2025 OCR GCSE Gateway Science - Biology A - J247 模擬試題連答案詳解

Enzymes are specific. The active site of a protease fits protein (lock and key) but not starch, so only protein is broken down. Enzymes are not used up and starch does not denature them.

B [1]. A, C and D are incorrect: enzymes are proteins, are not used up, and are not denatured by a substrate they do not act on. ALLOW no other answer.

Bacteria are prokaryotes: they have a cell wall and DNA free in the cytoplasm, with no nucleus. Root hair cells have a nucleus; red blood cells and sperm cells lack a cell wall.

A [1]. Only one answer is creditworthy.

The sugar solution has a lower water concentration than the potato cells, so water leaves the potato by osmosis and its mass falls.

C [1]. IGNORE any other option.

Surface area = \(6 \times 2 \times 2 = 24\ \text{cm}^2\). Volume = \(2 \times 2 \times 2 = 8\ \text{cm}^3\). Ratio = \(24 : 8 = 3 : 1\).

C [1]. Working (not required for the mark): surface area 24, volume 8, ratio 3 : 1.

The receptor detects the stimulus. A sensory neurone carries the impulse to a relay neurone in the CNS, then a motor neurone carries it to the effector.

A [1].

Evaporation of sweat takes energy from the skin, cooling the body. Constriction, hair raising and shivering are responses to cold.

A [1].

Benedict's reagent heated with a reducing sugar gives a colour change from blue to green, yellow, orange or brick-red. Starch is tested with iodine, protein with biuret and lipid with ethanol.

C [1].

Photosynthesis: carbon dioxide + water → glucose + oxygen, using light energy. Option A is aerobic respiration.

B [1].

Enzymes are specific because the active site has a shape that fits only its own substrate (lock-and-key model). The correct answer is B.

B [1]. A: enzymes are not chosen by substrate size alone. C: enzymes are not used up. D: describes temperature dependence, not specificity.

Benedict's reagent starts blue. On heating with a reducing sugar such as glucose it changes through green and yellow to brick red as the sugar concentration rises. The answer is D.

D [1]. Black is the iodine test positive result for starch; purple is the biuret test positive result for protein.

Surface area = \(6 \times 3 \times 3 = 54\ \text{cm}^2\). Volume = \(3 \times 3 \times 3 = 27\ \text{cm}^3\). Ratio = \(54:27 = 2:1\). The answer is C.

C [1]. Method: surface area 54 and volume 27 (not separately credited as this is a 1-mark item). A is the ratio for a 1 cm cube; D is the inverted ratio.

The receptor detects the stimulus. The impulse passes along a sensory neurone to a relay neurone in the CNS, then along a motor neurone to the effector (muscle). The answer is A.

A [1]. B, C and D have neurones or structures in the wrong order.

The pancreas secretes both insulin and glucagon to control blood glucose concentration. Insulin is not made by the pituitary, oestrogen is made by the ovary, and testosterone is made by the testes. The answer is D.

D [1]. A: insulin is made by the pancreas. B: adrenal glands make adrenaline. C: ovaries make oestrogen and progesterone.

When the body is too hot, more sweat is produced, and its evaporation transfers energy from the skin. Blood vessels near the skin widen (vasodilation) so more heat is lost by radiation. The answer is B.

B [1]. Both changes must be correct in the same option.

The sugar solution has a lower water concentration than the potato cells, so water moves out of the cells through the partially permeable membrane by osmosis. The cylinder loses mass. The answer is C.

C [1]. A describes a dilute external solution. B and D wrongly name sugar as the substance moving.

(a) Contamination is reduced by aseptic technique, for example flaming the inoculating loop or taping the lid on.

(b) Radius = 18 ÷ 2 = 9 mm. Area = 3.14 × 9² = 3.14 × 81 = 254.34 mm², which is 254 mm² to 3 significant figures.

(c) The clear zone for A (18 mm) is larger than for B (12 mm) and C (0 mm), so more bacteria were killed or prevented from growing.

(d) Some bacteria had a mutation making them resistant to antibiotic B, so they were not killed and reproduced to form colonies.

(a) [1] Any one from: keep the lid on the dish / tape the lid (loosely) / only lift the lid briefly; flame the loop or spreader before use; work close to a Bunsen flame; use sterilised agar and dish. IGNORE 'wear gloves' or 'wash hands' alone. DO NOT ALLOW 'sealing the dish completely' (anaerobic pathogens risk).

(b) [2] Correct final answer 254 (mm²) = 2 marks. If the answer is wrong or missing: 1 mark for radius = 9 (mm) OR for 3.14 × 9² / 254.34 shown. ALLOW 1 mark for ECF from an incorrect radius correctly squared and multiplied by 3.14, rounded to 3 s.f. Using diameter as radius gives 1018 mm²: award 0 for the radius mark but 1 mark ECF is NOT available unless working shown; award max 1 for 3.14 × 18² shown. ALLOW 254.3 or 254.34 for full marks (rounding to 3 s.f. not required if the unrounded value is correct). Unit mm² is given on the answer line.

(c) [1] Largest (clear) zone / 18 mm is bigger than the others (AW). Comparison with at least one other antibiotic is not required but a comparison must be implied. IGNORE 'kills the most bacteria' without reference to the zone.

(d) [1] The bacteria were resistant (to antibiotic B) / had a mutation / had an allele that gave resistance / survived and reproduced. IGNORE 'the bacteria adapted' or 'the bacteria became immune'.

(b) Mean = (20 + 18 + 22) / 3 = 60 / 3 = 20 mm.
Other parts are explained in the marking scheme.

(a) R does not kill / stop growth of the bacteria (AW) OR bacteria are resistant to R [1]. DO NOT ALLOW 'R is not an antibiotic'.

(b) Correct answer 20 [2]. If the answer is wrong, award 1 mark for (20 + 18 + 22) / 3 or for a total of 60. Unit mm is given; IGNORE missing unit.

(c) Q has the largest mean ring / 20 mm compared with 13 mm for P (AW) [1]. ALLOW ECF from (b). Must refer to data.

(d) Any two from: [2]
- 37 \(^\circ\)C is close to body temperature so would favour growth of human pathogens (AW);
- 25 \(^\circ\)C makes it less likely that harmful bacteria grow (AW);
- loose tape lets oxygen in so anaerobic pathogens do not grow (AW);
- tape stops lid being removed so harmful bacteria cannot escape / be exposed (AW).

(e) A cold is caused by a virus and antibiotics do not kill viruses (AW) [1].

(b) Total = 3 + 5 + 0 + 4 + 2 + 6 = 20; mean = 20 / 6 = 3.33 daisies per quadrat.
(c) Quadrat area = 0.5 x 0.5 = 0.25 m\(^2\). Number of quadrats that fit in the field = 800 / 0.25 = 3200. Estimated total = 3.33 x 3200 = about 10 667 daisies.

(a) To avoid bias / so that the sample is representative of the whole field (AW) [1]. DO NOT ALLOW 'to make it a fair test'.

(b) Correct answer 3.3 or 3.33 [2]. If incorrect, award 1 mark for 20 / 6 or total of 20.

(c) Correct answer 10 600 to 10 700 [3]. If incorrect, award:
- 1 mark for quadrat area = 0.25 m\(^2\);
- 1 mark for 800 / 0.25 = 3200 (or mean per m\(^2\) = 13.3);
- 1 mark for mean x 3200 (or 13.3 x 800).
ALLOW ECF from (b).

(d) Larger sample gives a more reliable / representative mean / reduces effect of anomalies or chance (AW) [1].

(c) Decrease = 4800 − 1200 = 3600. Percentage decrease = \( \frac{3600}{4800} \times 100 = 75\% \).

(a) lymphocyte(s) / B lymphocyte(s) [1]. ALLOW B cells. DO NOT ALLOW 'white blood cells' alone, phagocytes.
(b) Any three from: vaccine contains dead/weakened/inactive pathogen or its antigens (1); antigens stimulate lymphocytes / immune system to make specific antibodies (1); memory cells are produced / remain in the blood (1); on infection by live pathogen, antibodies produced faster / in greater amounts (1); pathogen destroyed before symptoms develop (1). IGNORE 'vaccine kills the virus'. [3]
(c) Correct answer 75 (%) = 2 marks. If incorrect: 4800 − 1200 / 3600 (1) or 3600 / 4800 × 100 (1). ECF from a wrong subtraction. [2]
(d) Any two from: many people not immune / unprotected (1); the virus is passed on to more people / more susceptible hosts (1); herd immunity is lost / too low (1); infected people are in contact with others and chains of transmission are not broken (1). [2]

(c) \( (36 + 40 + 38) \div 3 = 38 \) cm³.
(d) Rate = \( 25 \div 60 = 0.41666... \) = 0.42 cm³/s (2 s.f.).
(e) Volume increases as temperature increases up to 40 °C, then decreases at 50 °C because high temperature denatures the enzyme.

(a) temperature (of the water bath / solution) [1]
(b) Any two from: volume of hydrogen peroxide (1); concentration of hydrogen peroxide (1); mass / size / surface area of liver (1); time of collection (1); pH (1). ALLOW amount of liver. IGNORE 'same equipment', 'amount of enzyme' alone is ALLOWED as liver equivalent. [2]
(c) 38 [1]
(d) 0.42 = 2 marks. If incorrect: 25 ÷ 60 / 0.4166... (1) for method, or correct value not rounded to 2 s.f. ALLOW 0.41 for 1 mark only (truncation). [2]
(e) Volume increases as temperature increases up to (maximum at) 40 °C, then decreases (1); at 50 °C the enzyme is denatured / active site changes shape so substrate cannot fit (1). DO NOT ALLOW 'enzyme is killed' for the second mark. [2]

Carbohydrate digestion raises blood glucose. The pancreas detects this and secretes insulin, which is carried in the blood to the liver. The liver converts glucose to glycogen for storage, lowering blood glucose back to normal. In type 1 diabetes the pancreas does not produce enough insulin, so glucose remains high after eating. It is treated with insulin injections (with diet control), and the dose is matched to carbohydrate eaten.

(a) pancreas [1]
(b)* Level of response, 6 marks.
Level 3 (5–6 marks): Clear, detailed account covering BOTH strands, including the link between insulin and glucose uptake/glycogen storage in healthy person AND failure of insulin production with injections as treatment. Logically ordered, accurate scientific terms.
Level 2 (3–4 marks): Some correct description of both strands, or a detailed account of one strand with a brief mention of the other. Some linkage between ideas.
Level 1 (1–2 marks): Simple statements, e.g. 'insulin lowers glucose' or 'diabetics inject insulin', with little or no linkage.
Level 0 (0 marks): No relevant content.
Indicative content: glucose from digested carbohydrate absorbed so blood glucose rises; pancreas detects the rise; pancreas secretes insulin; insulin carried in blood; insulin causes cells (liver/muscle) to take up glucose; glucose converted to glycogen and stored in liver/muscles; blood glucose falls back to normal / negative feedback; type 1: pancreas does not produce enough/any insulin; blood glucose stays high / not removed from blood; treated by insulin injections; dose matched to food eaten / monitor blood glucose; avoid excess carbohydrate. ALLOW insulin pump. DO NOT ALLOW confusion of insulin with glucagon as the hormone lowering glucose; do not credit 'insulin turns glucose into glycogen' without cell uptake only as a Level 1–2 contribution.
(c) glycogen [1]. DO NOT ALLOW glucagon.

(a) Any one precaution, e.g. flame the loop, wash hands/disinfect the bench, tape the lid on the dish.

(b) Radius = 22 ÷ 2 = 11 mm. Area = \(\pi \times 11^2\) = \(\pi \times 121\) = 380.13... mm². To 3 s.f. = 380 mm².

(c) The clear zone for Q (22 mm) is the largest, so more bacteria were killed or prevented from growing around it. The water disc has no zone, which shows the zones are caused by the antibiotics.

(d) Bacteria vary because of random mutation. A mutation may make a bacterium resistant to the antibiotic. When the antibiotic is used, non-resistant bacteria are killed, but resistant bacteria survive, reproduce and pass on the resistance allele. Over generations the population is mostly resistant (natural selection). To slow this, doctors should prescribe antibiotics only for bacterial infections when needed, and patients should complete the full course. Good hygiene and isolating infected patients reduce spread.

(a) 1 mark. Any one from: flame/sterilise the loop or spreader; disinfect bench/wash hands; tape the lid (do not seal fully) / keep the lid on; use sterile agar/equipment. IGNORE 'wear gloves' alone. [1]

(b) Check the final answer first: 380 (mm²) = 3 marks. ALLOW 379 or 380.1 for 2 marks only if not given to 3 s.f.; 3 s.f. required for the final mark.
Otherwise: radius = 11 (mm) [1]; area = \(\pi \times 11^2\) = 380.13... [1]; rounded correctly to 3 s.f. = 380 [1]. ECF from an incorrect radius, e.g. 22 used as radius gives 1520 (mm²) for 2 marks. Unit mm² (not required for marks but DO NOT ALLOW if wrong unit with the final mark). [3]

(c) Q has the largest (diameter/area of) clear zone [1]. ALLOW 'it killed the most bacteria' AW. DO NOT ALLOW 'the zone is bigger' with no comparison to the others. [1]

(d)* Level of response, 5 marks. Indicative content:
- random mutation produces resistance allele in some bacteria
- antibiotic kills non-resistant bacteria
- resistant bacteria survive and reproduce / pass on the allele
- proportion of resistant bacteria increases (natural selection)
- doctors only prescribe antibiotics when necessary / not for viral infections
- patients finish the full course
- use of different/narrow-spectrum antibiotics; good hygiene/isolation of infected patients

Level 3 (5 marks, 4 for the lower end): clear and logical explanation of how resistance arises including selection AND at least two valid actions to slow the spread. Level 2 (3-2 marks): explains some of the mechanism OR gives actions with some explanation; may lack linkage. Level 1 (1 mark): one or two simple relevant points, e.g. 'bacteria become resistant' or 'finish the course', with no explanation. Level 0 (0 marks): no creditable response. Scientific coverage fixes the level; clarity and organisation of the answer set the mark within the level. IGNORE 'the bacteria get used to the antibiotic' / 'the person becomes resistant' (DO NOT ALLOW). [5]

Total 10 marks.

(a) Enzymes speed up reactions without being used up, so they are catalysts.
(b) Starch turns iodine blue-black. When no starch is left, the drop stays the orange/brown colour of iodine.
(c) \(1000 \div 90 = 11.11\), which is 11.1 to 1 d.p.
(d) Rate rises from 4.0 at pH 4 to 11.1 at pH 6 and 16.7 at pH 7, then falls to 2.5 at pH 9.
(e) At high pH the bonds holding the enzyme change, so the active site changes shape (denatured). Starch no longer fits the active site, so few reactions happen.
(f) pH 7 was only compared with values 1 or 2 units away, so the true optimum could be between 6 and 8. Testing pH 6.5, 7.5 etc. and repeating each pH would give a more reliable answer.

(a) catalyst / biological catalyst [1]. IGNORE protein. DO NOT ALLOW 'catalyse'/'catalysis' alone.

(b) iodine / drop stays orange / brown / does not turn blue-black [1]; because there is no starch left / starch has all been broken down [1]. ALLOW 'stays yellow'. Second mark depends on a correct colour idea.

(c) 11.1 [2]. If answer wrong, ALLOW 1000 ÷ 90 / 11.11 / 11 for [1]. ALLOW 11.2 for [1] only if working shown (rounding error).

(d) Any two from: rate increases from pH 4 to pH 7 / up to pH 7 [1]; highest rate at pH 7 [1]; then rate decreases / falls from pH 7 to pH 9 [1]. Use of data values to support can gain credit. IGNORE 'it goes up then down' without pH reference for second mark only. Max [2].

(e) enzyme is denatured / active site changes shape [1]; substrate / starch no longer fits (the active site) / fewer enzyme-substrate complexes form [1]. DO NOT ALLOW 'enzyme is killed' for the first mark. IGNORE 'enzyme is destroyed' alone.

(f) test more pH values / smaller intervals close to pH 7 / between pH 6 and 8 (e.g. 6.5, 7.5) [1]; repeat the investigation (at each pH) and calculate a mean [1]. ALLOW use of a pH meter/probe or more accurate timing (e.g. shorter sampling intervals) for [1] as an alternative to the repeat point. Max [2].

(a) The pancreas.
(b) Increase = 8.0 − 5.0 = 3.0 mmol/dm³. Percentage increase = \(3.0 \div 5.0 \times 100 = 60\%\).
(c) After glucose rises, the pancreas releases insulin. Insulin makes glucose move into body cells and the liver converts glucose to glycogen for storage, so blood glucose falls back towards normal.
(d) In type 1 diabetes the pancreas does not make enough insulin, so glucose stays in the blood. It is controlled by injecting insulin (often before meals), monitoring blood glucose, and managing carbohydrate intake and exercise. In type 2 diabetes body cells no longer respond properly to insulin, so injecting more insulin is not the main approach; reducing risk or controlling it relies on a balanced diet with less sugar, losing excess body mass and regular exercise.

(a) pancreas [1]. DO NOT ALLOW pancreases/pancreatic cells alone only if organ is clearly wrong; ALLOW 'islets/cells in the pancreas'.

(b) 60 (%) [2]. If answer wrong, 3.0 / 3 seen [1] or (3 ÷ 5) × 100 [1]. ALLOW ECF for a wrongly read increase from the table. Answer 160% scores [1].

(c) insulin is released (by the pancreas) in response to high glucose [1]; glucose moves / is taken up into cells / liver converts glucose to glycogen [1]. ALLOW 'glucose stored as glycogen'. DO NOT ALLOW 'insulin turns glucose into glycogen' without reference to the liver for second mark unless 'converted into glycogen' is stated. IGNORE glucagon.

(d)* Level of response, [6]. Indicative content:
- Type 1: pancreas does not produce enough/any insulin (cells destroyed); glucose is not taken up by cells / not converted to glycogen, so blood glucose remains high.
- Type 1 control: insulin injections / pump; taken before/after meals to match carbohydrate intake; monitoring blood glucose; controlling carbohydrate intake and exercise.
- Type 2: body cells do not respond properly to insulin (insulin resistance), so extra insulin is less effective as main approach.
- Type 2 risk reduction: balanced diet / less sugar and refined carbohydrate; lose excess body mass / avoid obesity; regular exercise; (medication in some cases).

Level 3 (5–6 marks): Clear, accurate explanation of the cause of high glucose in type 1, the insulin treatment, AND the type 2 lifestyle approach with a reason why insulin alone is not the main approach. Ideas are logically linked and well organised.
Level 2 (3–4 marks): Covers at least two of the three bullet strands with some correct linked explanation, or all three with limited detail/links. Mostly clear.
Level 1 (1–2 marks): Simple, mainly isolated statements about diabetes, e.g. 'insulin is injected' or 'eat less sugar', with little linking.
Level 0 (0 marks): No creditworthy response.

Best-fit marking: scientific coverage fixes the level; clarity and logical order of communication decide the position within the level. DO NOT ALLOW 'insulin cures type 1 diabetes'. IGNORE reference to glucagon unless incorrect.

The vaccine contains a harmless form of the virus, so the child makes memory cells. On re-infection these cells rapidly produce the specific antibodies, so the virus is destroyed before symptoms develop.

A (1 mark). B is wrong: vaccines do not contain antibiotics and antibiotics do not kill viruses. C is wrong: platelets help blood clotting. D is wrong: vaccines contain antigens, not antibodies.

Human pathogens grow best near body temperature (37 °C). Incubating at no more than 25 °C reduces the chance of growing harmful pathogens in school.

C (1 mark). A is wrong: many bacteria grow well at 37 °C. B is wrong: agar remains solid at 37 °C. D is not the reason given.

Aphids are the food source of ladybirds. With fewer aphids, ladybirds have less food, so more starve or fail to reproduce and numbers fall.

B (1 mark). A and D give incorrect reasoning about the food web; C is wrong because ladybirds in this web eat aphids, not plants.

Carbon dioxide is a reactant in photosynthesis. Raising its concentration removes a limiting factor, so more glucose is made and the plants grow more.

D (1 mark). A is wrong: carbon dioxide enters through the leaves. B and C are not effects of added carbon dioxide.

Both parents are Ff. Offspring: FF, Ff, Ff, ff. One in four is ff, so the probability is 25%.

C (1 mark). A would apply only if a parent were homozygous dominant; B is the probability of a carrier child; D is the probability of not having the disorder.

Mean = (3 + 5 + 4 + 8 + 5) ÷ 5 = 25 ÷ 5 = 5 per \(\text{m}^2\). Estimate = \(5 \times 200 = 1000\).

B (1 mark). A is the total of the counts; C multiplies the total by 200; D is the field area.

Antibiotic use kills non-resistant bacteria. Resistant ones survive, reproduce and pass on the allele, so the resistant proportion rises by natural selection.

A (1 mark). B, C and D wrongly suggest bacteria change deliberately or individually in response to the antibiotic.

The pancreas does not make enough insulin in type 1 diabetes. Without insulin, the liver does not convert glucose to glycogen, so blood glucose rises after the meal and stays high.

D (1 mark). A, B and C describe changes that would not occur without insulin.

A vaccine contains dead or weakened pathogens (antigens). These stimulate lymphocytes to produce antibodies and memory cells, so a faster response occurs if the real pathogen enters later. Option A is wrong because vaccines do not contain ready-made antibodies. Option C is wrong because antibiotics are not in vaccines and do not work on viruses. Option D is wrong because vaccines do not contain donor white blood cells.

B [1 mark]. Any other letter or more than one letter = 0.

Measles is caused by a virus. Malaria is caused by a protist, not a bacterium. Salmonella food poisoning is caused by a bacterium, not a protist. Athlete's foot is caused by a fungus, not a virus. Only A is correct.

A [1 mark]. Any other letter or more than one letter = 0.

With far fewer rabbits, less grass is eaten so the amount of grass increases. The foxes lose most of their food supply, so fewer survive and the fox population decreases. This is option A.

A [1 mark]. Any other letter or more than one letter = 0.

A larger clear zone means more bacterial growth was prevented around the disc. Antibiotic R has the largest clear zone (25 mm), so it is the most effective. Antibiotic P has no clear zone, so it had no effect on these bacteria.

C [1 mark]. Any other letter or more than one letter = 0.

Each parent is Ff. A genetic cross gives FF, Ff, Ff and ff in equal proportions. Only ff has the disorder, which is 1 in 4 = 25%.

B [1 mark]. Any other letter or more than one letter = 0.

Biological control uses a living organism rather than a chemical, so no toxic residues accumulate in food chains. It is usually slower than spraying, rarely removes a pest completely, and the introduced predator can affect other species.

B [1 mark]. Any other letter or more than one letter = 0.

Random mutation produced a few resistant bacteria. The antibiotic killed non-resistant bacteria, so the resistant ones survived, reproduced and passed on the allele. This is natural selection. Bacteria do not change deliberately or by getting used to antibiotics.

B [1 mark]. Any other letter or more than one letter = 0.

(b) Radius = 14 ÷ 2 = 7 mm. Area = \( \pi \times 7^2 = 153.94... \) mm² = 154 mm² (3 s.f.).
Other parts are recall/application as in the marking scheme.

(a) Any one from: flame the loop/neck of bottle; sterilise the loop before use; only lift the lid slightly / tape lid on; work near a Bunsen flame; use sterile spreader/agar. [1]

(b) Correct answer 154 (mm²) = 2 marks (ALLOW 153.9 or 153.94 if then rounded; ALLOW 150 for 1 mark only as 2 s.f.). If answer incorrect: radius = 7 mm (1 mark) OR \( \pi \times 7^2 \) shown (1 mark). Using 14 as radius gives 616 = 1 mark for correct use of formula (ECF). [2]

(c) It is a control / to show that water (or the disc) does not kill bacteria, so any clear zone is due to the extract. AW [1]

(d) Largest clear zone/diameter (14 mm) OR bigger than B and C. DO NOT ALLOW 'it is biggest' with no reference to zone. [1]

(e) Any one from: to avoid growing pathogens / harmful bacteria that grow best at body temperature (37 °C); for safety. IGNORE 'to keep it cooler' alone. [1]

(f) Influenza is caused by a virus / viruses are not killed by antibiotics / viruses live inside host cells. [1]

(b) Total = 3 + 5 + 4 + 6 + 2 + 4 = 24; mean = 24 ÷ 6 = 4.
(c) 4 snails in 0.5 m² = 8 per m². 8 × 600 = 4800 snails.

(a) grass → snail → song thrush → sparrowhawk with arrows in the correct direction AND song thrush as secondary consumer. [1]

(b) 4 [1]

(c) Correct answer 4800 = 2 marks. If incorrect: 4 ÷ 0.5 = 8 per m² (1 mark), OR 600 ÷ 0.5 = 1200 quadrats (1 mark). ECF from (b) allowed. [2]

(d) To avoid bias / to give a representative sample / so results are fair. AW [1]

(e) Grass decreases / is reduced (1). Fewer thrushes so snail numbers increase (as less predation), so more grass is eaten (1). Second mark dependent on a correct reference to snails. DO NOT ALLOW 'grass increases'. [2]

Level of response question. Good answers link the stimulus (glucose rise/fall), receptor/control (pancreas), hormone (insulin/glucagon), effector (liver, muscle) and the correcting change, and then explain negative feedback as reversing the original change.

(a) Pancreas. [1]

*(b) Level of response, 6 marks.

Level 3 (5–6 marks): Explains both situations with the correct hormones and the actions on liver/cells AND links to negative feedback. The answer is clear, logical and uses correct terms.
Level 2 (3–4 marks): Describes the response to one situation fully, or both situations partly (e.g. names insulin and glucagon with some effect), with limited or no link to negative feedback. Some structure.
Level 1 (1–2 marks): Simple statements, e.g. insulin lowers glucose / pancreas is involved. Little detail or logic.
Level 0 (0 marks): No relevant content.

Indicative scientific points:
- after meal, blood glucose rises and is detected by the pancreas
- pancreas releases insulin into the blood
- insulin makes liver (and muscle) cells take up glucose
- glucose converted to glycogen (for storage)
- blood glucose falls back to normal
- when fasting, glucose falls and the pancreas releases glucagon
- glucagon acts on the liver, converting glycogen into glucose which is released into the blood
- blood glucose rises back to normal
- negative feedback: a change in glucose causes a response that reverses the change, keeping the level near a set point

IGNORE 'sugar' for glucose if meaning is clear. DO NOT ALLOW insulin stored in the liver or converting glycogen to glucose. [6]

(a) Aseptic technique stops unwanted microbes contaminating the culture, which would affect the results, and stops pathogens escaping.
(b) Total = 14 + 11 + 14 = 39. Mean = 39 / 3 = 13 mm.
(c) Results: A (19 mm) > B (13 mm) > C (0 mm). Larger clear zone means more bacteria killed or inhibited, so A is the most effective against this bacterium. The conclusion is only partly supported: only one bacterium was tested, the discs may not have contained equal concentrations, and a larger zone may also reflect faster diffusion through the agar rather than greater killing. Improvements: use the same concentration and volume of each antibiotic, use identical discs and agar depth, keep temperature and incubation time constant, do more repeats, test other bacteria, and include a control disc soaked in sterile water. Resistance to C: bacteria vary because of mutation; a mutant bacterium may be resistant to C. When C is used, non-resistant bacteria die and resistant ones survive and reproduce, passing on the allele; the population becomes mainly resistant (natural selection). Patients' infections then cannot be treated with C, so other antibiotics are needed, and the infection may spread.

(a) [1] Prevent contamination of the culture by unwanted microbes/bacteria from the air or surroundings AW; OR stop harmful bacteria escaping/infecting the student. IGNORE 'to keep it clean' alone. DO NOT ALLOW 'to kill bacteria'.

(b) [2] Correct answer 13 (mm) = 2 marks. If the answer is wrong, 14 + 11 + 14 = 39 (or 39 / 3 seen) = 1 mark. ALLOW 13.0. Answer must appear in the table or on the answer line.

(c)* [6] Level of response. Indicative content:
Results and improvements:
- A has the largest clear zone (19 mm), B has a smaller zone (13 mm), C has no clear zone (0 mm)
- so A is the most effective of these three against this bacterium, and C is ineffective
- the conclusion is limited because only one species was tested / 'best for treating infections' is too general
- control variables: same concentration/volume of antibiotic, same size of disc, same agar depth, same temperature and incubation time
- repeat more times / calculate means / ignore anomalies
- use a control disc with no antibiotic (sterile water); test other bacteria
Resistance:
- bacteria vary / random mutation gives some bacteria resistance to C
- when antibiotic is used, non-resistant bacteria are killed; resistant bacteria survive and reproduce
- the resistance allele is passed on to offspring; the proportion of resistant bacteria increases (natural selection)
- infections become difficult/impossible to treat with that antibiotic; resistant strains may spread; need new antibiotics / use antibiotics only when necessary

Level 3 (5-6 marks): Detailed evaluation covering BOTH strands: uses the data accurately and gives at least two specific improvements, and gives a correct linked explanation of how resistance arises through mutation and selection with its consequence for patients. Clear, logical and well organised.
Level 2 (3-4 marks): Covers both strands but with limited detail, OR covers one strand in detail. Some links made. Mostly clear and relevant.
Level 1 (1-2 marks): Simple, isolated statements, such as 'A has the biggest zone' or 'bacteria become resistant'. Little or no link or structure.
Level 0 (0 marks): No response or no response worthy of credit.
Coverage determines the level; quality of communication determines the position within the level. ALLOW 'antibiotic resistance develops because the bacteria get used to it' only at Level 1 (misconception, no credit for the mechanism). DO NOT ALLOW 'bacteria adapt/change on purpose' or 'patients become resistant' for mechanism marks.

(a)(i) Sum = 14 + 16 + 15 = 45; 45 ÷ 3 = 15 mm.
(ii) P has the largest mean clear zone (15 mm compared with 8 mm for Q and 0 mm for R), so it killed or stopped the most bacteria.
(b) Use sterilised equipment and keep the lid on the dish as much as possible; secure the lid with tape.
(c) A mutation produces a resistant bacterium by chance. When the antibiotic is used, non-resistant bacteria are killed but the resistant one survives, reproduces and passes on the resistance gene, so the resistant strain becomes common (natural selection). Doctors should not prescribe antibiotics for viral infections or unnecessarily, and should use them only when needed; patients should complete the full course so that all bacteria are killed.

(a)(i) Correct answer 15 (mm) = 2 marks. If wrong, 14 + 16 + 15 = 45 (or ÷ 3 shown) = 1 mark (method). ALLOW ECF for a wrong sum.
(ii) P AND largest (clear) zone / most bacteria killed = 1 mark. DO NOT ALLOW P without a reason.
(b) Any two from: sterilise inoculating loop / equipment (by flame or autoclave); keep lid on / open as little as possible; tape the lid on (not fully sealed); disinfect bench / wash hands; use sterilised agar and dish. = 2 marks. IGNORE wear gloves / goggles on their own.
(c)* Levels of response (6 marks).
Level 3 (5-6 marks): Clear explanation of how resistance develops (mutation, survival of resistant bacteria, reproduction, passing on the allele) AND at least one correct action by doctors and one by patients. Logical, well-structured account.
Level 2 (3-4 marks): Some explanation of how resistance develops AND at least one correct action, but with gaps in the sequence, or a full explanation with no actions. Mostly clear.
Level 1 (1-2 marks): Simple statements, e.g. bacteria become resistant / antibiotics used too much. Little logical link.
Level 0 (0 marks): No creditable response.
Indicative content: random mutation in a bacterium; gives resistance to the antibiotic; antibiotic kills non-resistant bacteria; resistant bacteria survive (selection); they reproduce / divide; resistance gene passed on to offspring; population becomes mostly resistant; doctors do not prescribe antibiotics for viral infections / for minor illnesses; use only when necessary / use a narrow range; patients complete the whole course; do not demand antibiotics; do not share antibiotics; hygiene / isolating infected patients.
DO NOT ALLOW bacteria / people 'become used to' or 'build up immunity to' the antibiotic as the mechanism for Level 3. Marks within a level are set by quality of communication.

(a) The concentration of sucrose solution is the factor the student changes.
(b) Control the size of the cylinders, the volume of sucrose solution, the temperature and the time left in solution.
(c) Change = 2.75 - 2.50 = 0.25 g. Percentage change = 0.25 ÷ 2.50 × 100 = 10 %.
(d) At low concentrations the cylinders gain mass. As the concentration increases, the gain gets smaller, then the cylinders lose mass, with the biggest loss at 0.8 mol/dm³.
(e) The change is +1 % at 0.4 and -6 % at 0.6, so zero change is just above 0.4, about 0.43 mol/dm³.
(f) The sucrose solution is more concentrated (has a lower water concentration) than the solution in the potato cells. Water moves out of the cells by osmosis, down its concentration gradient, through the partially permeable cell membrane, so the mass falls.

(a) (Sucrose) concentration (of solution) = 1 mark.
(b) Any two from: length / mass / surface area / diameter of potato cylinder; volume of sucrose solution; temperature; time in solution; same potato / same type of potato; same method of drying. = 2 marks. IGNORE 'amount' of potato / solution without a measure. DO NOT ALLOW concentration of sucrose.
(c) Correct answer 10 (%) = 2 marks. If wrong: 0.25 (g) as the change = 1 mark; or 0.25 ÷ 2.50 (× 100) = 1 mark. ALLOW + 10. Answer of 0.1 scores 1 mark only.
(d) Two from: mass increases at low concentrations / 0.0 to 0.4 (1); mass decreases at high concentrations / 0.6 and above (1); as concentration increases the change in mass decreases / goes from gain to loss (1). Maximum 2 marks. ALLOW AW for gain / loss. DO NOT ALLOW 'mass decreases' on its own.
(e) Any value from 0.40 to 0.45 (mol/dm³) = 1 mark. DO NOT ALLOW values outside this range.
(f) Three from: sucrose solution is more concentrated than the solution in the cells / water concentration is lower outside (1); water moves out of the cells / potato (1); by osmosis, from a high to a low water concentration / down a concentration gradient (1); through a partially permeable membrane (1); so mass decreases (cylinder loses water) (1). Maximum 3 marks. DO NOT ALLOW sucrose moves out. IGNORE 'water moves from dilute to concentrated' without reference to a membrane or cells.

(a) Algae photosynthesise, so they are the producer.
(b) Minnows eat water fleas and tadpoles (primary consumers); diving beetles eat tadpoles. Both are secondary consumers.
(c) Fewer minnows means fewer water fleas are eaten, so water fleas increase. More water fleas eat more algae, so the amount of algae decreases.
(d) Total = 4 + 7 + 3 + 6 + 5 + 8 + 2 + 5 + 6 + 4 = 50. Mean per quadrat = 50 ÷ 10 = 5. Mean per m² = 5 ÷ 0.25 = 20. Estimated total = 20 × 150 = 3000.
(e) Random placement avoids bias, so the sample represents the whole pond surface.
(f) Energy is lost as heat from respiration and in uneaten parts or waste.

(a) Algae = 1 mark. IGNORE plants.
(b) Minnow OR diving beetle = 1 mark. DO NOT ALLOW heron, water flea or tadpole.
(c) Water fleas increase / go up (1); because fewer are eaten / less predation (1); algae decrease (1) because more water fleas eat them / more grazing (1). Maximum 3 marks. ECF is allowed for a reason matching the stated direction. IGNORE 'food chain is disrupted'.
(d) Correct answer 3000 = 3 marks. If wrong: mean per quadrat of 5 (1); 5 ÷ 0.25 = 20 per m² (1); 20 × 150 (1). ALLOW 3 marks for 3000 given with working. ALLOW ECF through the steps. An answer of 750 (5 × 150) scores 1 mark.
(e) To avoid bias / to give a representative sample / so every area has an equal chance of being sampled = 1 mark. IGNORE 'to make it a fair test'.
(f) Any two from: lost as heat / through respiration; used for movement / life processes; lost in waste / faeces / urine; some parts not eaten (bones, roots, shells) or not digested. = 2 marks. IGNORE 'used for growth' and 'lost' on its own.

(a)(i) Aseptic technique, e.g. flame the loop/spreader, disinfect the bench, lift the lid only slightly. (a)(ii) 25 C is lower than body temperature so pathogens that could infect humans are unlikely to grow well. (b)(i) Mean = (22 + 20 + 24) / 3 = 66 / 3 = 22 mm. (b)(ii) B has the largest mean clear area (22 mm, compared with 14 and 9 mm), so it killed or stopped the growth of the most bacteria; the results for the plates are close together (20 to 24) so are repeatable. (b)(iii) The water disc is a control: it shows that the liquid itself or the paper disc does not kill bacteria, so the clear areas are caused by the mouthwashes. (c) Within a bacterial population, random mutation produces some bacteria that are resistant to an antibiotic. When the antibiotic is used, non-resistant bacteria are killed but resistant ones survive, reproduce and pass on the resistance allele, so the resistant strain becomes more common and can spread between people. Infections caused by resistant strains cannot be treated with the usual antibiotic, so patients stay ill for longer, may need other drugs, and the infection can spread to others. Using antibiotics less often reduces this selection pressure.

(a)(i) One from: flame / sterilise the loop or spreader; disinfect the bench; keep the lid lifted only slightly / for a short time; use sterile agar and equipment [1]. IGNORE 'wear gloves' alone.
(a)(ii) Pathogens / harmful bacteria grow best at body temperature (37 C) / are less likely to grow at 25 C [1]. ALLOW to reduce the risk of growing harmful bacteria.
(b)(i) 22 [1]. Units are already given in the table; IGNORE a missing unit.
(b)(ii) Any two from: B has the largest mean clear area / diameter (22 mm) [1]; compared with A (14 mm) and C (9 mm) / the others [1]; results for B are close together / consistent / repeatable (20 to 24 mm) [1]. Max 2. ALLOW ECF from (b)(i).
(b)(iii) It is a control / to show that water (or the disc) does not kill bacteria / so that any clear area is due to the mouthwash [1]. DO NOT ALLOW 'to make it a fair test' alone.
(c)* Levels of response (6 marks). Scientific coverage sets the level; clear, relevant and logically organised communication sets the position within the level.
Level 3 (5-6 marks): Detailed, linked explanation covering BOTH strands: mutation produces resistant bacteria; antibiotic kills non-resistant bacteria (selection); resistant survive, reproduce and pass on resistance so the strain becomes more common / spreads; problems for treatment such as infections that cannot be treated, longer illness, need for new antibiotics. Logically structured.
Level 2 (3-4 marks): Explains most of the process of resistance developing OR gives some linked ideas on both strands, with some gaps in logic or detail.
Level 1 (1-2 marks): Simple, isolated statements, e.g. bacteria become resistant, or antibiotics will not work, with little or no explanation.
Level 0 (0 marks): No response or no response worthy of credit.
ALLOW 'immune' for resistant only at Level 1. DO NOT ALLOW the idea that the person or the bacteria 'get used to' the antibiotic or that bacteria change on purpose.
Total 12 marks.