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Pearson Edexcel · International A-Level · Biology

  • 1

    Molecules, Diet, Transport and Health

    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

    Water, carbohydrates and food tests

    Syllabus

    Pearson Edexcel IAL Biology Specification Issue 2 (February 2021), Topic 1 statements 1.1-1.4 with Core Practical 1 (spec pp.15-16). These are local teaching subdivisions of the official statements.

    Water as the transport solvent: dipole nature, hydrogen bonding between molecules, and why polar and ionic substances dissolve. Carbohydrates: mono/di/polysaccharides (glucose, fructose, galactose; maltose, sucrose, lactose; glycogen, amylose, amylopectin); condensation forming glycosidic bonds and hydrolysis splitting them; relating structure to energy supply and storage (alpha-glucose helix vs branched amylopectin/glycogen; cellulose excluded here). Core Practical 1: semi-quantitative Benedict's estimation of reducing-sugar concentration and iodine estimation of starch using colour standards.

    Source: Cambridge International syllabus

    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

    Lipids: triglycerides and ester bonds

    Syllabus

    Statement 1.5 (spec p.16): triglyceride synthesis by condensation of glycerol with three fatty acids forming ester bonds; saturated vs unsaturated fatty acids (no double bond vs one or more C=C, straight chains packing vs kinked chains, melting-point consequences). Links forward to CVD risk (1.5 subtopic) through dietary lipid chemistry and to the ethanol-emulsion test.

    Source: Cambridge International syllabus

    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

    Why animals need a pump: the heart and vessels

    Syllabus

    Statements 1.6-1.8 (spec pp.16-17): why multicellular animals need a mass-transport circulation (diffusion limits); structure-function of capillaries, arteries and veins (wall layers, lumen, valves, elastic and smooth muscle tissue); the mammalian heart's chambers, valves and major vessels; the cardiac cycle (atrial systole, ventricular systole, diastole) with pressure and volume changes and valve closure. Myogenic stimulation details are NOT needed at IAS (spec note).

    Source: Cambridge International syllabus

    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

    Haemoglobin and the oxygen journey

    Syllabus

    Statement 1.9 (spec p.17): haemoglobin's quaternary structure (four polypeptides, each with a haem group), cooperative binding shown by the sigmoid dissociation curve, transport of oxygen and carbon dioxide, and the Bohr effect - increasing carbon dioxide partial pressure shifts the curve right as hydrogen ions load haemoglobin. Compare myoglobin's hyperbolic curve where single-chain storage proteins appear.

    Source: Cambridge International syllabus

    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

    Cardiovascular disease: a chain of events

    Syllabus

    Statements 1.10-1.13 with Core Practical 2 (spec pp.17-18): the course of atherosclerosis (endothelial damage, inflammatory response, LDL deposition, plaque growth, aneurysm and thrombosis risk); the clotting cascade (thromboplastin release, prothrombin to thrombin, fibrinogen to fibrin); blood pressure (systolic/diastolic, hypertension as a risk factor) and Core Practical 2 heart-rate investigation (Daphnia or equivalent). Risk factors multiply rather than add.

    Source: Cambridge International syllabus

    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

    Diet, cholesterol and treating CVD

    Syllabus

    Statements 1.19-1.20 (spec p.18): how scientific knowledge about diet (energy balance, obesity measured by BMI and waist-to-hip ratio, salt, saturated vs unsaturated fat) informs choices; benefits and risks of CVD treatments - antihypertensives (beta-blockers, ACE inhibitors, diuretics), statins (LDL-receptor upregulation, muscle and liver side-effects), anticoagulants (warfarin, aspirin) and platelet-inhibitory drugs - alongside lifestyle intervention.

    Source: Cambridge International syllabus

    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

    Membranes and transport

    Syllabus

    Topic 2 statements 2.1-2.3 with Core Practical 3 (spec pp.19-20): properties of exchange surfaces (large surface area, thin, concentration gradient maintained); membrane structure from the fluid-mosaic model - phospholipid bilayer, intrinsic and extrinsic proteins, cholesterol, glycolipids and glycoproteins; membrane-model history (Overton to Singer-Nicolson) and Core Practical 3 membrane permeability investigation (beetroot at varying temperature or solvent concentration).

    Statements 2.4-2.5 (spec p.20): osmosis as movement of free water molecules from higher to lower water potential through a partially permeable membrane; passive transport (diffusion, facilitated diffusion through carrier and channel proteins) vs active transport (ATP, against the gradient); endocytosis and exocytosis for bulk material; Fick's law applied to surface area, concentration gradient and diffusion distance (rate proportional to SA x gradient / thickness).

    Source: Cambridge International syllabus

    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

    Proteins and enzymes

    Syllabus

    Statements 2.6-2.8 with Core Practical 4 (spec p.20): amino acid structure (amino group, carboxyl group, R group, peptide bond formation by condensation); protein levels of structure (primary sequence to hydrogen/ion/disulfide-bonded tertiary and quaternary); enzyme action - active site specificity, induced fit, temperature/pH/substrate concentration effects on rate, competitive and non-competitive inhibition; Core Practical 4 investigating enzyme activity (temperature or pH or [S] with e.g. amylase-starch). Immobilised and linked-enzyme applications.

    Source: Cambridge International syllabus

    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, replication and the code

    Syllabus

    Statements 2.9-2.12 (spec p.21): mononucleotide structure (pentose sugar, phosphate, nitrogenous base; deoxyribose vs ribose), DNA polymer formation by phosphodiester bonds, double helix with complementary base pairing and antiparallel strands; semi-conservative replication (helicase unwinding, DNA polymerase, free nucleotides, Meselson-Stahl evidence); a gene as a base sequence coding for an amino-acid sequence; the triplet code - non-overlapping, degenerate, universal, with start and stop codons.

    Source: Cambridge International syllabus

    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

    From gene to protein — and when it goes wrong

    Syllabus

    Statements 2.13-2.18 (spec pp.21-22): transcription (RNA polymerase, template strand, pre-mRNA splicing) and translation (mRNA codons, tRNA anticodons with amino acids, ribosomes); mutations from DNA replication errors - substitution, insertion, deletion with frameshift consequences; alleles, genotype, phenotype, dominance and sex linkage as far as WBI11 demands (cystic fibrosis as the model of a mutated gene's effect on CFTR chloride transport, mucus build-up, infection risk); genetic screening (carrier identification, pre-implantation genetic diagnosis, amniocentesis and chorionic villus sampling) with the ethical and social issues of screening programmes.

    Source: Cambridge International syllabus

    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

    Check yourself

    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

    Cells, Development, Biodiversity and Conservation

    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

    Cell ultrastructure and the microscope

    Syllabus

    Topic 3 statements 3.1-3.7 with Core Practical 5 (spec pp.22-23). Eukaryotic ultrastructure - nucleus and nucleolus, ribosomes, rough and smooth endoplasmic reticulum, Golgi apparatus and vesicles, mitochondria, chloroplasts, lysosomes, centrioles, cell-surface membrane - each organelle tied to its function, and prokaryotic ultrastructure (cell wall, capsule, plasmids, single circular DNA, 70S ribosomes, flagellum). Organisation into tissues, organs and systems (statement 3.2). Magnification vs resolution; light and electron microscopy (TEM/SEM); magnification calculations M = image/actual; the roles of staining. Core Practical 5 examines cells with microscopy.

    Source: Cambridge International syllabus

    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

    Meiosis: halving and shuffling

    Syllabus

    Statements 3.9-3.13 (spec p.23): loci and linkage; meiosis as two divisions producing four genetically different haploid cells; independent assortment of homologous pairs and crossing over of chromatids as the two sources of variation; mammalian gamete specialisation (sperm acrosome, midpiece mitochondria, egg cytoplasm and zona pellucida); fertilisation in mammals (acrosome reaction, cortical reaction blocking polyspermy) and in flowering plants (pollen tube, double fertilisation).

    Source: Cambridge International syllabus

    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

    Mitosis and the cell cycle

    Syllabus

    Statements 3.14-3.16 with Core Practical 6 (spec p.23): the cell cycle (interphase G1-S-G2, then mitosis and cytokinesis); prophase, metaphase, anaphase, telophase with what happens to chromosomes at each stage; mitosis producing genetically identical cells for growth, repair and asexual reproduction; calculating mitotic index (cells in mitosis over total cells) and its clinical use; Core Practical 6 prepares and observes root-tip squashes.

    Source: Cambridge International syllabus

    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

    Stem cells and switching genes on

    Syllabus

    Statements 3.17-3.21 (spec p.24): totipotent and pluripotent stem cells, morula and blastocyst stages; differentiation by differential gene expression; post-transcriptional changes to mRNA giving several proteins from one gene; interaction of genotype and environment in the phenotype; multiple alleles and polygenic inheritance producing continuous variation. Ethics of stem-cell use runs through the exam questions.

    Source: Cambridge International syllabus

    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

    Plant transport tissues

    Syllabus

    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).

    Statements 4.7-4.13 (spec pp.27-28): sustainability of plant fibres and starch; the importance of water and inorganic ions (nitrate for amino acids, calcium for wall structure and signalling, magnesium for chlorophyll); antimicrobial and therapeutic substances from plants (Core Practical 9 antimicrobial testing); bacterial growth conditions; the development of drug testing from historical to contemporary protocols including the stages of trials.

    Source: Cambridge International syllabus

    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

    Measuring life: classification and biodiversity

    Syllabus

    Statements 4.14-4.18 (spec p.28): classification organising life by shared characteristics, with the three-domain system (Archaea, Bacteria, Eukarya) and evidence from molecular sequences; species richness and index of diversity (D) calculations within a habitat; heterozygosity index within a species; endemism; the threats to biodiversity. Calculations with the D formula are examined every year.

    Source: Cambridge International syllabus

    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

    Niches, Hardy–Weinberg and conservation

    Syllabus

    Statements 4.19-4.21 (spec pp.28-29): the niche concept and adaptation examples (anatomical, physiological, behavioural); the Hardy-Weinberg equation p^2 + 2pq + q^2 = 1 for allele frequencies in a population and the conditions it assumes; evaluating zoo and seed-bank conservation methods, captive breeding and habitat corridors.

    Source: Cambridge International syllabus

    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

    Culturing and growing microbes

    Syllabus

    Statements 4.7-4.13 (spec pp.27-28): sustainability of plant fibres and starch; the importance of water and inorganic ions (nitrate for amino acids, calcium for wall structure and signalling, magnesium for chlorophyll); antimicrobial and therapeutic substances from plants (Core Practical 9 antimicrobial testing); bacterial growth conditions; the development of drug testing from historical to contemporary protocols including the stages of trials.

    Source: Cambridge International syllabus

    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

    Check yourself

    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

    Practical Skills in Biology I

    • 3.1 Planning investigations (Unit 3 spec, planning section)

      Learning program coming soon

    • 3.2 Implementation and measurements (Unit 3 spec, implementation section)

      Learning program coming soon

    • 3.3 Processing results, graphs and uncertainties (Unit 3 spec, processing section)

      Learning program coming soon

  • 4

    Energy, Environment, Microbiology and Immunity

    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

    Photosynthesis: two stages, two places

    Syllabus

    Topic 5 statements 5.1-5.8 with Core Practical 10 (spec pp.29-30). The overall photosynthesis reaction; light-dependent reactions (photolysis of water, photophosphorylation of ADP, reduction of NADP) in the thylakoid membranes; the light-independent reactions (Calvin cycle: carbon dioxide fixation by RuBP catalysed by rubisco, GALP production, ribulose bisphosphate regeneration) in the stroma; chloroplast structure matched to both stages; absorption and action spectra; chromatography of pigments with Rf values (Core Practical 10); limiting factors (light intensity, carbon dioxide concentration, temperature) and agricultural manipulation of them.

    Source: Cambridge International syllabus

    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

    Productivity and the short food chain

    Syllabus

    Statements 5.9-5.10 (spec p.30): gross primary productivity (GPP), net primary productivity (NPP = GPP minus respiratory losses) and the units used (kJ m-2 yr-1); biomass transfer efficiency between trophic levels and its calculation; why transfers lose energy (respiration, egestion, excretion, uneaten parts) and why food chains are short.

    Source: Cambridge International syllabus

    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

    Ecosystems, populations and succession

    Syllabus

    Statements 5.11-5.15 with Core Practical 11 (spec pp.30-31): the terms habitat, population, community and ecosystem; biotic and abiotic factors controlling distribution and abundance; the niche concept applied; predator-prey cycles; methods of measuring abundance (quadrats, transects, capture-mark-recapture); stages of succession from colonisation by pioneer species to climax community (Core Practical 11 surveys a habitat).

    Source: Cambridge International syllabus

    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

    Climate change: evidence, causes, effects

    Syllabus

    Statements 5.16-5.22 with Core Practical 12 (spec pp.31-32): evidence for climate change (temperature records, ice cores, dendrochronology, pollen in peat); greenhouse gases and the enhanced greenhouse effect (CO2, methane, nitrous oxide - sources, atmospheric lifetimes, global warming potential); the carbon cycle and methods to reduce atmospheric carbon; extrapolation of data and its reliability; effects of climate change on rainfall patterns, species distributions, enzyme-dependent ecosystems; Core Practical 12 investigates the effect of temperature on enzyme/habitat-relevant activity.

    Source: Cambridge International syllabus

    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

    Evolution and speciation

    Syllabus

    Statements 5.23-5.26 (spec p.32): evolution as a change in allele frequency brought about by mutation, selection (directional and stabilising), gene flow and drift; isolation reducing gene flow (geographical = allopatric; reproductive = sympatric) leading to speciation; interpreting controversial scientific conclusions (climate, evolution) by the standards of evidence; reforestation, sustainable resources and biofuels.

    Source: Cambridge International syllabus

    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

    Culturing microorganisms

    Syllabus

    Statements 6.1-6.4 with Core Practical 13 (spec p.33): culture media (nutrient agar, broth); aseptic technique in detail; methods of measuring microbial growth (cell counts, turbidity, viable counts by dilution plating); the bacterial growth curve's four phases (lag, log/exponential, stationary, death) and the exponential growth-rate constant k; Core Practical 13 investigates growth rate in liquid culture with a calibration curve.

    Source: Cambridge International syllabus

    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

    Pathogens and the body's barriers

    Syllabus

    Statements 6.5-6.7 (spec p.33): the structures of bacteria and viruses (nucleic acid, capsid, envelope) compared; infection routes (digestive, respiratory, sexual contact, wounds, vector); barriers to infection (skin, mucus, stomach acid, blood clotting); how Mycobacterium tuberculosis and HIV infect and damage the body, and how HIV worsens TB.

    Source: Cambridge International syllabus

    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

    The immune response

    Syllabus

    Statements 6.8-6.12 (spec p.34): non-specific responses (inflammation, lysozyme, interferon); antigens; the humoral response - B effector cells differentiating into plasma cells that secrete antibodies, B memory cells giving secondary response; the cellular response - T helper, T killer and T memory cells; the antibody structure (four polypeptide chains, variable region binding site) and antigen-antibody complex; natural and artificial, active and passive immunity; vaccination and herd immunity; the evolutionary race between pathogens and hosts.

    Source: Cambridge International syllabus

    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

    Antibiotics and hospital-acquired infections

    Syllabus

    Statements 6.13-6.15 with Core Practical 14 (spec p.34): bacteriostatic vs bactericidal antibiotics; how antibiotic resistance arises and spreads (selection); Core Practical 14 investigates antibiotic effects on bacteria; how an understanding of the causes of hospital-acquired infections (hygiene, invasive procedures, resistant strains) reduces their incidence.

    Source: Cambridge International syllabus

    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

    Decomposition, PCR, profiling and time of death

    Syllabus

    Statements 6.16-6.20 (spec pp.34-35): the role of microorganisms in decomposition and nutrient recycling; the polymerase chain reaction (denaturation, annealing of primers, extension) amplifying DNA; gel electrophoresis separating DNA fragments by length; DNA profiling for identification and genetic relationships; determining time of death from body temperature, rigor mortis, degree of muscle contraction and succession of insects on the body.

    Source: Cambridge International syllabus

    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

    Check yourself

    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

    Respiration, Internal Environment, Coordination and Gene Technology

    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

    Respiration: four stages, one purpose

    Syllabus

    Topic 7 statements 7.1-7.8 with Core Practicals 15-16 (spec pp.35-36). The overall aerobic reaction and respiration as a stepped, enzyme-controlled process; glycolysis in the cytoplasm (phosphorylation of hexoses, substrate-level phosphorylation of ATP, reduced NAD, pyruvate; lactate in anaerobic conditions); the link reaction and Krebs cycle in the mitochondrial matrix (decarboxylation, ATP, reduced NAD, reduced FAD); oxidative phosphorylation on the cristae - the electron transport chain pumping hydrogen ions into the intermembrane space and chemiosmosis through ATP synthase, oxygen as the final electron acceptor; lactate metabolism after anaerobic exercise; the respiratory quotient RQ = CO2 produced / O2 consumed for carbohydrate (1.0), lipid (0.7) and protein (0.9); Core Practicals 15-16 (respirometers and artificial hydrocarbonate-indicator respiration).

    Source: Cambridge International syllabus

    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

    Muscles: the sliding filament

    Syllabus

    Statements 7.9-7.11 (spec p.36): how muscles, tendons, the skeleton and ligaments interact in movement (muscle pulls bone across a joint; tendons join muscle to bone; ligaments hold bones together); the structure of a skeletal muscle fibre (fused multinucleate cells, myofibrils of actin and myosin); the sliding filament theory of contraction (calcium ions exposing binding sites, myosin heads with ATP attaching and pulling, actin sliding); fast-twitch and slow-twitch fibres and their adaptation to sprint and endurance work.

    Source: Cambridge International syllabus

    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

    The heart's own clock and the medulla's override

    Syllabus

    Statements 7.12-7.13, 7.15 with Core Practical 17 (spec p.36): the myogenic nature of cardiac muscle; how the heart's electrical activity begins at the sino-atrial node, passes to the atrio-ventricular node and along the Purkyne fibres; control of heart rate and ventilation rate by the cardiovascular control centre and ventilation centre in the medulla oblongata; cardiac output = stroke volume x heart rate and its calculation; variations in ventilation and cardiac output during exercise; Core Practical 17 uses spirometer traces to investigate the effects of exercise on tidal volume, breathing rate and minute ventilation.

    Source: Cambridge International syllabus

    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

    Homeostasis and feedback

    Syllabus

    Statements 7.14, 7.16-7.17 (spec p.36): homeostasis as maintaining the internal environment in dynamic equilibrium; negative feedback holding variables within narrow limits (and positive feedback amplifying change); the roles of the autonomic nervous system, adrenaline in the fight-or-flight response, and the hypothalamus in thermoregulation (vasodilation, sweating, shivering, vasoconstriction, hair erection).

    Source: Cambridge International syllabus

    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

    The kidney: filter, reabsorb, fine-tune

    Syllabus

    Statements 7.18-7.21 (spec pp.36-37): the gross and microscopic structure of the mammalian kidney (cortex, medulla, pelvis; nephron with glomerulus, Bowman's capsule, proximal convoluted tubule, loop of Henle, collecting duct); ultrafiltration (high pressure from the afferent arteriole, podocytes, basement membrane) and selective reabsorption in the PCT (glucose, amino acids, some salts and water); the loop of Henle and water potential gradient in the medulla; ADH from the pituitary increasing the collecting duct's water permeability; urea production in the liver from excess amino acids (deamination, ornithine cycle not required in detail).

    Source: Cambridge International syllabus

    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

    Switching genes: transcription factors and epigenetics

    Syllabus

    Statement 7.22 (spec p.37): how genes are switched on and off by transcription factors binding to DNA (including steroid hormones entering cells to act as transcription factors); epigenetics - DNA methylation and histone modification switching genes off and their consequences (e.g. tumour suppressor silencing, Prader-Willi imprinting); the lac operon as the model of gene regulation where appropriate.

    Source: Cambridge International syllabus

    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

    Nerve impulses and synapses

    Syllabus

    Statements 8.1-8.7 (spec pp.37-38): the structure of sensory, relay and motor neurones; the resting potential maintained by the sodium-potassium pump; how an action potential arises (depolarisation, the all-or-nothing threshold, repolarisation, refractory period); conduction along unmyelinated axons and saltatory conduction in myelinated axons; the structure and function of synapses (neurotransmitter release, receptor binding, unidirectionality); how drugs influence nerve impulses at synapses (agonists, inhibitors, SSRI reuptake blockade).

    Statements 8.8-8.10 (spec p.38): the organisation of the mammalian nervous system (CNS = brain and spinal cord; PNS, split into somatic and autonomic, sympathetic and parasympathetic); how receptors detect stimuli (rod and cone cells in the retina, photoreceptors, thermoreceptors, chemoreceptors, baroreceptors, proprioceptors, nociceptors); the spinal reflex arc (grey and white matter) and its three-neuron pathway; habituation as a decreased response to a repeated stimulus.

    Source: Cambridge International syllabus

    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

    Plants: light and growth switches

    Syllabus

    Statements 8.11-8.12 with Core Practical 18 (spec p.38): how phytochrome (Pr absorbing red light and Pfr absorbing far-red) controls flowering and germination; auxin (IAA) and gibberellins in growth and germination (cell elongation, enzyme induction in the aleurone layer); practical investigation of plant responses with Core Practical 18 and the recommended additional practical on habituation.

    Source: Cambridge International syllabus

    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

    The brain and its chemistry

    Syllabus

    Statements 8.13-8.16 (spec pp.38-39): coordination through nervous and hormonal systems; the location and functions of the cerebral hemispheres, hypothalamus, pituitary, cerebellum and medulla oblongata; how MRI, fMRI, CT and PET scan images are formed and their uses; imbalances in neurotransmitters (dopamine in Parkinson's, serotonin in depression) and how knowledge of brain chemistry underpins drug treatment (L-Dopa crossing the blood-brain barrier).

    Source: Cambridge International syllabus

    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

    Gene technology

    Syllabus

    Statements 8.17-8.19 (spec p.39): how drugs can be produced using genetically modified organisms (insulin from bacteria, human proteins from GM animals); producing recombinant DNA - restriction enzymes cutting at recognition sites with sticky ends, DNA ligase joining, plasmid vectors with marker genes; transforming host cells (heat shock, electroporation); marker gene selection.

    Statements 8.20-8.21 (spec p.39): microarrays to identify active genes (mRNA hybridisation); the term bioinformatics and its use in comparing DNA sequences between species and individuals; the risks and benefits of GM organisms in agriculture and medicine; applications such as recombinant human growth hormone and fluorescent zebrafish.

    Source: Cambridge International syllabus

    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

    Check yourself

    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

    Practical Skills in Biology II

    • 6.1 Planning A2 investigations (Unit 6 spec, planning section)

      Learning program coming soon

    • 6.2 Implementation and measurement critique (Unit 6 spec, implementation section)

      Learning program coming soon

    • 6.3 Analysis, statistics and justified conclusions (Unit 6 spec, analysis section)

      Learning program coming soon

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