2022 CCEA A-Level Biology 1010 模拟试题及答案详解

Glucose that enters the filtrate during ultrafiltration is reabsorbed back into the blood by active transport (co-transport with sodium ions) across the wall of the proximal convoluted tubule (PCT).

1 mark: proximal convoluted tubule. Reject 'kidney tubule' alone.

Blood enters the glomerulus via the afferent arteriole and leaves via the narrower efferent arteriole, generating the hydrostatic pressure needed for ultrafiltration.

1 mark: afferent arteriole.

B lymphocytes, once activated and having undergone clonal selection, differentiate into plasma cells that secrete large quantities of a specific antibody.

1 mark: B lymphocyte / B cell.

An antigen is a molecule, usually a protein found on the surface of a cell or pathogen, that is recognised as non-self and stimulates an immune response.

1 mark: antigen.

The myelin sheath electrically insulates the axon between the nodes of Ranvier, forcing the action potential to jump from node to node (saltatory conduction), greatly increasing conduction speed.

1 mark: myelin sheath (accept 'nodes of Ranvier' if clearly linked to saltatory conduction).

Acetylcholinesterase, present in the synaptic cleft, hydrolyses acetylcholine into choline and ethanoic acid, removing it from the receptors so the postsynaptic membrane is not continuously stimulated.

1 mark: breaks down / hydrolyses acetylcholine so postsynaptic membrane is not continuously stimulated.

Carrying capacity (K) is the maximum population size of a species that a given environment can sustain indefinitely, determined by the availability of resources such as food, water and space.

1 mark: maximum population size an environment can support/sustain (indefinitely, given available resources).

Primary succession begins on a substrate that has never previously supported a community, such as bare rock, sand dunes or newly cooled lava, with no soil initially present.

1 mark: primary succession.

The climax community is the final, relatively stable stage of succession, in equilibrium with the prevailing environmental conditions, showing little further change in species composition.

1 mark: climax community.

Productivity is a rate, so it is expressed as energy per unit area per unit time: \( \text{kJ m}^{-2}\text{ yr}^{-1} \).

1 mark: \( \text{kJ m}^{-2}\text{ yr}^{-1} \) (accept equivalent energy/area/time units).

Energy released during respiration is largely lost as heat to the surroundings rather than being converted into new biomass, so it is unavailable to the next trophic level.

1 mark: respiration (heat loss during respiration).

Nitrification is the oxidation of ammonium ions to nitrite ions (by Nitrosomonas) and then to nitrate ions (by Nitrobacter), both steps requiring aerobic conditions.

1 mark: nitrification.

(a) Glucose molecules are small enough to pass through the pores of the fenestrated capillary and the basement membrane during ultrafiltration, so they are not selectively filtered out and pass into the filtrate at the same concentration as the plasma. (b) All the filtered glucose is normally reabsorbed back into the blood by active transport (co-transport with sodium ions) across the epithelium of the proximal convoluted tubule, so under normal conditions none remains to appear in the urine.

(a) 1 mark: glucose molecules are small enough to pass through the pores of the basement membrane/fenestrated capillary during ultrafiltration. (b) 1 mark: all glucose is reabsorbed in the proximal convoluted tubule; 1 mark: by active transport / co-transport with sodium ions.

(a) Dehydration reduces the volume of water in the blood relative to solutes, so the water potential of the blood plasma decreases (becomes more negative). (b) This decrease is detected by osmoreceptors in the hypothalamus, which respond by triggering the posterior pituitary gland to release more antidiuretic hormone (ADH) into the blood. ADH increases the permeability of the collecting duct (and distal convoluted tubule) to water by inserting aquaporins into the cell membranes, so more water is reabsorbed from the filtrate by osmosis, producing a smaller volume of more concentrated urine.

(a) 1 mark: decreases / becomes more negative. (b) 1 mark: osmoreceptors in the hypothalamus detect the change and more ADH is released from the posterior pituitary; 1 mark: ADH increases permeability of the collecting duct (and DCT) to water so more water is reabsorbed by osmosis.

The secondary immune response has a much shorter lag (latent) period than the primary response and produces a greater concentration of antibody more rapidly, because memory B cells produced during the primary response are already present and can divide rapidly into antibody-secreting plasma cells on re-exposure to the antigen. Vaccination exposes the immune system to a safe form of the antigen, stimulating a primary response and the production of memory cells without causing disease; if the person later encounters the actual pathogen, a fast, large secondary response destroys it before symptoms of disease develop, providing immunity.

1 mark: secondary response shows a shorter lag period; 1 mark: higher/faster rise in antibody concentration, due to memory cells already present; 1 mark: vaccination pre-exposes the immune system to produce memory cells, so a future infection is cleared before illness develops.

T helper cells normally release cytokines that stimulate the clonal expansion and differentiation of B lymphocytes into antibody-secreting plasma cells, and that activate cytotoxic T cells and macrophages. With a greatly reduced number of T helper cells, both the humoral (antibody-mediated) response and the cell-mediated response are severely impaired, because the cells responsible for destroying pathogens and infected cells are not properly activated. As a result, pathogens that a healthy immune system would normally destroy are able to establish infections, making the patient highly susceptible to opportunistic infections.

1 mark: T helper cells normally stimulate B cells (antibody production) and cytotoxic T cells/macrophages via cytokines; 1 mark: reduced T helper cell numbers impair both humoral and cell-mediated immunity; 1 mark: pathogens/opportunistic infections are therefore not destroyed effectively.

Mechanical pressure deforms the lamellae of the Pacinian corpuscle, stretching the membrane of the sensory neurone inside it. This stretching opens stretch-mediated sodium channels, allowing Na⁺ ions to diffuse into the neurone down their electrochemical gradient, depolarising the membrane and producing a generator potential. If the generator potential is large enough to reach the threshold, it causes sufficient depolarisation to open voltage-gated sodium channels nearby, initiating an action potential that propagates along the axon.

1 mark: deformation/stretch opens stretch-mediated sodium channels; 1 mark: Na⁺ influx causes depolarisation, producing a generator potential; 1 mark: if threshold is reached, voltage-gated Na⁺ channels open, producing an action potential.

During the refractory period, the voltage-gated sodium channels responsible for the rising phase of the action potential are inactivated and cannot reopen immediately, regardless of any further stimulus. Because a new action potential cannot be generated until these channels have reset, a minimum time interval is imposed between successive impulses, which sets an upper limit on the frequency at which action potentials can be transmitted.

1 mark: voltage-gated Na⁺ channels are inactivated and cannot reopen immediately; 1 mark: no new action potential can be generated until the channels reset, limiting maximum frequency.

Transmission across a synapse is unidirectional, because neurotransmitter is released only from the presynaptic membrane and receptors are present only on the postsynaptic membrane, whereas an action potential can in principle travel in either direction along an isolated axon. Synaptic transmission is also slower than conduction along the axon because it depends on diffusion of a chemical neurotransmitter across the synaptic cleft (causing synaptic delay), whereas conduction along the axon is a self-propagating electrical event.

1 mark each for any two of: synaptic transmission is unidirectional (axonal conduction is not, in isolation); synaptic transmission is slower (synaptic delay); synaptic transmission involves diffusion of a chemical neurotransmitter rather than direct electrical conduction.

As the density of the rabbit population increases, intraspecific competition for the limited food supply intensifies, because each individual has access to a smaller average share of the available food. This reduces the birth rate (through poorer body condition and reduced reproductive success) and/or increases the death rate (through starvation), and crucially this negative effect becomes proportionally greater the higher the population density is. This density-dependent negative feedback tends to keep the population size close to the carrying capacity of the habitat, preventing indefinite growth.

1 mark: as density increases, intraspecific competition for food increases; 1 mark: this reduces birth rate and/or increases death rate; 1 mark: the effect is proportionally greater at higher densities (density-dependent), regulating the population around the carrying capacity.

An r-strategist species, such as the aphid, produces very large numbers of offspring, reaches sexual maturity quickly, provides little or no parental care, and is adapted to exploit unstable or unpredictable environments where rapid population growth is favoured. A K-strategist species, such as the elephant, produces few offspring per reproductive event, invests heavily in parental care, and maintains a population size close to the carrying capacity in a relatively stable environment.

1 mark: correct description of an r-strategist with a valid named example; 1 mark: correct description of a K-strategist with a valid named example.

Zonation is the distribution of different species in distinct horizontal bands or zones across a habitat. On a rocky shore, this pattern is caused mainly by a gradient in an abiotic factor, particularly the length of time each zone is exposed to air (and therefore desiccation risk and temperature extremes) between high and low tide, which different species tolerate to different degrees.

1 mark: distribution of species occurring in distinct bands/zones; 1 mark: caused by a gradient in an abiotic factor (e.g. exposure time/desiccation) up the shore.

A pioneer community exists under harsh abiotic conditions (e.g. little or no soil, extreme exposure), which only a small number of highly tolerant species can survive, so diversity is initially low. As pioneer species colonise and die, their organic remains build up soil and improve water/nutrient retention, moderating the abiotic conditions; this allows a progressively wider range of species, including less tolerant ones, to colonise and compete successfully, increasing species diversity as succession proceeds.

1 mark: pioneer/early conditions are harsh so only a few species can tolerate them; 1 mark: pioneers modify the abiotic environment (e.g. build soil), allowing more species to colonise as succession proceeds.

Suitable abiotic factors that could be measured include soil pH, light intensity, soil moisture content and temperature, any two of which could be correlated with the distribution of plant species recorded along the transect.

1 mark each for any two valid abiotic factors, e.g. soil pH, light intensity, soil moisture, temperature, wind speed.

(a) \( NPP = GPP - R = 20\,000 - 8\,000 = 12\,000 \text{ kJ m}^{-2}\text{ yr}^{-1} \). (b) Not all the plant material produced is eaten by primary consumers (e.g. roots and dead material are instead broken down by decomposers); of the material that is eaten, some is indigestible and is egested as faeces rather than absorbed; and of the energy that is absorbed, much is used in the consumer's own respiration and lost as heat rather than being converted into new biomass, so only a small fraction of NPP ends up incorporated into primary consumer biomass.

(a) 1 mark: \( 12\,000 \text{ kJ m}^{-2}\text{ yr}^{-1} \) (with working). (b) 1 mark: not all plant material is eaten, and some eaten material is egested (indigestible parts); 1 mark: energy absorbed is largely lost as heat via respiration before assimilation into new biomass.

Energy transfer between trophic levels is never 100% efficient because: not all of the organisms (or parts of organisms) at one trophic level are consumed by the next (e.g. roots, bones, feathers may remain uneaten); of the material that is consumed, not all is digested and absorbed, with a proportion being egested as faeces; and of the material that is absorbed, a large proportion is used in respiration to release energy for movement, growth and other life processes, and this energy is ultimately lost to the surroundings as heat rather than being incorporated into new biomass.

1 mark: not all of an organism/trophic level is eaten; 1 mark: not all eaten material is digested/absorbed (egestion as faeces); 1 mark: energy is lost as heat during respiration rather than being converted into new biomass.

Because energy is always lost between trophic levels, mainly as heat during respiration, the total amount of energy available must decrease at each successive trophic level; this means a pyramid of energy can never be inverted, as each bar must represent less energy than the one below it. A pyramid of numbers, in contrast, reflects the number of individual organisms rather than energy content, and can be inverted when a small number of large producers (e.g. a single tree) supports a large number of small primary consumers.

1 mark: energy is always lost between trophic levels (respiration/heat loss), so energy at each level must be less than the level below; 1 mark: this makes an inverted energy pyramid impossible, unlike a numbers pyramid, which depends on organism size rather than energy content.

Farmers can increase the efficiency of energy transfer by, for example, applying fertilisers and controlling pests/weeds to increase the productivity of crops, and by keeping livestock warm and confining their movement, which reduces the proportion of ingested energy lost through respiration (as heat and movement) rather than converted into biomass.

1 mark each for any two valid methods, e.g. fertiliser use, pest/weed control, housing/limiting movement of livestock, selective breeding for efficient growth.

Nitrogen-fixing bacteria such as Rhizobium convert atmospheric nitrogen gas \( (\text{N}_2) \), which most organisms cannot use directly, into ammonium ions/nitrogen-containing compounds that plants can absorb and use to synthesise amino acids and proteins. In the mutualistic (symbiotic) relationship found in the root nodules of legumes, the plant provides the bacteria with carbohydrates and a protected, low-oxygen environment, while the bacteria supply the plant with a source of usable nitrogen that it could not otherwise obtain from the atmosphere, so both organisms benefit.

1 mark: nitrogen-fixing bacteria convert \( \text{N}_2 \) gas into ammonium ions/nitrogen compounds usable by plants; 1 mark: bacteria live in root nodules and receive carbohydrates/organic compounds from the plant; 1 mark: the plant receives fixed nitrogen it could not otherwise obtain — a mutual benefit.

In waterlogged soil, anaerobic (low-oxygen) conditions favour denitrifying bacteria, which use nitrate ions as an alternative electron acceptor in respiration, converting them back into nitrogen gas \( (\text{N}_2) \). This nitrogen gas diffuses out of the soil into the atmosphere and is no longer available in a form that plant roots can absorb, reducing the pool of nitrate available for plant uptake.

1 mark: denitrifying bacteria convert nitrate to nitrogen gas under anaerobic conditions; 1 mark: nitrogen gas is lost to the atmosphere, reducing the nitrate available for plant uptake.

(a) \( \text{Net production} = 15\,000 - 6\,000 = 9\,000 \text{ kJ m}^{-2}\text{ yr}^{-1} \). (b) \( \dfrac{3\,600}{9\,000} \times 100 = 40.0\% \). (c) Biomass/energy available to primary consumers for growth = ingestion − egestion − respiration = \( 3\,600 - 900 - 1\,500 = 1\,200 \text{ kJ m}^{-2}\text{ yr}^{-1} \). Efficiency of transfer to secondary consumers = \( \dfrac{620}{1\,200} \times 100 = 51.7\% \).

(a) 1 mark: 9000 kJ m⁻² yr⁻¹. (b) 1 mark: correct method \( (3600/9000 \times 100) \); 1 mark: 40.0% (accept 40%). (c) 1 mark: correct calculation of primary consumer production available (3600−900−1500=1200) used as denominator; 1 mark: 51.7% (accept 51.6–51.7%, ECF from earlier values).

(a) The resting potential shown is −70 mV. It is maintained by the sodium–potassium pump, which actively transports 3 Na⁺ ions out of the neurone for every 2 K⁺ ions pumped in (using ATP), combined with the membrane being more permeable to K⁺ than Na⁺ via potassium 'leak' channels, so the inside of the axon remains negative relative to the outside. (b) Between 1 ms and 2 ms, depolarisation occurs: voltage-gated Na⁺ channels open in response to the stimulus, and Na⁺ ions diffuse into the neurone down their electrochemical gradient, making the inside progressively less negative and then positive, reaching a peak of +40 mV. Between 2 ms and 3 ms, repolarisation occurs: the Na⁺ channels close/inactivate and voltage-gated K⁺ channels open, so K⁺ ions diffuse out of the neurone, restoring the negative internal charge; because the K⁺ channels are slow to close, slightly too much K⁺ leaves the neurone, producing a brief hyperpolarisation to −80 mV before the sodium–potassium pump restores the resting potential.

(a) 1 mark: −70 mV; 1 mark: maintained by the Na⁺/K⁺ pump (3 Na⁺ out : 2 K⁺ in, active transport/ATP) and differential membrane permeability (K⁺ leak channels). (b) 1 mark: depolarisation caused by opening of voltage-gated Na⁺ channels and Na⁺ influx; 1 mark: repolarisation caused by Na⁺ channels closing and voltage-gated K⁺ channels opening, K⁺ efflux; 1 mark: hyperpolarisation as K⁺ channels close slowly, before the pump restores the resting potential.

(a) The percentage cover of marram grass increases from 5% at 0 m to a peak of 60% at 30 m from the sea, then decreases to 15% by 50 m; cover therefore shows a positive relationship with distance up to 30 m and a negative relationship beyond 30 m. (b) Beyond 30 m, soil moisture and stability are higher, which allows other, more competitive plant species to establish as succession proceeds; marram grass is adapted to the drier, more unstable conditions of the embryo/fore-dune and is progressively outcompeted for light, water and nutrients as later-successional species colonise the more stable, moister soil further inland. (c) Spearman's rank correlation coefficient; null hypothesis: there is no significant correlation between soil moisture and percentage cover of marram grass.

(a) 1 mark: cover increases from 0–30 m; 1 mark: cover decreases from 30–50 m (peak at 30 m). (b) 1 mark: increased competition from other, more competitive species as succession proceeds inland; 1 mark: marram grass is adapted to dry, unstable, low-nutrient conditions and is outcompeted in more stable/moist conditions. (c) 1 mark: Spearman's rank correlation coefficient with a correctly stated null hypothesis (no significant correlation).

(a) Urea molecules are small and diffuse freely down their concentration gradient across the partially permeable dialysis membrane, from the higher concentration in the blood to the lower concentration in the dialysis fluid; protein molecules, however, are far too large to pass through the pores of the membrane and so remain in the blood. (b) The glucose concentration of the dialysis fluid is set equal to that of normal blood plasma so that no concentration gradient exists for glucose; this prevents glucose, which the body needs, from diffusing out of the blood into the dialysis fluid and being lost. (c) Blood and dialysis fluid are made to flow in opposite directions through the dialysis machine; this maintains a concentration gradient for urea (and other wastes) along the entire length of the exchange membrane, rather than the gradient reducing progressively as in parallel flow, which maximises the total amount of urea removed from the blood in a given time.

(a) 1 mark: urea is small enough to diffuse through the partially permeable membrane pores; 1 mark: protein molecules are too large to pass through. (b) 1 mark: no concentration gradient for glucose, so it is not lost from the blood. (c) 1 mark: counter-current flow maintains a concentration gradient along the whole length of the membrane; 1 mark: this maximises diffusion/removal of urea compared with parallel flow.

(a) Decomposers, such as saprobiotic bacteria and fungi, break down dead organic matter and waste products through extracellular digestion and absorption; as they respire the absorbed organic molecules, they release carbon back into the atmosphere as \( \text{CO}_2 \), recycling carbon and making it available again to producers for photosynthesis. (b) Fossil fuels contain carbon that has been removed from the short-term carbon cycle and stored underground over millions of years. When fossil fuels are burned, this carbon is released as \( \text{CO}_2 \) at a much faster rate than it was originally removed by photosynthesis and geological processes, and faster than natural sinks such as oceans and forests can absorb it. This results in a net increase in the concentration of \( \text{CO}_2 \) in the atmosphere, disrupting the natural balance of the carbon cycle and enhancing the greenhouse effect.

(a) 1 mark: decomposers break down dead organic matter (via saprobiotic nutrition); 1 mark: respiration by decomposers releases CO₂ back to the atmosphere, recycling carbon. (b) 1 mark: fossil fuels represent carbon stored over millions of years, removed from the short-term cycle; 1 mark: combustion releases this carbon as CO₂ faster than natural removal by photosynthesis/sinks; 1 mark: this results in a net increase in atmospheric CO₂, disrupting the natural balance/enhancing the greenhouse effect.

A comprehensive answer should explain that the nervous system transmits information as fast-travelling electrical impulses (action potentials) along specific neurones to specific effectors, such as skeletal muscles or glands, producing a rapid but short-lived response; for example, in thermoregulation, the hypothalamus detects a fall in body temperature and sends nerve impulses to effectors such as skeletal muscles (causing shivering) and skin arterioles (causing vasoconstriction) to generate and conserve heat rapidly. By contrast, the endocrine system releases chemical messengers (hormones) into the blood, which travel more slowly to reach widespread target cells possessing specific complementary receptors, producing a slower-onset but often longer-lasting response; for example, blood glucose concentration is regulated by insulin and glucagon secreted from the islets of Langerhans in the pancreas, with insulin promoting glucose uptake and glycogenesis in liver and muscle cells, and glucagon promoting glycogenolysis, acting over a longer timescale than a nervous reflex. Both systems operate through negative feedback to restore a set point, but nervous responses are typically localised, fast and transient, while endocrine responses are typically widespread, slower to begin and more prolonged.

Level-based 3-band grid (12 marks). Level 1 (1–4 marks): basic, undeveloped points about nervous and/or endocrine coordination with little accurate detail; poor use of specialist terminology; weak QWC. Level 2 (5–8 marks): some relevant, accurate detail on both systems with at least one named example; reasonable structure; adequate QWC. Level 3 (9–12 marks): comprehensive, accurate, well-structured comparison of nervous and endocrine coordination covering named examples (e.g. thermoregulation, blood glucose regulation), explicit discussion of speed (fast electrical vs slower chemical), duration (short-lived vs prolonged) and mechanism (specific neurone-effector links vs hormone-receptor binding at widespread targets), with reference to negative feedback in both; accurate specialist vocabulary and coherent, logically sequenced prose.

Auxin is produced in the meristem at the tip of a shoot and is transported down the shoot, largely by diffusion and active transport, to the region of elongation just behind the tip. When light strikes a shoot from one direction only, auxin is transported laterally away from the illuminated side towards the shaded side of the shoot, so a higher concentration of auxin accumulates on the shaded side than on the illuminated side. Auxin stimulates cell elongation in the region behind the tip by loosening the cell wall (increasing wall extensibility, sometimes explained by the acid growth hypothesis), so the cells on the shaded side, having a higher auxin concentration, elongate more than those on the illuminated side. This unequal elongation causes the shoot to curve towards the light source, a response known as positive phototropism, which increases the plant's exposure to light for photosynthesis.

Level-based 3-band grid (6 marks). Level 1 (1–2 marks): basic statement that the shoot bends towards the light, with little or no reference to a mechanism. Level 2 (3–4 marks): auxin identified as the hormone responsible, with some reference to unequal distribution across the shoot. Level 3 (5–6 marks): clear, accurate explanation that auxin, produced at the shoot tip, is transported laterally to the shaded side in response to unilateral light; the resulting higher auxin concentration on the shaded side stimulates greater cell elongation there; this unequal elongation causes the shoot to curve towards the light (positive phototropism).

The Krebs cycle takes place in the mitochondrial matrix, following the link reaction that converts pyruvate to acetyl coenzyme A.

1 mark: mitochondrial matrix.

Glycolysis uses 2 ATP for phosphorylation of glucose but produces 4 ATP by substrate-level phosphorylation, giving a net yield of 2 ATP per glucose molecule.

1 mark: 2 ATP (net).

Photosystem II absorbs light and uses the energy to split water molecules (photolysis), releasing electrons, protons and oxygen.

1 mark: Photosystem II / PSII.

The light-dependent reaction produces reduced NADP (NADPH) and ATP, both of which are used in the light-independent reaction (Calvin cycle) to reduce glycerate 3-phosphate to triose phosphate.

1 mark: reduced NADP and ATP (both required for the mark).

DNA helicase breaks the hydrogen bonds between complementary base pairs, unwinding and separating the two strands of the double helix to expose them as templates.

1 mark: DNA helicase.

Hydrogen bonds hold complementary base pairs together in the DNA double helix and are broken by DNA helicase to separate the two strands.

1 mark: hydrogen bonds.

Restriction endonucleases recognise specific base sequences and cut the DNA at or near these sites, often leaving 'sticky ends' useful in genetic engineering.

1 mark: restriction endonuclease / restriction enzyme.

DNA ligase catalyses the formation of phosphodiester bonds between the sugar-phosphate backbones of adjacent DNA fragments, joining them together, for example when inserting a gene into a plasmid vector.

1 mark: DNA ligase.

An allele is one of two or more alternative versions of a gene, occupying the same locus on homologous chromosomes, that may produce different phenotypic effects.

1 mark: alternative form/version of a gene.

Crossing two heterozygotes (Aa × Aa) gives genotype ratio 1 AA : 2 Aa : 1 aa; since AA and Aa share the dominant phenotype, the phenotypic ratio is 3 dominant : 1 recessive.

1 mark: 3:1.

The Hardy-Weinberg principle states that, in a population at equilibrium, allele frequencies sum as \( p + q = 1 \), and genotype frequencies sum as \( p^2 + 2pq + q^2 = 1 \), where \( p \) and \( q \) are the frequencies of the dominant and recessive alleles respectively.

1 mark: both \( p + q = 1 \) and \( p^2 + 2pq + q^2 = 1 \) stated correctly.

Angiospermophyta is the phylum of flowering plants, distinguished from gymnosperms by producing seeds enclosed within an ovary that develops into a fruit.

1 mark: Angiospermophyta / flowering plants.

Monocotyledons have a single cotyledon in the seed embryo and typically parallel leaf venation, while dicotyledons have two cotyledons and typically net (reticulate) leaf venation.

1 mark: any valid distinguishing feature (e.g. number of cotyledons, leaf venation pattern, floral parts in multiples of three vs four/five).

Arthropoda is characterised by a segmented body, a hard chitinous exoskeleton, and paired jointed appendages, and includes insects, arachnids and crustaceans.

1 mark: Arthropoda.

Oxygen acts as the final electron acceptor at the end of the electron transport chain in the inner mitochondrial membrane, combining with electrons and H⁺ ions to form water. If oxygen is not available to accept these electrons, the electron transport chain stops functioning, so reduced coenzymes (NADH and FADH₂) cannot be reoxidised back to NAD⁺ and FAD. Because oxidative phosphorylation — the chemiosmotic synthesis of ATP via the proton gradient generated by the electron transport chain — produces the great majority of ATP in aerobic respiration, this process halts without oxygen, drastically reducing total ATP yield.

1 mark: oxygen is the final electron acceptor in the electron transport chain, forming water; 1 mark: without O₂, the electron transport chain stops and NAD⁺/FAD cannot be regenerated; 1 mark: oxidative phosphorylation (chemiosmosis/ATP synthase), which produces most ATP, therefore halts.

In anaerobic respiration in mammalian muscle, only glycolysis takes place in the cytoplasm, yielding a net 2 ATP per glucose molecule; pyruvate is converted to lactate to regenerate NAD⁺ so glycolysis can continue, rather than entering the mitochondrion. Because the Krebs cycle and oxidative phosphorylation, which together yield the great majority of ATP in aerobic respiration, do not take place, the overall ATP yield per glucose molecule is far lower than in aerobic respiration.

1 mark: only glycolysis (2 ATP net) occurs, pyruvate is converted to lactate; 1 mark: the Krebs cycle and oxidative phosphorylation (the major ATP-yielding stages) do not take place.

Respiratory quotient is calculated as \( RQ = \dfrac{\text{volume of CO}_2\text{ produced}}{\text{volume of O}_2\text{ consumed}} \). Fat molecules contain proportionally more carbon and hydrogen and less oxygen relative to carbohydrate molecules, so relatively more oxygen must be consumed to fully oxidise the additional hydrogen and carbon atoms in fats compared with the carbon dioxide produced. This gives fats a lower \( \text{CO}_2:\text{O}_2 \) ratio (RQ ≈ 0.7) than carbohydrates (RQ ≈ 1.0).

1 mark: fats contain relatively less oxygen (more C–H bonds) than carbohydrates; 1 mark: more O₂ is required per CO₂ produced during fat oxidation, giving a lower RQ (≈0.7 vs ≈1.0 for carbohydrate).

Light energy excites electrons in Photosystem II, which pass along a series of carriers forming an electron transport chain in the thylakoid membrane; the energy released as electrons move along the chain is used to actively pump protons (H⁺) from the stroma into the thylakoid lumen. Photolysis of water in the thylakoid lumen also releases protons directly into the lumen, further increasing the proton concentration there relative to the stroma, creating a steep proton gradient across the thylakoid membrane. Protons then diffuse back down their concentration gradient into the stroma through ATP synthase enzymes embedded in the membrane, and the energy released by this flow is used to phosphorylate ADP with inorganic phosphate to form ATP, a process known as chemiosmosis.

1 mark: the electron transport chain pumps H⁺ from the stroma into the thylakoid lumen; 1 mark: photolysis of water also releases H⁺ into the lumen, building a proton gradient; 1 mark: protons diffuse back through ATP synthase into the stroma, providing the energy for ATP synthesis (chemiosmosis).

When carbon dioxide concentration is very low, carbon dioxide — rather than light intensity — is the factor limiting the rate of the light-independent reaction (Calvin cycle), because there is insufficient substrate for carbon fixation regardless of how much light-dependent product (ATP and reduced NADP) is available. Since light intensity is not the limiting factor under these conditions, a further decrease in light intensity has little or no additional effect on the overall rate of photosynthesis.

1 mark: CO₂ is already the limiting factor at low concentration; 1 mark: light intensity is not limiting under these conditions, so reducing it further has little/no additional effect on rate.

DNA helicase unwinds and separates the two strands of the DNA double helix by breaking the hydrogen bonds between complementary base pairs, exposing each single strand to act as a template. Free DNA nucleotides in the nucleus pair by complementary base pairing with the exposed bases on each template strand, and DNA polymerase catalyses the formation of phosphodiester bonds joining these nucleotides together, synthesising a new complementary strand alongside each original strand. This produces two identical daughter DNA molecules, each consisting of one original (parental) strand and one newly synthesised strand — the semi-conservative model of replication.

1 mark: DNA helicase unwinds the double helix, breaking hydrogen bonds, exposing two template strands; 1 mark: free nucleotides pair by complementary base pairing with each template strand and are joined by DNA polymerase; 1 mark: each daughter DNA molecule consists of one original (parental) and one new strand.

If DNA replication were conservative, one round of replication would produce one band of fully heavy (¹⁵N/¹⁵N) DNA and one separate band of fully light (¹⁴N/¹⁴N) DNA, since the original double helix would remain intact and only entirely new molecules would be synthesised. Instead, a single band of intermediate density was observed, showing that every DNA molecule present contained a mixture of ¹⁵N and ¹⁴N; this is only consistent with each new molecule containing one original heavy strand and one newly synthesised light strand, as predicted by the semi-conservative model.

1 mark: conservative replication would predict two separate bands (heavy and light), which was not observed; 1 mark: a single intermediate band shows each molecule contains one old (¹⁵N) and one new (¹⁴N) strand, consistent only with semi-conservative replication.

The genetic code is degenerate (redundant), meaning that more than one triplet codon can code for the same amino acid, often differing only in the third base of the codon. If a base substitution mutation changes a codon into a different codon that happens to code for the same amino acid, the amino acid sequence of the resulting polypeptide is unchanged; this is known as a silent mutation, and because the primary structure of the protein is unaffected, its structure and function remain unchanged despite the change in the DNA base sequence.

1 mark: the genetic code is degenerate/redundant — more than one codon can code for the same amino acid; 1 mark: correct explanation that this can produce a silent mutation, where the substituted codon still specifies the same amino acid, so protein structure/function is unaffected.

A promoter is the specific DNA sequence recognised by RNA polymerase, allowing it to bind to the DNA and initiate transcription at the correct starting point. If a compatible promoter is not positioned correctly upstream of the inserted gene, RNA polymerase in the host bacterium will not bind there, and the gene will not be transcribed into mRNA or subsequently translated into protein.

1 mark: a promoter is required for RNA polymerase to bind and initiate transcription; 1 mark: without it, the inserted gene will not be transcribed/expressed in the host bacterium.

The DNA sample is first heated to around 95°C, causing denaturation as the hydrogen bonds between the two strands break and the double helix separates into single strands. The mixture is then cooled to around 50–65°C, allowing short, specific primers to anneal to complementary sequences flanking the target region on each single strand. The temperature is then raised to around 72°C, the optimum for the heat-tolerant Taq DNA polymerase, which extends new complementary strands from each primer using free nucleotides present in the reaction mixture. This three-step cycle of denaturation, annealing and extension is repeated many times (typically 25–35 cycles), with the amount of the target DNA sequence doubling with each cycle, resulting in exponential amplification.

1 mark: denaturation at high temperature separates the DNA strands (breaks hydrogen bonds); 1 mark: annealing — primers bind to complementary flanking sequences at a lower temperature; 1 mark: extension — Taq polymerase synthesises new strands from the primers at ~72°C, and the cycle is repeated to amplify the DNA exponentially.

(a) The heterozygous long-winged parent is Ll and the vestigial-winged parent is ll (vestigial wing is recessive, so this fly must be homozygous recessive). (b) The Ll parent produces gametes L and l in equal proportions; the ll parent produces only l gametes. Combining these gametes in a Punnett square gives offspring genotypes Ll, Ll, ll and ll, i.e. a 1:1 ratio of Ll to ll. Since Ll is phenotypically long-winged and ll is vestigial-winged, the expected phenotypic ratio of the offspring is 1 long-winged : 1 vestigial-winged.

(a) 1 mark: Ll and ll. (b) 1 mark: correct genetic diagram/Punnett square showing gametes and offspring genotypes (Ll, Ll, ll, ll); 1 mark: correct ratio 1:1 long-winged : vestigial-winged.

Epistasis occurs when the alleles present at one gene locus mask or otherwise influence the phenotypic expression of alleles at a different, independently assorting gene locus. For example, in mice, coat colour depends on a gene with alleles for black (B) and brown (b) fur, but a separate gene controls whether pigment is deposited at all (C, pigment present) or not (c, albino); mice with the genotype cc are albino regardless of their genotype at the colour locus, because the cc genotype masks the effect of the B/b gene. As a result, the standard 9:3:3:1 dihybrid ratio expected from two independently assorting genes is modified, for example to a 9:3:4 ratio, because all genotypes containing cc (which would otherwise fall into two separate phenotypic classes) are combined into a single albino phenotypic class.

1 mark: epistasis defined as one gene masking/influencing the expression of another gene at a different locus; 1 mark: a valid named example given (e.g. coat colour/pigment gene interaction in mice); 1 mark: correct explanation of how the standard 9:3:3:1 ratio is modified (e.g. to 9:3:4) because of the masking effect.

\( q^2 = 0.16 \), so \( q = \sqrt{0.16} = 0.4 \). Since \( p + q = 1 \), \( p = 1 - 0.4 = 0.6 \).

1 mark: \( q = \sqrt{0.16} = 0.4 \); 1 mark: \( p = 1 - 0.4 = 0.6 \).

Xylem vessel elements are dead cells with no cell contents (cytoplasm, nucleus) and no end walls between adjacent cells, forming a long, continuous, unobstructed tube that offers minimal resistance to the flow of water. Their walls are thickened and impregnated with lignin, giving them the strength to withstand the strong negative pressure (tension) generated during transpiration pull, without the vessel collapsing inwards. Small, unlignified regions called pits are present in the walls, allowing lateral (sideways) movement of water between adjacent vessels and into surrounding living cells.

1 mark each for any three of: dead cells with no cytoplasm/cross walls removed, forming a continuous tube; lignified walls providing strength/preventing collapse under tension; pits allowing lateral water movement; narrow diameter aiding capillarity/reducing risk of air blockage — any three valid, correctly explained adaptations.

Angiosperms have generally more efficient reproductive strategies than gymnosperms: many have co-evolved with animal pollinators, allowing more targeted and efficient pollen transfer than the largely wind-based pollination relied on by most gymnosperms, and their fruits provide an effective means of seed dispersal by attracting animals. Angiosperms also typically have shorter life cycles from pollination to seed maturity and produce a nutritive endosperm following double fertilisation, improving seedling survival, giving them a competitive advantage that has allowed them to dominate a wider range of habitats than gymnosperms.

1 mark each for any two of: more efficient/co-evolved pollination and seed dispersal mechanisms (e.g. via animals, fruit); faster reproductive cycle; double fertilisation producing a nutritive endosperm, improving seedling survival.

Organisms are classified within the phylum Chordata if they possess, at some stage of their development, the following features: a notochord — a flexible, rod-like structure that provides skeletal support along the length of the body; a dorsal, hollow nerve cord running along the back of the body; pharyngeal slits or pouches in the region of the throat; and a post-anal tail extending beyond the anus.

1 mark each for any three of: notochord present at some life stage; dorsal hollow nerve cord; pharyngeal slits/pouches present at some stage; post-anal tail present at some stage.

(a) A control respirometer, set up identically but without a respiratory substrate (or with dead/boiled yeast), accounts for any change in gas volume caused by factors other than respiration, such as fluctuations in atmospheric temperature or pressure during the investigation; the change recorded in the control can be subtracted from the experimental readings to give a corrected, valid measure of the rate of respiration due to the substrate alone. (b) Yeast has little or no lactase enzyme, so lactose cannot be readily hydrolysed into glucose and galactose for use in glycolysis and respiration; with less usable substrate available, the rate of oxygen consumption (respiration) is much lower than with glucose, which can be respired directly. (c) Temperature should be controlled (e.g. using a water bath), as it affects the rate of enzyme-catalysed reactions in respiration; other valid answers include yeast concentration/volume and pH.

(a) 1 mark: the control accounts for changes in gas volume not due to respiration (e.g. temperature/pressure changes); 1 mark: this allows the change to be subtracted, giving a corrected/valid measure of respiration rate. (b) 1 mark: yeast has little or no lactase enzyme to hydrolyse lactose; 1 mark: so less glucose/usable substrate is available for respiration, lowering O₂ uptake. (c) 1 mark: any valid controlled variable (e.g. temperature, yeast concentration, pH).

(a) Distance from the lamp can be measured precisely and reproducibly using a ruler or metre rule, and because light intensity is inversely proportional to the square of the distance from the source, varying distance provides a controllable, indirect way of varying light intensity without needing a light meter. (b) According to the inverse square law, light intensity is proportional to \( 1/d^2 \), so intensity falls very steeply as distance increases from a small value, but the fall becomes progressively smaller at greater distances; this matches the data, in which the rate of photosynthesis (bubbles per minute) decreases sharply between 10 cm and 30 cm (where light intensity is dropping fastest and is likely the limiting factor), then decreases more gradually between 30 cm and 50 cm. (c) A limitation is that the volume of individual bubbles produced may not be consistent, so counting the number of bubbles does not give an accurate measure of the actual volume of oxygen produced; this could be improved by collecting the gas produced in a set time using a gas syringe or capillary tube and measuring its volume directly.

(a) 1 mark: distance can be measured precisely/reproducibly and relates to light intensity via the inverse square law. (b) 1 mark: light intensity decreases with the square of distance (inverse square law); 1 mark: correct link to the pattern of decreasing rate of photosynthesis with increasing distance (steep decline at short distance, levelling at long distance). (c) 1 mark: valid limitation (e.g. bubble size varies, affecting reliability of the volume estimate); 1 mark: valid improvement (e.g. use a gas syringe/capillary tube to measure the actual volume of O₂ produced).

(a) Since \( A = T = 30.0\% \), and \( A + T + G + C = 100\% \), \( G + C = 100 - 60 = 40\% \); since \( G = C \), \( C = 20.0\% \). (b) Chargaff's rule shows that the percentage of adenine always equals the percentage of thymine, and the percentage of guanine always equals the percentage of cytosine, in any organism's DNA; this equal pairing supports the idea that in the double helix, adenine always pairs specifically with thymine (via two hydrogen bonds) and guanine always pairs specifically with cytosine (via three hydrogen bonds), a fixed complementary base-pairing arrangement, rather than bases pairing randomly. (c) Organism C is likely to have DNA with a higher melting temperature, because it has a higher percentage of guanine-cytosine base pairs (35% G, and therefore 35% C) than organism A; G–C base pairs are held together by three hydrogen bonds, compared with only two hydrogen bonds in an A–T pair, so DNA richer in G–C pairs requires more energy (a higher temperature) to separate the two strands.

(a) 1 mark: 20.0%. (b) 1 mark: equal A=T and G=C proportions support fixed complementary base pairing; 1 mark: correct explanation that A always pairs with T and G always pairs with C. (c) 1 mark: organism C identified; 1 mark: correct explanation — C has more G–C pairs, which form 3 hydrogen bonds (vs 2 for A–T), requiring more energy/a higher temperature to break.

(a) The expected phenotypic ratio for a dihybrid cross between two individuals heterozygous for two independently assorting genes is 9:3:3:1. (b) Since the calculated \( \chi^2 \) value (0.47) is less than the critical value (7.81) at \( p = 0.05 \), the null hypothesis — that there is no significant difference between the observed and expected results — is accepted; the differences between the observed and expected numbers of offspring are due to chance, supporting the conclusion that the two genes assort independently. (c) If the genes for seed shape and seed colour were closely linked on the same chromosome, they would tend to be inherited together rather than independently, and the offspring ratio would be skewed strongly towards the parental combinations (round yellow and wrinkled green), with far fewer recombinant offspring (round green and wrinkled yellow) than expected under independent assortment. Because the observed ratio closely matches the expected 9:3:3:1 ratio (confirmed by the chi-squared test), this indicates that the genes assort independently, consistent with them being located on different chromosomes, or far enough apart on the same chromosome that crossing over occurs frequently enough to produce effectively independent assortment.

(a) 1 mark: 9:3:3:1. (b) 1 mark: calculated value is less than the critical value; 1 mark: null hypothesis accepted — no significant difference between observed and expected, supporting independent assortment. (c) 1 mark: linked genes would be inherited together, producing a ratio skewed towards the parental phenotypes rather than 9:3:3:1; 1 mark: the observed near-9:3:3:1 ratio indicates independent assortment, consistent with the genes being on separate chromosomes (or effectively unlinked).

(a) The xerophyte's thicker cuticle forms a more effective waterproof barrier over the leaf surface, reducing evaporative water loss directly through the cuticle (cuticular transpiration). The xerophyte's much lower stomatal density (20 per mm² compared with 150 per mm² in the mesophyte) means there are far fewer stomata available for water vapour to diffuse out of the leaf, reducing the total water lost through stomatal transpiration. (b) A smaller leaf surface area reduces the total area of the leaf exposed to heat, sunlight and moving air, all of which promote evaporation; by minimising this exposed surface, the xerophyte reduces its overall rate of transpiration and therefore its water loss, which is a significant advantage in a water-scarce (xeric) environment.

(a) 1 mark: thicker cuticle reduces cuticular water loss/evaporation; 1 mark: lower stomatal density reduces water lost via transpiration through stomata. (b) 1 mark: smaller surface area reduces exposure to conditions promoting evaporation (heat/air movement); 1 mark: this reduces the overall rate of transpiration/water loss, an advantage in a water-scarce environment.

(a) Based on the table, Arthropoda can be distinguished from Annelida by the presence of jointed appendages (limbs), which Annelida lack, and by the presence of a hard exoskeleton, which Annelida also lack. (b) Segmentation reflects an underlying, repeated body plan that can indicate shared evolutionary ancestry; although it occurs in more than one phylum, when considered alongside other distinguishing features — such as the presence or absence of jointed appendages and an exoskeleton — it contributes to building a fuller, more reliable picture of the true evolutionary relationships between organisms, rather than being used as a classification criterion in isolation.

(a) 1 mark: jointed appendages present in Arthropoda, absent in Annelida; 1 mark: hard exoskeleton present in Arthropoda, absent in Annelida. (b) 1 mark: segmentation indicates a shared, repeated body plan; 1 mark: classification is more reliable when several features are considered together (not segmentation alone), helping to reveal true evolutionary relationships.

During prophase I of meiosis, crossing over can occur between non-sister chromatids of homologous chromosomes, exchanging sections of DNA and creating new combinations of alleles on each chromatid that were not present on either parental chromosome. During metaphase I and anaphase I, homologous chromosome pairs line up and separate independently of one another (independent assortment), so that the combination of maternal and paternal chromosomes distributed to each gamete is random, producing a very large number of possible chromosome combinations. Random fertilisation then means that any one of the many genetically varied gametes produced by one parent can fuse with any one of the many genetically varied gametes produced by the other parent, further multiplying the number of possible genotype combinations in the resulting offspring.

Level-based 3-band grid (6 marks). Level 1 (1–2 marks): basic idea that meiosis/fertilisation produce variation, with little detail. Level 2 (3–4 marks): identifies crossing over and/or independent assortment with limited explanation. Level 3 (5–6 marks): clear explanation of crossing over (exchange of DNA between non-sister chromatids of homologous chromosomes in prophase I, creating new allele combinations), independent assortment (random orientation/separation of homologous pairs in meiosis I, producing many possible chromosome combinations in gametes), and random fertilisation (any sperm can fuse with any egg, multiplying possible genotype combinations).

Genetic variation within a population arises through mutation, which creates new alleles, and, in sexually reproducing species, through the processes of meiosis (crossing over and independent assortment) and random fertilisation, which generate new combinations of existing alleles. When a selection pressure is present in the environment — such as the presence of an antibiotic, a predator, or a change in climate — individuals whose alleles confer a phenotype better suited to that pressure are more likely to survive and successfully reproduce than individuals without those alleles; this is known as differential survival and reproductive success. Because these better-adapted individuals pass their advantageous alleles on to their offspring, the frequency of the advantageous allele increases within the population's gene pool over successive generations, while less advantageous alleles decrease in frequency; this directional shift in allele frequency over time is the mechanism of evolution by natural selection, and, given enough time and sufficient divergence, can lead to the formation of new species. Strong supporting evidence for this process includes the well-documented rise in antibiotic-resistant bacteria following widespread antibiotic use, in which resistant strains, initially rare, come to dominate a bacterial population because non-resistant strains are selectively killed, directly demonstrating a measurable shift in allele frequency driven by a selection pressure; similarly, industrial melanism in the peppered moth showed a shift in allele frequency for wing colour that correlated closely with changes in predation pressure linked to pollution levels.

Level-based 3-band grid (12 marks). Level 1 (1–4 marks): basic/simplistic account, e.g. 'the strongest survive', with little scientific detail; weak QWC. Level 2 (5–8 marks): explains that variation arises through mutation/recombination and that a selection pressure favours certain phenotypes, with a named example, but limited discussion of allele frequency change over generations; QWC generally sound. Level 3 (9–12 marks): comprehensive, well-structured account covering the source of variation (mutation, and meiosis/recombination in sexual reproduction), identification of a selection pressure and the resulting differential survival and reproductive success of better-adapted individuals, explanation that advantageous alleles increase in frequency across generations, and discussion of a valid named example (e.g. antibiotic resistance, peppered moth) with supporting evidence; high-quality, accurate, logically sequenced prose with correct terminology throughout.

Working near a Bunsen burner flame using aseptic technique, transfer 1 cm³ of the original bacterial culture into a test tube containing 9 cm³ of sterile broth or diluent and mix thoroughly, producing a 10⁻¹ dilution. Using a fresh sterile pipette, transfer 1 cm³ of this 10⁻¹ dilution into a further tube containing 9 cm³ of sterile diluent to produce a 10⁻² dilution. This process is repeated, using a new sterile pipette at each stage to avoid transferring residual, more concentrated culture, until the 10⁻⁵ dilution has been produced.

2 marks: correct serial 1:10 dilution technique described (1 cm³ into 9 cm³ diluent, repeated through to 10⁻⁵) with reference to using a fresh sterile pipette at each stage/aseptic technique; 1 mark awarded if the method is broadly correct but aseptic technique/fresh pipette is not mentioned.

Using a fresh sterile pipette at each dilution stage prevents contamination between samples and avoids carrying over a significant volume of the more concentrated preceding dilution, which would otherwise make the resulting dilution series inaccurate.

1 mark: prevents contamination and/or avoids carry-over of the more concentrated solution, which would make the dilution inaccurate.

An appropriate colour filter is selected, chosen to be a colour complementary to the colour of the test solution, to maximise the sensitivity of absorbance readings. A cuvette filled only with the solvent (or blank solution), containing no coloured solute, is then placed in the colorimeter, and the instrument is set to zero absorbance (100% transmission) using this blank, before any test samples are measured.

1 mark: an appropriate filter is selected (a colour complementary to the solution being tested); 1 mark: the colorimeter is zeroed/calibrated using a blank cuvette of solvent/distilled water before readings are taken.

Different cuvettes may vary slightly in wall thickness or have scratches, both of which can affect the amount of light absorbed independently of the sample's concentration; using the same or matched cuvettes throughout ensures that absorbance readings are directly comparable and any differences reflect only differences in sample concentration.

1 mark: avoids errors caused by variation in cuvette thickness/imperfections, ensuring readings are directly comparable.

The shoot is cut underwater at a slant to prevent air from entering the xylem vessels at the cut surface; the potometer is then assembled with the cut shoot inserted into the apparatus while still underwater, to avoid introducing air locks into the system. All joints in the apparatus are sealed with waterproof (petroleum) jelly to prevent air entering or water leaking out. A single air bubble is introduced into the capillary tube (for example, by briefly lifting the end of the tube out of the water reservoir), and its rate of movement along the graduated scale is timed to give a measure of the rate of water uptake by the shoot.

1 mark for any two of: shoot cut underwater at a slant; apparatus assembled underwater/joints sealed with waterproof jelly to prevent air entry/leaks. 1 mark: an air bubble is introduced into the capillary tube and its movement along the scale timed to measure the rate of water uptake.

It is assumed that the rate of water uptake measured by the potometer is approximately equal to the rate of water loss by transpiration from the leaves, even though a small proportion of the water taken up is actually used in photosynthesis or to maintain cell turgidity rather than being lost through transpiration.

1 mark: rate of water uptake is assumed approximately equal to rate of transpirational water loss (small amounts used in photosynthesis/turgidity are considered negligible).

Two tape measures are laid at right angles along two sides of the field to create a coordinate system; a random number generator (or random number table) is used to generate pairs of coordinates. The quadrat is placed at each randomly generated coordinate, and the percentage cover of the species of interest within the quadrat frame is estimated (for example, using a grid of sub-squares). This is repeated for a sufficient number of quadrats (e.g. at least 20–30) at different random locations to obtain a representative, unbiased sample, and the mean percentage cover is calculated.

1 mark: random coordinates generated (e.g. using random numbers along two measuring tapes) to avoid bias in quadrat placement; 1 mark: percentage cover estimated within the quadrat at each random point, repeated for a sufficient number of samples, with a mean calculated.

Random sampling avoids any conscious or unconscious bias in the placement of quadrats (for example, a tendency to place them in more visually interesting or accessible areas), and ensures that every part of the habitat has an equal chance of being included in the sample, making the resulting data more representative of the whole habitat.

1 mark: avoids observer bias in quadrat placement and/or ensures every part of the habitat has an equal chance of being sampled.

A series of standard solutions of known concentration is prepared, for example by serial dilution of a stock solution of the pigment; the absorbance of each standard is measured using a calibrated colorimeter (with an appropriate filter, zeroed against a blank) and plotted against concentration to produce a calibration curve (line or curve of best fit). The absorbance of the unknown sample is then measured under identical conditions, and its concentration is determined by reading off the calibration curve (or by calculation from the equation of the line) at the corresponding absorbance value.

1 mark: standards of known concentration prepared and their absorbance measured to plot a calibration curve of absorbance vs concentration; 1 mark: the absorbance of the unknown sample is measured under the same conditions and its concentration determined by reference to (interpolation from) the calibration curve.

If conditions such as the filter used, the cuvette, or the calibration of the colorimeter differ between measuring the standards and the unknown sample, the absorbance readings would not be directly comparable, making any concentration read from the calibration curve invalid or inaccurate.

1 mark: ensures readings are directly comparable, so the concentration determined from the calibration curve is valid/accurate.

A small, concentrated spot of leaf pigment extract is applied repeatedly, allowing it to dry between applications, to a pencil line (the origin) drawn near the bottom of a strip of chromatography paper. The paper is then placed in a suitable solvent inside a sealed container, ensuring the solvent level remains below the pencil origin line/pigment spot. As the solvent moves up the paper by capillary action, it carries the different pigments with it at different rates, depending on their relative solubility in the solvent and their adsorption to the paper, separating them into distinct spots. Once the solvent front has moved a suitable distance (and is marked before evaporating), the Rf value of each separated spot is calculated as the distance travelled by the pigment divided by the distance travelled by the solvent front, and each pigment is identified by comparing its Rf value with known reference Rf values.

1 mark: pigment extract spotted on an origin line (pencil, above the solvent level) and the paper run in solvent, pigments separating by capillary action according to solubility; 1 mark: Rf value calculated (distance moved by pigment ÷ distance moved by solvent) and compared with known reference values to identify each pigment.

The Rf value of a substance is defined as the distance it has travelled up the chromatography paper, divided by the distance travelled by the solvent front, measured from the same origin: \( R_f = \dfrac{\text{distance travelled by pigment}}{\text{distance travelled by solvent front}} \).

1 mark: correct definition/formula — distance travelled by pigment ÷ distance travelled by solvent front.

Wearing appropriate personal protective equipment (e.g. gloves) and working on a dissecting board or tray with the specimen securely pinned in place, the operculum (gill cover) is carefully lifted and cut using scissors or a scalpel, cutting away from the body and hands at all times, to expose the gills beneath it. The gill arches, filaments and lamellae are then examined, for example using a hand lens or a dissecting microscope, and their structural features (such as large surface area, thin epithelium and a rich blood supply) relevant to efficient gaseous exchange are noted.

1 mark: appropriate, safe technique described (specimen secured, careful use of scissors/scalpel to open the operculum, cutting away from the body, PPE considered); 1 mark: gills correctly exposed and observed (e.g. using a hand lens/microscope), identifying filaments/lamellae.

When using a scalpel, the cutting motion should always be directed away from the body and hands to avoid injury, and any used blades should be disposed of safely and immediately in a designated sharps container.

1 mark: any valid safety precaution, e.g. cut away from the body/hands; dispose of blades in a sharps bin.

(a) \( \text{Mean} = \dfrac{3.2+3.4+3.0+3.6+3.2+3.4+3.0+3.2+3.4+3.6}{10} = \dfrac{33.0}{10} = 3.3 \) eyepiece units. (b) \( 3.3 \times 2.5 = 8.25 \ \mu\text{m} \). (c) Measuring more cells reduces the impact of any single anomalous or atypical measurement and accounts for natural variation between cells, giving a more representative and reliable estimate of the true mean diameter.

(a) 1 mark: 3.3 eyepiece units (correct working shown). (b) 1 mark: 8.25 μm (ECF from (a)). (c) 1 mark: reduces the impact of anomalies/natural variation, giving a more representative, reliable mean.

(a) \( \text{Rate} = \dfrac{15}{60} = 0.25 \text{ cm}^3\text{s}^{-1} \). (b) The results suggest an optimum temperature of approximately 30°C, since this produced the greatest volume of oxygen in the given time. (c) At temperatures above the optimum, the increased kinetic energy of the molecules disrupts the hydrogen and ionic bonds that maintain the tertiary structure of the enzyme catalase; this changes the shape of the active site so that hydrogen peroxide molecules can no longer bind as effectively (or at all), reducing the rate of reaction and hence the volume of oxygen produced.

(a) 1 mark: 0.25 cm³ s⁻¹ (with working). (b) 1 mark: approximately 30°C. (c) 1 mark: the enzyme denatures at higher temperature — bonds maintaining tertiary structure/active site shape are disrupted, reducing substrate binding and rate of reaction.

(a) \( \text{Concentration} = \dfrac{85}{1 \times 10^{-4}} = 8.5 \times 10^{5} \text{ cells cm}^{-3} \). (b) A source of error is that cells lying on the boundary lines of the counting square may be inconsistently double-counted or missed altogether; this can be reduced by applying a consistent counting rule, such as counting only cells that touch the top and left boundary lines of each square and ignoring those touching the bottom and right lines.

(a) 1 mark: correct method (85 ÷ 0.0001); 1 mark: 8.5 × 10⁵ cells cm⁻³ (850,000 cells cm⁻³). (b) 1 mark: a valid source of error (e.g. boundary cell double-counting) with a valid, matched method of reduction (e.g. a consistent counting rule for boundary cells).

(a) The rate of reaction increases roughly proportionally with substrate concentration at lower concentrations, but the rate of increase slows and the rate eventually levels off (plateaus) at higher substrate concentrations. (b) At high substrate concentrations, the active sites of the fixed number of enzyme molecules present become saturated, meaning nearly all active sites are occupied by substrate at any given moment; adding further substrate cannot increase the rate any further because enzyme concentration, rather than substrate concentration, has become the limiting factor, so the rate reaches a maximum value (Vmax).

(a) 1 mark: rate increases (roughly proportionally) then plateaus/levels off at higher concentration. (b) 1 mark: enzyme active sites become saturated with substrate; 1 mark: enzyme (not substrate) becomes the limiting factor, so the rate cannot increase further, reaching Vmax.

(a) Radius \( = \dfrac{17}{2} = 8.5 \text{ mm} \); Area \( = \pi r^2 = \pi \times 8.5^2 = \pi \times 72.25 = 227 \text{ mm}^2 \) (3 s.f.). (b) The zone of inhibition is the clear area surrounding the disc where no bacterial growth has occurred, showing that the antibiotic, diffusing outward from the disc, has inhibited or killed bacteria within that region; a larger zone of inhibition at a given concentration indicates that the bacteria are more sensitive to (or the antibiotic is more effective against) that particular strain.

(a) 1 mark: correct method (radius = 8.5 mm, area = πr²); 1 mark: 227 mm² (3 s.f., accept 226–227). (b) 1 mark: the zone shows the region where bacterial growth was inhibited by the diffusing antibiotic; a larger zone indicates greater antibiotic effectiveness/bacterial sensitivity.

(a) \( 24 \times 0.02 = 0.48 \text{ cm}^3 \). (b) \( \dfrac{0.48}{5} = 0.096 \text{ cm}^3\text{min}^{-1} \). (c) Soda lime absorbs the carbon dioxide produced by the respiring seeds; without it, the \( \text{CO}_2 \) released would replace, in volume, the \( \text{O}_2 \) consumed, masking any change in gas volume, so soda lime ensures that the only change in gas volume recorded is due to oxygen uptake.

(a) 1 mark: 0.48 cm³. (b) 1 mark: 0.096 cm³ min⁻¹ (ECF from (a)). (c) 1 mark: soda lime absorbs the CO₂ produced, so only the volume change due to O₂ uptake is measured.

(a) Student's t-test is the appropriate statistical test for comparing the means of two independent, normally distributed samples such as these. (b) The null hypothesis states that there is no significant difference between the mean height of Variety X and the mean height of Variety Y (any observed difference is due to chance). (c) Since the calculated test statistic exceeds the critical value at \( p = 0.05 \), the null hypothesis is rejected; there is a statistically significant difference between the mean heights of the two wheat varieties, and the observed difference is unlikely to be due to chance alone.

(a) 1 mark: Student's t-test. (b) 1 mark: correctly stated null hypothesis (no significant difference between the two means). (c) 1 mark: null hypothesis rejected — significant difference between the varieties' mean heights.

(a) \( CV = \dfrac{SD}{\text{mean}} \times 100 = \dfrac{2.4}{15.0} \times 100 = 16.0\% \). (b) The coefficient of variation, being a percentage relative to the mean, allows a fair comparison of the relative variability (spread) between two data sets that have different means or are measured in different units, which cannot be done fairly by comparing raw standard deviation values alone, as standard deviation is affected by the scale of the mean.

(a) 1 mark: correct method (2.4/15.0 × 100); 1 mark: 16.0%. (b) 1 mark: coefficient of variation allows comparison of relative variability between data sets with different means/units, unlike raw standard deviation.

The student should set up a range of buffer solutions at different pH values (e.g. pH 4, 5, 6, 7 and 8), covering a suitable range around the expected optimum for amylase. At each pH, a fixed volume and concentration of amylase solution is mixed with a fixed volume and concentration of starch solution and the appropriate buffer, keeping other variables such as temperature constant across all trials. At regular time intervals, a sample is removed and tested with iodine solution, and the time taken for the iodine to no longer turn blue-black (indicating complete starch digestion) is recorded, or a colorimeter is used to measure the decreasing blue-black colour intensity over time. This procedure is repeated at each pH value (with repeat trials for reliability), and the rate of reaction (1/time) is calculated at each pH, allowing a valid quantitative comparison of enzyme activity across the range of pH values tested.

1 mark: an appropriate range of buffered pH values is used, with other variables (temperature, enzyme/substrate concentration) controlled; 1 mark: a valid method of monitoring reaction progress is described (e.g. iodine test at intervals/colorimetry) with a defined end-point; 1 mark: rate is calculated (e.g. 1/time) and used to allow valid comparison across pH values, with repeats for reliability.

The investigation has low reliability, because taking only a single reading for each plant species provides no way of checking whether the result was reproducible or of identifying and excluding an anomalous reading. In addition, environmental conditions such as light intensity, humidity and air movement may have varied over the course of the single day, and since these were not controlled or monitored throughout the investigation, any such variation could have influenced the results independently of the species being tested; a single reading could also be undetectably affected by an air leak or an irregularity in the potometer. To improve reliability, several repeat readings should be taken for each species (with a mean calculated and any clear anomalies discarded), ideally under controlled or continuously monitored environmental conditions.

1 mark: a single reading gives no indication of reproducibility/cannot identify anomalies, reducing reliability; 1 mark: environmental conditions not controlled/monitored across a single day could introduce uncontrolled variation; 1 mark: a valid improvement is suggested (e.g. multiple repeats per species, mean calculated, environmental conditions controlled/monitored).

Temperature must be controlled, for example by placing a water bath or a heat-absorbing filter between the lamp and the plant to prevent heat from the lamp affecting the water temperature. If temperature were not controlled, moving the lamp closer to the plant to increase light intensity would also raise the temperature of the water surrounding the plant, due to heat radiating from the lamp; because the rate of the enzyme-catalysed reactions of photosynthesis is temperature-dependent, this uncontrolled rise in temperature would independently increase the rate of photosynthesis regardless of light intensity. This would mean any observed increase in bubble rate could not be validly attributed to light intensity alone, since temperature would be acting as a confounding variable, undermining the validity of the investigation.

1 mark: a valid variable is identified (e.g. temperature) with a valid method of control described; 1 mark: explanation that failing to control it (e.g. temperature rising as the lamp moves closer) would independently affect the rate of photosynthesis; 1 mark: explanation that this would confound the investigation, meaning changes in rate could not be validly attributed to light intensity alone.

The student should use an appropriate, consistent and unbiased sampling method at both sites, such as randomly placed quadrats generated using coordinate-based random numbers (or a belt transect if a clear environmental gradient is present). Crucially, the same quadrat size and the same total number of quadrats (the same overall sampling effort) should be used at both sites, to ensure the comparison between them is fair and valid. At each quadrat, the number of individuals of each species present should be identified and recorded, giving the species richness and abundance data required to calculate Simpson's Index of Diversity for each site.

1 mark: a consistent, appropriate, unbiased sampling method is used (e.g. random quadrats) at both sites; 1 mark: the same sampling effort/quadrat size/number of quadrats is used at both sites, ensuring a fair comparison; 1 mark: species richness and abundance of each species are recorded at each quadrat, providing the data required to calculate Simpson's Index.

A pilot study allows the student to identify a suitable range of salt concentrations to use in the main investigation, in particular ensuring that the range includes concentrations that produce both a mass increase and a mass decrease in the potato chips, bracketing the isotonic point at which no net mass change occurs. It also allows practical or methodological issues to be identified and resolved in advance, such as determining an appropriate time period long enough to produce a measurable but not excessive mass change, or ensuring the blotting technique used to remove surface water before weighing is consistent and reliable. A pilot study additionally helps the student determine an appropriate sample size or number of repeats needed at each concentration to obtain reliable results, improving the overall quality and validity of the main investigation.

1 mark: allows an appropriate range of concentrations to be identified/refined (e.g. to bracket the isotonic point); 1 mark: allows practical/methodological issues to be identified and resolved (e.g. suitable time period, consistent technique); 1 mark: helps determine a suitable sample size/number of repeats for reliability in the main investigation.