus sampling). Screening questions always carry ethics: who is tested, who is told, what happens to a positive result.

    1.12

    自我检测

    1. Explain why amylopectin releases glucose faster than amylose, referring to bonds and ends of chains.
    2. Calculate the BMI of a 91 kg man 1.83 m tall and classify him.
    3. During ventricular systole, which valves are open and which are shut, and why?
    4. The oxygen saturation at 4 kPa falls from about 60 % to about 34 % when CO₂ rises. Explain the molecular cause.
    5. Describe the clotting cascade in order, from endothelial damage to fibrin.
    6. A study reports that people who eat more yoghurt have fewer heart attacks. List three checks you would make before believing a causal claim.
    7. Explain why a competitive inhibitor lowers the rate at low substrate concentration but not at very high concentration.
    8. Meselson and Stahl ruled out conservative replication. Which observation, after one round, did the job?
    9. A frameshift mutation is usually worse than a substitution. Explain why, using the words triplet and degenerate.
    10. Name two benefits and two risks of genetic screening for cystic fibrosis carriers.

    Answers: 1 many 1,6-branch points give many chain ends for enzymes to attack simultaneously; 2 $91/1.83^2 = 27.2$, overweight; 3 atrio-ventricular shut, semilunar open, because ventricular pressure exceeds aortic but not atrial-from-above pressure; 4 hydrogen ions from CO₂ bind haemoglobin, changing its shape and lowering oxygen affinity (Bohr effect); 5 damage exposes collagen, platelets and damaged cells release thromboplastin, thromboplastin + Ca²⁺ converts prothrombin to thrombin, thrombin converts fibrinogen to fibrin, fibrin mesh traps cells; 6 sample size and selection, confounders such as age/exercise/diet, blinded comparison, mechanism or dose–response; 7 the inhibitor competes for the active site, so at very high substrate concentration nearly every collision with the site is substrate; 8 after one round all DNA was a single intermediate band — conservative predicts one heavy and one light band; 9 a frameshift changes every triplet after the mutation, so the whole protein from that point is a different amino-acid sequence, while degeneracy lets many substitutions change nothing; 10 benefits — informed family planning, earlier treatment or lifestyle change; risks — anxiety, discrimination by insurers or employers, false positives or negatives.

  • 2

    细胞、发育、生物多样性和保护

    2.1

    From one cell to a living world

    You began as one fertilised egg cell. This unit explains how that cell divided, how the copies were shuffled and halved to make gametes, how identical genes became different tissues, and how the variety that produces is counted, classified and — increasingly — defended. It is the unit of microscopes, calculations and definitions done precisely.

    WBI12 Cells, Development, Biodiversity and Conservation is the second IAS unit: 1 hour 30 minutes, 80 marks, all compulsory. It assesses Topic 3 (Cell Structure, Reproduction and Development) and Topic 4 (Plant Structure and Function, Biodiversity and Conservation). Expect labelling diagrams, microscope and magnification calculations, the index of diversity and Hardy–Weinberg questions, and long answers on conservation evaluation.

    2.1

    细胞超微结构和显微镜技术(陈述3.1–3.7,核心实践5)

    教学大纲

    主题 3 陈述 3.1-3.7 结合核心实验 5(见第22-23页)。真核细胞超微结构——细胞核与核仁、核糖体、粗面与滑面内质网、高尔基体与囊泡、线粒体、叶绿体、溶酶体、中心体、细胞表面膜——每种细胞器与其功能相关联;原核细胞超微结构(细胞壁、荚膜、质粒、单个环状DNA、70S核糖体、鞭毛)。组织、器官和系统的构成(陈述 3.2)。放大倍数与分辨率;光学与电子显微镜(TEM/SEM);放大倍数计算公式 M = 图像/实际大小;染色的作用。核心实验 5 使用显微镜观察细胞。

    来源:Cambridge International 教学大纲

    A eukaryotic 真核 cell is a set of specialised compartments:

    Structure Function
    Nucleus and nucleolus DNA stored behind a double membrane; nucleolus builds ribosomes
    Ribosomes (80S) site of protein synthesis
    Rough endoplasmic reticulum folds and transports proteins made on its ribosomes
    Golgi apparatus modifies, packages and ships proteins in vesicles
    Mitochondrion site of aerobic respiration, ATP production
    Chloroplast (plants) site of photosynthesis
    Lysosome digestive enzymes for worn organelles
    Cell-surface membrane controls transport, recognition

    Cells organise into tissues 组织 (similar cells), organs 器官 (cooperating tissues) and organ systems 器官系统. A prokaryotic 原核 cell has none of the membrane-bound organelles: a cell wall, a slime capsule 荚膜, one circular DNA molecule plus plasmids 质粒, 70S ribosomes, and sometimes a flagellum 鞭毛.

    Magnification 放大倍数 is how much bigger the image is; resolution 分辨率 is the smallest separation still seen as two. Light microscopes resolve about 200 nm; electron beams, far shorter in wavelength, resolve to a few nanometres — but need thin stained sections (TEM) or gold-coated surfaces (SEM). $M = \text{image size} / \text{actual size}$, keeping units consistent.

    Worked check. A mitochondrion measures 14 mm on a photograph labelled ×20 000. Actual length $= 14\,000\ \mu\text{m} / 20\,000 = 0.7\ \mu$m. A nucleus that must be seen with its double membrane needs an electron microscope: the two membranes lie closer than a light microscope's resolution.

    2.2

    减数分裂、配子和受精(陈述3.9–3.13)

    教学大纲

    陈述 3.9-3.13(见第23页):基因座与连锁;减数分裂作为两次分裂产生四个遗传上不同的单倍体细胞;同源染色体对的独立分配与染色单体交叉互换作为变异的两个来源;哺乳动物配子的特化(精子顶体、中段线粒体、卵细胞质和透明带);哺乳动物的受精(顶体反应、皮层反应阻止多精入卵)及开花植物的受精(花粉管、双受精)。

    来源:Cambridge International 教学大纲

    Meiosis 减数分裂 is two successive divisions after one DNA replication, producing four genetically different haploid 单倍体 cells from one diploid 二倍体 cell. Two processes create the variation:

    • Independent assortment 独立分配 — each homologous pair lines up and separates at random in meiosis I; each gamete gets one of each pair, from either parent at random ($2^n$ combinations).
    • Crossing over 交叉互换 — in prophase I, homologous chromatids swap sections, mixing alleles within chromosome pairs.

    Mammalian gametes are specialised for delivery and provisioning: the sperm 精子 carries an acrosome 顶体 (digestive enzymes), a midpiece of mitochondria and a flagellum; the egg 卵子 contributes nearly all cytoplasm and a zona pellucida 透明带 that locks out extra sperm after the cortical reaction 皮质反应. In flowering plants the pollen tube delivers two male nuclei: one fertilises the egg, the other the polar nuclei — double fertilisation 双受精.

    DNA content per cell through the cell cycle, both meiotic divisions, gametes and fertilisation. Note the doubling in S phase and the two halvings.
    2.3

    有丝分裂、细胞周期和有丝分裂指数(陈述3.14–3.16,核心实践6)

    教学大纲

    陈述 3.14-3.16 结合核心实验 6(见第23页):细胞周期(间期 G1-S-G2,然后是有丝分裂和胞质分裂);前期、中期、后期、末期各阶段染色体的行为;有丝分裂产生遗传上相同的细胞以用于生长、修复和无性繁殖;计算有丝分裂指数(处于有丝分裂期的细胞数除以总细胞数)及其临床应用;核心实验 6 制备和观察根尖压片。

    来源:Cambridge International 教学大纲

    Mitosis 有丝分裂 produces two genetically identical diploid cells: growth, repair, asexual reproduction. Interphase ($G_1$, S — DNA replication, $G_2$) fills most of the cycle; mitosis itself is prophase (chromosomes condense, spindle forms), metaphase (chromosomes line at the equator), anaphase (spindle fibres pull chromatids to poles), telophase (nuclear membranes reform), then cytokinesis 胞质分裂.

    The mitotic index 有丝分裂指数 $=$ cells in mitosis $\div$ total cells counted. High values in a tissue biopsy mean rapid division — a cancer signature.

    Worked check. A root-tip squash shows 36 of 450 cells in mitosis: index $= 36/450 = 0.08 = 8\%$. If the cycle takes 20 hours, mitosis occupies $0.08 \times 20 = 1.6$ hours.

    2.4

    干细胞、分化和表型(陈述3.17–3.21)

    教学大纲

    陈述 3.17-3.21 (考纲 p.24):全能和多能干细胞、桑椹胚和囊胚阶段;通过差异基因表达进行分化;mRNA的转录后修饰使一个基因产生多种蛋白质;基因型与环境在表型中的相互作用;复等位基因和多基因遗传导致连续变异。干细胞使用的伦理问题贯穿整个考试题目。

    来源:Cambridge International 教学大纲

    A totipotent 全能性 stem cell can form a whole organism (the first divisions of the zygote); a pluripotent 多能性 cell (inner blastocyst 胚泡) can form any tissue but not the whole organism. Differentiation happens by differential gene expression 差异基因表达: every cell keeps the whole genome, but each type switches on only its own subset. One gene can yield several proteins by post-transcriptional modification 转录后修饰 — RNA splicing choices of the pre-mRNA. The final phenotype also reflects the environment (height needs both genes and nutrition) and, for many traits, multiple genes acting together (polygenic 多基因 inheritance giving continuous variation).

    2.5 2.6

    植物细胞结构、木质部和韧皮部(陈述4.1–4.6,核心实践7–8)

    教学大纲

    Topic 4 statements 4.1-4.6 with Core Practicals 7-8 (spec pp.26-27): plant cell ultrastructure (cellulose cell wall, plasmodesmata, chloroplasts, amyloplasts, vacuole); starch and cellulose structure contrasted with storage vs support roles; cellulose microfibrils and secondary thickening giving tensile strength; xylem vessels (lignified, dead, water transport and support) and phloem sieve tubes with companion cells (translocation); distribution in root, stem and leaf; Core Practical 7 investigates plant tissue structure and Core Practical 8 water transport (potometer work).

    陈述 4.7-4.13 (考纲 pp.27-28):植物纤维和淀粉的可持续性;水和无机盐的重要性(硝酸盐用于合成氨基酸,钙用于细胞壁结构和信号传导,镁用于合成叶绿素);来自植物的抗菌和治疗物质(核心实验 9 抗菌测试);细菌生长条件;从历史到现代方案的药物测试发展,包括试验的各个阶段。

    来源:Cambridge International 教学大纲

    Plant cells add a cellulose cell wall 细胞壁 joined to neighbours by plasmodesmata 胞间连丝. Cellulose chains bundle into microfibrils 微纤维 whose criss-cross laying gives tensile strength; secondary thickening with lignin strengthens and waterproofs.

    Tissue Built from Function
    Xylem 木质部 dead hollow cells, lignified walls water and mineral transport; support
    Phloem 韧皮部 living sieve-tube elements + companion cells translocation of sugars
    Schematic cross-sections: in the root the xylem forms a central star with phloem between its arms; in the stem, vascular bundles arrange phloem outside xylem in a ring.

    Water and ions matter beyond drinking: nitrate 硝酸盐 builds amino acids, calcium 钙离子 cross-links pectin in walls and signals inside cells, magnesium 镁离子 sits at the heart of chlorophyll — a shortage yellows the leaves.

    2.7

    分类和测量生物多样性(陈述4.14–4.18)

    教学大纲

    陈述 4.14-4.18 (考纲 p.28):分类学根据共同特征组织生命,采用三域系统(古菌、细菌、真核生物),并有分子序列证据;物种丰富度和多样性指数(D)在生境内的计算;种内杂合度指数;特有性;对生物多样性的威胁。每年考试都会考查使用D公式的计算题。

    来源:Cambridge International 教学大纲

    Classification arranges life by shared characteristics; the three-domain system 三域系统 (Bacteria, Archaea, Eukarya) rests on molecular evidence — ribosomal RNA sequences — that revealed Archaea are closer to us than to bacteria. Within a habitat, diversity has two measures:

    • Species richness 物种丰富度 — how many species.
    • Index of diversity 多样性指数 — $D = \dfrac{N(N-1)}{\sum n(n-1)}$, where $N$ is all individuals and $n$ each species' count. Higher $D$, more diverse.

    Worked check. Habitat A: 3 species with counts 40, 35, 25 ($N = 100$). $D = 100 \times 99 / (40 \times 39 + 35 \times 34 + 25 \times 24) = 9900 / (1560 + 1190 + 600) = 9900/3350 = 2.96$. Habitat B: 3 species, counts 90, 6, 4: $D = 9900/(8010 + 30 + 12) = 1.23$. Same richness, very different balance — $D$ sees what richness cannot.

    Within one species, the heterozygosity index 杂合度指数 $H =$ (number of heterozygotes) $\div$ (number of individuals) measures genetic variety. A species found nowhere else is endemic 特有种 — and its island home makes it both precious and fragile.

    2.8

    生态位、哈迪-温伯格平衡和保护(陈述4.19–4.21)

    教学大纲

    陈述 4.19-4.21 (考纲 pp.28-29):生态位概念及适应实例(解剖学、生理学、行为学);哈迪-温伯格方程 p^2 + 2pq + q^2 = 1 用于种群中的等位基因频率及其假设条件;评估动物园和种子库保护方法、圈养繁殖和栖息地走廊。

    来源:Cambridge International 教学大纲

    A niche 生态位 is a species' way of life: its role, its requirements, its tolerances. Adaptations come in three registers — anatomical (thick fur), physiological (enzyme variants), behavioural (migration timing).

    The Hardy–Weinberg equation 哈迪–温伯格方程 tracks allele frequencies in a large, randomly breeding population with no selection, migration or mutation:

    $$p^2 + 2pq + q^2 = 1$$

    $q^2$ = frequency of the homozygous recessive genotype. If 4 % of a population shows the recessive phenotype, $q^2 = 0.04$, so $q = 0.2$, $p = 0.8$, carriers $2pq = 0.32$.

    Worked check. Of 500 foxes, 45 are homozygous recessive: $q^2 = 0.09$, $q = 0.3$, $p = 0.7$. Expected carriers $= 2pq \times 500 = 0.42 \times 500 = 210$.

    Conservation questions want balance. Zoos and seed banks 种子库 keep species against extinction and run captive breeding, but cost space, lose natural behaviour and serve few individuals; habitat protection and corridors 走廊 preserve ecology at scale but need land, money and political will. Credit the specific argument, not the sentiment.

    2.6

    植物产物、矿物质离子和药物检测(陈述4.7–4.13)

    教学大纲

    陈述 4.7-4.13 (考纲 pp.27-28):植物纤维和淀粉的可持续性;水和无机盐的重要性(硝酸盐用于合成氨基酸,钙用于细胞壁结构和信号传导,镁用于合成叶绿素);来自植物的抗菌和治疗物质(核心实验 9 抗菌测试);细菌生长条件;从历史到现代方案的药物测试发展,包括试验的各个阶段。

    来源:Cambridge International 教学大纲

    For the microbiology here (and Unit 4's): grow bacteria on nutrient agar or in broth, using aseptic technique 无菌技术 — sterilise loops by flaming, flame bottle necks, work near an updraught, seal plates, incubate below body temperature so human pathogens cannot grow. Count colonies on spread or pour plates (each colony from one cell or a clump: colony-forming units), or count cells with a haemocytometer, or follow turbidity 浊度 in broth.

    A microbial growth curve on a log scale: lag, exponential (log), stationary and death phases.

    Worked check. In the exponential phase, the growth-rate constant $k = (\log_{10} N_t - \log_{10} N_0)/(0.301 \times t)$. From $1 \times 10^4$ to $4 \times 10^7$ cells in 4 hours: $\log$ difference $= 3.602$, so $k = 3.602/(0.301 \times 4) \approx 3.0$ divisions per hour — about a 20-minute generation time.

    2.6

    自我检测

    1. Calculate the actual width of a nucleus that appears 24 mm wide at ×4 000. Give your answer in µm.
    2. Name the two processes in meiosis that create genetic variation, and state when each happens.
    3. A cell has 2 a.u. of DNA in G1. Sketch how DNA content changes through S phase, meiosis I and meiosis II.
    4. Explain why all cells of a blastocyst can become any tissue, but a mature muscle cell cannot.
    5. Explain two ways xylem is adapted to its function, one structural and one chemical.
    6. Two habitats each contain 5 plant species. Explain why the index of diversity can still differ between them.
    7. In a population of 1 000, 16 people show a recessive condition. Calculate q, p and the expected number of carriers.
    8. A species of snail is endemic to one island. Explain why endemic species face higher extinction risk.
    9. State three conditions a population must meet for the Hardy–Weinberg equation to hold.
    10. Give one benefit and one limitation of seed banks compared with protecting habitat.

    Answers: 1 $24\,000/4\,000 = 6\ \mu$m; 2 crossing over in prophase I, independent assortment in metaphase/anaphase I (and II); 3 rises to 4 in S, halves to 2 after meiosis I, halves to 1 after meiosis II, restored to 2 at fertilisation; 4 blastocyst cells are pluripotent — all genes still switchable; a muscle cell has switched off other pathways permanently in ordinary conditions; 5 hollow, lignified dead cells form continuous water columns with no cytoplasm to resist flow; lignin waterproofs and strengthens walls against collapse; 6 D weighs the evenness of abundance, not just the species count; 7 $q^2 = 0.016$, $q \approx 0.126$, $p \approx 0.874$, carriers $2pq \approx 0.22$, about 220 people; 8 a single local event (storm, disease, introduced predator, habitat loss) removes the whole species at once; 9 large population, random mating, no selection/mutation/migration; 10 seed banks store genetic diversity cheaply and safely against habitat loss (benefit) but species exist out of their ecology — no interactions, no evolution in place, and recolonisation is uncertain (limitation).

  • 3

    生物学实践技能 I

    • 3.1 规划调查(单元3大纲,规划部分)

      学习计划即将上线

    • 3.2 实施与测量(单元3大纲,实施部分)

      学习计划即将上线

    • 3.3 数据处理、图表与不确定性(单元3大纲,数据处理部分)

      学习计划即将上线

  • 4

    能量、环境、微生物学和免疫学

    4.1

    Energy in, energy through, energy lost

    A wheat field intercepts sunlight, fixes carbon, feeds a food chain — and loses energy at every step. This unit follows the energy from photon to ecosystem, then turns to the smallest participants: the microbes that recycle, the pathogens that take, and the immune system that defends. WBI14 is also the most calculation-heavy paper after Unit 1: expect NPP, growth constants, diversity indices and time-of-death arithmetic.

    WBI14 Energy, Environment, Microbiology and Immunity is the first IA2 unit: 1 hour 30 minutes, 90 marks, all compulsory. It assesses Topic 5 (Energy Flow, Ecosystems and the Environment) and Topic 6 (Microbiology, Immunity and Forensics).

    4.1

    光合作用与叶绿体色素(知识点5.1–5.8,核心实践10)

    教学大纲

    主题 5 陈述 5.1-5.8 及核心实验 10 (考纲 pp.29-30)。光合作用的总反应式;光依赖反应(水的光解、ADP的光磷酸化、NADP的还原)发生在类囊体膜上;光非依赖反应(卡尔文循环:RuBP催化固定二氧化碳、GALP生成、核酮糖二磷酸再生)发生在基质中;叶绿体结构与这两个阶段的对应关系;吸收光谱和作用光谱;色素的纸层析法及Rf值(核心实验 10);限制因子(光照强度、二氧化碳浓度、温度)及其农业应用调控。

    来源:Cambridge International 教学大纲

    The overall reaction: $6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}$, driven by light energy. In the thylakoid 类囊体 membranes, the light-dependent reactions 光反应 photolyse water (releasing O₂ and H⁺), photophosphorylate 光合磷酸化 ADP and reduce NADP. In the stroma 基质, the light-independent reactions 暗反应 (Calvin cycle) fix carbon dioxide onto ribulose bisphosphate with rubisco, reduce the products using the ATP and reduced NADP from stage one, and make GALP — some to glucose, most to regenerate RuBP.

    Structure matches function: thylakoids stacked for absorbing surface; stroma packed with enzymes for the cycle.

    Pigments and spectra

    Chlorophyll a and b absorb red and blue light; carotenoids widen the range and protect. The absorption spectrum 吸收光谱 shows what each pigment takes up; the action spectrum 作用光谱 shows what rate of photosynthesis each wavelength drives — their agreement is evidence that these pigments do the work. Chromatography separates pigments by solubility; $\mathrm{Rf} = \text{distance moved by pigment} / \text{distance moved by solvent}$.

    Absorption spectra of the chloroplast pigments, with the shaded action spectrum tracking their combined absorption.

    Limiting factors 限制因素: light intensity, CO₂ concentration, temperature. A greenhouse raises the weakest of the three — and exam answers say which and why.

    4.2

    生产力与能量传递(知识点5.9–5.10)

    教学大纲

    陈述 5.9-5.10 (考纲 p.30):总初级生产力(GPP)、净初级生产力(NPP = GPP减去呼吸消耗)及所用单位(kJ m-2 yr-1);营养级间的生物量传递效率及其计算;能量传递损失的原因(呼吸作用、排遗、排泄、未食用部分)以及食物链为何较短。

    来源:Cambridge International 教学大纲

    Gross primary productivity 总初级生产力 (GPP) is the energy fixed by photosynthesis per area per time (kJ m⁻² yr⁻¹). Plants spend much of it on their own respiration; what remains is net primary productivity 净初级生产力:

    $$\mathrm{NPP} = \mathrm{GPP} - R$$

    Only about 10 % of the energy at each trophic level passes upward — lost to respiration as heat, egestion, excretion and parts never eaten. That is why chains rarely exceed four or five levels and why top predators are rare.

    Energy leaving a food chain: GPP split by plant respiration, then roughly a tenth passing at each step upward.

    Worked check. A grassland fixes 45 000 kJ m⁻² yr⁻¹ and respirers 19 000. NPP $= 26\,000$ kJ m⁻² yr⁻¹. Cattle eating that grass keep about 10 %: 2 600 kJ m⁻² yr⁻¹. Humans eating the cattle keep 260 — the energy arithmetic behind "eating lower on the chain feeds more people".

    4.3

    生态系统、种群与演替(知识点5.11–5.15,核心实践11)

    教学大纲

    陈述 5.11-5.15 及核心实验 11 (考纲 pp.30-31):生境、种群、群落和生态系统的术语;控制分布和数量的生物和非生物因素;生态位概念的应用;捕食者-猎物周期;测量密度的方法(样方、样线、标志重捕法);从先锋物种定殖到顶级群落的演替阶段(核心实验 11 调查生境)。

    来源:Cambridge International 教学大纲

    Four definitions, precisely: habitat 栖息地 (where), population 种群 (one species' individuals), community 群落 (all populations), ecosystem 生态系统 (community + environment). Abiotic factors (light, water, temperature, pH) and biotic factors (competition, predation, disease) set who lives where and how abundantly — together, the niche 生态位.

    Predator and prey cycle with a delay: hares rise, lynx follow, hares fall, lynx starve — a lag of about a quarter cycle.

    A classic predator-prey cycle: the lynx population tracks the hare population with a delay.

    Succession 演替 begins when pioneer species colonise bare ground; each stage changes the soil and microclimate, enabling the next, until the climax community 顶极群落 stabilises. Measuring abundance uses quadrats and transects (random sampling for estimates, systematic along a gradient) and capture-mark-recapture for animals ($N \approx \text{first catch} \times \text{second catch} / \text{marked recaptures}$).

    4.4

    气候变化的证据、成因及影响(知识点5.16–5.22,核心实践12)

    教学大纲

    陈述 5.16-5.22 及核心实验 12 (考纲 pp.31-32):气候变化的证据(温度记录、冰芯、树木年轮、泥炭中的花粉);温室气体及温室效应增强(CO2、甲烷、氧化亚氮——来源、大气寿命、全球变暖潜能值);碳循环及降低大气碳含量的方法;数据外推及其可靠性;气候变化对降雨模式、物种分布、酶依赖性生态系统的影响;核心实验 12 探究温度对酶/相关生境活性的影响。

    来源:Cambridge International 教学大纲

    Evidence comes from direct temperature records, ice cores (trapped air giving ancient CO₂), tree rings (dendrochronology 树轮年代学) and pollen layers in peat. The enhanced greenhouse effect 温室效应: outgoing long-wave radiation is absorbed by CO₂, methane and nitrous oxide and re-emitted downward. The gases differ — methane is potent but short-lived; CO₂ lasts centuries; nitrous oxide combines both — so policy weighs global warming potential 全球增温潜势 against lifetime.

    Extrapolating a trend beyond its data is a judgement, not a fact: state the assumption (the trend continues) and its weakness. Effects already examined: shifting rainfall and species ranges, and — close to this course — temperature's effect on enzyme rates in cold-blooded organisms and whole ecosystems.

    Worked check. Methane's warming potential is about 28× CO₂'s per kg, but it lasts ~12 years against centuries. A one-tonne methane leak matters 28× a one-tonne CO₂ leak this decade, but fades while the CO₂ persists — the arithmetic of "warming potential vs cumulative load".

    4.5

    进化与物种形成(知识点5.23–5.26)

    教学大纲

    陈述5.23-5.26(大纲页32):进化作为由突变、选择(定向选择和稳定化选择)、基因流和遗传漂变引起的等位基因频率变化;隔离减少基因流(地理隔离=异域物种形成;生殖隔离=同域物种形成)导致物种形成;根据证据标准解读有争议的科学结论(气候、进化);重新造林、可持续资源和生物燃料。

    来源:Cambridge International 教学大纲

    Evolution is a change in allele frequency 等位基因频率. It needs variation (mutation, meiosis) and a filter: directional selection 定向选择 shifts the mean when the environment moves; stabilising selection 稳定选择 trims the extremes in a stable one. Drift changes frequencies by chance in small populations; gene flow (migration) mixes them back.

    Speciation 物种形成 needs isolation: allopatric 异域的 (a physical barrier — allopatric = different homeland) or sympatric 同域的 (reproductive isolation in the same place — behavioural, temporal or genetic). Isolated populations diverge until they can no longer interbreed.

    4.6

    微生物培养与生长曲线(知识点6.1–6.4,核心实践13)

    教学大纲

    陈述6.1-6.4及核心实验13(大纲页33):培养基(营养琼脂、肉汤);无菌技术的细节;测量微生物生长的方法(细胞计数、浊度、通过稀释涂布法的活菌计数);细菌生长曲线的四个阶段(延滞期、对数/指数期、稳定期、衰亡期)及对数增长速率常数k;核心实验13利用校准曲线研究液体培养物中的生长速率。

    来源:Cambridge International 教学大纲

    Aseptic technique protects you and the culture: flame the loop to red heat, flame bottle necks, lift lids briefly, seal plates with tape, incubate at 25 °C (never body temperature, so human pathogens cannot multiply). Measure growth by direct cell count (haemocytometer), turbidity 浊度 in broth, or viable counts by dilution plating 稀释涂布 (colonies counted × dilution factor).

    The growth curve: lag 延迟期 (enzymes synthesising), log 对数期 (exponential doubling), stationary 稳定期 (deaths = divisions), death 衰亡期. The exponential constant:

    $$k = \frac{\log_{10} N_t - \log_{10} N_0}{0.301 \times t}$$

    Worked check. From $2\times10^4$ to $1.6\times10^7$ cells in 5 hours: log difference $= 2.903$; $k = 2.903/(0.301\times5) = 1.93$ generations per hour — a doubling time of about 31 minutes.

    4.7

    病原体、感染途径及结核病/HIV(知识点6.5–6.7)

    教学大纲

    陈述6.5-6.7(大纲页33):比较细菌和病毒的结构(核酸、衣壳、包膜);感染途径(消化道、呼吸道、性接触、伤口、媒介);感染屏障(皮肤、黏液、胃酸、血液凝固);结核分枝杆菌和HIV如何感染并损害人体,以及HIV如何加重结核病。

    来源:Cambridge International 教学大纲

    Bacteria: cell wall, plasmids, ribosomes, circular DNA, some with capsule and flagellum. Viruses: nucleic acid (DNA or RNA) in a capsid 衣壳, some with an envelope — no organelles, no metabolism of their own; they hijack host cells. Mycobacterium tuberculosis 结核分枝杆菌 infects lungs, surviving inside macrophages; HIV 人类免疫缺陷病毒 destroys T helper cells, so the immune system collapses and TB reactivates — the classic AIDS-defining illness.

    Barriers come first: skin, mucus and cilia, stomach acid, lysozyme in tears, and the clotting cascade sealing wounds.

    4.8

    免疫反应、抗原与抗体(知识点6.8–6.12)

    教学大纲

    陈述6.8-6.12(大纲页34):非特异性反应(炎症、溶菌酶、干扰素);抗原;体液反应——B效应细胞分化为分泌抗体的浆细胞,B记忆细胞介导二次反应;细胞反应——T辅助细胞、T杀伤细胞和T记忆细胞;抗体结构(四条多肽链,可变区结合位点)及抗原-抗体复合物;天然与人工、主动与被动免疫;疫苗接种及群体免疫;病原体与宿主之间的进化军备竞赛。

    来源:Cambridge International 教学大纲

    Non-specific: inflammation (vasodilation and fluid leak bringing phagocytes), phagocytosis, interferon. Specific — antigens 抗原 trigger it:

    • Humoral 体液: B effector cells → plasma cells → antibodies 抗体; B memory cells B记忆细胞 remain for the faster, stronger secondary response.
    • Cellular 细胞: T helper cells coordinate, T killer cells T杀伤细胞 destroy infected cells, T memory cells persist.

    An antibody is four polypeptide chains (two heavy, two light) with a variable region 可变区 forming one specific binding site — the shape that fits one antigen. Immunity splits two ways: active 主动 (your own antibodies — natural infection or vaccination) vs passive 被动 (given antibodies — mother to baby, antiserum); natural vs artificial in each. Vaccination works at the scale of populations too (herd immunity 群体免疫) — but pathogens evolve back: the evolutionary race.

    Worked check. A vaccine's second dose produces antibody levels ten times the first and within days rather than weeks: clonal selection finds the memory B cells, which divide rapidly into plasma cells — the secondary response the primary response built.

    4.9

    抗生素与医院获得性感染(知识点6.13–6.15,核心实践14)

    教学大纲

    陈述6.13-6.15及核心实验14(大纲页34):抑菌剂与杀菌剂类抗生素;抗生素耐药性的产生与传播(自然选择);核心实验14探究抗生素对细菌的作用;如何通过了解医院获得性感染的成因(卫生条件、侵入性操作、耐药菌株)来降低其发生率。

    来源:Cambridge International 教学大纲

    Bacteriostatic 抑菌的 antibiotics stop bacterial growth; bactericidal 杀菌的 ones kill. Resistance arises by mutation and spreads by selection whenever exposure kills the susceptible and spares the resistant — finishing a course and avoiding unnecessary use both slow it. Hospital-acquired infections thrive where vulnerable patients, invasive devices and resistant strains meet; hand hygiene, sterile procedure and isolating carriers break the chain.

    4.10

    分解作用、PCR、DNA指纹鉴定与死亡时间推断(知识点6.16–6.20)

    教学大纲

    陈述 6.16-6.20 (考纲 pp.34-35):微生物在分解和养分循环中的作用;聚合酶链式反应(变性、引物退火、延伸)扩增DNA;凝胶电泳按长度分离DNA片段;用于识别和遗传关系的DNA指纹鉴定;根据体温、尸僵程度、肌肉收缩程度及昆虫演替顺序推断死亡时间。

    来源:Cambridge International 教学大纲

    Decomposers recycle carbon and nitrogen — the ecosystem's waste system. The polymerase chain reaction 聚合酶链式反应 cycles denaturation (95 °C), primer annealing (50–60 °C) and extension (72 °C), doubling the target DNA each cycle: $n$ cycles give $2^n$ copies. Gel electrophoresis 凝胶电泳 drags negatively charged DNA fragments through gel — short fragments run furthest. The profile 电泳图谱 of band positions identifies individuals (enough loci differ between unrelated people) and measures relatedness (shared bands).

    Time of death narrows by four clocks: body temperature (falls ~1 °C per hour, modified by size, clothing, air), rigor mortis 尸僵 (sets then passes), decomposition degree, and insect succession on the body — each stage's arrivals time-stamp the interval.

    Worked check. A body found at 22 °C core temperature in a 15 °C room: $(37-22)/(1\ ^\circ\text{C per hour}) \approx 15$ hours — but state the assumptions (still air, average build) before trusting it.

    4.10

    自我检测

    1. Name the products of the light-dependent reactions used by the Calvin cycle.
    2. A crop fixes 60 000 kJ m⁻² yr⁻¹ and respirers 20 000. Calculate NPP and the energy reaching a third trophic level at 10 % transfer each step.
    3. Explain why the action spectrum is evidence that chlorophyll carries out photosynthesis.
    4. Distinguish directional and stabilising selection with one example of each.
    5. Give two reasons a food chain rarely exceeds five trophic levels.
    6. A culture grows from $5\times10^3$ to $4\times10^6$ in 6 hours. Calculate k and the doubling time.
    7. Explain why a second vaccine dose raises antibody titre far faster than the first.
    8. State the difference between bacteriostatic and bactericidal, and why the distinction matters for a patient with a weak immune system.
    9. Three PCR cycles from one double-stranded template: how many copies?
    10. A body's core is 30 °C. Estimate the time since death under standard assumptions and name one factor that would extend the true interval.

    Answers: 1 ATP and reduced NADP; 2 NPP $= 40\,000$; third level $= 400\,\text{kJ m}^{-2}\text{yr}^{-1}$ (10 % of 4 000); 3 wavelengths chlorophyll absorbs best are the wavelengths that drive photosynthesis best; 4 directional — antibiotic resistance shifting the mean; stabilising — human birth weights; 5 energy lost at each step to respiration/heat/egestion, so little remains; producers also lose to respiration before the first transfer; 6 log difference 2.903, $k = 2.903/(0.301\times6) = 1.61$ h⁻¹, doubling ≈ 37 min; 7 memory B cells persist; the second encounter selects clones that divide rapidly into plasma cells — the secondary response; 8 bacteriostatic stops growth (immune system must clear), bactericidal kills outright — a weak immune system needs bactericidal support; 9 $2^3 = 8$ copies; 10 $(37-30)/1 = 7$ hours; heavy clothing or a warm room slows cooling, so true time is longer.

  • 5

    呼吸、内环境、协调与基因技术

    5.1

    Inside the body: energy, balance and control

    Sprint for a bus and your body performs a symphony: muscles burn glucose faster, the heart doubles its output, ventilation deepens, sweat starts, and the kidney re-tunes its water recovery — all controlled without a conscious thought. This unit follows that machinery: respiration at molecular scale, muscle and heart at tissue scale, nerves and hormones as the controllers, and gene technology as the modern window onto it.

    WBI15 Respiration, Internal Environment, Coordination and Gene Technology is the second IA2 unit: 1 hour 45 minutes, 90 marks. It assesses Topic 7 (Respiration, Muscles and the Internal Environment) and Topic 8 (Coordination, Response and Gene Technology), and one 20-mark question is built on a pre-released scientific article — you will have studied it in advance.

    5.1

    呼吸作用:从糖酵解到氧化磷酸化(知识点7.1–7.8,核心实践15–16)

    教学大纲

    主题7陈述7.1-7.8及核心实验15-16(大纲页35-36)。总需氧反应及呼吸作用作为分级、受酶控制的过程;细胞质中的糖酵解(己糖磷酸化、底物水平磷酸化生成ATP、还原型NAD、丙酮酸;无氧条件下生成乳酸);线粒体基质中的连接反应和克雷布斯循环(脱羧、ATP、还原型NAD、还原型FAD);嵴上的氧化磷酸化——电子传递链将氢离子泵入膜间隙并通过ATP合酶进行化学渗透,氧气作为最终电子受体;无氧运动后的乳酸代谢;呼吸商RQ = 产生的CO2 / 消耗的O2,针对碳水化合物(1.0)、脂质(0.7)和蛋白质(0.9);核心实验15-16(呼吸计和人工碳酸氢盐指示剂呼吸法)。

    来源:Cambridge International 教学大纲

    Respiration releases energy in steps, each catalysed by a specific enzyme. Aerobic respiration's summary:

    $$\mathrm{C_6H_{12}O_6} + 6\mathrm{O_2} \rightarrow 6\mathrm{CO_2} + 6\mathrm{H_2O}\ (+\ \text{ATP})$$
    Stage Where What happens Net yield
    Glycolysis 糖酵解 cytoplasm hexose phosphorylated then split; pyruvate formed 2 ATP, 2 reduced NAD
    Link reaction 丙酮酸氧化脱羧 mitochondrial matrix pyruvate oxidised and decarboxylated to acetyl 1 CO₂, 1 reduced NAD per pyruvate
    Krebs cycle 三羧酸循环 matrix acetyl fully oxidised; substrates regenerate CO₂, ATP, reduced NAD, reduced FAD
    Oxidative phosphorylation 氧化磷酸化 cristae electron transport chain pumps H⁺; chemiosmosis through ATP synthase; O₂ the final acceptor most ATP

    Anaerobic conditions stop the chain — reduced NAD cannot be recycled — so pyruvate accepts its hydrogen and becomes lactate 乳酸 in muscle. After exercise, lactate is carried to the liver and rebuilt into glucose (needing oxygen: the oxygen debt 氧债).

    Respiratory quotient 呼吸商: $\mathrm{RQ} = \mathrm{CO_2}\ \text{produced} / \mathrm{O_2}\ \text{consumed}$. Carbohydrate 1.0, protein 0.9, lipid ≈ 0.7 — an organism's RQ reveals what it is burning.

    Left: respiration rate peaks at the enzyme optimum near body temperature and collapses as proteins denature. Right: respiratory quotient by substrate.

    Worked check. A respirometer shows 60 cm³ CO₂ produced and 60 cm³ O₂ consumed in the same interval: RQ = 1.0 — carbohydrate. A germinating seed with RQ 0.7 is burning lipid reserves.

    5.2

    肌肉与运动(知识点7.9–7.11)

    教学大纲

    陈述7.9-7.11(大纲页36):肌肉、肌腱、骨骼和韧带如何在运动中相互作用(肌肉牵拉骨骼跨过关节;肌腱连接肌肉与骨骼;韧带固定骨骼);骨骼肌纤维的结构(融合的多核细胞,由肌动蛋白和肌球蛋白组成的肌原纤维);滑行丝理论(钙离子暴露结合位点,带ATP的肌球蛋白头部附着并拉动,肌动蛋白滑动);快缩纤维与慢缩纤维及其对短跑和耐力运动的适应。

    来源:Cambridge International 教学大纲

    A tendon 肌腱 joins muscle to bone; a ligament 韧带 joins bone to bone across a joint; muscles work in antagonistic pairs. A skeletal muscle fibre is a fused, multinucleate cell packed with myofibrils 肌原纤维 of actin 肌动蛋白 and myosin 肌球蛋白.

    Contraction is the sliding filament cycle: an action potential releases Ca²⁺; calcium exposes actin's binding sites; the energised myosin head (its ATP split) attaches, pulls the actin, then detaches when new ATP binds — thousands of heads per second, each stroke a few nanometres.

    Fast-twitch fibres: thick, few mitochondria, quick and powerful, fatigue fast (sprinting). Slow-twitch: many mitochondria, rich blood supply, myoglobin — endurance.

    5.3

    心脏活动与呼吸控制(知识点7.12–7.15,核心实践17)

    教学大纲

    陈述 7.12-7.13,7.15 结合核心实验 17(规格页码 p.36):心肌的肌源性本质;心脏电活动如何起自窦房结,传导至房室结并沿浦肯野纤维传播;延髓中心的心血管中枢和呼吸中枢对心率和呼吸频率的控制;心输出量 = 每搏输出量 × 心率及其计算;运动期间呼吸量和心输出量的变化;核心实验 17 利用肺活量计曲线研究运动对潮气量、呼吸频率和分钟通气量的影响。

    来源:Cambridge International 教学大纲

    Cardiac muscle is myogenic 自主性的: it beats without nerve input. The sino-atrial node 窦房结 sets the pace; the wave spreads over the atria to the atrio-ventricular node, then along Purkyne fibres to the ventricles — atria contract first, ventricles a fraction later.

    The cardiovascular control centre and the ventilation centre in the medulla oblongata 延髓 adjust both systems to demand: chemical receptors sense CO₂ and pH, baroreceptors pressure.

    $$\text{cardiac output} = \text{stroke volume} \times \text{heart rate}$$

    Worked check. At rest: 70 cm³ × 72 min⁻¹ ≈ 5.0 dm³ min⁻¹. During exercise: 120 cm³ × 150 min⁻¹ = 18 dm³ min⁻¹ — the table the exam loves. Ventilation rises in step, read from spirometer traces (tidal volume × breathing rate = minute ventilation 每分钟通气量).

    5.4

    稳态、反馈调节与体温调节(知识点7.14,7.16–7.17)

    教学大纲

    陈述 7.14,7.16-7.17(规格页码 p.36):稳态作为维持内环境动态平衡;负反馈将变量保持在狭窄范围内(正反馈放大变化);自主神经系统、肾上腺素在“战斗或逃跑”反应中的作用,以及下丘脑在体温调节中的功能(血管舒张、出汗、寒战、血管收缩、立毛)。

    来源:Cambridge International 教学大纲

    Homeostasis 稳态 holds the internal environment in dynamic equilibrium — not stillness, but variables oscillating inside narrow limits. Negative feedback 负反馈 corrects deviations in either direction (the thermostat pattern); positive feedback 正反馈 amplifies them (a spiral, not a loop — as in an action potential or fever runaway).

    Thermoregulation runs through the hypothalamus 下丘脑: heat loss by vasodilation and sweating; heat conservation by vasoconstriction, shivering and hair erection. Adrenaline 肾上腺素 prepares the fight-or-flight response — heart rate, blood glucose and airways all respond together.

    5.5

    肾脏结构与渗透压调节(知识点7.18–7.21)

    教学大纲

    陈述 7.18-7.21(规格页码 pp.36-37):哺乳动物肾脏的宏观与微观结构(皮质、髓质、肾盂;肾单位包括肾小球、鲍曼囊、近曲小管、髓袢、集合管);超滤过(入球小动脉的高压、足细胞、基膜)及近曲小管的选择性重吸收(葡萄糖、氨基酸、部分盐类和水分);髓袢与髓质中的水势梯度;垂体分泌的抗利尿激素增加集合管对水的通透性;肝脏中由过量氨基酸产生的尿素(脱氨基作用,鸟氨酸循环无需详细掌握)。

    来源:Cambridge International 教学大纲

    Blood enters the nephron 肾单位's glomerulus at high pressure — the afferent arteriole is wider than the efferent — and small molecules are forced through podocytes and the basement membrane into Bowman's capsule 肾小囊 (ultrafiltration: everything except cells and large proteins). The proximal convoluted tubule 近曲小管 reabsorbs all glucose and amino acids plus most salts and water (active transport then osmosis). The loop of Henle 亨勒环 builds a water-potential gradient down the medulla by countercurrent multiplication. The collecting duct 集合管 makes the final decision: ADH 抗利尿激素 from the pituitary inserts aquaporins so more water is reabsorbed — concentrated urine when dehydrated, dilute when not.

    Water potential of the filtrate falls steeply through the loop of Henle; the collecting duct's permeability — set by ADH — decides the final urine concentration.

    Excess amino acids are deaminated 脱氨 in the liver; the toxic ammonia becomes urea 尿素 for excretion.

    Worked check. ADH present → collecting duct walls permeable → water leaves down the gradient into the medullary blood → small volume, concentrated urine. Alcohol suppresses ADH — large volume, dilute.

    5.6

    基因开关与转录因子(知识点7.22)

    教学大纲

    陈述 7.22(规格页码 p.37):转录因子如何通过与DNA结合来开启或关闭基因(包括进入细胞作为转录因子的类固醇激素);表观遗传学——DNA甲基化和组蛋白修饰关闭基因及其后果(例如抑癌基因沉默、普瑞德-威利综合征印记);乳糖操纵子作为基因调控的模型(视情况而定)。

    来源:Cambridge International 教学大纲

    A transcription factor 转录因子 binds a gene's promoter and switches transcription on or off. Steroid hormones diffuse through the membrane and act as transcription factors themselves — one gene can serve several tissues. Epigenetics 表观遗传学 silences genes without changing the base sequence: methylation 甲基化 of DNA and modification of histones (acetylation) pack the DNA out of the transcription machinery's reach. Failures matter: a methylated tumour-suppressor gene cannot hold cell division in check; imprinting errors explain Prader–Willi syndrome.

    5.7 5.8

    神经元、神经冲动与突触(知识点8.1–8.7)

    教学大纲

    陈述 8.1-8.7(规格页码 pp.37-38):感觉神经元、联络神经元和运动神经元的结构;钠钾泵维持静息电位;动作电位的产生过程(去极化、全或无阈值、复极化、不应期);无髓鞘轴突上的传导与有髓鞘轴突上的跳跃式传导;突触的结构与功能(神经递质释放、受体结合、单向传递);药物如何在突触处影响神经冲动(激动剂、抑制剂、SSRI再摄取阻断)。

    陈述 8.8-8.10(规格页码 p.38):哺乳动物神经系统的组成(中枢神经系统 = 脑和脊髓;周围神经系统分为躯体神经系统和自主神经系统,后者又分交感神经和副交感神经);感受器如何检测刺激(视网膜中的视杆细胞和视锥细胞、光感受器、温度感受器、化学感受器、压力感受器、本体感受器、伤害感受器);脊髓反射弧(灰质和白质)及其三元神经元通路;习惯化是对重复刺激反应减弱的现象。

    来源:Cambridge International 教学大纲

    The resting neurone pumps Na⁺ out and K⁺ in (the sodium-potassium pump 钠钾泵), holding an interior at about −70 mV. A stimulus past threshold 阈值 (about −55 mV) opens sodium channels: an all-or-nothing 全或无 action potential spikes to +30 mV, then potassium channels repolarise; the refractory period 不应期 prevents backward travel and sets a maximum frequency.

    The action potential: depolarisation as sodium enters, repolarisation as potassium leaves, then the refractory period.

    In myelinated 有髓 neurones the impulse jumps node to node (saltatory conduction 跳跃传导) — far faster. At the synapse 突触, calcium entry triggers vesicles of neurotransmitter to fuse with the presynaptic membrane; the transmitter diffuses and binds postsynaptic receptors; enzymes then clear the gap. Drugs act here: SSRIs block serotonin re-uptake, so the signal persists; lidocaine blocks sodium channels; ecstasy (MDMA) floods the cleft with serotonin then depletes it.

    The nervous system divides: CNS (brain, spinal cord) and PNS — somatic (voluntary) and autonomic (involuntary; sympathetic accelerates, parasympathetic calms). Receptors are transducers: rods (dim light, many per bipolar cell — sensitivity), cones (colour, one-to-one — acuity), plus thermoreceptors, chemoreceptors, baroreceptors, nociceptors 伤害感受器 (pain). The spinal reflex arc 反射弧 passes sensory → relay → motor neurone in the cord's grey matter — fast, involuntary, protective. Habituation 习惯化 is a decreasing response to a repeated harmless stimulus.

    5.9

    植物反应与光敏色素(陈述8.11–8.12,核心实验18)

    教学大纲

    陈述 8.11-8.12 结合核心实验 18(规格页码 p.38):光敏色素(吸收红光的Pr和吸收远红光的Pfr)如何控制开花和萌发;生长素(IAA)和赤霉素在生长和萌发中的作用(细胞伸长、胚乳层酶诱导);使用核心实验 18 及推荐的额外关于习惯化的实践调查植物反应。

    来源:Cambridge International 教学大纲

    Phytochrome 光敏色素 flips between Pr (absorbs red) and Pfr (absorbs far-red); sunlight's red bias leaves Pfr dominant, controlling germination and flowering by day length. Auxin 生长素 elongates cells by acidifying walls; gibberellins 赤霉素 induce enzymes (amylase in the aleurone layer) that mobilise starch in germination.

    5.10

    大脑、成像与脑化学(陈述8.13–8.16)

    教学大纲

    陈述8.13-8.16(大纲页38-39):神经系统和激素系统的协调;大脑半球、下丘脑、垂体、小脑和延髓的位置与功能;MRI、fMRI、CT和PET扫描图像的形成原理及其用途;神经递质失衡(帕金森病中的多巴胺、抑郁症中的血清素);脑化学知识如何支撑药物治疗(L-Dopa跨越血脑屏障)。

    来源:Cambridge International 教学大纲

    Landmarks and jobs: cerebral hemispheres 大脑半球 (voluntary, thought), cerebellum 小脑 (coordination, balance), hypothalamus and pituitary 垂体 below them (homeostasis, hormones), medulla oblongata (heart, breathing). MRI shows structure; fMRI shows activity via oxygenated-blood flow; CT fast structural scans; PET metabolism with radioactive tracers.

    Chemistry first: Parkinson's is a dopamine 多巴胺 deficit — treated with L-Dopa, which crosses the blood-brain barrier and is converted to dopamine in the brain. Low serotonin links to depression — the target of SSRIs.

    5.11 5.12

    重组DNA技术与药物生产(陈述8.17–8.19)

    教学大纲

    陈述8.17-8.19(大纲页39):如何利用转基因生物生产药物(细菌生产胰岛素、转基因动物生产人源蛋白);重组DNA的制备——限制性内切酶在识别位点切割产生粘性末端,DNA连接酶连接,带有标记基因的质粒载体;转化宿主细胞(热激、电穿孔);标记基因筛选。

    陈述8.20-8.21(大纲页39):微阵列用于鉴定活跃基因(mRNA杂交);生物信息学术语及其在物种间和个体间比较DNA序列中的应用;农业和医学中转基因生物的风险与益处;应用实例包括重组人生长激素和荧光斑马鱼。

    来源:Cambridge International 教学大纲

    Recombinant DNA 重组DNA: a restriction enzyme 限制性内切酶 cuts DNA at its recognition sequence, leaving sticky ends 黏性末端; the human gene (made without introns from mRNA) is spliced into a plasmid 质粒 with DNA ligase 连接酶; the plasmid transforms host bacteria; marker genes (antibiotic resistance or fluorescence) select the cells that took it up. Scale up in a fermenter and the bacteria secrete human insulin 胰岛素 — cleaner and safer than extracting from pigs. GM animals make human proteins in milk; risks and benefits of GM crops (resistance, yields versus gene flow to wild relatives) are the standard evaluation.

    Microarrays 微阵列 identify active genes: sample mRNA hybridises to complementary probes on the chip; spots that fluoresce mark genes being transcribed. Bioinformatics 生物信息学 compares sequences across species and individuals — how relatedness is now measured.

    5.11 5.12

    自我检测

    1. Name the stage of respiration occurring in the mitochondrial matrix and one of its inputs.
    2. A runner's muscles produce lactate. Explain why, and what happens to the lactate afterwards.
    3. Myosin heads cannot detach without ATP. Explain what this means for muscles after death.
    4. Calculate cardiac output for stroke volume 85 cm³ and heart rate 140 min⁻¹.
    5. Explain why the afferent arteriole is wider than the efferent arteriole.
    6. State two differences between negative and positive feedback, with one example of each.
    7. Why does myelination increase conduction speed?
    8. An axon's threshold is −55 mV. A stimulus moves it to −60 mV. What happens, and why?
    9. Red light converts phytochrome to which form?
    10. Explain why L-Dopa is given instead of dopamine itself.
    11. A microarray spot fluoresces strongly. What does that show?
    12. Why must the human insulin gene be made from mRNA rather than from genomic DNA when inserting into bacteria?

    Answers: 1 the link reaction (or Krebs cycle) — input pyruvate (or acetyl); 2 oxygen is short so reduced NAD is recycled by reducing pyruvate to lactate; afterwards lactate goes to the liver and is converted back to glucose, using oxygen; 3 without ATP the myosin stays bound to actin — rigor mortis; 4 $85 \times 140 = 11\,900\ \text{cm}^3\,\text{min}^{-1} = 11.9\ \text{dm}^3\,\text{min}^{-1}$; 5 the pressure difference drives ultrafiltration of plasma into Bowman's capsule; 6 negative returns a variable toward its set point (thermoregulation); positive pushes it further away (fever rising); 7 the impulse jumps between nodes of Ranvier instead of depolarising the whole membrane — fewer, bigger steps; 8 nothing — threshold is not reached, so no action potential (all-or-nothing); 9 Pfr; 10 dopamine cannot cross the blood-brain barrier but L-Dopa can, and is converted to dopamine inside the brain; 11 that gene was being actively transcribed — its mRNA was present and hybridised; 12 bacteria cannot splice introns, so the gene must come from mature mRNA that already lacks them.

  • 6

    生物学实践技能 II

    • 6.1 规划A2探究(单元6大纲,规划部分)

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    • 6.2 实施与测量评估(单元6大纲,实施部分)

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    • 6.3 分析、统计与有根据的结论(单元6大纲,分析部分)

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IGCSE、A-Level 与 AP