2025 CCEA A-Level Biology 1010 模拟试题及答案详解

(a) The sodium-potassium pump actively transports Na⁺ out of the neurone and K⁺ into the neurone (3 Na⁺ out : 2 K⁺ in), using ATP; the membrane is more permeable to K⁺ than Na⁺ at rest (more K⁺ leak channels), so K⁺ diffuses back out; this results in the inside of the neurone being negative relative to the outside, at approximately −70 mV.
(b) The arrival of the action potential depolarises the presynaptic membrane; this opens voltage-gated calcium channels, and Ca²⁺ diffuses into the presynaptic knob; the influx of Ca²⁺ causes synaptic vesicles (containing the neurotransmitter, e.g. acetylcholine) to move to and fuse with the presynaptic membrane; the neurotransmitter is released into the synaptic cleft by exocytosis; the neurotransmitter diffuses across the cleft and binds to specific complementary receptors on the postsynaptic membrane, opening ligand-gated Na⁺ channels; Na⁺ ions diffuse into the postsynaptic neurone, causing a local depolarisation — the EPSP.
(c) Neurotransmitter is only stored in, and released from, synaptic vesicles in the presynaptic neurone/knob; the specific receptors for the neurotransmitter are only located on the postsynaptic membrane; therefore the signal can only pass from the presynaptic to the postsynaptic neurone, not in reverse.

(a) [3] any three of: Na⁺/K⁺ pump; active transport using ATP; 3 Na⁺ out : 2 K⁺ in; membrane more permeable to K⁺ (more leak channels); resulting potential difference ≈ −70 mV, inside negative. (b) [5] any five of: depolarisation of presynaptic membrane; voltage-gated Ca²⁺ channels open; Ca²⁺ influx into presynaptic knob; vesicles move to/fuse with presynaptic membrane; exocytosis of neurotransmitter (accept named, e.g. acetylcholine); diffusion across the synaptic cleft; binding to specific/complementary receptors on postsynaptic membrane; ligand-gated Na⁺ channels open; Na⁺ influx causes depolarisation/EPSP of postsynaptic membrane. (c) [2] neurotransmitter/vesicles only present in presynaptic knob; receptors only present on postsynaptic membrane.

(a) Blood entering the glomerulus is at high (hydrostatic) pressure; this pressure forces water and small solute molecules (e.g. glucose, amino acids, urea, ions) out of the capillary; filtration occurs through the fenestrations (pores) of the capillary endothelium, the basement membrane (which acts as the filter, retaining large molecules) and the gaps between the podocytes of the Bowman's capsule wall; large molecules such as plasma proteins and blood cells are too large to pass through and remain in the blood.
(b) Glucose and amino acids. Both are small enough to be filtered into the glomerular filtrate, but under normal conditions they are almost completely reabsorbed back into the blood by selective reabsorption (active transport/co-transport with Na⁺) in the proximal convoluted tubule, so almost none remains in the urine.
(c) The afferent arteriole is wider than the efferent arteriole; this creates a resistance to outflow, raising the hydrostatic (blood) pressure within the glomerular capillaries; this elevated pressure is what forces fluid out of the capillaries and drives ultrafiltration.

(a) [4] any four of: high hydrostatic/blood pressure in glomerulus; forces water and small solutes out; through fenestrations/pores of capillary wall; through basement membrane (acts as filter); through gaps between podocytes; large molecules (proteins, cells) retained in blood. (b) [3] B1 glucose; B1 amino acids; B1 reason — selectively/actively reabsorbed in the proximal convoluted tubule. (c) [3] wider afferent than efferent; creates resistance to outflow / narrower exit; raises hydrostatic pressure within the glomerulus, driving filtration.

(a) A muscle fibre contains many sarcomeres in series, each bounded by Z-lines/discs; myosin (thick) filaments and actin (thin) filaments overlap; myosin heads attach to specific binding sites on the actin filaments, forming cross-bridges; ATP hydrolysis provides energy for a change in orientation (tilting) of the myosin heads; this tilting pulls the actin filaments inward, sliding them further over the myosin filaments; the myosin head detaches (requires ATP binding), returns to its original position, and reattaches further along the actin filament, repeating the cycle; as a result, the actin filaments from each end of the sarcomere are pulled towards the centre, the sarcomere (and I-band/H-zone) shortens, but the length of the filaments themselves does not change.
(b) When an action potential reaches the muscle fibre, calcium ions are released from the sarcoplasmic reticulum into the sarcoplasm; Ca²⁺ ions bind to troponin, causing a change in its shape/conformation; this moves the attached tropomyosin strand, which was previously blocking the myosin-binding sites on the actin filament; this exposes the binding sites, allowing the myosin heads to attach to actin and cross-bridge cycling (and hence contraction) to begin.

(a) [7] any seven of: sarcomere bounded by Z-lines; actin and myosin filaments overlap; myosin heads bind to actin forming cross-bridges; ATP hydrolysed for energy; myosin heads change orientation/tilt; actin filaments pulled/slide over myosin; heads detach (ATP binds) and reattach further along; cycle repeats; sarcomere shortens; filament lengths themselves unchanged. (b) [4] Ca²⁺ released from sarcoplasmic reticulum; Ca²⁺ binds to troponin; troponin changes shape, moving tropomyosin; myosin-binding sites on actin exposed, allowing cross-bridge formation.

(a) Lag phase — population size remains low/constant initially, as cells adjust/synthesise enzymes needed to metabolise the new medium; exponential (log) phase — population increases rapidly, as resources (nutrients, space) are not yet limiting and the rate of reproduction greatly exceeds the death rate; deceleration phase — growth rate slows as resources begin to become limiting/competition increases; stationary phase — population size levels off and remains roughly constant, as the birth rate becomes equal to the death rate, i.e. the carrying capacity has been reached.
(b) Carrying capacity is the maximum population size (of a species) that a particular environment/habitat can sustain indefinitely, given the resources available. Factor 1: Depletion of nutrients/food in the broth — as nutrients become scarce, growth rate decreases and death rate increases, limiting further population growth. Factor 2: Accumulation of toxic metabolic waste products (e.g. ethanol, CO₂) in the fixed volume of broth — this increasingly inhibits reproduction and increases mortality, preventing further increase in population size.

(a) [6] B1 each named phase in correct sequence (lag, exponential/log, deceleration, stationary — accept 3 or 4 phases); B1/B2 correct description of what limits/drives growth in each named phase (up to 2 marks for detailed causal descriptions). (b) [4] B1: carrying capacity defined as maximum sustainable population size for the environment/resources available; B1 + B1: two valid limiting factors named; B1: at least one factor clearly explained in context (e.g. nutrient depletion or waste accumulation in the broth).

(a) Decomposers first break down proteins/organic nitrogenous waste to release ammonium ions; nitrifying bacteria then oxidise this ammonium — one group (e.g. Nitrosomonas) oxidises ammonium to nitrite; a second group (e.g. Nitrobacter) oxidises nitrite to nitrate; nitrate is the form of nitrogen most readily absorbed and used by plant roots.
(b) Denitrifying bacteria, active in anaerobic/waterlogged soil conditions, convert nitrate (and nitrite) back into gaseous nitrogen (N₂), which returns to the atmosphere; this represents a loss of usable nitrogen from the soil, reducing the nitrate available for uptake by plants and therefore reducing soil fertility.
(c) Nitrogen-fixing bacteria (e.g. free-living Azotobacter or symbiotic Rhizobium in legume root nodules) convert atmospheric nitrogen gas (N₂), which most organisms cannot use directly, into ammonium/organic nitrogen compounds that can be used to build amino acids and proteins; this makes nitrogen available to plants (and hence the rest of the food chain), replenishing nitrogen removed from the soil by processes such as denitrification and harvesting.

(a) [4] decomposers release ammonium from organic matter; nitrifying bacteria oxidise ammonium; ammonium → nitrite (e.g. Nitrosomonas); nitrite → nitrate (e.g. Nitrobacter) — nitrate usable by plants. (b) [3] denitrifying bacteria active in anaerobic conditions; convert nitrate/nitrite to N₂ gas, returned to atmosphere; reduces nitrate available to plants, lowering soil fertility. (c) [3] converts atmospheric N₂ (unusable by most organisms) into ammonium/organic nitrogen; usable to build amino acids/proteins; named example (Rhizobium/root nodules or Azotobacter); replenishes soil nitrogen.

(a) Osmoreceptors in the hypothalamus detect a decrease in the water potential of the blood (i.e. blood becomes more concentrated); this stimulates the synthesis of ADH by the hypothalamus; ADH is stored in, and released from, the posterior pituitary gland into the bloodstream; ADH travels to the kidney and increases the permeability of the walls of the distal convoluted tubule and collecting duct to water (by inserting aquaporins into the membrane); this increases the reabsorption of water from the filtrate back into the blood by osmosis; producing a smaller volume of more concentrated urine, and restoring blood water potential towards normal (negative feedback).
(b) Drinking a large volume of water increases the water potential of the blood (blood becomes more dilute); this is detected by osmoreceptors in the hypothalamus, which reduces the synthesis and release of ADH from the posterior pituitary; with less ADH circulating, the distal convoluted tubule and collecting duct become less permeable to water; less water is reabsorbed from the filtrate, so a larger volume of more dilute urine is produced, which removes the excess water and restores blood water potential to normal.

(a) [6] any six of: osmoreceptors in hypothalamus detect fall in blood water potential; stimulates ADH synthesis; ADH stored in/released from posterior pituitary; travels in blood to kidney; increases permeability of DCT/collecting duct to water (aquaporins); increases water reabsorption by osmosis; smaller volume/more concentrated urine produced; restores blood water potential (negative feedback). (b) [5] blood water potential increases/blood more dilute; detected by hypothalamic osmoreceptors; ADH secretion decreases; DCT/collecting duct less permeable to water; less water reabsorbed; larger volume of more dilute urine produced.

(a) Antibody-mediated immunity involves the division of B-lymphocytes following exposure to a foreign antigen, forming plasma cells, which synthesise and secrete specific antibodies, and memory cells, which provide long-term immunity; there is a delay before sufficient antibody is produced, meaning the infected individual can show symptoms of disease during this time. Cell-mediated immunity involves T-lymphocytes, which are sensitised by viral antigens, abnormal self-antigens (e.g. on tumour cells) or transplanted foreign tissue antigens, and divide to form several types of T-lymphocyte: killer T-cells, which directly and enzymatically destroy cells bearing the foreign antigen; helper T-cells, which co-operate with B-cells in antibody formation; memory T-cells, providing long-term immunity; and suppressor T-cells, which deactivate/switch off the immune response of both B- and T-cells once the infection has been dealt with.
(b) Antibodies bind to complementary antigens on the pathogen surface, causing agglutination (clumping) of the pathogens, forming an antigen-antibody complex; this makes the pathogens easier for phagocytic cells (polymorphs) to locate and engulf, by phagocytosis; once engulfed, the pathogen is contained within a phagocytic vacuole, which fuses with lysosomes; lysosomal enzymes then digest and destroy the pathogen intracellularly.

(a) [6] any six of: antibody-mediated involves B-lymphocytes; division forms plasma cells; plasma cells secrete specific antibodies; also forms memory cells (long-term immunity); delay in antibody-mediated response; cell-mediated involves T-lymphocytes, sensitised by viral/abnormal self/transplant antigens; forms killer T-cells (direct destruction); helper T-cells (aid antibody production); memory T-cells; suppressor T-cells (deactivate response). (b) [4] agglutination — antigen-antibody complex forms; phagocytosis by polymorphs; engulfed pathogen in vacuole fuses with lysosomes; intracellular digestion by lysosomal enzymes.

(a) NPP = GPP − respiration = 20 000 − 12 000 = 8000 kJ m⁻² yr⁻¹.
(b) % NPP consumed = (ingested ÷ NPP) × 100 = (3000 ÷ 8000) × 100 = 37.5%.
(c) Assimilation efficiency = (assimilated ÷ ingested) × 100 = (1800 ÷ 3000) × 100 = 60%.
(d) Not all energy assimilated is converted into new biomass (production); a large proportion of assimilated energy is used in respiration to release ATP for life processes (movement, maintaining body temperature, etc.) and is ultimately lost as heat; some assimilated energy may also be lost in excretory products (e.g. urine); therefore production (biomass gained, 800 kJ m⁻² yr⁻¹ here) is always less than the energy assimilated (1800 kJ m⁻² yr⁻¹), giving a production efficiency below 100% (here, 800/1800 × 100 = 44.4%).

(a) [2] M1: GPP − R; A1: 8000 kJ m⁻² yr⁻¹ (units required). (b) [2] M1: 3000/8000 × 100; A1: 37.5%. (c) [3] M1: 1800/3000 (× 100); M1: correct method; A1: 60%. (d) [3] most assimilated energy lost via respiration (for ATP/life processes); further lost as heat; some lost in excretory products — so production < assimilation, giving efficiency < 100%.

(a) Indicative content: a fall in blood glucose is detected by alpha cells in the islets of Langerhans of the pancreas (acting as both receptor and effector in this hormonal pathway); alpha cells respond by secreting the hormone glucagon into the blood; glucagon travels in the bloodstream to target cells, mainly in the liver, where it stimulates glycogenolysis (breakdown of glycogen to glucose) and gluconeogenesis (production of glucose from non-carbohydrate sources), raising blood glucose back towards normal (negative feedback); this is a hormonal/endocrine response — the sympathetic nervous system can also be involved, e.g. stimulating adrenaline release from the adrenal medulla in a more severe or rapid fall in blood glucose (a 'fight or flight'-linked response), which also raises blood glucose via glycogenolysis.
Comparison: nervous coordination (electrical impulses along neurones, chemical transmission only at synapses) acts much faster than hormonal coordination (hormones diffuse/travel via the bloodstream, which is comparatively slow); nervous responses tend to be short-lived/of short duration (e.g. a single muscle twitch), whereas hormonal responses tend to be longer-lasting, as hormones remain in the blood and continue to act until broken down or removed; nervous responses are highly specific, acting only on the specific effector(s) connected by neurones, whereas hormonal responses may be more widespread, as the hormone travels throughout the whole body in the blood but only affects cells with the complementary receptor.
(b) The tip of the shoot produces auxin (e.g. indoleacetic acid, IAA), which is transported down the shoot; when light shines on the shoot from one side only, auxin becomes unequally distributed, with a higher concentration accumulating on the shaded side of the shoot than on the illuminated side; auxin promotes cell elongation in shoot cells; because there is more auxin on the shaded side, the cells on the shaded side elongate more than the cells on the illuminated side; this unequal growth causes the shoot to bend/curve towards the light source (positive phototropism), maximising the light available for photosynthesis.

[18 total] Part (a) [12]: indicative points include — fall in blood glucose detected by alpha cells of islets of Langerhans; glucagon secreted; travels in blood to liver; stimulates glycogenolysis; stimulates gluconeogenesis; raises blood glucose (negative feedback); adrenaline/sympathetic nervous system involvement in rapid/severe falls; nervous system faster than hormonal system; nervous responses shorter-lived/localised via specific neural connections; hormonal responses longer-lasting/more widespread via the blood; correct use of specialist terms throughout. Band 3 (9–12 marks): ≥9 valid indicative points, wide-ranging and accurate, balanced coverage of both coordination systems and a clear comparison, accurate QWC. Band 2 (5–8 marks): 5–8 valid points, reasonable coverage but may be unbalanced (e.g. little comparison) or contain minor inaccuracies, competent QWC. Band 1 (1–4 marks): ≤4 valid points, basic/fragmented answer, limited specialist vocabulary, basic QWC.
Part (b) [6]: indicative points — auxin produced in the shoot tip; transported down the shoot; unequal/lateral distribution under unilateral light, more auxin accumulates on the shaded side; auxin promotes cell elongation; greater elongation of cells on the shaded side than the illuminated side; shoot bends/curves towards the light (positive phototropism). Award up to 6 marks for accurate, well-sequenced coverage of these points; cap at 3 marks if the direction of auxin distribution or the direction of the growth response is reversed/incorrect.

(a) Any two of: no mutation occurring; mating is random (no selective mate choice); no selection (all genotypes equally likely to survive and reproduce); no migration into or out of the population (a closed population); the population is very large (so chance/genetic drift is negligible).
(b) Frequency of the recessive condition = q² = 1/2500 = 0.0004. So q = √0.0004 = 0.02. Since p + q = 1, p = 1 − 0.02 = 0.98.
(c) Frequency of heterozygotes = 2pq = 2 × 0.98 × 0.02 = 0.0392. As a percentage: 0.0392 × 100 = 3.92%.
(d) Out of 1000 people: 1000 × 0.0392 = 39.2 ≈ 39 people are expected to be carriers.
(e) Any two of: mutation (creating new alleles, changing allele frequency); non-random mating/selective mate choice; natural selection (one genotype has a survival/reproductive advantage or disadvantage); migration (gene flow) into or out of the population; genetic drift (in a small population).

(a) [2] B1 each for any two valid Hardy–Weinberg conditions (max 2). (b) [3] M1: q² = 0.0004; M1: q = √0.0004; A1: q = 0.02 and p = 0.98 (both required, ft). (c) [3] M1: 2pq substitution with their p, q; M1: correct evaluation; A1: 3.92% (ft their p, q). (d) [1] B1 ft: 39 (or 39.2), from 1000 × their (c). (e) [2] B1 each for any two valid factors (mutation, non-random mating, selection, migration, genetic drift) — max 2.

(a) A dihybrid cross between two heterozygotes (RrYy × RrYy) gives the classic 9:3:3:1 phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. Out of 160 offspring (16 parts): round yellow = 9/16 × 160 = 90; round green = 3/16 × 160 = 30; wrinkled yellow = 3/16 × 160 = 30; wrinkled green = 1/16 × 160 = 10.
(b) Null hypothesis: there is no significant difference between the observed and expected (9:3:3:1) numbers of offspring of each phenotype; any difference is due to chance.
(c) χ² = Σ(O−E)²/E.
Round yellow: (86−90)²/90 = 16/90 = 0.178.
Round green: (32−30)²/30 = 4/30 = 0.133.
Wrinkled yellow: (34−30)²/30 = 16/30 = 0.533.
Wrinkled green: (8−10)²/10 = 4/10 = 0.400.
Σ = 0.178+0.133+0.533+0.400 = 1.244 (χ² ≈ 1.24).
(d) Degrees of freedom = number of categories − 1 = 4 − 1 = 3. The calculated χ² (1.24) is less than the critical value (7.815) at p = 0.05 with 3 degrees of freedom. Therefore the null hypothesis is accepted — there is no significant difference between the observed and expected numbers, and the data support a 9:3:3:1 ratio (independent assortment).

(a) [4] B1: correct 9:3:3:1 ratio stated; B1 each for two of the four correct expected numbers (90, 30, 30, 10), max 3 further marks for all four correct (accept as: 1 mark ratio + 3 marks for all four expected values correct). (b) [1] B1: valid null hypothesis referencing no significant difference between observed and expected. (c) [4] M1: correct (O−E)² term for at least two categories; M1: correct division by E; A1: individual values correct (or ft); A1: total χ² = 1.24 (accept 1.2–1.25). (d) [2] B1 ft: correct comparison of their χ² to 7.815 with 3 d.f.; B1 ft: correct conclusion (accept H0, no significant difference / data fit 9:3:3:1 ratio) consistent with their χ² value.

(a) Carrier female genotype = XᴮXᵇ; normal male genotype = XᴮY. Cross: XᴮXᵇ × XᴮY. Gametes from female: Xᴮ, Xᵇ. Gametes from male: Xᴮ, Y. Offspring: XᴮXᴮ (normal female); XᴮXᵇ (normal, carrier female); XᴮY (normal male); XᵇY (pale male). Ratio 1 : 1 : 1 : 1 — i.e. all female offspring have normal plumage (half being carriers), and of the male offspring, half have normal plumage and half have pale plumage.
(b) Because the gene is located on the X chromosome, males (XY) only possess one copy of this gene (they are hemizygous), so if they inherit the recessive allele (Xᵇ) they will always show the pale plumage phenotype, as there is no second allele that could be dominant and mask it; females (XX) possess two X chromosomes, so a female showing pale plumage must inherit the recessive allele on both X chromosomes (XᵇXᵇ), which is a much rarer combination — a female with just one copy (XᴮXᵇ) is an unaffected carrier.

(a) [5] B1: correct parental genotypes XᴮXᵇ × XᴮY; B1: correct gametes identified; B1: all four correct offspring genotypes shown; B1: correct phenotypes assigned to each genotype; B1: correct 1:1:1:1 ratio stated (or equivalent, e.g. all females normal, males 1 normal:1 pale). (b) [2] B1: males only have one X chromosome/hemizygous, so a single recessive allele is always expressed; B1: females need two copies (homozygous recessive) to show the phenotype, which is rarer.

(a) Ferns are multicellular plants that are well differentiated, with true roots, stems and leaves, unlike simpler plants such as mosses; they possess a vascular system (xylem and phloem) for the transport of water/minerals and organic solutes; they possess a waterproof cuticle and stomata with fine control (able to open and close), reducing uncontrolled water loss; support is provided by turgor pressure within cells and by the woody xylem vessels and other strengthening tissue of the vascular bundles; they disperse by spores, which germinate in moist conditions but are only partially resistant to desiccation.
(b) Angiosperms possess all the water-retention and support features seen in ferns, but these are generally more highly evolved; for example, angiosperms (especially trees) develop more extensive woody xylem tissue, giving greater structural support and allowing growth to a large size/height; many angiosperms also show specific xerophytic adaptations (e.g. reduced leaf surface area, sunken stomata, thick cuticle) that allow survival in a wider range of habitats, including very dry ones, which ferns (restricted mostly to moist habitats) cannot tolerate.
(c) Seeds have a tough outer coat (testa) that allows them to withstand desiccation and remain dormant until conditions are favourable for germination (moist conditions), whereas fern spores are only partially resistant to drying out and are more vulnerable to desiccation; this means seed dispersal is more successful than spore dispersal in dry terrestrial environments, or in surviving an unfavourable dry period before germinating.

(a) [6] any six of: well-differentiated with true roots/stems/leaves; possess vascular system (xylem/phloem); waterproof cuticle; fine control of stomata (open/close); support by turgor within cells; support from woody xylem/vascular bundle strengthening; disperse by spores; spores germinate in moist conditions; spores only partially resistant to desiccation. (b) [4] angiosperms possess same features as ferns, more highly evolved; more extensive/advanced xylem tissue (wood) for support; allows growth in trees/larger size; xerophytic adaptations named (e.g. reduced leaf area, sunken stomata, thick cuticle); allows survival in a wider range of/drier habitats than ferns. (c) [3] seeds have tough outer coat/testa; resistant to desiccation, allowing dormancy in dry conditions; therefore more successful/reliable dispersal and survival in dry terrestrial environments than spores.

(a) W = Phylum Annelida (e.g. earthworm) — metameric segmentation, hydrostatic skeleton from segmental body cavities. X = Phylum Arthropoda (e.g. insect) — jointed limbs, segments grouped into regions. Y = Phylum Cnidaria (e.g. hydra/jellyfish) — radially symmetrical, hydrostatic skeleton from the fluid-filled enteron. Z = Phylum Platyhelminthes (e.g. planarian/liver fluke) — flattened, single gut opening.
(b) The jointed limbs of arthropods have allowed the basic body plan to be modified in many different directions during evolution — e.g. for walking, swimming, digging or (in insects) flight; this adaptability of a single basic body plan for many different functions/habitats has allowed arthropods, and particularly insects, to become the most successful animal group in terms of both the number of species and the total number of individuals.
(c) Platyhelminthes (Animal Z) lack a specialised skeletal system/body cavity providing a hydrostatic skeleton, unlike Annelida (Animal W), which have a hydrostatic skeleton formed from fluid-filled segmental body cavities. Consequence: Annelida can generate more effective, localised support and directional movement (e.g. peristaltic locomotion) using their segmented hydrostatic skeleton, whereas Platyhelminthes rely only on the support provided by their body tissue, restricting them to simpler, less powerful movement and generally a flattened body form (which also aids gas exchange by diffusion, given the lack of a transport system).

(a) [4] B1 each for correct phylum: W — Annelida; X — Arthropoda; Y — Cnidaria; Z — Platyhelminthes. (b) [3] jointed limbs allow modification of the basic body plan; for multiple functions/habitats (e.g. flight, named); has led to arthropods/insects being most successful in species number and/or individual number. (c) [3] B1: Platyhelminthes lack a hydrostatic skeleton/specialised body cavity that Annelida possess; B1: Annelida's hydrostatic skeleton (from segmental cavities) allows more effective support/powerful, directional movement; B1: Platyhelminthes rely on body tissue only for support, limiting movement/body form to flattened shape.

(a) In Vessel A, the KOH absorbs any CO₂ produced by the respiring peas, so the only factor causing a change in gas volume (and hence manometer fluid movement) is the uptake of oxygen by the peas for respiration; therefore the volume decrease in Vessel A (32 mm³) represents the volume of oxygen consumed by the peas in 5 minutes.
(b) In Vessel B (no KOH), the volume change reflects the net effect of O₂ consumed AND CO₂ produced: decrease = O₂ consumed − CO₂ produced. So CO₂ produced = O₂ consumed − decrease in Vessel B = 32 − 4 = 28 mm³.
(c) RQ = volume of CO₂ produced ÷ volume of O₂ consumed = 28 ÷ 32 = 0.875.
(d) An RQ of 1.0 indicates carbohydrate is the sole respiratory substrate, while an RQ of around 0.7 indicates fat (or protein) is being used. An RQ of 0.875, between these two values, suggests that the peas are respiring using a mixture of substrates — mainly carbohydrate, but with some fat and/or protein also being respired.

(a) [2] B1: KOH absorbs CO₂ produced; B1: so the volume change reflects only O₂ consumed/uptake. (b) [3] M1: recognise decrease in B = O₂ consumed − CO₂ produced (or equivalent); M1: 32 − 4; A1: 28 mm³. (c) [3] M1: RQ = CO₂ produced ÷ O₂ consumed; M1: correct substitution 28/32 (ft); A1: 0.875 (ft). (d) [2] B1: RQ = 1.0 for carbohydrate only, RQ ≈ 0.7 for fat/protein only, as reference points; B1: RQ of 0.875 (between these) indicates a mixture of respiratory substrates, mainly carbohydrate with some fat/protein.

(a) Light intensity directly affects the rate of the light-dependent stage of photosynthesis (e.g. rate of photolysis of water, generation of ATP and reduced NADP); at low light intensities, light is the limiting factor — as light intensity increases, the rate of the light-dependent reactions (and hence overall photosynthesis) increases proportionally/linearly, because more light energy is available to drive these reactions.
(b) Eventually another factor becomes limiting instead of light — here, carbon dioxide concentration (which limits the light-independent stage/rate of carbon fixation) becomes the limiting factor, so further increases in light intensity no longer increase the rate of photosynthesis, producing a plateau. Since both graphs are recorded at the same, constant temperature, and the initial linear (light-limited) sections have the same slope, the light intensity at which CO₂ (rather than light) first becomes limiting is the same for both graphs — it is determined by the fixed CO₂ concentration/temperature relative to light, not by which CO₂ treatment is used, up until the point CO₂ becomes limiting.
(c) With a higher CO₂ concentration, more CO₂ is available for the light-independent stage (fixation by RuBisCO onto RuBP), meaning CO₂ does not become limiting until a higher light intensity/rate of the light-dependent stage is reached; therefore the overall rate of photosynthesis can rise further before CO₂ availability limits it, giving a higher plateau rate than at the lower CO₂ concentration.
(d) ATP and reduced NADP (both produced during the light-dependent stage on the thylakoid membranes, and used to reduce/fix carbon dioxide during the light-independent stage in the stroma).

(a) [3] light intensity affects rate of light-dependent reactions; light is the limiting factor at low intensities; rate increases proportionally/linearly as more light energy available. (b) [3] another factor (CO₂) becomes limiting; light-independent stage/CO₂ fixation cannot proceed faster without more CO₂; same temperature and slope in both, so CO₂ becomes limiting at the same light intensity in each case. (c) [2] more CO₂ available for fixation/light-independent stage; delays CO₂ becoming limiting, allowing a higher rate/plateau before CO₂ limits further increase. (d) [2] B1: ATP; B1: reduced NADP.

(a) The DNA double helix unwinds and the hydrogen bonds between complementary base pairs break, catalysed by the enzyme DNA helicase, separating the two strands; each original (parental) strand then acts as a template; free DNA nucleotides in the nucleus pair with their complementary exposed bases on each template strand (A with T, C with G) according to the base-pairing rule; these nucleotides are joined together by the enzyme DNA polymerase, forming a new sugar-phosphate backbone; this produces two new DNA double helices, each consisting of one original (parental) strand and one newly synthesised strand — hence 'semi-conservative'.
(b) The triplet code refers to the fact that a sequence of three consecutive bases (a triplet/codon) on DNA (or mRNA) codes for one specific amino acid. During transcription: the enzyme RNA polymerase binds to and unwinds a region of the DNA double helix at the gene to be transcribed; hydrogen bonds between the two strands break, exposing the bases; one strand (the template/antisense strand) is used as a template; free RNA nucleotides align opposite their complementary exposed bases on the template strand (A pairs with U in RNA, and T, C, G pair as usual with A, G, C respectively); RNA polymerase joins these RNA nucleotides together to form a strand of pre-mRNA/mRNA with a base sequence complementary to the template DNA strand.

(a) [5] any five of: helicase unwinds double helix/breaks hydrogen bonds; two strands separate; each strand acts as a template; free nucleotides pair by complementary base pairing (A-T, C-G); DNA polymerase joins nucleotides/forms new strand; two double helices produced; each with one original and one new strand (semi-conservative). (b) [5] B1: triplet code = sequence of three bases codes for one amino acid; then any four of: RNA polymerase binds to/unwinds DNA at the gene; hydrogen bonds break, strands separate; one strand acts as template; free RNA nucleotides pair by complementary base pairing (A-U, T-A, C-G, G-C); RNA polymerase joins nucleotides to form mRNA strand complementary to the template strand.

(a) A restriction enzyme (endonuclease) recognises and cuts DNA at a specific, complementary base sequence (restriction site); the same restriction enzyme is used to cut both the DNA containing the desired human gene and the plasmid vector; because the same enzyme is used, this produces complementary, single-stranded overhangs, or 'sticky ends', on both the gene and the cut plasmid; the human gene (with its sticky ends) can then base-pair with the complementary sticky ends of the cut plasmid; the enzyme DNA ligase is used to join the sugar-phosphate backbones together, sealing the gene into the plasmid to form recombinant DNA; the recombinant plasmid is then introduced into a bacterium (e.g. by heat-shock or electroporation), which can then use its own transcription and translation machinery to express the human gene and produce the human protein.
(b) Indicative content: for a population to remain in Hardy–Weinberg equilibrium (allele frequencies unchanged generation to generation), several conditions must be met — no mutation, random mating, no selection, no migration, and a very large population size (negligible genetic drift); natural selection changes allele frequencies when individuals of one genotype/phenotype have a survival or reproductive advantage (are better adapted to the environment) over other genotypes; these better-adapted individuals are more likely to survive to reproductive age and to produce more offspring, passing on their alleles more frequently to the next generation; over many generations, this causes the frequency of the advantageous allele to increase in the population (and the disadvantageous allele to decrease), which directly violates the 'no selection' condition required for equilibrium; mutation introduces entirely new alleles into the gene pool (or changes existing ones) that were not previously present, and so directly violates the 'no mutation' condition; while an individual mutation is rare, mutation is the ultimate source of all new genetic variation upon which natural selection can then act; other factors, such as non-random mating (e.g. selective mate choice) and migration (gene flow of individuals, and their alleles, into or out of the population), can likewise change allele frequencies and move a population away from equilibrium; a conclusion could note that natural selection acts on the variation that mutation provides, so together they are central to the process of evolutionary change in allele frequencies over time.

[18 total] Part (a) [6]: indicative points — restriction enzyme/endonuclease cuts DNA at specific sequence; same enzyme used to cut plasmid and gene-containing DNA; produces complementary sticky ends; gene and plasmid ends base-pair/anneal; DNA ligase joins/seals sugar-phosphate backbones (forms recombinant DNA); plasmid inserted into bacterium (named method optional, e.g. heat-shock); bacterium transcribes/translates the gene to produce the protein. Award up to 6 marks for accurate, logically sequenced coverage of these points.
Part (b) [12]: indicative points include — Hardy–Weinberg conditions listed (no mutation, random mating, no selection, no migration, large population); natural selection defined — differential survival/reproduction due to advantageous allele/phenotype; advantageous allele passed on more frequently; allele frequency changes/increases over generations; violates 'no selection' condition; mutation creates new alleles/genetic variation; violates 'no mutation' condition; mutation is the ultimate source of variation for selection to act on; additional valid factors (non-random mating, migration/gene flow, genetic drift) and their link to specific Hardy–Weinberg conditions; a reasoned link/conclusion connecting mutation (source of variation) and selection (mechanism of change). Band 3 (9–12 marks): ≥9 valid indicative points, wide-ranging, accurate, clear links to Hardy–Weinberg conditions, sound concluding discussion, accurate QWC and specialist vocabulary. Band 2 (5–8 marks): 5–8 valid points, generally accurate but may lack clear linkage to Hardy–Weinberg conditions or balance between selection and mutation, competent QWC. Band 1 (1–4 marks): ≤4 valid points, basic/fragmented, limited specialist vocabulary, basic QWC.

(a) Each dilution step halves the concentration (5 cm³ stock + 5 cm³ water = double the volume, half the concentration). After 1 dilution: 50 mmol dm⁻³. After 2 dilutions: 25 mmol dm⁻³. After 3 dilutions: 12.5 mmol dm⁻³.
(b) To make a 1/4 dilution directly: take 2.5 cm³ of the stock solution (10 ÷ 4 = 2.5) and add 7.5 cm³ of distilled water (10 − 2.5 = 7.5), then mix thoroughly, giving 10 cm³ of solution at 1/4 the original concentration.
(c) Use a graduated pipette or volumetric/graduated glassware (rather than an unmarked container) to measure the volumes accurately, or use the same pipette/glassware for each measurement to reduce/standardise systematic error.
(d) Any two of: same type/size of cuvette (test tube) used each time; same wavelength/colour filter used on the colorimeter; same volume of solution in the cuvette each time; colorimeter zeroed/calibrated with a blank (distilled water) before use.

(a) [2] M1: correct halving method shown (e.g. 100→50→25→12.5); A1: 12.5 mmol dm⁻³. (b) [2] M1: 2.5 cm³ stock identified; A1: made up to 10 cm³ with 7.5 cm³ distilled water (accept ratio stated correctly). (c) [1] B1: valid method to improve volume-measuring accuracy (e.g. graduated/volumetric pipette). (d) [2] B1 each for any two valid controlled variables (max 2).

(a) Null hypothesis: there is no significant difference between the mean leaf length of leaves grown in full sun and leaves grown in shade (any difference is due to chance).
(b) Mean (sun) = (5.8+6.1+6.5+6.0+6.9+6.2+6.6+6.3)/8 = 50.4/8 = 6.3 cm. Mean (shade) = (8.4+7.9+8.2+7.7+8.5+8.0+7.8+8.3)/8 = 64.8/8 = 8.1 cm.
(c) \( t = \dfrac{|6.3 - 8.1|}{\sqrt{\frac{0.35^2}{8} + \frac{0.29^2}{8}}} = \dfrac{1.8}{\sqrt{0.0153 + 0.0105}} = \dfrac{1.8}{\sqrt{0.0258}} = \dfrac{1.8}{0.1607} \approx 11.2 \).
(d) Degrees of freedom = n₁ + n₂ − 2 = 8 + 8 − 2 = 14. The calculated t value (≈11.2) is much greater than the critical value (2.145) at p = 0.05 with 14 degrees of freedom. Therefore the null hypothesis is rejected — there IS a statistically significant difference between the mean leaf lengths of sun and shade leaves (shade leaves are significantly longer).

(a) [1] B1: valid null hypothesis (no significant difference in mean leaf length). (b) [2] B1: mean sun = 6.3 cm; B1: mean shade = 8.1 cm. (c) [3] M1: correct substitution into the formula given; M1: correct evaluation of the expression under the square root; A1: t ≈ 11.1–11.2 (accept ft from their means). (d) [2] B1 ft: correct comparison of their t to 2.145 (14 d.f.); B1 ft: correct conclusion — reject H0, significant difference, shade leaves longer.

(a) Null hypothesis: there is no significant difference between the observed distribution of dandelion plants across the four zones and an even distribution; any difference is due to chance.
(b) Expected number per zone (if evenly distributed) = 96 ÷ 4 = 24.
(c) χ² = Σ(O−E)²/E.
North: (18−24)²/24 = 36/24 = 1.50.
South: (32−24)²/24 = 64/24 = 2.67.
East: (25−24)²/24 = 1/24 = 0.04.
West: (21−24)²/24 = 9/24 = 0.375.
Σ = 1.50+2.67+0.04+0.375 = 4.58.
(d) Degrees of freedom = 4 − 1 = 3. The calculated χ² (4.58) is less than the critical value (7.815) at p = 0.05 with 3 degrees of freedom, so the null hypothesis is accepted — there is no significant difference from an even distribution across the four zones.

(a) [1] B1: valid null hypothesis referencing no significant difference from even distribution. (b) [1] B1: 24 (all four zones). (c) [4] M1: correct (O−E)² for at least two zones; M1: correct division by E; A1: individual values correct (or ft); A1: total χ² = 4.58 (accept 4.5–4.6). (d) [1] B1 ft: accept H0 — χ² < 7.815, no significant difference (consistent with their χ²).

(a) Lincoln index: estimated population size, N = (number marked and released, n₁ × total number captured in second sample, n₂) ÷ number of marked individuals recaptured in the second sample, m₂. N = (40 × 50) ÷ 8 = 2000 ÷ 8 = 250 woodlice.
(b) Any three of: the population is 'closed' between the two sampling occasions (no significant immigration, emigration, births or deaths); marked individuals mix/redistribute randomly and fully throughout the population before the second sample is taken; marking does not affect the survival, behaviour or 'catchability' of the marked individuals; the marks remain visible/do not fall off between sampling occasions; sampling method/effort is the same on both occasions.
(c) If marking made woodlice more visible to predators, marked woodlice would be preferentially eaten between the release and the second sampling; this would reduce the proportion of marked individuals recaptured in the second sample (a lower m₂ than would otherwise occur); since N is inversely proportional to m₂, this would cause the population size to be overestimated.

(a) [3] B1: correct formula N = (n₁ × n₂)/m₂ (or equivalent, correctly described); M1: correct substitution 40×50/8; A1: 250. (b) [3] B1 each for any three valid assumptions (max 3), from: closed population; random mixing/redistribution of marked individuals; marking does not affect survival/behaviour/catchability; marks not lost; consistent sampling method. (c) [2] B1: fewer marked individuals survive to be recaptured/lower m₂; B1: since N is inversely proportional to m₂ (or equivalent reasoning), the population estimate would be too high/an overestimate.

(a) Rf = distance travelled by pigment ÷ distance travelled by solvent front = 6.7 ÷ 9.6 = 0.6979 ≈ 0.70.
(b) Carotene is the most soluble in the solvent, because it has the highest Rf value (9.1/9.6 = 0.95), i.e. it travelled furthest relative to the solvent front, indicating it spent proportionally more time dissolved in/carried by the moving solvent (mobile phase) than adsorbed to the paper (stationary phase).
(c) If the original spot is too large or dilute, the pigments will diffuse/spread out further as they run, causing the separated spots to be larger, more spread out, and more likely to overlap with each other; this makes it difficult to accurately measure the distance travelled by each pigment, reducing the accuracy/reliability of the calculated Rf values.
(d) A sealed/covered container prevents the volatile solvent from evaporating out of the container and maintains a atmosphere saturated with solvent vapour inside; this ensures the solvent runs consistently and pigments separate reproducibly, giving accurate and repeatable Rf values (an unsaturated atmosphere would cause uneven/faster evaporation of solvent from the paper, distorting the results).

(a) [2] M1: 6.7/9.6; A1: 0.70 (2 d.p.). (b) [2] B1: carotene; B1: correct reason — highest Rf/travelled furthest relative to solvent front. (c) [2] B1: prevents excessive diffusion/spreading of the spot; B1: link to more accurate/distinct measurement of distance travelled (accurate Rf). (d) [2] B1: prevents solvent evaporation/maintains saturated atmosphere; B1: ensures consistent/reproducible solvent run and accurate Rf values.

(a) Mean = (24+27+22+26+29+25+23+28+26+24) ÷ 10 = 254 ÷ 10 = 25.4 cm.
(b) Deviations from the mean (x − x̄): −1.4, 1.6, −3.4, 0.6, 3.6, −0.4, −2.4, 2.6, 0.6, −1.4. Squared deviations: 1.96, 2.56, 11.56, 0.36, 12.96, 0.16, 5.76, 6.76, 0.36, 1.96. Sum of squared deviations = 44.4. Divide by (n−1) = 9: 44.4 ÷ 9 = 4.933. Square root: √4.933 ≈ 2.22 cm.
(c) A small standard deviation indicates that the data values are clustered closely around the mean, i.e. there is little variation/spread in the data set.
(d) The range only uses the two most extreme values (highest and lowest) in the data set, so it can be strongly distorted by a single unusually high or low value (an outlier/anomaly); the standard deviation takes every value in the data set into account, so it gives a more representative/reliable measure of the overall spread of the data around the mean, and is less affected by a single outlier.

(a) [1] B1: 25.4 cm. (b) [3] M1: correct deviations (x − x̄) calculated/squared; M1: Σ(x−x̄)² ÷ (n−1) = 44.4/9 (ft their mean); A1: √(their value) ≈ 2.22 cm (ft). (c) [1] B1: data clustered close to the mean/little variation. (d) [2] B1: range only uses two extreme values, easily distorted by an outlier; B1: standard deviation uses all values, giving a more representative/reliable measure of spread.

(a) Independent variable: temperature (of the water bath/reaction mixture). Dependent variable: volume of oxygen gas produced in 60 seconds. Controlled variable (any one): concentration and volume of hydrogen peroxide used; concentration and volume of catalase enzyme used; pH of the reaction mixture.
(b) The anomalous high reading at 40°C, not repeated at nearby temperatures, is more likely to be due to a random experimental/measurement error on that particular run (e.g. the gas syringe/measuring cylinder was misread, gas escaped/was collected inconsistently, or the water bath temperature briefly fluctuated) rather than a real biological effect, since a genuine effect of temperature on enzyme activity would be expected to follow a smooth trend, and would likely also be reflected (at least partially) in the neighbouring 35°C or 45°C readings.
(c) Repeat the investigation at each temperature (e.g. three times) and calculate a mean volume of oxygen produced at each temperature; this reduces the impact of random error/anomalous results on the overall trend/mean, and allows anomalous results to be identified and potentially discarded or repeated, making the results more reliable/reproducible.

(a) [3] B1: IV = temperature; B1: DV = volume of oxygen produced (in a fixed time); B1: valid controlled variable named (e.g. H₂O₂ or enzyme concentration/volume, pH). (b) [2] B1: identifies likely random/measurement error (not a genuine trend), e.g. gas leak, misreading, timing error; B1: valid reasoning, e.g. real effect would be expected to follow a smooth trend / would be reflected in neighbouring readings too. (c) [3] B1: repeat the investigation at each temperature; B1: calculate a mean at each temperature; B1: explanation — reduces impact of random error/anomalies on the trend, improving reliability/reproducibility.

(a) Mean = (15+20+10+25+30+15+20+10+25+20) ÷ 10 = 190 ÷ 10 = 19%.
(b) If the investigator chose quadrat positions themselves, they might (consciously or unconsciously) select positions that appear to have particularly high or low clover cover, introducing bias into the sample; using randomly-generated coordinates (e.g. from random number tables/a random number generator, used to set distances along two perpendicular tape measures/a grid) ensures every point in the field has an equal chance of being sampled, giving a representative, unbiased sample of the whole field.
(c) Increasing the number of quadrat samples reduces the impact of natural variation/any anomalous individual readings on the overall mean, since the sample as a whole is less likely to be skewed by a small number of unusually high or low values; a larger sample is more likely to be representative of percentage cover across the whole field, so the resulting population estimate is more reliable/more likely to be close to the true value.
(d) It can be very difficult to distinguish where one individual clover plant ends and another begins, as clover spreads vegetatively and its leaves/stems overlap, so percentage cover is a subjective/estimated measure rather than a precise count of individuals (accept: percentage cover does not give information about the actual number, size, or age structure of plants).

(a) [2] M1: sum ÷ 10; A1: 19%. (b) [2] B1: avoids investigator bias (conscious or unconscious selection of favourable areas); B1: every point has an equal chance of selection, giving a representative sample. (c) [2] B1: reduces effect of natural variation/anomalous individual quadrat readings on the mean; B1: sample more likely to be representative of the whole field/estimate closer to the true value. (d) [1] B1: valid limitation, e.g. difficulty distinguishing individual plants/overlapping vegetative growth, or percentage cover gives no information on plant number/size.