2022 CCEA GCSE Biology 1010 模擬試題連答案詳解

The cell membrane is partially permeable and controls which substances enter and leave the cell. The cytoplasm is a jelly-like substance in which most of the chemical reactions of the cell, controlled by enzymes, take place.

1 mark: cell membrane controls entry/exit of substances (partially permeable); 1 mark: cytoplasm is site of most metabolic/enzyme reactions.

The nucleus contains the cell's genetic material (DNA/chromosomes) and controls the cell's activities, including which proteins are made. Ribosomes are the organelles at which the process of protein synthesis actually occurs.

1 mark: nucleus; 1 mark: ribosome (ribosomes).

Photosynthesis is summarised by the word equation: carbon dioxide + water, in the presence of light energy and chlorophyll, produces glucose + oxygen. The pigment that absorbs the light energy needed to drive this reaction is chlorophyll, found in chloroplasts.

1 mark: correct word equation (carbon dioxide + water → glucose + oxygen, light/chlorophyll as conditions); 1 mark: chlorophyll.

The rate of photosynthesis can be limited by light intensity (too little light means less energy is available), carbon dioxide concentration (too little CO2 restricts the supply of raw material) and temperature (too low a temperature slows the enzyme-controlled reactions).

1 mark each for any two of: light intensity; carbon dioxide concentration; temperature. Max 2 marks.

Amylase, present in saliva, catalyses the breakdown of starch into the sugar maltose.

1 mark: amylase; 1 mark: maltose.

An enzyme is a biological catalyst: a substance that increases the rate of a specific chemical reaction without itself being changed or used up. All enzymes are proteins, each with a specifically shaped active site that fits only its complementary substrate molecule(s).

1 mark: biological catalyst that speeds up a specific reaction without being used up; 1 mark: protein.

Oxygen moves from the alveoli into the blood by diffusion, down a concentration gradient. The alveoli are adapted to make this diffusion rapid: they have a very large total surface area, walls that are only one cell thick (a short diffusion distance), an extensive capillary network (good blood supply) and a moist lining that allows gases to dissolve.

1 mark: diffusion; 1 mark: any one of large surface area / thin wall (one cell thick) / good blood supply / moist lining.

Aerobic respiration is summarised by the word equation: glucose + oxygen → carbon dioxide + water, with energy released for use by the cell.

1 mark: correct reactants (glucose + oxygen); 1 mark: correct products (carbon dioxide + water).

Nervous coordination sends electrical impulses along neurones, producing a very fast response that is short-lived. Hormonal coordination relies on chemical messengers carried in the blood, so the response is much slower to start but tends to last much longer.

1 mark: nervous response is faster than hormonal response; 1 mark: hormonal response lasts longer than nervous response (accept converse statements).

Insulin is produced by the pancreas. It lowers blood glucose concentration by causing cells, particularly in the liver and muscles, to take up glucose from the blood and convert it to glycogen for storage.

1 mark: pancreas; 1 mark: lowers/decreases blood glucose concentration.

Iodine solution (iodine dissolved in potassium iodide solution) is added to a food sample. If starch is present, the colour changes from orange-brown to blue-black.

1 mark: iodine solution (iodine/potassium iodide solution); 1 mark: orange-brown to blue-black.

A food chain is a diagram showing the feeding relationships between organisms, and the transfer of energy along it, starting with a producer and passing through a series of consumers. Energy is lost between each trophic level because some energy is released as heat during respiration, some is used for movement, and some is lost in materials that are not eaten or not digested (egested).

1 mark: shows transfer of energy/biomass between organisms starting with a producer; 1 mark: any one valid reason for energy loss (heat from respiration / movement / undigested material egested).

The table shows that as the distance from the lamp increases from 10 cm to 40 cm, the number of bubbles produced per minute falls steadily, from 48 to 6, because light intensity decreases with distance and light intensity was limiting the rate of photosynthesis over this range. Between 40 cm and 50 cm the rate stays constant at 6 bubbles per minute. This shows that light intensity is no longer the limiting factor at this low intensity; instead, a different factor, such as carbon dioxide concentration or temperature, must now be limiting the rate of photosynthesis, so reducing the light intensity further has no additional effect.

1 mark: rate of bubble production decreases as distance from the lamp increases (light intensity decreases); 1 mark: rate is constant/unchanged between 40 and 50 cm; 1 mark: because light intensity is no longer limiting – another factor (CO2 concentration or temperature) is now limiting the rate.

At light intensity A, the rate of photosynthesis rises from 4 to 8 units as CO2 concentration increases from 0.01% to 0.04%, but then stays at 8 units even though CO2 concentration continues to rise. This shows that above 0.04% CO2, light intensity (which is low) has become the limiting factor, so increasing CO2 further cannot increase the rate. At light intensity B, more light energy is available, so the rate continues to rise with CO2 concentration, from 4 to 19 units, up to 0.06%, at which point it also plateaus – showing that at the higher light intensity, CO2 concentration itself remains the limiting factor for longer, until 0.06%, after which it is presumably some other factor (e.g. temperature) that limits the rate.

1 mark: rate at A plateaus at 0.04% CO2 because light intensity becomes limiting; 1 mark: rate at B continues to rise past 0.04% because more light is available/light is not limiting at B; 1 mark: correct use of data values from the table to support the explanation.

The data show that pepsin activity is greatest at pH 2 (9 units) and falls steadily as pH increases, reaching 0 units at pH 8; amylase activity, in contrast, is low at pH 2 (1 unit), rises to a peak of 10 units at pH 7, then falls again at pH 8. Each enzyme has an active site with a specific shape that only functions correctly, binding efficiently to its substrate, within a narrow range of pH around its optimum. Away from this optimum, changes in pH alter the shape of the enzyme's active site (and, at extremes, denature it), reducing activity. Pepsin's optimum is acidic because it functions in the stomach, where hydrochloric acid keeps the pH low; amylase's optimum is close to neutral because it functions in the mouth and small intestine, which are not strongly acidic.

1 mark: pepsin activity highest at low/acidic pH and decreases as pH rises; 1 mark: amylase activity highest around neutral pH (pH 7) and lower either side; 1 mark: each enzyme's active site shape is suited to the pH of the environment in which it normally works (accept reference to denaturation/shape change away from optimum).

The volume of oxygen produced per minute rises steadily from 6 cm³ at 10°C to a maximum of 27 cm³ at 40°C, as increasing temperature gives the enzyme and substrate molecules more kinetic energy, increasing the frequency of successful collisions between the enzyme's active site and its substrate. Between 40°C and 50°C the rate falls sharply, to 9 cm³/min. This is because 40°C was close to the enzyme's optimum temperature; above this, the excess heat energy breaks the bonds holding the enzyme's tertiary structure together, changing the shape of its active site. The enzyme is denatured, so the substrate (hydrogen peroxide) can no longer bind to the active site, and the rate of reaction falls.

1 mark: rate increases from 10°C to 40°C (peak/maximum at 40°C); 1 mark: rate falls sharply at 50°C; 1 mark: because the enzyme is denatured above its optimum temperature – active site changes shape and substrate can no longer bind.

The data show that breathing rate rises from 15 to 28 breaths per minute and tidal volume rises from 500 cm³ to 950 cm³ after exercise. During exercise, muscle cells respire at a faster rate to release the extra energy needed for contraction, so they use oxygen more quickly and produce carbon dioxide more quickly. An increase in both the rate and depth of breathing increases the volume of air exchanged with the alveoli each minute (ventilation rate), which increases the rate of oxygen diffusion into the blood and carbon dioxide diffusion out of the blood, meeting the muscles' increased demand and removing the extra waste CO2.

1 mark: breathing rate increases after exercise (15 to 28 breaths/min); 1 mark: tidal volume increases after exercise (500 to 950 cm³); 1 mark: explanation linking increased muscle respiration/oxygen demand and CO2 production to the need for increased gas exchange.

In the healthy person, blood glucose concentration rises from 4.5 to a peak of 7.0 mmol/dm³ one hour after the meal, then falls back to close to its starting value (4.5 mmol/dm³) by three hours. In the untreated diabetic, blood glucose concentration rises much higher, to 14.0 mmol/dm³ by two hours, and remains high (13.0 mmol/dm³) at three hours rather than falling. This difference occurs because, in the healthy person, the rise in blood glucose after eating stimulates the pancreas to secrete insulin, which causes cells (particularly liver and muscle cells) to take up glucose from the blood and store it as glycogen, bringing blood glucose back down towards its normal level. In a person with untreated type 1 diabetes, the pancreas produces little or no insulin, so glucose is not taken up efficiently by cells and blood glucose concentration remains abnormally high.

1 mark: healthy person's blood glucose rises then falls back towards normal within the 3 hours; 1 mark: diabetic's blood glucose rises higher and stays high/does not fall back; 1 mark: explanation referring to insulin secretion (healthy) versus lack of effective insulin (diabetic) causing cells to take up glucose.

The table shows that fat contains 37 kJ of energy per gram, more than twice the energy content of carbohydrate or protein, both of which contain 17 kJ per gram. This means that a relatively small mass of fatty food supplies a large amount of energy, which is useful for an athlete who needs to consume a lot of energy without having to eat a very large volume of food. Conversely, because fat is so energy-dense, it is easy to take in more energy than the body uses simply by eating fatty foods; when energy intake exceeds energy expenditure, the excess energy is stored in the body as fat, causing an increase in body mass. A person trying to lose body mass should therefore limit fat intake to help ensure that overall energy intake does not exceed energy expenditure.

1 mark: fat has the highest energy content per gram (37 kJ/g, more than carbohydrate/protein at 17 kJ/g); 1 mark: this explains why fat is useful for an athlete needing high energy in a small mass of food; 1 mark: explanation that limiting energy-dense fat helps to avoid an excess energy intake, which would otherwise be stored as body fat and increase body mass.

The data show a large decrease in the energy available at each successive trophic level: from 20 000 kJ/m²/year in the grass, to 1 800 kJ/m²/year in the rabbits (herbivores), to only 160 kJ/m²/year in the foxes (carnivores) – each level holding roughly a tenth or less of the energy of the level below it. This happens for several reasons: a large proportion of the energy taken in at each level is released as heat during respiration and lost to the surroundings; energy is also used by the organisms for movement and other life processes rather than being stored in new biomass; and not all of the biomass of one trophic level is eaten by the next, and of what is eaten, not all is digested and absorbed – some is egested as faeces. As a result, only a small proportion of the energy at one trophic level becomes available to the next.

1 mark: describes the decrease in energy available at each successive trophic level (using data); 1 mark and 1 mark: any two valid reasons for this decrease, e.g. energy lost as heat in respiration; energy used in movement/other life processes; not all organisms/biomass eaten or digested (egestion). Max 3 marks.

Increase in breathing rate = 35 − 14 = 21 breaths per minute. Percentage increase = (increase ÷ original) × 100 = (21 ÷ 14) × 100 = 1.5 × 100 = 150%. Check by a second route: 14 breaths/min increased by 150% is 14 × 2.5 = 35 breaths/min, which matches the given value, confirming the answer. The percentage increase in breathing rate is 150%.

1 mark: correct increase calculated (35 − 14 = 21); 1 mark: correct formula used (increase ÷ original × 100); 1 mark: correct substitution (21 ÷ 14 × 100); 1 mark: correct final answer, 150%, with % sign.

Increase in rate = 24 − 8 = 16 cm³/min. Percentage increase = (increase ÷ original) × 100 = (16 ÷ 8) × 100 = 2 × 100 = 200.0%. Check by a second route: 8 cm³/min increased by 200% is 8 × 3 = 24 cm³/min, which matches the given value, confirming the answer. The percentage increase in the rate of reaction is 200.0%.

1 mark: correct increase calculated (24 − 8 = 16); 1 mark: correct formula used (increase ÷ original × 100); 1 mark: correct substitution (16 ÷ 8 × 100); 1 mark: correct final answer to 1 d.p., 200.0%, with % sign.

The ileum is adapted in several ways to absorb digested food molecules efficiently. First, the ileum itself is long, which provides a large overall surface area over which absorption can occur. Second, its inner wall is folded, which increases the surface area available for absorption still further. Third, these folds are covered with millions of villi, finger-like projections that greatly increase the surface area of the ileum wall in contact with digested food, allowing more molecules to be absorbed at once. Looking at the structure of an individual villus, the outer wall is only one cell thick (a single layer of surface/epithelial cells), giving digested molecules only a very short distance to diffuse across, which speeds up absorption. Each villus also contains an extensive network of blood capillaries; blood is continually carried away from the villus, keeping the concentration of absorbed nutrients such as glucose and amino acids inside the villus low, which maintains a steep concentration gradient and so maintains a fast rate of diffusion into the blood. Finally, each villus contains a lacteal, a small vessel of the lymphatic system, which absorbs digested fats in the form of fatty acids and glycerol and transports them away in lymph. Together, the large surface area, short diffusion distance, good blood supply and presence of a lacteal mean that the ileum is very well adapted to absorb digested food molecules efficiently. In summary: the ileum's length, folded wall and villi give it a large surface area; each villus has a wall only one cell thick, an extensive capillary network and a lacteal, so together these features maximise the rate of diffusion and absorption of digested nutrient molecules into the blood and lymphatic system.

Band A (5–6 marks): answer refers to at least five relevant, accurate points using appropriate specialist terms (e.g. villi, epithelium, capillary network, lacteal, diffusion, concentration gradient), clearly and coherently organised. Band B (3–4 marks): at least three relevant points made, with reasonable use of specialist terms; may lack full coherence. Band C (1–2 marks): at least one relevant point made; answer may be simplistic, poorly expressed or lacking specialist terms. Band D (0 marks): no relevant content / response not creditworthy. Indicative content: ileum is long, providing a large surface area; inner wall is folded, increasing surface area further; folds covered in villi, greatly increasing surface area; villus wall is a single layer of (epithelial) cells / thin, giving a short diffusion distance; villus has an extensive capillary network / good blood supply, maintaining a steep concentration gradient by removing absorbed nutrients; villus contains a lacteal, which absorbs digested fats (fatty acids and glycerol); overall combination of features maximises rate of diffusion/absorption.

Auxin is a plant hormone produced at the tip of the shoot, from where it moves down the shoot. When light reaches the shoot from only one side, auxin becomes unevenly distributed: more auxin accumulates on the shaded side of the shoot than on the side facing the light. Because auxin causes cell elongation, the cells on the shaded side, which contain more auxin, elongate more than the cells on the illuminated side. This differential (unequal) growth on the two sides causes the shoot to bend towards the light.

1 mark: auxin is produced at the shoot tip and moves down the shoot; 1 mark: light causes an uneven/unequal distribution of auxin, with more accumulating on the shaded side; 1 mark: greater auxin concentration on the shaded side causes greater cell elongation there, causing the shoot to bend towards the light.

Auxin is produced only at the tip of a shoot. In Seedling A, the tip is intact, so auxin is produced and, under one-sided light, becomes unequally distributed, accumulating on the shaded side. This causes greater cell elongation on the shaded side than the illuminated side, so the shoot bends towards the light. In Seedling B, the tip (the site of auxin production) has been removed, so no auxin is produced. Without auxin, there can be no unequal distribution of the hormone between the two sides of the shoot, and so no difference in the rate of cell elongation on each side; the seedling therefore grows straight upwards rather than bending.

1 mark: auxin is produced only at the shoot tip; 1 mark: Seedling A's intact tip allows auxin production and unequal distribution under one-sided light, causing bending; 1 mark: Seedling B has no tip/no auxin production, so there is no unequal distribution and no bending.

The seedling would be predicted to grow straight upwards rather than bending towards the light. The tip of the shoot is the region that detects the direction of light and is also the site of auxin production. Because the cap is opaque, it prevents light from reaching the tip, so the tip cannot detect that light is coming from one side only. As a result, auxin produced at the tip is distributed evenly down both sides of the shoot rather than accumulating more on one side. With no unequal distribution of auxin, there is no difference in the rate of cell elongation between the two sides of the shoot, so the seedling grows straight upwards instead of bending.

1 mark: seedling grows straight upwards / does not bend towards the light; 1 mark: cap prevents the tip from detecting the direction of light; 1 mark: without directional light reaching the tip, auxin remains evenly distributed, so there is no differential elongation/bending.

(a) The seedling was predicted to bend, curving away from the side on which the agar block had been placed. (b) Auxin diffused out of the agar block and into the side of the cut stem that it was covering, raising the auxin concentration on that side compared with the other side. Because auxin promotes cell elongation, the cells on the side with the higher auxin concentration (under the block) elongated more than the cells on the other side, causing the shoot to curve away from that side. Crucially, this bending occurred even though the seedling was kept in complete darkness, so light itself cannot have been responsible for the unequal growth. This demonstrates that it is the unequal distribution/concentration of the hormone auxin across the shoot, rather than light acting directly on the cells, that is the actual cause of the differential growth and bending response; light's role is simply to cause auxin to become unequally distributed in an intact seedling.

(a) 1 mark: seedling bends/curves, away from the side with the agar block. (b) 1 mark: auxin diffuses from the block into the adjacent side of the stem, raising its auxin concentration; 1 mark: higher auxin concentration causes greater cell elongation on that side, producing curvature; 1 mark: because bending occurred in darkness, this shows auxin concentration/distribution itself (not light directly) causes the bending response.

The genome of an organism is the complete set of genetic material (all of its DNA), including all of its genes. In a human cell, most of the genome is contained within the nucleus, packaged as chromosomes (a small amount is also found in mitochondria, but this is not the main location).

1 mark: genome is the entire genetic material/all the DNA of an organism; 1 mark: nucleus (as chromosomes).

An allele is one of two or more alternative versions of a gene, which may differ slightly in their DNA base sequence. An individual that has two different alleles for a particular gene (one on each chromosome of a homologous pair) is described as heterozygous for that gene.

1 mark: allele is an alternative form/version of a gene; 1 mark: heterozygous.

Gametes are produced by meiosis, a type of cell division that halves the chromosome number, so that each gamete contains only one chromosome from each homologous pair. Human body (somatic) cells are diploid and normally contain 46 chromosomes (23 pairs); gametes therefore contain 23 chromosomes each.

1 mark: meiosis; 1 mark: 46.

Osmosis is a special case of diffusion involving only water molecules. It is defined as the diffusion of water molecules from a dilute solution (higher water concentration) to a more concentrated solution (lower water concentration), through a selectively/partially permeable membrane.

1 mark: diffusion of water molecules from a dilute to a more concentrated solution; 1 mark: through a selectively/partially permeable membrane.

When a plant cell is placed in a concentrated sugar solution, the external solution has a lower water concentration than the cell's cytoplasm/vacuole, so water leaves the cell by osmosis. As the cell loses water, the cytoplasm shrinks and the cell membrane pulls away from the rigid cell wall. A cell in this state is described as plasmolysed.

1 mark: water leaves the cell by osmosis (cell membrane/cytoplasm pulls away from the cell wall); 1 mark: plasmolysed.

When a plant cell is placed in pure water, the external solution has a higher water concentration than the cell's cytoplasm/vacuole, so water enters the cell by osmosis. The cell swells, but the rigid cell wall exerts an inward pressure that prevents the cell from bursting. A cell in this firm, swollen state is described as turgid.

1 mark: water enters the cell by osmosis, cell swells but does not burst (cell wall prevents bursting); 1 mark: turgid.

Phagocytes are a type of white blood cell that defend the body against pathogens by engulfing and digesting them. This process of engulfing and digesting a pathogen (or other particle) is called phagocytosis.

1 mark: phagocyte; 1 mark: phagocytosis.

Penicillin is naturally produced by a fungus (mould) of the genus Penicillium. A substance, such as penicillin, that kills bacteria or inhibits/slows their growth is called an antibiotic.

1 mark: fungus/mould (accept Penicillium); 1 mark: antibiotic.

A person's risk of developing cardiovascular disease is increased by lifestyle factors such as a poor diet high in saturated fat and sugar, a lack of regular physical exercise, smoking (tobacco use) and drinking excessive amounts of alcohol.

1 mark each for any two of: poor diet (excess fat/sugar); lack of exercise; smoking; excessive alcohol consumption. Max 2 marks.

The left ventricle has thick, muscular walls and pumps oxygenated blood at high pressure into the aorta, which carries it to the rest of the body. Deoxygenated blood returning from the body enters the heart through the vena cava.

1 mark: left ventricle; 1 mark: vena cava.

Arteries carry blood away from the heart at high pressure. They have thick walls containing muscle and elastic fibres, which allow the artery to withstand this high pressure without bursting, and to stretch and recoil, helping to maintain blood flow between heartbeats.

1 mark: correct structural feature (e.g. thick/muscular/elastic wall) linked correctly to its function (withstanding high pressure/maintaining blood flow). Accept other valid feature-function pairs, e.g. narrow lumen to maintain pressure.

Capillaries are the site of exchange between blood and tissues. Their walls are only one cell thick, which minimises the diffusion distance and allows substances such as oxygen, glucose, carbon dioxide and waste products to diffuse efficiently between the blood and surrounding cells.

1 mark: correct structural feature (wall one cell thick, narrow diameter/large total surface area, or leaky/permeable wall); 1 mark: correctly related to efficient diffusion/exchange of substances between blood and tissues.

Variation refers to the differences that exist between individuals of the same species. Some variation is genetic (inherited), but variation can also be caused by environmental factors that are not inherited, such as diet, climate, or the amount of exercise or sunlight an individual is exposed to.

1 mark: variation is the differences between individuals of the same species; 1 mark: any one valid environmental (non-inherited) cause of variation, e.g. diet, climate, sunlight exposure.

The contraceptive pill (and the contraceptive implant or injection) are hormonal methods of contraception, which work by using hormones to prevent ovulation or otherwise disrupt the menstrual cycle. The condom and the diaphragm are non-hormonal, barrier methods, which physically prevent sperm from reaching an egg.

1 mark: valid hormonal method (pill, implant or injection); 1 mark: valid barrier/non-hormonal method (condom or diaphragm).

(a) Both parents are heterozygous, Ff. Each parent can produce gametes carrying either the F allele or the f allele. Combining these gametes in a Punnett square:
F f
F FF Ff
f Ff ff
This gives offspring genotypes in the ratio 1 FF : 2 Ff : 1 ff.
(b) Of the four possible genotype combinations, three (FF, Ff, Ff) give an unaffected phenotype and one (ff) gives cystic fibrosis, so the expected phenotype ratio is 3 unaffected : 1 affected.
(c) The probability of a child being ff (affected) is 1 in 4, which as a percentage is 25%.
(d) The 25% probability is a long-term average that applies to a very large number of offspring (or many such families); it does not guarantee that exactly 1 in 4 children in any one family will be affected. Each pregnancy is an independent event with the same 1-in-4 chance, so, just as tossing a coin four times will not always give exactly two heads, it is entirely possible by chance for all four children in this family to inherit the unaffected combination of alleles.

(a) 1 mark: correct gametes shown (F and f from each parent); 1 mark: correct offspring genotypes shown (FF, Ff, Ff, ff). (b) 1 mark: 3 unaffected : 1 affected (accept 3:1). (c) 1 mark: 25%. (d) 1 mark: each pregnancy/child is an independent event with the same probability; 1 mark: the ratio/probability is a long-term average (applies over many offspring) and does not guarantee the exact outcome in a small number of children, so all four could by chance be unaffected.

(a) Because short is the recessive phenotype, the short parent must be homozygous recessive, genotype tt. The tall parent is stated to be heterozygous, genotype Tt.
(b) The heterozygous tall parent (Tt) produces gametes T and t; the short parent (tt) produces only gametes t. Combining these in a Punnett square:
T t
t Tt tt
t Tt tt
This gives offspring genotypes in the ratio 2 Tt : 2 tt, i.e. 1 Tt : 1 tt.
(c) Since Tt is tall and tt is short, the expected phenotype ratio is 1 tall : 1 short. In practice, results from a real cross often differ from this expected ratio because which gamete fuses with which at fertilisation is a matter of chance, and the number of offspring produced in any one cross is relatively small; with a small sample, random chance can cause the observed ratio to deviate noticeably from the theoretical ratio. A much larger sample of offspring (or repeating the cross many times) would be expected to give a ratio closer to the theoretical 1:1.

(a) 1 mark: Tt (tall parent); 1 mark: tt (short parent). (b) 1 mark: correct gametes identified (T and t from the tall parent; t only from the short parent); 1 mark: correct offspring genotypes shown (Tt, Tt, tt, tt). (c) 1 mark: expected ratio 1 tall : 1 short; 1 mark: explanation referring to chance at fertilisation and/or small sample size causing deviation from the expected ratio.

The data show that as sucrose concentration increases from 0.0 to 0.8 mol/dm³, the percentage change in mass decreases steadily, from +12% down to -15%, changing from a gain in mass to a loss in mass somewhere between 0.2 and 0.4 mol/dm³. In pure water (0.0 mol/dm³), the external solution has a much higher water concentration than the solution inside the potato cells, so water moves into the cells by osmosis, down the water potential/concentration gradient, causing the cylinder to gain mass (+12%). In the 0.8 mol/dm³ sucrose solution, the external solution is more concentrated (has a lower water concentration) than the solution inside the potato cells, so water moves out of the cells by osmosis, causing the cylinder to lose mass (-15%).

1 mark: as sucrose concentration increases, % change in mass decreases; 1 mark: change in mass goes from positive to negative between 0.2 and 0.4 mol/dm³; 1 mark: in pure water, external solution more dilute than cell contents so water moves into cells by osmosis, mass increases; 1 mark: in 0.8 mol/dm³ solution, external solution more concentrated than cell contents so water moves out of cells by osmosis, mass decreases; 1 mark: correct use of osmosis terminology (dilute/concentrated solution, partially permeable membrane) throughout.

Over 15 minutes, the bubble moved 24 mm in still air, a rate of 24 ÷ 15 = 1.6 mm/min, compared with 47 mm in moving air, a rate of 47 ÷ 15 ≈ 3.1 mm/min – almost twice as fast. This is because moving air continually removes/disperses the water vapour that builds up in the layer of air close to the leaf surface, maintaining a steeper concentration (water potential) gradient between the air spaces inside the leaf and the air outside. This steeper gradient increases the rate of diffusion of water vapour out through the stomata, increasing the rate of transpiration. Because water is drawn up through the xylem to replace water lost in transpiration, a faster rate of transpiration produces a faster rate of water uptake by the shoot, as shown by the potometer readings.

1 mark: rate of water uptake (bubble movement) is greater in moving air than in still air, with reference to data (e.g. 47 mm vs 24 mm in 15 minutes); 1 mark: correct calculation/comparison of rate (approximately 3.1 mm/min vs 1.6 mm/min, almost double); 1 mark: moving air removes water vapour from the leaf surface more quickly; 1 mark: this maintains/steepens the diffusion (water potential/concentration) gradient out of the stomata; 1 mark: increasing the rate of transpiration, and therefore the rate of water uptake, to replace the water lost.

The data show a positive relationship between antibiotic use and bacterial resistance: as prescriptions rise from 420 to 560 per 1000 patients between 2018 and 2021, the percentage of resistant bacterial samples also rises, from 6% to 19%. This pattern can be explained by natural selection. Within any bacterial population there is natural genetic variation, and some individual bacteria may already carry a mutation that makes them resistant to a particular antibiotic. When the antibiotic is used, non-resistant bacteria are killed, but resistant bacteria survive. These surviving resistant bacteria then reproduce, passing the resistance allele on to their offspring, so the proportion of resistant bacteria in the population increases. The more frequently the antibiotic is prescribed, the stronger this selection pressure becomes, killing off non-resistant bacteria more often and allowing resistant strains to become increasingly common — matching the upward trend in resistance shown in the table.

1 mark: positive relationship described (as prescriptions increase, % resistant increases), with reference to data; 1 mark: natural variation exists in the bacterial population, some bacteria already resistant due to mutation; 1 mark: antibiotic use kills non-resistant bacteria; 1 mark: resistant bacteria survive and reproduce, passing on the resistance allele (differential survival/reproduction); 1 mark: greater/more frequent antibiotic use increases selection pressure, so resistance increases over time, consistent with the data.

Group A had a lower mean resting heart rate (62 bpm) than Group B (78 bpm), and a much shorter mean recovery time (3 minutes) than Group B (9 minutes). Both differences indicate that Group A, the regular exercisers, have greater cardiovascular fitness than Group B. This effect occurs because regular exercise causes the heart muscle to strengthen and, in particular, the muscle of the left ventricle to thicken, increasing the volume of blood the heart can pump with each contraction (the stroke volume). Because more blood is pumped per beat, fewer beats per minute are needed to circulate the same volume of blood around the body, lowering resting heart rate. A stronger, more efficient heart is also able to reduce heart rate back down to its resting value more quickly once the increased demand for oxygen created by exercise has been met, giving a shorter recovery time.

1 mark: Group A has a lower resting heart rate than Group B, using data (62 vs 78 bpm); 1 mark: Group A has a shorter recovery time than Group B, using data (3 vs 9 minutes); 1 mark: conclusion that Group A (exercisers) has greater cardiovascular fitness; 1 mark: regular exercise strengthens the heart muscle and increases stroke volume; 1 mark: explains why this produces a lower resting heart rate and/or faster recovery.

After the low-salt diet, mean systolic pressure fell from 158 mmHg to 142 mmHg, and mean diastolic pressure fell from 98 mmHg to 88 mmHg — both measures of blood pressure decreased. A lower salt intake means a lower concentration of sodium ions in the blood plasma. Because water tends to follow sodium (by osmosis, and because the kidneys reabsorb more water when sodium concentration is higher), a lower sodium concentration results in less water being retained in the blood, reducing the total volume of blood in the circulatory system. A smaller volume of blood exerts less force against the walls of the blood vessels, which lowers blood pressure.

1 mark: systolic pressure decreased (158 to 142 mmHg) after the diet; 1 mark: diastolic pressure decreased (98 to 88 mmHg) after the diet; 1 mark: lower salt intake reduces sodium ion concentration in the blood, reducing water retention/reabsorption; 1 mark: this reduces blood volume; 1 mark: lower blood volume reduces the pressure/force exerted on blood vessel walls, lowering blood pressure.

Arteries carry blood at the highest pressure of the three vessels (about 100 mmHg), as it has just been pumped from the heart; their thick, muscular and elastic walls allow them to withstand this high pressure without bursting and to stretch and recoil to help maintain blood flow. Capillaries carry blood at a much lower pressure (about 30 mmHg); because their walls do not need to withstand high pressure, they can be only one cell thick, which minimises the diffusion distance and allows efficient exchange of substances between blood and tissues. Veins carry blood at the lowest pressure of all (about 10 mmHg), so a thinner, less muscular wall is sufficient to contain the blood; because the pressure is so low, veins also contain valves, which prevent blood flowing backwards and help return blood to the heart, often assisted by the squeezing action of surrounding skeletal muscles.

1 mark: artery has highest pressure and thick/muscular/elastic wall to withstand/accommodate this; 1 mark: capillary has much lower pressure and a wall one cell thick, permitting efficient diffusion; 1 mark: vein has lowest pressure and a thinner, less muscular wall; 1 mark: valves in veins prevent backflow at this low pressure; 1 mark: overall correct link between decreasing pressure (artery > capillary > vein) and the structural adaptation of each vessel, using the data given.

The data show that as daily UV exposure increases from 0 to 45 minutes, the number of mutated cells per cm² of skin increases, from 2 to 29, with the increase becoming steeper at higher exposure durations. UV radiation is a mutagen: it damages the structure of DNA in skin cells, for example by causing incorrect bonds to form between bases. When this damaged DNA is copied during cell division, errors can be introduced into the base sequence, producing mutations. Because a greater duration of UV exposure causes more DNA damage overall, it results in a greater number of mutated cells, consistent with the upward trend shown in the table.

1 mark: number of mutated cells increases as UV exposure increases, with reference to data; 1 mark: UV radiation is a mutagen that damages/alters the structure of DNA in skin cells; 1 mark: this damage causes errors when DNA is copied/replicated, producing mutations; 1 mark: greater UV exposure causes more DNA damage and therefore more mutated cells, consistent with the data.

Before pollution, tree bark was light in colour, so light-coloured moths were well camouflaged against it and less likely to be seen and eaten by predators such as birds, while dark-coloured moths stood out and were more likely to be eaten; this is reflected in the data, with light moths making up 95% of the population. After pollution coated the bark in dark soot, the situation reversed: dark-coloured moths were now better camouflaged, while light-coloured moths were more visible to predators. Dark moths therefore had a survival advantage over light moths in the polluted environment, were more likely to survive to reproduce, and passed on the allele for dark colouration to their offspring. Because this advantageous allele was passed on generation after generation, while light moths were increasingly predated, the proportion of dark moths in the population increased sharply, to 85%, while light moths decreased to 15%. This is an example of natural selection: the environment (through predation) selected for the phenotype/allele best suited to it, changing the frequency of alleles in the population over time.

1 mark: before pollution, light moths more common (95%) because light bark provided camouflage from predators; 1 mark: dark moths more visible/more likely to be eaten before pollution; 1 mark: after pollution, dark bark meant dark moths better camouflaged, light moths more visible; 1 mark: dark moths had greater survival/reproductive success after pollution, passing on the dark-colour allele; 1 mark: population change over time (dark moths becoming more common) correctly identified as an example of natural selection.

Change in mass = 4.14 − 4.80 = −0.66 g. Percentage change = (change ÷ original mass) × 100 = (−0.66 ÷ 4.80) × 100 = −13.75%, which to 3 significant figures is −13.8%. Check by a second route: 4.80 g reduced by 13.75% is 4.80 × (1 − 0.1375) = 4.80 × 0.8625 = 4.14 g, which matches the given final mass, confirming the calculation. This is a decrease in mass, because the cylinder lost water to a sucrose solution that was more concentrated than the cell contents.

1 mark: correct change in mass calculated (4.14 − 4.80 = −0.66 g); 1 mark: correct formula used (change ÷ original × 100); 1 mark: correct substitution (−0.66 ÷ 4.80 × 100); 1 mark: correct final answer to 3 s.f., −13.8% (allow 13.8% decrease); 1 mark: correctly states this is a decrease in mass, with a valid reason (water lost to a more concentrated external solution).

Ultraviolet radiation is a mutagen, meaning it increases the rate of mutation, and overexposure to UV radiation from the sun can damage the DNA within skin cells, altering the base sequence of genes. If this damage affects a gene that normally controls the rate of cell division, the mutation can cause the affected cell to lose control of its cell cycle and begin to divide repeatedly and uncontrollably. This uncontrolled division produces a growing mass of abnormal cells, known as a tumour. If the tumour is malignant (cancerous), the abnormal cells can invade and damage surrounding healthy tissue, and in more advanced cases cells may break away and spread to other parts of the body, forming secondary tumours. Because DNA damage accumulates with exposure, the greater or more prolonged a person's exposure to UV radiation, the greater their risk of developing mutations of this kind and therefore of developing skin cancer; reducing UV exposure, for example using sunscreen or avoiding intense sun, reduces this risk. In summary: UV radiation is a mutagen that damages the DNA of skin cells; if this mutation affects a gene that controls cell division, the cell may divide uncontrollably, forming a tumour, which is cancerous (malignant) if it invades surrounding tissue or spreads to other parts of the body; greater or more prolonged UV exposure increases the risk of this happening.

Band A (5–6 marks): at least five relevant, accurate points made using appropriate specialist terms (e.g. mutagen, mutation, gene, cell division, tumour, malignant), clearly and coherently organised. Band B (3–4 marks): at least three relevant points, with reasonable use of specialist terms. Band C (1–2 marks): at least one relevant point; answer may be simplistic or lack specialist terms. Band D (0 marks): no relevant content / not creditworthy. Indicative content: UV radiation is a mutagen; it damages/alters the DNA (base sequence) of skin cells; if a gene controlling cell division is mutated, the cell may divide uncontrollably; uncontrolled cell division produces a tumour; a malignant tumour can invade surrounding tissue and/or spread to other parts of the body (secondary tumours); risk increases with greater/more prolonged UV exposure; reducing exposure (e.g. sunscreen) reduces risk.

Rate of reaction = volume of gas produced ÷ time taken = 12 cm³ ÷ 60 s = 0.20 cm³/s. Check by a second route: at 2% concentration, 24 cm³ ÷ 60 s = 0.40 cm³/s, matching the value already given in the table, confirming the method is being applied correctly, so 12 cm³ ÷ 60 s = 0.20 cm³/s is correct for 1%.

1 mark: correct working shown (12 ÷ 60); 1 mark: correct answer, 0.20 cm³/s (accept 0.2 cm³ s-1).

Total number of daisies counted = 6 + 8 + 5 + 9 + 7 + 6 + 8 + 7 + 9 + 5 = 70. Mean = total ÷ number of quadrats = 70 ÷ 10 = 7.0. Check by a second route: adding the values in a different order/pairing (6+5=11, 8+9=17, 5+7=12, 9+6=15, 7+8=15; 11+17+12+15+15=70) gives the same total of 70, confirming the sum, so the mean is 70 ÷ 10 = 7.0 daisies per quadrat.

1 mark: correct total (70); 1 mark: correct formula used (total ÷ number of quadrats); 1 mark: correct mean to 1 d.p., 7.0.

The graph should have hydrogen peroxide concentration (%) on the x-axis and rate of reaction (cm³/s) on the y-axis, each with a suitable linear scale using at least half of the available grid, and each of the four points (1, 0.20), (2, 0.40), (3, 0.55) and (4, 0.60) plotted accurately and joined with a single smooth curve. (a) The curve rises steadily from 1% to 4% concentration, but the increase in rate becomes smaller for each 1% increase in concentration (0.20 to 0.40 is a rise of 0.20, but 0.40 to 0.55 is a rise of only 0.15, and 0.55 to 0.60 is a rise of only 0.05), so the graph starts to level off/plateau at higher concentrations. (b) Reading from the curve at x = 2.5%, the value lies between the plotted points for 2% (0.40) and 3% (0.55); interpolating along the curve gives an estimated rate of approximately 0.47 cm³/s. Check by a second route: a straight-line interpolation between (2, 0.40) and (3, 0.55) gives 0.40 + 0.5 x (0.55-0.40) = 0.475 cm³/s, consistent with the estimate read from the curve, so approximately 0.47 cm³/s (accept 0.45-0.50 cm³/s) is correct.

Graph (2 marks): 1 mark for a suitable scale on each axis (using at least half the available grid) with axes correctly labelled with quantity and unit; 1 mark for all points plotted accurately (±½ small square) and joined with a single smooth curve (not point-to-point with sharp kinks). (a) 1 mark: rate of reaction increases as hydrogen peroxide concentration increases, but the rate of increase becomes smaller at higher concentrations (curve levels off). (b) 1 mark: value read from the candidate's own graph, consistent with their plotted curve (own figure rule); 1 mark: value within the range 0.45-0.50 cm³/s.

The bar chart should have two categories on the x-axis (mown area and unmown area), a suitable linear scale on the y-axis labelled 'mean number of daisy plants per quadrat', and two bars of equal width, accurately drawn to heights of 7.0 and 2.4 respectively. (a) The mean number of daisy plants per quadrat is much higher in the mown area (7.0) than in the unmown area (2.4) – almost three times as many (7.0 ÷ 2.4 ≈ 2.9). (b) Daisies are low-growing plants. In the unmown area, taller grasses and other plants can grow unchecked and shade the daisies, competing with them for light (and potentially also for water, nutrients and space); with less light available for photosynthesis, fewer daisies survive/grow well. In the mown area, regular mowing keeps the surrounding vegetation short, reducing this competition and allowing more light to reach the daisies, so more of them can grow and reproduce successfully, giving the higher abundance observed.

Bar chart (3 marks): 1 mark for a suitable scale and correctly labelled y-axis with units; 1 mark for two bars of equal width plotted accurately to the given values (7.0 and 2.4); 1 mark for clearly labelled categories (mown/unmown) with an appropriate chart title. (a) 1 mark: mown area has a much higher (approximately three times) mean daisy abundance than the unmown area, using the given values. (b) 1 mark: taller vegetation in the unmown area shades/outcompetes daisies for light (or another valid resource); 1 mark: mowing reduces this competition, allowing more daisies to grow successfully in the mown area.

The independent variable is the variable that is deliberately changed by the investigator. In Task 1, the student deliberately varied the concentration of hydrogen peroxide used, so this is the independent variable.

1 mark: hydrogen peroxide concentration.

To ensure that any difference in the rate of reaction is caused only by the change in hydrogen peroxide concentration, other variables that could also affect the rate must be kept the same each time. These include the total volume of hydrogen peroxide solution used, the mass or amount of catalase (for example, the mass of potato pieces or volume of enzyme extract used), the temperature at which the reaction is carried out, and the length of time over which gas is collected.

1 mark each for any two of: volume of hydrogen peroxide solution; mass/amount of catalase (e.g. mass of potato); temperature; time of gas collection. Max 2 marks.

If a different gas syringe or a different total volume of reaction mixture were used for different concentrations, this could itself affect the volume of gas collected or the rate measured, independently of the hydrogen peroxide concentration being tested. Using the same gas syringe and the same total reaction volume for every concentration tested means that hydrogen peroxide concentration is the only variable that changes between readings. This makes it a fair test, so any difference observed in the rate of reaction can be confidently attributed to the change in hydrogen peroxide concentration, allowing the results for different concentrations to be validly compared.

1 mark: standardising the apparatus/reaction volume removes it as a possible source of variation between readings; 1 mark: this ensures the test is fair, so any difference in rate can be attributed only to the change in hydrogen peroxide concentration (the independent variable).

With only ten quadrats sampled in each area, the sample may not fully capture the natural variation in daisy distribution across the whole field; an unusually high or low count in just one or two quadrats can have a disproportionately large effect on the calculated mean, reducing the reliability of the result. This could be improved by using a much larger number of randomly placed quadrats in each area (and/or repeating the whole investigation), which would give a more representative sample of each area and produce a mean that more reliably reflects the true daisy abundance.

1 mark: valid reason why a small number of quadrats may reduce reliability (e.g. small sample may not represent variation across the field / anomalous counts have a larger effect on the mean); 1 mark: valid improvement (e.g. use a greater number of randomly placed quadrats / repeat the investigation).

Random placement of quadrats is used so that every part of an area has an equal chance of being sampled, giving a sample that fairly represents the whole area. If quadrats were placed only along the edge of the field, this would introduce bias, because conditions at the edge (for example, more light where there is less shading from surrounding trees or buildings, more trampling from footpaths, or different drainage) may differ systematically from conditions in the rest of the field. The results would then reflect only the edge conditions rather than the field as a whole, so any conclusion drawn about daisy abundance across the whole field could be inaccurate or unrepresentative.

1 mark: sampling only at the edge introduces bias, because conditions there may differ from the rest of the field; 1 mark: this means the sample is not representative of the whole field, so conclusions about overall daisy abundance may be inaccurate.

The catalase used in the investigation was present in a fixed amount. As hydrogen peroxide concentration increases, more substrate molecules are available to collide with and bind to the enzyme's active sites, increasing the rate of reaction. However, once the concentration of hydrogen peroxide is high enough, the active sites of all of the available catalase molecules become saturated with substrate almost all of the time; the enzyme is then working at, or close to, its maximum possible rate. Adding more hydrogen peroxide beyond this point makes little additional difference to the rate, because the amount of enzyme, not the amount of substrate, has become the limiting factor. This explains why the rate levels off/increases by progressively smaller amounts at higher hydrogen peroxide concentrations.

1 mark: at high substrate concentration, the active sites of the (fixed amount of) enzyme become saturated with substrate; 1 mark: the enzyme is working at/close to its maximum rate, so the amount of enzyme (not substrate) becomes the limiting factor, and further increases in concentration have little additional effect.

Daisy plants grow and flower most abundantly when light intensity and temperature are higher, which typically occurs in mid-summer rather than early spring. In early spring, lower light intensity and lower temperatures mean plants have had less time and less favourable conditions for growth, so overall daisy numbers would be expected to be lower in both the mown and unmown areas than the mid-summer counts recorded. However, the mown area would still be expected to show relatively higher daisy abundance than the unmown area, for the same underlying reason as in mid-summer: even in spring, taller grasses in the unmown area would tend to shade and outcompete daisies for light, while mowing continues to reduce this competition in the mown area, so the general pattern (more daisies where mown) would likely still be observed, just with lower absolute numbers in both areas.

1 mark: daisy abundance in both areas expected to be lower in early spring than in mid-summer; 1 mark: valid reason (e.g. lower light intensity/temperature in spring means less growth/flowering has occurred); 1 mark: mown area still expected to show relatively higher abundance than unmown area, for the same competition-based reason as in summer.

A syringe (or a pipette) is used to measure small, precise volumes of liquid. Compared with a measuring cylinder, a syringe has finer graduations, so it can measure a given volume more precisely and accurately, improving the precision of the investigation.

1 mark: syringe or pipette; 1 mark: valid advantage, e.g. more precise/accurate/finer graduations than a measuring cylinder.

A gas syringe is connected to the reaction vessel and collects the gas produced; as gas enters, it pushes the plunger of the syringe outwards, allowing its volume to be read directly from the scale on the syringe, normally in cm³.

1 mark: gas syringe; 1 mark: cm³ (cubic centimetres).

An electronic (top-pan) balance is used to accurately measure the mass of a solid sample such as a piece of potato.

1 mark: (electronic/top-pan) balance.

The size of the potato cylinders used (their length and diameter, and hence their surface area and starting mass) should be kept the same for every sucrose concentration tested. This is important because a larger cylinder has a greater surface area and volume, so it could gain or lose more water than a smaller cylinder even in the same solution, independently of the effect of sucrose concentration. Keeping cylinder size constant ensures that any difference observed in the change in mass between concentrations is caused only by the difference in sucrose concentration, making it a fair test.

1 mark: size (length/diameter/surface area/mass) of potato cylinders should be kept the same across all concentrations tested; 1 mark: explains that differing cylinder size would independently affect water gained/lost, making comparisons between concentrations invalid/unfair.

A quadrat is a square frame of known area that is placed on the ground to sample and count the plants (or slow-moving animals) present within that defined area.

1 mark: quadrat.

Aseptic technique involves working close to a lit Bunsen burner and sterilising equipment such as the inoculating loop in the flame before and after use, to kill any unwanted microorganisms on it before it touches the agar or culture. The lid of the Petri dish is only partially opened, rather than fully removed, while inoculating the plate, in order to minimise the length of time that the agar surface is exposed to the air; this reduces the chance of unwanted microorganisms from the air settling on the agar and contaminating the culture.

1 mark: valid aseptic technique described, e.g. sterilising the inoculating loop in a Bunsen flame / working close to a lit Bunsen burner; 1 mark: partial opening minimises exposure of the agar to the air, reducing the risk of airborne contamination.

Sealing the Petri dish with tape in a criss-cross pattern, rather than sealing it completely around the edge, still allows a small amount of air exchange (preventing anaerobic conditions building up inside), while holding the lid securely in place to prevent it from being removed and to reduce the risk of the culture being spilled or escaping. The dish is incubated at a maximum of about 25°C rather than at human body temperature (37°C) because 37°C favours the growth of microorganisms adapted to living in the human body, which are more likely to include species that are pathogenic (harmful) to humans; incubating at a lower temperature reduces this risk while still allowing the culture to grow.

1 mark: criss-cross taping allows some air exchange while still securing the lid/reducing risk of spillage; 1 mark: lower incubation temperature (around 25°C) reduces the risk of growing microorganisms pathogenic to humans, compared with 37°C.

The purpose of destarching is to ensure that no starch is already present in a plant's leaves before an investigation into the need for light for photosynthesis begins. This means that if starch is later detected in a leaf that has been exposed to light during the investigation, it must have been produced by photosynthesis during the investigation itself, rather than being left over from before, making the investigation a valid test of the need for light. A plant is destarched by placing it in complete darkness (for example, in a dark cupboard) for at least 24-48 hours; without light, no further photosynthesis can occur, and any starch reserves already present are converted back to sugars and used or transported away by the plant, leaving the leaves starch-free.

1 mark: destarching ensures no starch is already present before the investigation, so a later positive result can be attributed to photosynthesis during the investigation; 1 mark: correct method described, i.e. leaving the plant in complete darkness for at least 24-48 hours.

The test tube containing the leaf and ethanol is heated indirectly by standing it in a water bath, rather than being heated directly over a naked flame, because ethanol is flammable.

1 mark: water bath.

The concentration of the solution placed inside each piece of Visking tubing (and the volume used) should be the same for every tube tested, only the concentration of the external (surrounding) solution should be varied. This is important because if the internal solution's concentration also varied between tubes, this would independently affect the amount of water entering or leaving each tube by osmosis, alongside any effect of the external solution. Keeping the internal solution the same for every tube ensures the investigation is a fair test of the effect of external solution concentration alone.

1 mark: concentration (and volume) of solution inside the Visking tubing should be the same for every tube tested; 1 mark: explains that varying the internal concentration too would confound the effect of the external solution, making the test unfair/invalid.

(a) Total magnification = eyepiece lens magnification x objective lens magnification = 10 x 40 = x400. (b) Real size = image size ÷ magnification = 32 mm ÷ 400 = 0.08 mm. Converting to micrometres (1 mm = 1000 um): 0.08 x 1000 = 80 um. Check by a second route: image size = real size x magnification = 80 um x 400 = 32 000 um = 32 mm, which matches the given image length, confirming the answer of 80 um.

(a) 1 mark: x400 (10 x 40). (b) 1 mark: correct method shown (image size ÷ magnification, i.e. 32 ÷ 400); 1 mark: correct final answer with correct unit conversion, 80 um.

(a) Total magnification with the x40 objective lens = 10 x 40 = x400. Image size = real size x magnification = 0.05 mm x 400 = 20 mm. Check by a second route: real size = image size ÷ magnification = 20 mm ÷ 400 = 0.05 mm, which matches the given real length, confirming the answer. (b) The x4 objective lens gives the lowest total magnification of the three options (x40, compared with x100 for the x10 objective and x400 for the x40 objective). A lower magnification shows a wider field of view, so more complete cells can be seen within it at once; a higher magnification shows a narrower field of view containing fewer, larger cells, some of which may be cut off at the edges.

(a) 1 mark: correct method shown (real size x magnification, i.e. 0.05 x 400); 1 mark: correct final answer, 20 mm. (b) 1 mark: x4 objective lens; 1 mark: correct reason, i.e. lowest magnification gives the widest field of view, showing the greatest number of complete cells.

(a) The results 12, 13 and 12 cm³ are close together, but 4 cm³ is much lower than the other three and does not fit the pattern, so it is the anomalous result. (b) A result this much lower than the others could be caused by a leak in the apparatus (so some gas escaped without being measured), a gas bubble being lost when connecting the apparatus, or a timing error (for example, starting the stopwatch late or stopping it early). (c) An anomalous result should not be included when calculating a mean, as it would distort the mean away from the true value; the student should exclude/discard this result and calculate the mean using only the remaining (reliable) repeats, ideally repeating the reading to replace it.

(a) 1 mark: 4 cm³. (b) 1 mark: valid possible cause, e.g. leak in apparatus / gas bubble lost / timing error. (c) 1 mark: exclude/discard from the mean, calculating the mean from the remaining results (accept: repeat to replace it).

The student's method mixes the enzyme and substrate solutions immediately after placing them in the water bath, before either solution has had time to actually reach the required temperature; the reaction would therefore begin at a temperature closer to room temperature (or whatever temperature the solutions started at) rather than the intended test temperature. This would make the measured rate an inaccurate/invalid measure of the reaction rate at the intended temperature. The method should be improved by leaving the two solutions separately in the water bath for several minutes (to allow them to fully reach the required temperature) before mixing them and starting timing, so that the reaction genuinely takes place at the intended temperature throughout.

1 mark: improvement identified, e.g. pre-incubate/leave both solutions in the water bath separately for several minutes before mixing; 1 mark: explains that immediate mixing means the reaction does not actually start at the intended temperature; 1 mark: explains that this makes the measured rate an inaccurate/invalid representation of the effect of that temperature.

A potometer works because water taken up by the shoot at one end pulls the marker bubble along the capillary tube; the distance the bubble moves in a given time is used to measure the rate of water uptake. Air, unlike water, is compressible. If additional air bubbles are trapped elsewhere in the apparatus (for example, in the rubber tubing or joints), these bubbles would compress and expand slightly as pressure in the system changes, absorbing some of the pressure change that should instead move the marker bubble along the tube. As a result, the marker bubble would move a different (smaller/less consistent) distance than it should for a given amount of water actually taken up, so its movement would no longer give an accurate, reliable measure of the true rate of water uptake by the shoot.

1 mark: trapped air bubbles are compressible, unlike water; 1 mark: they absorb/interfere with the pressure changes that should move the marker bubble; 1 mark: this means the marker bubble's movement no longer accurately/reliably represents the true rate of water uptake by the shoot.

Flaming (sterilising) an inoculating loop between uses is intended to destroy any microorganisms remaining on it, so that only the intended bacterial culture, applied in a controlled and consistent way, is transferred to each agar plate. If the loop is not flamed between transferring bacteria to two different plates, unwanted microorganisms (for example, from the air or a previous culture) could be carried over onto the second plate, contaminating it, or an inconsistent amount of the original culture could be transferred. Either possibility means that any difference later observed between the two plates (for example, in the size of an inhibition zone) could be caused by this contamination or inconsistency, rather than purely by the effect of the antibiotic being tested, so the comparison between the plates would no longer be fair, making the results invalid.

1 mark: not flaming the loop risks transferring unwanted microorganisms between plates (cross-contamination); 1 mark: or risks an inconsistent amount of the original culture being transferred; 1 mark: either possibility means any difference between plates cannot be confidently attributed only to the antibiotic being tested, making the comparison invalid.

To check whether the plant is fully destarched, the student could remove and test one leaf for starch, using the standard iodine test (boiling the leaf in ethanol to remove chlorophyll, then adding iodine solution), before beginning the main investigation. If this leaf shows a negative result (no blue-black colour, remaining orange-brown), this confirms the plant contains no starch and is fully destarched, so the main investigation can proceed; if it still shows a positive result, the plant should be left in the dark for longer. This check is important because if the plant were not fully destarched, any starch detected in a leaf at the end of the main investigation could simply be starch that was already present before the investigation began, rather than starch newly produced by photosynthesis during the investigation. This would make it impossible to draw a valid conclusion about whether photosynthesis (and therefore starch production) had actually occurred as a result of the conditions being tested.

1 mark: check described, i.e. remove and starch-test (iodine test) one leaf before starting the main investigation; 1 mark: a negative result on this leaf confirms the plant is fully destarched; 1 mark: explains that without this check, a positive result later in the main investigation could be due to starch left over from before, rather than starch produced during the investigation, invalidating the conclusion.

(a) Counting bubbles as a measure of the rate of photosynthesis assumes that every bubble produced has the same (or a very similar) volume. (b) In reality, bubbles produced are unlikely to all be exactly the same size; some may be larger or smaller than others, so counting the number of bubbles may not accurately reflect the actual total volume of gas produced in a given time, making bubble counting an imprecise/inaccurate measure of the true rate of photosynthesis. (c) A more accurate method would be to collect all of the gas produced over a set period of time in a suitable graduated container, such as a gas syringe (or an inverted measuring cylinder/funnel arrangement), and measure its actual volume directly. This measures the true volume of gas produced rather than relying on the assumption that all bubbles are the same size, giving a more accurate and reliable measure of the rate of photosynthesis.

(a) 1 mark: assumption that all bubbles are the same/a similar size. (b) 1 mark: bubbles are unlikely to actually be the same size, so counting them is an imprecise/inaccurate measure of gas volume. (c) 1 mark: valid alternative method, e.g. collect gas in a gas syringe/graduated container and measure its volume directly; 1 mark: correct explanation of why this is more accurate (measures actual volume rather than relying on an assumption about bubble size).

(a) Any single reading is affected by random experimental error and natural variation, so it may not accurately represent the true rate of reaction. By repeating the reading three times and calculating a mean, the effect of any one unusually high or low reading, caused by random error, is reduced (averaged out), so the mean is more likely to be close to the true rate of reaction than any single reading would be; this makes the result more reliable and more likely to be reproducible if the investigation were repeated. (b) A repeat result that differs greatly from the other two repeats (an anomalous result) is likely to have been affected by an error in carrying out that particular repeat, for example a leak in the apparatus, a timing error, or a misread scale, rather than reflecting a genuine difference in the rate of reaction. If this anomalous value were included when calculating the mean, it would pull the mean away from the true rate, making the mean less accurate. The student should therefore identify and exclude such anomalous results (ideally repeating that reading to replace it) before calculating the mean, so that the mean is based only on reliable, consistent data.

(a) 1 mark: repeating reduces the effect of random error/anomalies present in any one single reading; 1 mark: the mean is therefore more likely to be close to the true value, making the result more reliable/reproducible. (b) 1 mark: an anomalous result is likely to be due to an error made during that particular repeat; 1 mark: including it would distort the mean away from the true value, so it should be excluded (and ideally repeated) before calculating the mean.

Several hazards are present in this procedure, along with ways of reducing the risk associated with each. Ethanol is highly flammable, so there is a hazard of it catching fire if it comes into contact with a naked flame; this risk is reduced by heating the ethanol indirectly, using a water bath, rather than a Bunsen burner flame directly, and by ensuring no naked flames are in use nearby. Ethanol vapour is also flammable and can be harmful if inhaled in large quantities, so the risk is reduced by carrying out the procedure in a well-ventilated room, for example near an open window or under a fume cupboard, and by keeping the test tube away from flames. The water bath itself involves boiling water, which presents a scalding/burn hazard; this risk is reduced by using tongs or a test-tube holder to handle hot apparatus, rather than bare hands, and by taking care when placing the test tube into or removing it from the hot water. Iodine solution can irritate the skin and eyes if it comes into contact with them; the risk is reduced by wearing eye protection (safety goggles or glasses) throughout the procedure, avoiding skin contact, and washing off any spills immediately with plenty of water. Finally, glassware such as test tubes and boiling tubes may crack, particularly with rapid changes in temperature; this risk is reduced by checking glassware for existing cracks before use, handling it carefully, and allowing very hot glassware to cool before handling it directly. Together, these precautions reduce the risk associated with each identified hazard: ethanol is flammable, so it should be heated indirectly in a water bath, away from naked flames, in a well-ventilated room; boiling water presents a scalding hazard, so hot apparatus should be handled with tongs/a test-tube holder; iodine solution can irritate skin and eyes, so eye protection should be worn and spills avoided; glassware may crack with sudden temperature change, so it should be checked and handled carefully.

Band A (5-6 marks): at least five relevant hazard-and-control pairs identified, using appropriate specialist terms (e.g. flammable, indirect heating, scalding, irritant), clearly and coherently organised. Band B (3-4 marks): at least three relevant hazard-and-control pairs identified, with reasonable use of specialist terms. Band C (1-2 marks): at least one relevant hazard-and-control pair identified; answer may be simplistic or lack detail. Band D (0 marks): no relevant content / not creditworthy. Indicative content: ethanol is flammable - heat indirectly using a water bath, away from naked flames; ethanol vapour flammable/harmful if inhaled - use a well-ventilated area; boiling water in the water bath is a scalding/burn hazard - use tongs/test-tube holder to handle hot apparatus; iodine solution can irritate skin/eyes - wear eye protection, avoid contact, wash off spills; glassware may crack (especially with sudden temperature change) - check for cracks and handle carefully.

(a) Change in mass = 9.65 - 8.20 = 1.45 g. Percentage change = (change ÷ original mass) x 100 = (1.45 ÷ 8.20) x 100 = 17.68...%, which to 3 significant figures is 17.7%. Check by a second route: 8.20 g increased by 17.7% is 8.20 x 1.177 = 9.65 g (to 3 s.f.), matching the given final mass, confirming the answer. (b) There was a net movement of water into the Visking tubing, by osmosis, because the 25% sucrose solution inside the tubing had a lower water concentration (was more concentrated) than the distilled water outside, so water moved from the dilute solution (distilled water) to the more concentrated solution (inside the tubing) through the selectively permeable membrane.

(a) 1 mark: correct change in mass calculated (9.65 - 8.20 = 1.45 g); 1 mark: correct formula/substitution (1.45 ÷ 8.20 x 100); 1 mark: correct final answer to 3 s.f., +17.7% (accept 17.7% increase). (b) 1 mark: correct direction (net movement of water into the tubing) with a correct reason (sucrose solution inside more concentrated than distilled water outside).

Change in mass = 7.61 - 9.40 = -1.79 g. Percentage change = (change ÷ original mass) x 100 = (-1.79 ÷ 9.40) x 100 = -19.04...%, which to 3 significant figures is -19.0%. Check by a second route: 9.40 g decreased by 19.0% is 9.40 x (1 - 0.190) = 9.40 x 0.810 = 7.61 g (to 3 s.f.), matching the given final mass, confirming the answer. This is a decrease in mass, because water moved out of the tubing, by osmosis, from the dilute distilled water inside into the more concentrated 25% sucrose solution outside.

1 mark: correct change in mass calculated (7.61 - 9.40 = -1.79 g); 1 mark: correct formula/substitution (-1.79 ÷ 9.40 x 100); 1 mark: correct final answer to 3 s.f., -19.0%; 1 mark: correctly states this is a decrease in mass, with a valid reason (water moved out into the more concentrated external sucrose solution by osmosis).

(a) Antibiotic C produced the largest inhibition zone, with a diameter of 22 mm, compared with 14 mm for Antibiotic A and 8 mm for Antibiotic B. A larger inhibition zone means that the antibiotic diffused outward through the agar and prevented (or killed) bacterial growth over a larger surrounding area, indicating that Antibiotic C was the most effective of the three against the bacterium being tested. (b) Because part of Antibiotic C's inhibition zone extended beyond the edge of the agar, the zone could not actually grow to its full, true size within the dish (its edge was cut off by the boundary of the agar, not by the extent of the antibiotic's effect), so the diameter measured is smaller than the true size of the zone that would have formed with more space available. This makes the measured value an underestimate, so it cannot be validly compared with the fully-formed zones measured for Antibiotics A and B. To obtain a valid result, the student should repeat the test with Antibiotic C's disc placed centrally on a fresh agar plate, with enough space around it for the inhibition zone to form completely within the agar.

(a) 1 mark: Antibiotic C identified as most effective; 1 mark: correct explanation that a larger inhibition zone diameter (22 mm, greater than A and B) indicates greater inhibition of bacterial growth/greater effectiveness. (b) 1 mark: explains that the true zone size could not be measured because it extended beyond the edge of the agar, so the recorded value is an underestimate/not comparable; 1 mark: valid improvement, e.g. repeat with the disc placed centrally, with sufficient space for the zone to form fully.

(a) The total clear zone has a diameter of 18 mm, so its radius is \( 18 \div 2 = 9 \text{ mm} \). Area \( = \pi r^2 = 3.14 \times 9^2 = 3.14 \times 81 = 254.34 \text{ mm}^2 \), which to 3 significant figures is 254 mm². (b) The disc has a diameter of 6 mm, so its radius is \( 6 \div 2 = 3 \text{ mm} \). Area \( = \pi r^2 = 3.14 \times 3^2 = 3.14 \times 9 = 28.26 \text{ mm}^2 \), which to 3 significant figures is 28.3 mm². (c) Area of inhibited bacterial growth = total clear zone area minus disc area \( = 254.34 - 28.26 = 226.08 \text{ mm}^2 \), which to 3 significant figures is 226 mm². Check by a second route: 254 (3 s.f. total) − 28.3 (3 s.f. disc) = 225.7, which rounds to 226 mm², consistent with the unrounded calculation, confirming the answer.

(a) 1 mark: correct substitution (radius = 9 mm, \( \text{area} = 3.14 \times 9^2 \)); 1 mark: correct final answer to 3 s.f., 254 mm². (b) 1 mark: correct final answer to 3 s.f., 28.3 mm² (radius = 3 mm, \( \text{area} = 3.14 \times 3^2 = 28.26 \)). (c) 1 mark: correct method (total clear zone area minus disc area); 1 mark: correct final answer to 3 s.f., 226 mm² (allow ecf from candidate's own values in (a) and (b)).