2024 CCEA GCSE Biology 1010 模擬試題連答案詳解

Plant cells have additional structures beyond those shared with animal cells: a cellulose cell wall (for support), a large permanent vacuole (containing cell sap, for support/storage) and chloroplasts (containing chlorophyll, for photosynthesis). Answer: cellulose cell wall, large permanent vacuole, chloroplasts.

1 mark each for any of: cellulose cell wall, large permanent vacuole, chloroplasts (max 3).

Although both have a cell wall, a bacterial cell wall is not made of cellulose (unlike a plant cell wall). Bacterial cells have no true nucleus — their genetic material lies free in the cytoplasm rather than being enclosed by a nuclear membrane. Bacteria may also contain plasmids, small circular pieces of DNA separate from the main chromosome, which are not found in plant cells. Any two of these differences are creditworthy.

1 mark each for any two valid differences: cell wall not made of cellulose; no true/membrane-bound nucleus (DNA free in cytoplasm); presence of plasmids.

(a) Palisade mesophyll cells are column-shaped, tightly packed and contain many chloroplasts, and are positioned close to the upper surface of the leaf where light intensity is greatest, maximising light absorption for photosynthesis.
(b) Stomata are pores whose size is controlled by guard cells, which can open or close the pore; this regulates the diffusion of carbon dioxide into, and oxygen and water vapour out of, the leaf.
(c) The spongy mesophyll layer contains large, interconnected air spaces between its cells, giving a large internal surface area over which gases (carbon dioxide and oxygen) can diffuse to and from the mesophyll cells.
Answer: palisade cells are chloroplast-rich and near the top surface for light absorption; stomata/guard cells open and close to control gas exchange and water loss; air spaces give a large surface area for gas diffusion.

1 mark for each correctly explained adaptation (a), (b) and (c).

Biuret solution turns lilac/purple in the presence of protein, so this positive result shows the food sample contains protein. Iodine solution turns blue-black in the presence of starch; because no colour change occurred (it remained yellow-brown), this is a negative result, showing the sample does not contain starch. To test for reducing sugar, the student would need to carry out Benedict's test (adding Benedict's solution and heating), which gives a brick-red precipitate if reducing sugar is present. Answer: protein is present; starch is absent; Benedict's test would confirm reducing sugar.

1 mark: protein present (from positive Biuret test). 1 mark: starch absent (from negative iodine test). 1 mark: Benedict's test named.

Digestive enzymes are specific to the substrate they break down: carbohydrase (specifically amylase, for starch) breaks starch down into sugars; protease breaks proteins down into amino acids; lipase breaks fats down into fatty acids and glycerol. Answer: (a) carbohydrase/amylase, (b) protease, (c) lipase.

1 mark each: (a) carbohydrase/amylase, (b) protease, (c) lipase.

The many millions of alveoli give the lungs a very large total surface area for gas exchange. Each alveolus has thin walls, just one cell thick, minimising the diffusion distance for gases. Each alveolus is surrounded by a dense network of capillaries, giving a good blood supply that maintains a steep concentration gradient by constantly bringing blood low in oxygen/high in carbon dioxide and removing blood high in oxygen. The alveoli also have a moist lining, which allows gases to dissolve before diffusing across the exchange surface. Any three of these features are creditworthy.

1 mark each for any three of: large surface area, thin walls/short diffusion distance, good blood supply, moist lining (max 3).

The cornea is the transparent front part of the eye that carries out most of the refraction (bending) of light entering the eye. The iris is a ring of muscle that controls the diameter of the pupil, and therefore the amount of light entering the eye. The retina lines the back of the eye and contains light-sensitive receptor cells, which respond to light by generating nerve impulses that travel to the brain via the optic nerve. Answer: cornea — refracts light; iris — controls amount of light entering (pupil size); retina — detects light/generates nerve impulses.

1 mark each: (a) cornea refracts light, (b) iris controls light entering by changing pupil size, (c) retina detects light/contains light-sensitive cells that generate impulses.

A population is all the organisms of a single species found in a particular area at a given time. A community is made up of all the different populations (of different species) living together and interacting within a particular area. A habitat is the specific place in which an organism lives. Answer as above.

1 mark each for correct definitions of (a) population, (b) community and (c) habitat.

The producer is the organism that makes its own food by photosynthesis — here, the oak tree (leaves). The primary consumer is the first organism to feed on the producer — the caterpillar, which eats the oak leaves. The secondary consumer is the next organism along the chain, which eats the primary consumer — the blue tit, which eats the caterpillar. (The sparrowhawk, which eats the blue tit, would be the tertiary consumer.) Answer: producer = oak leaves; primary consumer = caterpillar; secondary consumer = blue tit.

1 mark each: producer = oak leaves/oak tree; primary consumer = caterpillar; secondary consumer = blue tit.

First convert the image size to micrometres: \( 15\ \text{mm} = 15000\ \mu\text{m} \). Rearranging the equation for the size of the real object: \( \text{size of real object} = \frac{\text{size of image}}{\text{magnification}} = \frac{15000\ \mu\text{m}}{600} = 25\ \mu\text{m} \). Answer: 25 μm.

1 mark: image size correctly converted to μm (15000 μm). 1 mark: correct rearrangement of the magnification equation. 1 mark: correct final answer, 25 μm.

(a) Rate = number of bubbles ÷ time taken: \( \frac{45}{5\ \text{min}} = 9 \) bubbles/minute.
(b) Between 10 and 20 units the rate of bubble production rises steeply (15 to 45 bubbles in 5 minutes), showing light intensity is limiting the rate of photosynthesis over this range. Between 20 and 30 units, the increase is much smaller (45 to 48 bubbles), so the rate levels off; this suggests that, above 20 units, light is no longer the limiting factor, and some other factor — such as carbon dioxide concentration or temperature — has become limiting instead, so further increases in light intensity produce little extra increase in rate.
Answer: rate at 20 units = 9 bubbles/minute; rate rises steeply then levels off, suggesting CO2 concentration or temperature becomes the limiting factor above 20 units.

(a) 1 mark: 9 bubbles/minute. (b) 1 mark: correctly describes the trend (rises steeply, then levels off/rises only slightly). 1 mark: valid alternative limiting factor suggested (CO2 concentration or temperature) with correct reasoning.

Temperature rise = 45 °C − 20 °C = 25 °C. Energy transferred to the water: \( E = 50\ \text{g} \times 4.2 \times 25\ \text{°C} = 5250\ \text{J} \). Energy released per gram of peanut: \( \frac{5250\ \text{J}}{0.5\ \text{g}} = 10500\ \text{J/g} \). (In practice, the true energy content of peanuts is considerably higher than this, because heat is lost to the surroundings rather than all being transferred to the water — this is a typical limitation of this experimental method.) Answer: 10500 J/g.

1 mark: correct temperature rise (25 °C). 1 mark: correct energy transferred to water (5250 J). 1 mark: correctly divides by the mass of peanut burned. 1 mark: correct final answer, 10500 J/g (accept ecf).

(a) Rate = \( \frac{1}{\text{time}} = \frac{1}{40\ \text{s}} = 0.025\ \text{s}^{-1} \).
(b) As temperature rises from 10 °C to 40 °C, the time taken for starch to disappear falls from 300 s to 40 s, i.e. the rate of reaction increases, because both the enzyme and substrate molecules have more kinetic energy and move faster, increasing the frequency of successful collisions between the enzyme's active site and the starch molecules. Above 40 °C, the trend reverses sharply — the time increases again (to 300 s by 60 °C), showing the rate falls. This is because 40 °C is close to the enzyme's optimum temperature; above this, the extra energy breaks the bonds holding the enzyme's tertiary structure together, irreversibly changing the shape of its active site (denaturation), so it can no longer bind starch effectively and catalysis is lost. Answer: rate at 40 °C = 0.025 s⁻¹; rate rises up to 40 °C due to increased collision frequency, then falls sharply above 40 °C as the enzyme denatures.

(a) 1 mark: 0.025 s⁻¹. (b) 1 mark: correctly describes rate increasing up to 40 °C with correct explanation (more kinetic energy/collisions). 1 mark: correctly describes rate decreasing above 40 °C. 1 mark: correct explanation in terms of denaturation/active site shape change.

(a) The data show an inverse relationship: as light intensity increases from 1 to 5 units, pupil diameter steadily decreases, from 7.0 mm to 3.0 mm.
(b) Receptors in the retina detect the increase in light intensity, and nerve impulses trigger a reflex response in the muscles of the iris: the circular muscles contract while the radial muscles relax, reducing the diameter of the pupil. This reduces the amount of light entering the eye, which prevents too much light reaching (and potentially damaging) the light-sensitive cells of the retina. (c) Controlling the amount of light entering the eye in this way prevents excessive light from reaching and damaging the light-sensitive cells of the retina, and helps ensure a clear, well-focused image can still form across a wide range of light conditions.
Answer: pupil diameter decreases as light intensity increases, because a reflex causes the iris's circular muscles to contract, narrowing the pupil to limit the amount of light entering the eye and protect the retina.

(a) 1 mark: correctly describes the inverse relationship (pupil diameter decreases as light intensity increases). (b) 1 mark: circular muscles of the iris contract (radial muscles relax), making the pupil smaller. 1 mark: correct reasoning that this reduces light entering the eye/protects the retina. (c) 1 mark: valid reason, e.g. prevents damage to the retina from too much light, or helps maintain clear vision across different light levels.

A diet that is high in saturated fat and cholesterol encourages the build-up of fatty deposits on the inner walls of arteries, a process that narrows the arteries and restricts blood flow through them. If this happens in the coronary arteries, which supply the heart muscle itself with oxygenated blood, the heart muscle may not receive enough oxygen, increasing the risk of a heart attack. A diet high in salt tends to raise blood pressure, which places additional strain on the heart and blood vessels and further increases the risk of cardiovascular disease. Reducing the amount of saturated fat and salt eaten (for example, by choosing lower-fat foods, or eating more fibre, fruit and vegetables) can help lower this risk. Answer: high saturated fat/cholesterol narrows arteries via fatty deposits, restricting blood flow to the heart and increasing heart attack risk; high salt raises blood pressure; reducing saturated fat and salt intake lowers this risk.

1 mark: saturated fat/cholesterol causes fatty deposits to build up, narrowing arteries. 1 mark: correct consequence — restricted blood flow to the heart, increasing heart attack risk. 1 mark: salt linked to raised blood pressure (or another valid additional risk factor). 1 mark: valid dietary change that would reduce risk.

The lock-and-key model describes how an enzyme's active site has a specific three-dimensional shape that is complementary to the shape of one particular substrate molecule (or a small group of very similar substrates) — just as a key must have the right shape to fit a specific lock. Because only a substrate with the matching shape can bind to the active site, each enzyme is specific and can only catalyse the breakdown (or formation) of that particular substrate, not other molecules.

When an enzyme is heated well above its optimum temperature, the additional kinetic energy causes the atoms within the enzyme molecule to vibrate more vigorously, which breaks the weak bonds holding the enzyme's tertiary (3D) structure in shape. This causes the active site to change shape — the enzyme is denatured. Because the substrate can no longer fit into the altered active site (like trying to fit a key into a lock that has been bent out of shape), the enzyme can no longer catalyse the reaction, and its activity falls; this loss of activity is permanent/irreversible on cooling.

Answer: an enzyme's active site shape is complementary to one specific substrate (lock and key), explaining its specificity; heating well above the optimum denatures the enzyme, permanently changing the active site's shape so the substrate can no longer bind, reducing enzyme activity.

1 mark: active site has a shape complementary to a specific substrate. 1 mark: correct link — only the matching substrate can bind, giving specificity. 1 mark: excess heat breaks bonds maintaining the enzyme's shape/structure, changing the active site's shape (denaturation). 1 mark: correct consequence — substrate can no longer bind, so activity is lost/decreases.

After eating a meal, glucose absorbed from digested food raises the concentration of glucose in the blood above its normal level. This rise is detected, and the pancreas responds by releasing more insulin into the blood. Insulin travels in the blood to target cells, particularly in the liver and muscles, and causes these cells to take up more glucose from the blood; the glucose taken up is converted into glycogen for storage in the liver and muscles (and some is used directly in respiration). As glucose is removed from the blood in this way, blood glucose concentration falls back towards its normal level. Once it returns to normal, less insulin is released. This is an example of negative feedback: a change away from the normal level (a rise in blood glucose) is detected, and triggers a response (insulin release and glucose uptake) that acts to reverse the change and restore the normal level, rather than amplifying it further.
Answer: rising blood glucose triggers insulin release, causing cells to take up glucose (stored as glycogen), lowering blood glucose back to normal — this reversal of the original change is negative feedback.

1 mark: rise in blood glucose is detected, causing the pancreas to release more insulin. 1 mark: insulin causes cells (liver/muscle) to take up glucose from the blood. 1 mark: glucose is converted to/stored as glycogen (or used in respiration), lowering blood glucose. 1 mark: correct explanation/use of the term negative feedback (change is detected and reversed).

During exercise, muscles contract more frequently and more forcefully, requiring more energy; this energy is released by an increased rate of (aerobic) respiration in the muscle cells, which uses up oxygen faster and produces carbon dioxide faster than at rest. The resulting change in blood gas levels (a fall in oxygen and/or rise in carbon dioxide) is detected, triggering an increase in both the rate and depth of breathing. Breathing faster and more deeply increases the volume of air exchanged with the lungs each minute, increasing the rate of gas exchange across the alveoli: more oxygen diffuses into the blood to be delivered to the working muscles for respiration, and more carbon dioxide diffuses out of the blood to be exhaled, preventing a harmful build-up of carbon dioxide (which would otherwise lower blood pH). Answer: exercising muscles respire faster, using more oxygen and producing more carbon dioxide; breathing rate and depth increase to supply the extra oxygen needed and remove the extra carbon dioxide produced.

1 mark: exercising muscles respire faster, using more oxygen/producing more carbon dioxide. 1 mark: breathing rate and depth both increase. 1 mark: this increases the rate of gas exchange, supplying more oxygen to the muscles. 1 mark: correctly explains that this also removes carbon dioxide faster, preventing its build-up.

Only a fraction of the energy present in the organisms at one trophic level is transferred to and becomes new biomass in the organisms at the next trophic level. Energy is lost at each stage in several ways: much is released as heat during respiration; some is lost in egestion, as undigested material passed out as faeces; some is lost in excretion, as waste products removed from the body; and some remains in parts of the organism that the next consumer does not eat, such as bones, roots or fur. As a result of these combined losses, typically only around 10% of the energy present at one trophic level is transferred into new biomass at the next trophic level — the efficiency of energy transfer between trophic levels is low. Because so much energy is lost at every step, the total amount of energy available falls very rapidly as you move up a food chain; after four or five trophic levels, so little energy remains that there is not enough to support a viable population of organisms at a further trophic level, which is why food chains are rarely longer than this.
Answer: only around 10% of the energy at each trophic level is transferred to the next, due to losses as heat (respiration), egestion, excretion and uneaten parts, so too little energy remains to support more than about four or five trophic levels.

1 mark: energy is lost as heat during respiration. 1 mark: energy is lost in egestion (undigested material/faeces). 1 mark: energy is lost in excretion and/or uneaten structures. 1 mark: correct statement that only around 10% of energy is transferred to the next trophic level (efficiency is low). 1 mark: correct conclusion that this limits food chains to about four or five trophic levels, since too little energy remains further up the chain.

As the number of individuals in a population increases, the demand placed on the limited resources available in the habitat also increases. For animals, these resources include food, water, territory and mates; for plants, they include water, light, space and mineral ions. Because these resources are finite, individuals must compete with one another to obtain enough of them. As competition intensifies, some individuals will be less successful in obtaining sufficient resources; they may be more likely to die (for example, from starvation) or fail to breed successfully, reducing the population's overall rate of survival and reproduction. This acts to slow down or limit further growth of the population, eventually stabilising it at a level the habitat's resources can support.

For example, as a population of rabbits in a field grows, the amount of grass available per rabbit decreases; competition for the remaining grass increases, and rabbits that are less successful at finding enough food are more likely to starve or have too little energy to reproduce successfully, limiting further increase in the rabbit population.

Answer: a growing population increases competition for limited resources (e.g. food, space); individuals that lose this competition are more likely to die or fail to reproduce, limiting further population growth — e.g. rabbits competing for a limited supply of grass as their population grows.

1 mark: growing population increases demand for a limited resource. 1 mark: this increases competition between individuals. 1 mark: less successful individuals are more likely to die/fail to reproduce. 1 mark: valid, clearly explained example (e.g. rabbits competing for grass) correctly linked to limiting population growth.

The pancreas continuously monitors the concentration of glucose in the blood. If blood glucose concentration rises above its normal level, for example after a meal, the pancreas responds by releasing more of the hormone insulin into the blood. Insulin travels to target cells, particularly in the liver and muscles, and causes them to take up glucose from the blood and convert it into glycogen for storage; some of the glucose taken up is also used directly in respiration. This removes glucose from the blood, bringing its concentration back down towards normal.

If blood glucose concentration falls below normal, for example between meals or during exercise, the pancreas instead releases the hormone glucagon. Glucagon travels to the liver and causes stored glycogen to be broken down back into glucose, which is then released into the blood, raising blood glucose concentration back towards normal.

This control system is an example of negative feedback: in each case, a change in blood glucose concentration away from the normal level is detected, and this triggers a hormonal response (insulin or glucagon) that acts to reverse the change and return the concentration to normal, rather than allowing it to drift further away from normal.

Maintaining a constant blood glucose concentration matters for two main reasons. First, cells throughout the body, especially in the brain, rely on a continuous and stable supply of glucose as a respiratory substrate to release the energy they need; large swings in glucose availability could starve cells of energy or (if too high) draw water out of cells osmotically. Second, the enzymes that control the body's metabolic reactions only function efficiently within a narrow range of internal conditions; large, uncontrolled fluctuations in blood glucose (and the related changes in cell water content) can disrupt the precise three-dimensional shape enzymes need to bind their substrates, reducing the efficiency of essential metabolic reactions throughout the body.

Answer: the pancreas releases insulin when glucose is high (causing cells to remove/store glucose) and glucagon when glucose is low (causing the liver to release stored glucose); because each response reverses the original change, this is negative feedback; stable blood glucose matters because cells (e.g. in the brain) need a steady glucose supply, and enzymes require stable internal conditions to keep their active site functional.

Band A (5-6 marks): accurate, detailed and logically sequenced account covering detection and correction of both a rise (insulin, glucose uptake/storage as glycogen) and a fall (glucagon, glycogen breakdown), a correct explanation of negative feedback, and a clear, accurate link between stable glucose and both cell function (respiration) and enzyme activity (active site shape). Wide and accurate use of specialist terms (e.g. glucagon, glycogen, negative feedback) with few SPG errors.
Band B (3-4 marks): reasonably accurate account of at least one correction mechanism (insulin or glucagon) and a partial or correct explanation of negative feedback; some reference to why constant glucose matters, though possibly generic (e.g. 'cells need energy') rather than linked clearly to enzymes; some errors, generally competent scientific language.
Band C (1-2 marks): basic, fragmented statements, e.g. 'insulin controls blood sugar' or 'the body keeps things constant', with little development or accurate terminology.
Band D (0 marks): no relevant content / not creditworthy.

A gene is a length of DNA, located at a particular position on a chromosome, that acts as a functional unit controlling a characteristic. An allele is one particular version of that gene — different alleles of the same gene can produce different versions of the characteristic (e.g. different alleles of an eye-colour gene). Answer: gene = section of DNA/chromosome controlling a characteristic; allele = one version/form of a gene.

1 mark: correct definition of gene. 1 mark: correct definition of allele, distinguishing it from a gene.

Cardiovascular disease risk is increased by several lifestyle factors, including smoking (tar, nicotine and carbon monoxide all contribute), a poor diet high in saturated fat and/or salt, lack of physical exercise, high stress levels and excess alcohol consumption. Any three of these are creditworthy.

1 mark each for any three valid lifestyle factors (max 3): smoking, poor diet/saturated fat/salt, lack of exercise, stress, excess alcohol.

Osmosis is a special case of diffusion that applies specifically to water molecules, moving down their own concentration gradient (from a region of higher water/more dilute solution to a region of lower water/more concentrated solution), and requires the two solutions to be separated by a selectively (partially) permeable membrane. Answer: diffusion of water from a dilute to a more concentrated solution through a selectively permeable membrane.

1 mark: diffusion of water molecules down a concentration gradient (dilute to concentrated). 1 mark: through a selectively/partially permeable membrane.

Arteries carry blood under high pressure, so they have thick walls containing more muscle and elastic fibres, able to withstand and maintain this pressure, and a relatively narrow lumen (central channel). Veins carry blood under much lower pressure, so their walls are thinner and contain less muscle/elastic tissue, and they have a wider lumen; because blood flows through veins under low pressure, veins also contain valves at intervals to prevent blood from flowing backwards. Answer: arteries have thicker, more muscular/elastic walls, a narrower lumen, and (usually) no valves, compared with veins.

1 mark each for any three of: arteries have thicker walls/more muscle and elastic fibres; arteries have a narrower lumen; veins have valves (arteries do not).

Red blood cells contain haemoglobin, which binds oxygen, so their function is oxygen transport around the body. White blood cells are involved in defending the body against disease-causing microorganisms (e.g. by engulfing pathogens or producing antibodies). Platelets play a role in blood clotting, converting the soluble protein fibrinogen into insoluble fibrin, which forms a mesh that traps blood cells and forms a clot/scab. Answer: (a) oxygen transport, (b) defence against disease, (c) blood clotting.

1 mark each: (a) oxygen transport, (b) defence against disease, (c) blood clotting (fibrinogen to fibrin).

The biconcave (dimpled disc) shape of a red blood cell increases its surface area relative to its volume, allowing oxygen to diffuse into and out of the cell more efficiently. Red blood cells lack a nucleus, leaving more internal space to be packed with haemoglobin. Haemoglobin, an iron-containing protein, binds reversibly to oxygen, allowing the cell to carry oxygen from the lungs to respiring tissues. Answer: biconcave shape, absence of nucleus, presence of haemoglobin (containing iron).

1 mark each for any three of: biconcave shape (large surface area), absence of a nucleus (more room for haemoglobin), contains haemoglobin/iron for binding oxygen.

In double circulation, blood passes through the heart twice for each full circuit of the body. Deoxygenated blood is pumped from the heart along the pulmonary artery to the lungs, where it picks up oxygen and releases carbon dioxide; the now oxygenated blood returns to the heart via the pulmonary vein. The heart then pumps this oxygenated blood out along the aorta to the rest of the body, where oxygen is delivered to respiring tissues and carbon dioxide is picked up; the deoxygenated blood returns to the heart via the vena cava, completing the circuit. Answer: heart -> pulmonary artery -> lungs -> pulmonary vein -> heart -> aorta -> body -> vena cava -> heart.

1 mark: correct pulmonary circuit (heart -> lungs -> heart, via pulmonary artery/vein). 1 mark: correct systemic circuit (heart -> body -> heart, via aorta/vena cava). 1 mark: correct overall sequence showing blood passes through the heart twice (double circulation).

Fertility treatments for infertility include: giving hormones to stimulate a woman's ovaries to produce multiple eggs (ova) in one cycle; in vitro fertilisation (IVF), in which eggs are collected and fertilised by sperm outside the body, in a laboratory dish; and the transfer of several fertilised embryos into the uterus, to increase the chance that at least one successfully implants. Answer: hormone stimulation of multiple ova; in vitro fertilisation (IVF); transfer of several embryos into the uterus.

1 mark each for any three of: hormone treatment to produce multiple ova; in vitro fertilisation (IVF); transfer of several embryos into the uterus.

Continuous variation occurs when a characteristic can take any value across a range, such as height; this is normally displayed using a histogram. Discontinuous variation occurs when a characteristic falls into a small number of distinct categories, such as the ability (or inability) to roll the tongue, or left/right hand dominance; this is normally displayed using a bar chart. Answer: continuous — height (histogram); discontinuous — tongue-rolling/hand dominance (bar chart).

1 mark: valid example of continuous variation. 1 mark: valid example of discontinuous variation. 1 mark: both correct graph types identified (histogram for continuous, bar chart for discontinuous).

Variation between individuals of the same species can arise from genetic causes — differences in the alleles inherited from parents — and from environmental causes, such as differences in diet, climate, exercise or other lifestyle factors experienced during life. For most characteristics in humans (e.g. body mass, exam performance), variation results from a combination of both genetic and environmental influences acting together, rather than either cause alone. Answer: genetic and environmental variation; most human characteristics are affected by a combination of both.

1 mark: genetic variation. 1 mark: environmental variation. 1 mark: correctly states that most characteristics result from a combination of both genetic and environmental factors.

(a) A child develops Huntington's disease if they inherit at least one copy of the dominant H allele; from this cross, the only genotype possible that includes H is Hh (since the mother can only contribute h). So an affected child has genotype Hh.
(b) The father (Hh) produces gametes H and h; the mother (hh) produces only gametes h.

H h
h Hh hh
h Hh hh

This gives possible offspring genotypes Hh, Hh, hh, hh — a 1:1 ratio of Hh to hh.
(c) Two of the four possible outcomes are Hh (affected, since H is dominant); the probability of an affected child is 2/4 = 1/2 = 50%.
(d) Because Huntington's disease is caused by a dominant allele, only one copy of the allele (as in genotype Hh) is enough to cause the condition — there is no need to inherit two copies, unlike a recessive condition. This means a parent who is heterozygous (Hh) — and therefore themselves affected, since H is dominant — can pass the allele on to any of their children; on average, half of their children will inherit the H allele and develop the disease, even if their partner does not carry the allele at all, because only one parent needs to supply the dominant allele for it to be expressed in the child.
Answer: affected child's genotype is Hh; offspring ratio Hh:hh is 1:1; probability of the disease = 50%; the condition can appear from just one carrying parent because only one copy of the dominant allele is needed to cause it.

(a) 1 mark: Hh.
(b) 1 mark: correct gametes shown (H, h from father; h, h from mother). 1 mark: correct offspring genotypes Hh, Hh, hh, hh shown.
(c) 1 mark: 50% (accept 1/2, 1 in 2); accept ecf from an internally consistent Punnett square.
(d) 1 mark: correct explanation that only one copy of a dominant allele is needed to cause the condition, so a heterozygous parent can pass it on to a child regardless of the other parent's genotype.

(a) For their child to have genotype ff, each parent must have contributed one f allele, so both parents must carry at least one copy of f. However, neither parent has cystic fibrosis themselves, so each must also carry the dominant allele F (to mask the effect of f). Therefore both parents must be heterozygous carriers, genotype Ff.
(b) Crossing Ff x Ff:

F f
F FF Ff
f Ff ff

This gives offspring genotypes FF, Ff, Ff, ff. Only the ff genotype has cystic fibrosis, so the probability for this (or any) child of this couple is 1/4 = 25%. This probability applies independently to each pregnancy — it is not affected by the genotype of the first child.
Answer: both parents are Ff (carriers); the probability that the second child has cystic fibrosis is 25%, independent of the first child's genotype.

(a) 1 mark: both Ff. 1 mark: correct reasoning (must carry f to have an affected child, but must also carry F since unaffected).
(b) 1 mark: correct Punnett square/genotypes (FF, Ff, Ff, ff). 1 mark: 25% (accept 1 in 4), with recognition that this probability is independent for each pregnancy.

(a) The mother (XHXh) produces gametes XH and Xh; the father (XHY) produces gametes XH and Y.

XH Xh
XH XHXH XHXh
Y XHY XhY

This gives possible offspring: XHXH (unaffected daughter), XHXh (unaffected carrier daughter), XHY (unaffected son) and XhY (affected son).
(b) Of the two possible genotypes for sons (XHY and XhY), one (XhY) has haemophilia; the probability that a son has haemophilia is therefore 1/2 = 50%.
(c) Because the allele for haemophilia is carried on the X chromosome, and males have only one X chromosome (plus a Y chromosome, which does not carry a corresponding allele), a male needs only one copy of the recessive allele to be affected — there is no second X chromosome present to carry a dominant, masking allele. Females, however, have two X chromosomes, so they need to inherit the recessive allele on both X chromosomes (making them homozygous recessive) to be affected; if they inherit just one copy, the other X chromosome's dominant allele masks it and they are an unaffected carrier. Because two copies are much less likely to be inherited together than one, haemophilia is much more common in males than in females.
Answer: daughters are XHXH or XHXh (both unaffected); sons are XHY (unaffected) or XhY (affected); probability a son is affected = 50%; haemophilia is more common in males because they need only one copy of the recessive allele (having only one X chromosome), while females need two.

(a) 1 mark: correct gametes from each parent. 1 mark: all four correct offspring genotypes shown (XHXH, XHXh, XHY, XhY).
(b) 1 mark: 50% (accept 1 in 2), based on sons only.
(c) 1 mark: correct explanation referring to males having only one X chromosome (so one recessive allele is sufficient), compared with females needing two copies.

(a) Because Huntington's disease is dominant, an unaffected person cannot carry the H allele, so individual 2 must be homozygous recessive: hh.
(b) Individual 1 is affected, so his genotype is either HH or Hh. However, individual 4 (his son) is unaffected — genotype hh. If individual 1 were HH, every one of his children would receive an H allele from him and, since H is dominant, would be affected; this would make it impossible for individual 4 to be unaffected. Therefore individual 1 cannot be HH, and must instead be heterozygous, Hh — this allows him to (by chance) pass on his recessive h allele to individual 4, who also receives h from the unaffected mother (hh), giving individual 4 the unaffected genotype hh.
(c) Since individual 1 is Hh and individual 2 is hh, the only possible offspring genotypes are Hh or hh (an HH child is not possible, as the mother can only ever contribute h). Individual 3 is affected, so she cannot be hh; her genotype must therefore be Hh.
Answer: individual 2 is hh; individual 1 is Hh (proven by individual 4 being unaffected, which would be impossible if individual 1 were HH); individual 3 is Hh.

(a) 1 mark: hh.
(b) 1 mark: Hh. 1 mark: correct reasoning using individual 4's unaffected phenotype to rule out HH for individual 1.
(c) 1 mark: Hh, with reasoning (or clearly derived from the cross Hh x hh) that HH is not possible from this cross.

Statins are drugs that reduce the concentration of cholesterol in the blood (by reducing its production in the liver); since cholesterol contributes to the fatty deposits that build up in artery walls, taking statins slows this build-up and reduces the risk of arteries narrowing or becoming blocked. Aspirin reduces the tendency of blood to clot (it has an anti-clotting/anti-platelet effect); this lowers the risk that a blood clot will form on a narrowed, fatty-deposit-lined artery and block it completely, which could otherwise trigger a heart attack or stroke.

Smoking increases cardiovascular disease risk in several ways: nicotine increases heart rate and raises blood pressure, putting extra strain on the heart and blood vessels, while carbon monoxide binds to haemoglobin in place of oxygen, reducing the blood's oxygen-carrying capacity and forcing the heart to work harder to deliver enough oxygen to tissues. Lack of exercise contributes to obesity and means the heart muscle is not regularly strengthened by exercise, both of which increase blood pressure and the likelihood of fatty deposits building up in the arteries.

Answer: statins lower blood cholesterol, slowing fatty deposit build-up; aspirin reduces blood clotting, lowering the risk of a blocked artery; smoking raises heart rate/blood pressure and reduces blood oxygen capacity, while lack of exercise contributes to obesity and a weaker heart, both increasing cardiovascular disease risk.

1 mark: statins reduce blood cholesterol, slowing fatty deposit build-up in arteries. 1 mark: aspirin reduces blood clotting, lowering the risk of a blocked artery. 1 mark: correct explanation of how smoking increases risk (nicotine raises heart rate/blood pressure, or CO reduces oxygen-carrying capacity). 1 mark: correct explanation of how lack of exercise increases risk (obesity/weaker heart muscle). 1 mark: for a complete, clearly reasoned answer covering both treatments and both lifestyle factors.

A heart attack occurs when a blood vessel supplying the heart muscle itself (a coronary artery) becomes blocked — often by a blood clot forming where fatty deposits have narrowed the artery. This cuts off the supply of oxygenated blood to the region of heart muscle beyond the blockage; without oxygen, the heart muscle cells in that region cannot respire aerobically and die, reducing the heart's ability to pump blood effectively (and potentially causing the heart to stop working properly).

A stroke occurs when a blood vessel supplying part of the brain becomes blocked (or, in some cases, bursts). This cuts off the supply of oxygenated blood to the affected region of the brain; the brain cells in that region are deprived of oxygen and die. Because different parts of the brain control different functions (such as movement, speech or memory), the death of brain cells in the affected region results in a loss or impairment of whichever function(s) that part of the brain controlled.

Answer: a heart attack is caused by a blocked coronary artery, cutting off oxygen to heart muscle, causing muscle cells to die and reducing pumping ability; a stroke is caused by a blocked (or burst) blood vessel in the brain, cutting off oxygen and causing brain cells to die, resulting in loss of the functions that part of the brain controlled.

1 mark: heart attack — blockage of a coronary artery cuts off blood/oxygen supply to heart muscle. 1 mark: heart muscle cells die, reducing the heart's pumping ability. 1 mark: stroke — blockage (or bursting) of a blood vessel supplying the brain cuts off blood/oxygen supply. 1 mark: brain cells in the affected area die. 1 mark: correctly links this to loss/reduction of the function controlled by that part of the brain.

Cancer results from the uncontrolled division of abnormal cells, which can form a tumour. A benign tumour remains localised — it is typically encapsulated (contained within a membrane) and does not spread to other parts of the body, so although it can still cause problems (e.g. by pressing on nearby organs), it is generally less dangerous and easier to treat, often by surgical removal. A malignant tumour, by contrast, is capable of spreading (metastasis): cells can break away from the original tumour and travel via the blood or lymphatic system to establish secondary tumours in other parts of the body, making the cancer much more difficult to treat and more dangerous.

Screening programmes (for example for breast, cervical, testicular or skin cancer) aim to detect cancers at an early stage, often before any symptoms are noticeable to the patient. Detecting a cancer while the tumour is still small and has not yet had the chance to spread means that treatment (such as surgical removal, or more localised radiotherapy) is much more likely to remove or destroy all of the cancerous cells successfully. If a cancer is instead only detected once it has grown large or already spread (metastasised) to other organs, it is much harder to treat completely, so survival rates are lower. Early detection through screening therefore significantly improves the chances of successful treatment and survival.

Answer: benign tumours stay localised/encapsulated and don't spread; malignant tumours can spread (metastasise) to form secondary tumours, which is far more dangerous; screening detects cancer early, when it is small and localised, making treatment much more likely to succeed and improving survival rates.

1 mark: benign — remains localised/encapsulated, does not spread. 1 mark: malignant — capable of spreading (metastasis) to form secondary tumours. 1 mark: screening detects cancer at an early stage/before symptoms appear. 1 mark: a small, localised (unspread) tumour is more likely to be successfully treated. 1 mark: for a clear, correctly reasoned overall link between early detection and improved survival rate.

Within any large population of bacteria there is natural variation, arising from random mutations; by chance, a small number of bacteria may already carry a mutation that gives them some resistance to a particular antibiotic. When that antibiotic is used — especially if it is overused, or a course is not completed — the non-resistant bacteria in the population are killed, but the resistant bacteria survive, since they are less affected by the antibiotic; with less competition for resources, these survivors reproduce successfully, and because resistance is genetically determined, they pass the resistance allele/gene on to their offspring. Over many generations of bacterial reproduction (which can be very rapid) and repeated antibiotic use, the proportion of resistant bacteria in the population increases — this is natural selection acting on the bacterial population, favouring resistant individuals. Eventually, a population (or strain) that is highly resistant, such as MRSA, can dominate.

This is a growing public health concern because infections caused by antibiotic-resistant bacteria are much harder — and in some cases currently impossible — to treat effectively with existing antibiotics, increasing the risk of serious illness, prolonged hospital stays and death; resistant strains can also spread between patients, particularly in hospital settings, and relatively few new antibiotics are currently being developed to replace those that are becoming ineffective.

Answer: random mutation gives some bacteria resistance; antibiotic use kills non-resistant bacteria but resistant bacteria survive and reproduce (natural selection), so resistance becomes more common over generations, especially with overuse; resulting resistant strains (e.g. MRSA) are much harder to treat, threatening public health.

1 mark: random mutation gives some bacteria in the population natural resistance. 1 mark: antibiotic exposure kills non-resistant bacteria, but resistant bacteria survive and reproduce. 1 mark: resistance allele is passed to offspring, so resistant bacteria become more common over successive generations (natural selection). 1 mark: correctly links repeated/incomplete antibiotic use to accelerating this process. 1 mark: correct explanation of the public health concern (infections harder to treat, risk of serious illness/spread).

Osmosis is the diffusion of water molecules from a dilute solution to a more concentrated solution, through a selectively permeable membrane. When a plant cell is placed in a concentrated sugar solution — more concentrated than the solution inside the cell — water moves out of the cell, across the cell membrane, by osmosis, down the water concentration gradient. As the cell loses water, its cytoplasm and vacuole shrink, and the cell membrane pulls away from the rigid cellulose cell wall; this is called plasmolysis. The cell wall itself keeps its original shape and does not collapse, because it is rigid and not affected by the loss of water from the cytoplasm.

An animal cell, such as a red blood cell, has a cell membrane but no cell wall. When it is placed in the same concentrated solution, water similarly moves out of the cell by osmosis, but because there is no rigid outer wall for the shrinking membrane to pull away from, the whole cell instead shrinks in size and its surface becomes shrivelled/crenated, rather than showing the distinct membrane-pulled-away-from-wall appearance of plasmolysis.

Answer: water leaves both cells by osmosis into the concentrated solution; the plant cell's membrane pulls away from its rigid cell wall (plasmolysis), while the animal cell, lacking a wall, simply shrinks and shrivels.

1 mark: correct definition of osmosis (water moves from dilute to concentrated solution, through a selectively permeable membrane). 1 mark: water moves out of the plant cell by osmosis into the concentrated solution. 1 mark: the cell membrane pulls away from the rigid cell wall — this is plasmolysis. 1 mark: the animal cell has no cell wall, so it simply shrinks/shrivels instead. 1 mark: for a clear, complete comparison of both cell types.

Water is absorbed from the soil into root hair cells, which have a large surface area, by osmosis (since the soil water is more dilute than the cell contents). This water then passes into the xylem, a tissue made of long, hollow tubes formed from dead cells joined end to end with no internal cell contents obstructing flow, which offers very little resistance to the upward movement of water and dissolved minerals.

In the leaves, water evaporates from the moist surfaces of the mesophyll cells into the internal air spaces of the leaf, and then diffuses out of the leaf through the stomata into the surrounding air — this loss of water vapour from the plant is called transpiration. As water is continually lost from the leaf in this way, it creates a 'pull' on the water remaining in the xylem: because water molecules are attracted to one another (cohesion), they form a continuous, unbroken column within the narrow xylem vessels, so as water is lost at the top, more water is drawn up from below to replace it, all the way from the roots. This continuous, one-way flow of water from the roots, through the xylem, and out through the leaves is called the transpiration stream.

Answer: root hairs absorb water by osmosis into the xylem, whose hollow, dead-cell structure allows easy upward flow; evaporation and diffusion of water out through the leaf stomata (transpiration) pulls a continuous column of water up through the xylem, via cohesion between water molecules, from the roots to the leaves.

1 mark: root hair cells absorb water from the soil by osmosis (large surface area). 1 mark: xylem structure (hollow tubes of dead cells) allows water to flow up with little resistance. 1 mark: water evaporates from mesophyll cells and diffuses out through stomata (transpiration). 1 mark: this water loss draws more water up through the xylem, via cohesion between water molecules. 1 mark: for a clear, complete description of the transpiration stream from roots to leaves.

Transpiration is the loss of water vapour from a leaf by evaporation (from mesophyll cell surfaces) followed by diffusion out through the stomata; like any diffusion process, its rate depends on the concentration gradient and the surface area available.

A leaf with a larger surface area has more mesophyll cell surface from which water can evaporate, and more stomata through which water vapour can diffuse out, so a greater surface area increases the rate of transpiration.

Higher wind speed continually moves the more humid air immediately surrounding the leaf away, replacing it with drier air; this maintains a steeper concentration gradient of water vapour between the air spaces inside the leaf and the air immediately outside it, which increases the rate of diffusion of water vapour out of the leaf, and therefore increases the rate of transpiration.

Higher humidity means the air surrounding the leaf already contains a relatively high concentration of water vapour; this reduces the concentration gradient between the (still saturated) air spaces inside the leaf and the air outside, which reduces the rate of diffusion of water vapour out of the leaf, and so decreases the rate of transpiration.

Answer: greater surface area increases transpiration (more area for evaporation/diffusion); higher wind speed increases transpiration (maintains a steep water vapour concentration gradient); higher humidity decreases transpiration (reduces the concentration gradient).

1 mark: greater surface area increases the rate (larger area for evaporation/diffusion). 1 mark: higher wind speed increases the rate (removes water vapour, maintaining a steep concentration gradient). 1 mark: higher humidity decreases the rate (reduces the concentration gradient). 1 mark: correct, consistent use of 'diffusion/concentration gradient' reasoning across at least two factors. 1 mark: for a complete, clearly explained answer covering all three factors.

Before watering, a wilted plant's cells have lost water and are flaccid (not turgid), so they no longer provide enough support to hold the plant upright. When the plant is watered, the water surrounding the root hair cells becomes more dilute than the contents of the root hair cells, so water moves into the root hair cells (and subsequently into other plant cells, including those in the leaves) by osmosis, from a region of higher water concentration to a region of lower water concentration, across the selectively permeable cell membrane. As each cell absorbs water, its vacuole swells, and the cell contents push outward against the cellulose cell wall; because the cell wall is relatively rigid and does not stretch much, it resists this outward pressure, and the cell becomes firm — this state is called turgor. Many turgid cells pressing against one another throughout the plant's tissues provide the mechanical support needed to hold the stem and leaves upright, so the plant recovers its firm, unwilted form.
Answer: watering allows water to enter root hair (and leaf) cells by osmosis; as vacuoles swell, cell contents press against the cell wall, making cells turgid; turgid cells support the plant, so it recovers from wilting.

1 mark: correct explanation that water enters root hair (and other plant) cells by osmosis after watering. 1 mark: vacuole/cell contents swell, pushing outward against the cell wall. 1 mark: the (relatively rigid) cell wall resists this, making the cell turgid. 1 mark: turgid cells provide support, allowing the plant to stand upright/recover from wilting.

Within any population of bacteria, there is natural variation between individuals, arising largely from random mutations in their DNA. By chance, a small number of bacteria in the original population may already carry a mutation that gives them some degree of resistance to a particular antibiotic, even before that antibiotic has ever been used against them.

When the population is exposed to the antibiotic, the non-resistant bacteria are killed or have their growth stopped. The resistant bacteria, however, are much less affected and survive; because they now face less competition for resources (nutrients, space) from the non-resistant bacteria that have been killed, the survivors are able to reproduce successfully. As bacterial reproduction is asexual and rapid, the resistance allele is passed on directly to all of the offspring of the resistant survivors.

Over many generations — and bacteria can reproduce extremely quickly, so many generations can occur in a short time — the proportion of resistant bacteria in the overall population increases, because resistant individuals consistently out-survive and out-reproduce non-resistant ones whenever the antibiotic is present. This effect is made worse by overuse of antibiotics or patients not completing a full prescribed course, since either allows some bacteria (including less-resistant ones) to survive and potentially develop or spread resistance further. Eventually, especially where antibiotic use is frequent (for example, in hospitals), the bacterial population can become dominated by highly resistant strains, such as MRSA, which show resistance to multiple antibiotics.

This is a serious and growing concern for public health because infections caused by such resistant bacteria become very difficult, or in some cases currently impossible, to treat using existing antibiotics. This increases the risk of serious illness, longer hospital stays and higher mortality, particularly for vulnerable patients, and resistant strains can spread readily between people, especially in healthcare settings; relatively few new antibiotics are currently being developed to replace those becoming ineffective, making this an urgent, ongoing challenge.

Answer: random mutation gives some bacteria natural resistance; antibiotic exposure kills non-resistant bacteria while resistant bacteria survive and reproduce, passing on resistance; over successive generations resistant bacteria become increasingly common (natural selection), and dominant resistant strains such as MRSA make infections much harder to treat, posing a growing public health threat.

Band A (5-6 marks): detailed, accurate, logically sequenced account covering random mutation producing initial variation in resistance, differential survival of resistant bacteria under antibiotic exposure, transmission of resistance to offspring over successive generations (with reference to natural selection), and a clear, accurate explanation of the public health concern. Wide and accurate use of specialist terms (e.g. mutation, resistance, natural selection, strain) with few SPG errors.
Band B (3-4 marks): reasonably accurate account of most stages (e.g. resistant bacteria survive and reproduce, becoming more common over time), but with some gaps (e.g. no clear reference to random mutation as the origin of variation, or limited development of the public health concern); some errors, generally competent scientific language.
Band C (1-2 marks): basic, fragmented statements, e.g. 'the bacteria become resistant to the antibiotic' or 'MRSA is hard to treat', with little development or accurate terminology.
Band D (0 marks): no relevant content / not creditworthy.

The independent variable is the one deliberately changed by the experimenter — here, the light intensity reaching the pondweed, controlled by changing the distance between the lamp and the pondweed. The dependent variable is the one measured to see the effect of this change — here, the rate of photosynthesis, measured indirectly as the number of oxygen bubbles produced in one minute. To make this a fair test, other variables that could also affect the rate of photosynthesis must be kept constant (controlled), such as the water temperature, the length or mass of pondweed used, or the concentration of sodium hydrogencarbonate solution (which provides a carbon dioxide source) in the water.
Answer: independent variable = light intensity/distance from lamp; dependent variable = rate of photosynthesis (bubbles/minute); controlled variable = e.g. water temperature, pondweed length, CO2/sodium hydrogencarbonate concentration.

1 mark: correct independent variable. 1 mark: correct dependent variable. 1 mark: any one valid controlled variable.

A suitable table lists the independent variable in the left-hand column, with repeat readings of the dependent variable and a mean in the columns to the right, with units given once in each column heading (not repeated in every cell):

Distance of lamp | Number of bubbles per minute | Mean number of
from pondweed / cm | Repeat 1 | Repeat 2 | Repeat 3 | bubbles per minute
-------------------|----------|----------|----------------|-------------------
10 | | | |
20 | | | |
30 | | | |

Answer: a table with the independent variable (distance / cm) in the first column, three repeat readings of bubbles per minute, and a mean column, all headed with correct quantities and units.

1 mark: independent variable (distance) correctly headed with units in its own column. 1 mark: repeat readings of the dependent variable and a mean column, correctly headed with units (bubbles per minute).

Burning a crisp involves an open flame and hot apparatus, so appropriate safety precautions include: wearing safety goggles to protect the eyes; using a heatproof mat to protect the work surface; using tongs or a mounted needle to hold the crisp while it burns, rather than fingers; tying back long hair and keeping loose clothing/sleeves away from the flame; and working in a well-ventilated area, away from flammable materials. Any three of these are creditworthy.

1 mark each for any three valid safety points (max 3): eye protection, heatproof mat, tongs/mounted needle to hold the crisp, hair/clothing precautions, ventilation/away from flammables.

To compare crisps fairly, the same conditions must apply to every test: the same mass of crisp should be burned each time; the same volume/mass of water should be heated; the water should start at the same temperature each time; and the distance between the burning crisp and the boiling tube should be kept the same. Any two of these are creditworthy.

1 mark each for any two of: mass of crisp burned; volume/mass of water; starting temperature of water; distance between flame and boiling tube.

Temperature rise = 71 °C − 21 °C = 50 °C. Energy transferred to the water: \( E = 25\ \text{g} \times 4.2 \times 50\ \text{°C} = 5250\ \text{J} \). Since exactly 1.0 g of crisp was burned, the energy released per gram is \( \frac{5250\ \text{J}}{1.0\ \text{g}} = 5250\ \text{J/g} = 5.25\ \text{kJ/g} \). Answer: 5.25 kJ/g.

1 mark: correct temperature rise (50 °C). 1 mark: correct energy transferred to water (5250 J). 1 mark: correctly expressed as energy per gram in kJ/g (5.25 kJ/g), accept ecf.

This method underestimates the true energy content because not all of the heat released by the burning crisp is transferred to the water: much of it is lost to the surrounding air by convection and radiation, some heat is absorbed by the glass boiling tube and clamp/stand rather than the water, and the crisp may not combust completely, releasing less energy than its full chemical energy content. Any two of these reasons are creditworthy.

1 mark each for any two of: heat lost to surrounding air; heat lost to/absorbed by the boiling tube or apparatus; incomplete combustion of the crisp.

On a graph of temperature against time, the gradient (slope) at any point represents the rate of change of temperature with time, i.e. the rate at which the water is being heated, in units such as °C per second or °C per minute. While the crisp is burning and releasing heat energy, the graph rises with a positive gradient. Once the crisp has finished burning, no further heat is being supplied to the water by combustion, so the temperature stops rising — the graph becomes flat (a gradient of approximately zero); over a longer time period it might even begin to slope downward slightly, as the water, now hotter than the surrounding air, starts to lose heat to its surroundings and cool.
Answer: the gradient represents the rate of heating (temperature rise per unit time); once burning stops, the graph levels off (or slowly falls) as no more heat is supplied.

1 mark: gradient represents the rate of temperature increase (rate of heating). 1 mark: correctly states the graph levels off/becomes flat once burning stops (no more heat supplied). 1 mark: additionally recognises the graph could then slowly fall as the water cools/loses heat to the surroundings.

Gradient = change in temperature ÷ change in time: \( \frac{68\ \text{°C} - 20\ \text{°C}}{60\ \text{s}} = \frac{48}{60} = 0.8\ \text{°C/s} \). Answer: 0.8 °C/s.

1 mark: correct change in temperature (48 °C) used. 1 mark: correct final answer, 0.8 °C/s.

A major source of inaccuracy in this method is heat being lost from the burning crisp and boiling tube to the surrounding air, rather than being transferred to the water, so the measured temperature rise (and calculated energy value) is lower than the true energy content. Surrounding the apparatus with a draught excluder or heat shield, or using a more enclosed calorimeter design, would reduce heat losses to the air and draughts, so a greater proportion of the energy released by the burning crisp would be transferred to the water; this would give a larger, more accurate temperature rise and a calculated energy value closer to the crisp's true energy content.
Answer: use a heat shield/more enclosed calorimeter to reduce heat loss to the surroundings, so more of the released energy is transferred to and measured in the water, giving a more accurate result.

1 mark: valid improvement (e.g. heat shield/draught excluder, more enclosed calorimeter). 1 mark: correct reasoning that this reduces heat loss to the surrounding air. 1 mark: correct explanation that this results in a more accurate (closer to true) measurement of energy content.

A single reading may be affected by random errors, such as slight variation in exactly how the crisp burns, small draughts, or timing/reading errors, which could make that one result unrepresentative (anomalous). By repeating the test several times and calculating a mean of the (concordant) results, the effect of any single anomalous reading is reduced, giving a value that is more reliable and more representative of the crisp's true energy content.

1 mark: repeating reduces the effect of random errors/anomalous results. 1 mark: calculating a mean gives a more reliable/representative result.

The value calculated from this simple experiment is almost always lower than the value on the packaging (which is determined using more precise laboratory calorimetry equipment) because much of the heat energy released as the crisp burns escapes into the surrounding air, is absorbed by the glass boiling tube, clamp and stand, or is lost before the flame is directly beneath the tube, rather than being transferred to and measured by the water. To reduce this difference, the student could insulate the boiling tube (e.g. with cotton wool or foil, avoided directly over the flame), use a heat shield or draught excluder around the apparatus, or use a more enclosed calorimeter, all of which would capture more of the released heat and give a value closer to the packaging's stated energy content.
Answer: much of the heat released is lost to the surroundings rather than transferred to the water; insulating the apparatus/using a more enclosed calorimeter would capture more of this heat and reduce the difference.

1 mark: valid reason for underestimate (heat lost to surroundings/apparatus, not fully transferred to water). 1 mark: valid improvement (insulation, heat shield, more enclosed calorimeter). 1 mark: correct reasoning linking the improvement to reduced heat loss and a more accurate result.

The student could repeat the test for that sample again, under the same conditions, and compare the new reading with the original set of repeat readings; if a result stands out as very different from the other, more consistent (concordant) readings, it can be identified as anomalous, and should be discarded (not included when calculating the mean) or the test repeated further to confirm.

1 mark: repeats the test and compares with other readings. 1 mark: identifies an anomalous result as one that differs greatly from the other (concordant) readings, and should be discarded/repeated.

(a) Disc B produced by far the largest clear zone (28 mm, compared with 12 mm for A and 6 mm for C). A larger clear zone shows that the antibiotic was still effective at inhibiting bacterial growth over a greater distance/area from the disc, so disc B was the most effective antibiotic against this bacterium.
(b) As the antibiotic diffuses outward from the disc into the surrounding agar, its concentration is high enough near the disc to kill the bacteria or prevent them from growing, leaving a visibly clear area with no bacterial colonies present. Further from the disc, the antibiotic has become too dilute (its concentration has fallen too low through diffusion) to have an effect, so bacteria are able to grow normally there, giving a visible boundary between the clear zone and normal growth.
(c) To make this a fair test allowing valid comparison between antibiotics, other variables should be kept constant, such as the bacterial species and starting density used, the size of the agar plate and antibiotic discs, and the incubation temperature and time.
Answer: disc B was most effective (largest clear zone); the clear zone forms because the antibiotic diffusing from the disc kills/inhibits bacteria nearby but becomes too dilute further away; controlled variables should include bacterial species/density, disc/plate size, and incubation temperature and time.

(a) 1 mark: disc B, with correct reasoning (largest clear zone). (b) 1 mark: antibiotic diffuses out and kills/inhibits bacteria near the disc. 1 mark: further away the antibiotic is too dilute to be effective, so bacteria grow normally. (c) 1 mark: any one valid controlled variable.

(a) The greatest volume of CO2 (18 cm³) was produced at 30 °C, so this is the temperature at which the rate of respiration was greatest.
(b) Respiration is controlled by enzymes, which have an optimum temperature at which they work fastest — here, close to 30 °C. Above this optimum, the extra thermal energy breaks the bonds holding the enzymes' tertiary structure together, changing the shape of their active sites (denaturation); denatured enzymes can no longer bind their substrates effectively, so the rate of the enzyme-catalysed reactions involved in respiration falls sharply, resulting in much less CO2 being produced at 50 °C.
Answer: rate of respiration was greatest at 30 °C; the sharp fall between 40 °C and 50 °C occurs because the enzymes controlling respiration become denatured above their optimum temperature.

(a) 1 mark: 30 °C. (b) 1 mark: correctly identifies that respiratory enzymes are denatured above the optimum temperature. 1 mark: correctly links denaturation (active site shape change) to the sharp fall in the rate of respiration/CO2 production.

(a) Mean % cover: \( \frac{20+15+30+10+25+5+20+15+10+30}{10} = \frac{180}{10} = 18\% \).
(b) If the mean percentage cover from the sample is representative of the whole field, then 18% of the total field area is covered by daisies: \( 18\% \times 800\ \text{m}^2 = 0.18 \times 800\ \text{m}^2 = 144\ \text{m}^2 \).
Answer: mean percentage cover = 18%; estimated area covered by daisies = 144 m².

(a) 1 mark: correct total (180). 1 mark: correct mean, 18%. (b) 1 mark: correct method (mean % x total area). 1 mark: correct answer, 144 m² (accept ecf from the candidate's own mean).

(a) The data show a clear negative correlation: as sucrose concentration increases from 0.0 M to 0.8 M, the percentage change in mass steadily decreases, from a gain of +12% (at 0.0 M, distilled water) to a loss of -18% (at 0.8 M). The chips gain mass at low concentrations (0.0-0.2 M) but lose mass at higher concentrations (0.4 M and above), so the value crosses from a gain to a loss somewhere between 0.2 M and 0.4 M.
(b) Using linear interpolation between the two nearest readings, (0.2 M, +5%) and (0.4 M, -3%): the change in mass falls by 8 percentage points over an increase of 0.2 M in concentration, so the concentration at which the change would be 0% is approximately \( 0.2 + \left(\frac{5}{8} \times 0.2\right) \approx 0.33\ \text{M} \), i.e. approximately 0.3 M (this could equally be estimated by reading the x-intercept from a graph of the data). At this sucrose concentration, the external solution has the same effective solute concentration as the potato cells' own cell sap, so there is no net movement of water into or out of the cells by osmosis (water moves into and out of the cells at equal rates), and so no net change in mass. This value can therefore be used as an estimate of the solute concentration inside the potato cells.
Answer: percentage change in mass decreases as sucrose concentration increases (negative correlation), crossing from a gain to a loss between 0.2 M and 0.4 M; the zero-change concentration is approximately 0.3 M, which estimates the potato cells' own solute concentration.

(a) 1 mark: correctly describes the negative correlation/decreasing trend. 1 mark: correctly identifies the change from gain to loss occurring between 0.2 M and 0.4 M. (b) 1 mark: a reasonable estimate (0.3-0.35 M, from interpolation or graph-reading). 1 mark: correct explanation that this value represents the potato cells' own solute concentration, since there is no net osmotic water movement at this point.

(a) Increase in pulse rate = pulse rate immediately after exercise − resting pulse rate = 150 − 70 = 80 bpm.
(b) Recovery time is the time taken, after exercise stops, for pulse rate to return to its normal resting value. From the table, pulse rate first returns to the resting value of 70 bpm at 5 minutes after exercise, so the recovery time is 5 minutes.
(c) Regular exercise strengthens the heart muscle, increasing its stroke volume (the volume of blood pumped per beat); a fitter person's heart can therefore deliver the oxygen and glucose needed by the muscles, and remove waste products such as carbon dioxide, using fewer, more powerful beats rather than needing to beat as rapidly. As a result, a fitter person's heart rate does not need to rise or stay elevated for as long during and after exercise, so their pulse rate returns to its resting value more quickly.
Answer: increase in pulse rate = 80 bpm; recovery time = 5 minutes; a fitter person's more efficient (higher stroke volume) heart returns pulse rate to resting level faster.

(a) 1 mark: 80 bpm. (b) 1 mark: correct definition of recovery time. 1 mark: 5 minutes, correctly read from the data. (c) 1 mark: correct explanation referring to a fitter heart's greater efficiency/stroke volume needing fewer/less prolonged beats to meet demand.

(a) For a cube of side length L, surface area = 6L² and volume = L³. For a 10 mm cube: surface area = 6 x 10² = 600 mm²; volume = 10³ = 1000 mm³. Surface area to volume ratio = \( \frac{600}{1000} = 0.6 \), i.e. 0.6 : 1 (equivalently, \( \frac{6}{L} = \frac{6}{10} = 0.6 \)).
(b) Using the same method, the 5 mm cube has a surface area to volume ratio of \( \frac{6}{5} = 1.2 \), while the 20 mm cube has a ratio of \( \frac{6}{20} = 0.3 \) — as cube size increases, the surface area to volume ratio decreases. A smaller surface area to volume ratio means that, relative to the total volume the acid must diffuse through to reach the centre, there is less surface area available for the acid to enter, and a greater internal distance/volume to cross. This is consistent with the data: the time increases much faster than the side length (multiplying the side length by 4, from 5 mm to 20 mm, multiplies the time by 16, from 40 s to 640 s), showing diffusion becomes disproportionately slower as size increases and the surface area to volume ratio falls.
Answer: 10 mm cube SA:V = 0.6:1; the 20 mm cube has a much smaller SA:V ratio (0.3) than the 5 mm cube (1.2), meaning relatively less surface area and a greater distance for acid to diffuse to the centre, so it takes much longer.

(a) 1 mark: correct SA:V ratio, 0.6 (or 0.6:1, or 6:10). (b) 1 mark: correctly calculates or states that SA:V ratio decreases as cube size increases. 1 mark: correct explanation linking the smaller SA:V ratio in the larger cube to a longer diffusion time (less relative surface area / greater distance to the centre).

(a) Total magnification = eyepiece magnification x objective magnification = \( 10 \times 40 = 400 \), i.e. ×400.
(b) Actual width = number of eyepiece units x real size per unit: \( 100 \times 25\ \mu\text{m} = 2500\ \mu\text{m} \). Converting to mm: \( 2500\ \mu\text{m} = 2.5\ \text{mm} \).
(c) The real distance that one eyepiece graticule unit represents depends on the total magnification being used — specifically, on the objective lens in use, since the eyepiece stays the same. At ×40 objective, 1 unit = 25 μm; because real size per eyepiece unit is inversely proportional to objective magnification, switching to a ×10 objective (four times lower magnification) means 1 eyepiece unit now actually represents four times the real distance, i.e. 100 μm, not 25 μm. If the student continued to use the old calibration (25 μm per unit) after switching to the ×10 objective without recalibrating, any measurement taken would be calculated as only a quarter of the object's true size.
Answer: total magnification = ×400; actual cell width = 2500 μm (2.5 mm); switching to a lower-power (×10) objective without recalibrating means each eyepiece unit now represents 4 times more real distance, so the old calibration would make the calculated width 4 times too small.

(a) 1 mark: ×400. (b) 1 mark: correct method (100 x 25 μm). 1 mark: correct answer, 2500 μm and 2.5 mm. (c) 1 mark: correctly identifies that the real size represented by one eyepiece unit depends on/changes with the objective magnification used. 1 mark: correctly quantifies the resulting error (a 4x underestimate) from using the ×40 calibration at ×10 objective.

(a) Rate = distance moved ÷ time taken: \( \frac{45\ \text{mm}}{3\ \text{min}} = 15\ \text{mm/min} \).
(b) Volume of water taken up = distance moved x cross-sectional area: \( 45\ \text{mm} \times 0.5\ \text{mm}^2 = 22.5\ \text{mm}^3 \).
(c) For a potometer to give accurate readings of water uptake, the apparatus must be airtight, with no unwanted air bubbles in the tubing (other than the single bubble being timed), since a leak or an extra bubble would allow air uptake to be mistaken for, or to interfere with, water uptake; all joints should be sealed (e.g. with Vaseline/petroleum jelly), and the shoot should be cut underwater (or its cut end kept underwater) when setting up, to prevent air being drawn into the xylem vessels, which would block water movement.
Answer: rate = 15 mm/min; volume taken up = 22.5 mm³; the apparatus must be airtight (no stray air bubbles/leaks), with the shoot cut underwater to prevent air entering the xylem.

(a) 1 mark: 15 mm/min. (b) 1 mark: correct method. 1 mark: correct answer, 22.5 mm³. (c) 1 mark: any one valid precaution (airtight joints/no stray bubbles, or shoot cut underwater).

(a) Mean rate = volume of gas produced ÷ time taken: \( \frac{24\ \text{cm}^3}{30\ \text{s}} = 0.8\ \text{cm}^3/\text{s} \).
(b) Doubling the mass of chopped potato roughly doubles the amount (and total number of active sites) of the enzyme catalase available in the reaction mixture. With more active sites present, more hydrogen peroxide molecules can bind to an active site and be broken down at the same time, so more successful enzyme-substrate collisions occur per second, initially increasing the rate at which oxygen gas is produced (this increase would continue until some other factor, such as the concentration of hydrogen peroxide itself, becomes the limiting factor).
Answer: mean rate = 0.8 cm³/s; doubling the potato/catalase would increase the initial rate of reaction, since more enzyme active sites allow more hydrogen peroxide to be broken down per second.

(a) 1 mark: correct method. 1 mark: correct answer, 0.8 cm³/s. (b) 1 mark: correctly predicts the rate would increase. 1 mark: correct explanation in terms of more available enzyme/active sites increasing successful collisions.

(a) Volume of gas produced = 35 cm³ − 5 cm³ = 30 cm³. Mean rate = volume produced ÷ time taken: \( \frac{30\ \text{cm}^3}{5\ \text{min}} = 6\ \text{cm}^3/\text{min} \).
(b) In the absence of sufficient oxygen, yeast cells respire anaerobically, breaking down glucose incompletely to release energy, producing ethanol (alcohol) and carbon dioxide as waste products: glucose -> ethanol + carbon dioxide + energy. This process is called anaerobic respiration, or fermentation, in yeast.
Answer: mean rate = 6 cm³/min; glucose -> ethanol + carbon dioxide + energy, in the process of anaerobic respiration (fermentation).

(a) 1 mark: correct volume produced (30 cm³). 1 mark: correct rate, 6 cm³/min. (b) 1 mark: correct word equation (glucose -> ethanol + carbon dioxide (+ energy)). 1 mark: correctly names the process as anaerobic respiration/fermentation.

The starch solution and the amylase solution should first be placed in separate test tubes and left in a water bath set to the required temperature for a few minutes, so that both solutions reach that temperature before they are mixed (rather than starting the reaction at room temperature). The two solutions are then mixed together and a timer started immediately. At regular intervals (e.g. every 30 seconds), a drop of the reaction mixture is removed and tested with iodine solution on a spotting tile; the reaction is judged to be complete once the iodine no longer turns blue-black (showing all the starch has been broken down), and this time is recorded.

To ensure the investigation is valid (a fair test), only the temperature should be deliberately varied between tests; the volume and concentration of both the starch solution and the amylase solution, the volume of iodine solution used for each test, and the sampling interval should all be kept the same for every temperature tested. A suitable range of temperatures should be tested (e.g. 10 °C, 20 °C, 30 °C, 40 °C, 50 °C, 60 °C) to identify the enzyme's optimum.

To ensure the results are reliable, each temperature should be tested at least three times, and a mean time calculated; this reduces the effect of random errors (such as slight timing inaccuracies or an anomalous individual reading) on the final result.

Answer: pre-incubate starch and amylase separately to each temperature, then mix, time, and test with iodine at regular intervals until no blue-black colour forms; keep all other variables (concentrations, volumes, sampling method) constant, and repeat each temperature at least three times to calculate a reliable mean.

1 mark: pre-incubates starch and amylase separately to the test temperature before mixing. 1 mark: correct method for timing/testing (regular iodine sampling until no colour change, or equivalent valid method). 1 mark: identifies a suitable range of temperatures to test. 1 mark: correctly identifies at least two variables to keep constant for a fair test. 1 mark: repeats each temperature (at least three times) and calculates a mean for reliability.

Using quadrats to sample daisy cover has several strengths: it is a relatively quick, simple and inexpensive method that does not disturb or destroy the plants being sampled, and if quadrat positions are chosen randomly (for example, using random number coordinates rather than the student's own judgement) this reduces bias in where samples are taken, helping the sample to be more representative of the field as a whole.

However, the method also has weaknesses. If only a small number of quadrats are sampled, and daisies happen to be distributed unevenly (patchily) across the field — for example, clustered in one sunnier corner — the sample may not accurately represent the true overall population, giving a misleading estimate. Estimating percentage cover by eye can also be somewhat subjective, and different students may estimate the same quadrat differently, introducing further inconsistency (a form of measurement/observer error).

One clear improvement would be to increase the number of quadrats sampled across the field. A larger sample size reduces the impact that any one unusually patchy or atypical quadrat has on the overall mean, making the estimate of daisy population/cover more reliable and more representative of the field as a whole.

Answer: quadrats are quick and, if randomly placed, give an unbiased sample; but a small, patchy sample may not represent the whole field well, and percentage cover estimation can be subjective; increasing the number of quadrats sampled would improve the reliability of the result.

1 mark: valid strength (e.g. quick/simple, or random placement reduces bias). 1 mark: valid weakness relating to sample size/patchy distribution. 1 mark: valid weakness relating to subjectivity of estimating cover. 1 mark: valid improvement (increasing number of quadrats). 1 mark: correct reasoning for why the improvement helps (more representative/reliable estimate).

Several aseptic techniques should be used to avoid contaminating the bacterial culture and to keep the student safe. The inoculating loop (or spreader) used to transfer bacteria should be sterilised by passing it through a Bunsen flame both before and after use; this kills any microorganisms already present on it, preventing unwanted microorganisms from contaminating the culture, and preventing the bacteria being studied from being spread elsewhere. Work should be carried out close to a lit Bunsen burner, since the flame creates a rising convection current of hot air that carries airborne microorganisms upward and away from the working area, reducing the chance of contamination landing on the culture. The lid of the Petri dish should be secured with a few short pieces of tape around the edge, rather than sealed completely; this limits (without fully preventing) the entry of unwanted microorganisms from the air, while still allowing enough air exchange to prevent anaerobic conditions developing, which could otherwise favour the growth of certain harmful bacteria. The culture should be incubated at a controlled, relatively low temperature, typically around 25 °C in school laboratories, rather than close to human body temperature (37 °C), to reduce the risk of encouraging the growth of microorganisms that could be pathogenic to humans. Finally, once the investigation is complete, the culture (and any used equipment) should be disposed of safely by autoclaving — sterilising with high-pressure steam — to kill all microorganisms before disposal, preventing the release of potentially harmful bacteria.
Answer: sterilise the loop in a flame before and after use; work near a Bunsen flame; tape (not seal) the Petri dish lid; incubate at a controlled low temperature (~25 °C); dispose of cultures by autoclaving — each technique reduces contamination risk or the risk of growing/releasing harmful microorganisms.

1 mark: sterilising the loop in a flame before and after use, with correct reasoning. 1 mark: working near a Bunsen flame, with correct reasoning (updraught reduces airborne contamination). 1 mark: taping (not sealing) the lid, with correct reasoning. 1 mark: incubating at a controlled, relatively low temperature (~25 °C), with correct reasoning. 1 mark: safe disposal by autoclaving, with correct reasoning.

For an investigation to be a valid, fair test, only the independent variable being investigated — here, temperature — should be deliberately changed between tests, with every other variable that could affect the outcome kept constant. Because different groups used different concentrations of amylase, more than one variable differed between groups' results at once: both the temperature tested and the enzyme concentration used may have varied. This means that if one group's results differ from another's, it is impossible to be certain whether this difference was caused by the temperature being investigated, by the difference in enzyme concentration, or by some combination of both — so the class results cannot be validly compared or combined with confidence.

To allow valid comparisons between groups, every group should have used exactly the same concentration and volume of amylase solution, and the same concentration and volume of starch solution, so that only the temperature varied between the tests carried out by different groups. This would ensure that any difference observed between groups' results could be confidently attributed to the difference in temperature alone, making the combined class data a valid, fair test.

Answer: differing amylase concentrations mean more than one variable changed at once, so any difference in results cannot be confidently attributed to temperature alone; every group should have used the same concentration and volume of amylase (and starch), varying only temperature, to allow valid comparison.

1 mark: correctly identifies that more than one variable differed between groups. 1 mark: correctly explains that this means differences in results cannot be confidently attributed to temperature alone. 1 mark: states that all groups should use the same amylase concentration/volume. 1 mark: states that all groups should use the same starch concentration/volume (or other consistent controlled variable). 1 mark: correct overall conclusion that only temperature should have varied, to allow valid comparison.

A control experiment is set up in exactly the same way as the main investigation, with every variable kept the same, except for the one factor being tested — here, whether the disc contains an antibiotic. Using a disc soaked only in water (rather than an antibiotic) shows what the bacterial growth pattern looks like when no antibiotic effect is present at all.

If the bacteria are observed to grow normally, right up to the edge of the water-soaked disc, with no clear zone forming around it, this demonstrates that it is specifically the presence of an antibiotic — and not some other factor, such as the physical presence of the disc itself, general conditions on the agar plate, or an unintended contaminant — that is responsible for any clear zones seen around the true antibiotic discs.

Without this control, it would not be possible to rule out these alternative explanations with confidence; a clear zone might, in principle, be caused by something other than the antibiotic being tested. Including a water-soaked control disc is therefore a necessary part of making the investigation a valid test of the antibiotics' effect specifically, and gives confidence that any differences observed between antibiotic discs are genuinely due to differences in antibiotic effectiveness.

Answer: a water-soaked control disc shows bacterial growth with no antibiotic present; if bacteria grow normally up to this disc (no clear zone), it confirms that clear zones around the true antibiotic discs are genuinely caused by the antibiotic, making the control a necessary part of a valid investigation.

1 mark: correctly describes what the control disc is (soaked in water, all else identical). 1 mark: correctly states the expected result if the control works as intended (bacteria grow normally up to the disc, no clear zone). 1 mark: correctly explains that this rules out other explanations for a clear zone (e.g. the disc itself, general conditions). 1 mark: correctly explains why this makes the investigation a valid test of the antibiotic specifically. 1 mark: for a clear, well-reasoned, complete evaluation.

To investigate the effect of sucrose concentration on percentage change in mass, the student should first prepare a suitable range of sucrose concentrations, for example 0.0 M (distilled water), 0.2 M, 0.4 M, 0.6 M, 0.8 M and 1.0 M, made up accurately using a measuring cylinder or serial dilution from a stock solution.

For each concentration, an equal-sized potato chip should be prepared, using a cork borer of the same diameter and cutting each chip to the same length, ideally all taken from the same potato (or potatoes of the same variety), to reduce variation in the cells' own starting solute concentration between chips. Each chip should be gently blotted dry with paper towel and weighed on a balance before being placed into its test tube of sucrose solution, and left for a fixed period of time, such as 30 minutes. At the end of this time, each chip should be removed, blotted dry again in the same way, and reweighed; the percentage change in mass can then be calculated for each chip using the formula \( \%\ \text{change} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100 \).

To make this a valid, fair test, every variable other than sucrose concentration should be kept constant across all chips: the size and shape of the chips, the volume of solution each chip is placed in, the length of time left in the solution, and the temperature at which the investigation is carried out, since temperature also affects the rate of osmosis. To obtain reliable results, the investigation should be repeated (e.g. at least three chips tested at each concentration), and a mean percentage change in mass calculated for each concentration, reducing the effect of any individual anomalous reading (for example, caused by a chip not being blotted consistently, or an inaccurate balance reading).

The mean percentage change in mass at each concentration can then be plotted on a graph, with sucrose concentration on the x-axis and percentage change in mass on the y-axis, and a line of best fit drawn. At low sucrose concentrations, where the surrounding solution is more dilute than the potato cells' own cell sap, water moves into the cells by osmosis and mass increases; at high sucrose concentrations, where the surrounding solution is more concentrated than the cell sap, water moves out of the cells and mass decreases. The concentration at which the graph crosses zero percentage change (the x-intercept) represents the point where the external sucrose solution has the same effective solute concentration as the potato cells themselves, so there is no net movement of water into or out of the cells by osmosis; this concentration can therefore be used as an estimate of the solute concentration inside the potato cells.

Answer: prepare a range of sucrose concentrations and equal-sized potato chips; measure mass before and after a fixed time in each concentration to calculate % change in mass, keeping chip size, solution volume, time and temperature constant, and repeating for reliability; the concentration at which the graph of results crosses 0% change estimates the potato cells' own solute concentration, since no net osmotic water movement occurs at that point.

Band A (5-6 marks): a detailed, accurate, well-sequenced method covering preparation of a suitable range of concentrations, correct procedure for measuring mass change (blotting, weighing before/after, % change calculation), at least two correctly identified controlled variables, repeats/means for reliability, and a correct, clearly explained method for estimating the potato cells' solute concentration from the zero-change point of the results. Wide and accurate use of specialist terms (e.g. osmosis, solute concentration, controlled variable) with few SPG errors.
Band B (3-4 marks): a reasonably complete method covering most of the above, but with some gaps (e.g. missing repeats, or an incomplete/vague explanation of how the zero-change point relates to the cells' solute concentration); some errors, generally competent scientific language.
Band C (1-2 marks): basic, fragmented statements, e.g. 'put chips in different sugar solutions and weigh them', with little development or accurate terminology.
Band D (0 marks): no relevant content / not creditworthy.