2025 OCR GCSE Twenty First Century Science - Biology B - J257 模擬試題連答案詳解

Ignore 9.9 cm. Sum of remaining values = 4.2 + 4.8 + 4.5 + 4.5 = 18.0 cm. Mean = 18.0 ÷ 4 = 4.5 cm. Option B (5.6 cm) is the mean if the anomaly is included.

Award 1.5 marks for A. Method: anomaly excluded (idea of 4 values, 18.0 total) then divided by 4. No marks for B, C or D.

The data show that the two variables rise together (a positive correlation). A correlation does not by itself show that one variable causes the other, as other factors could be involved, so only C is justified.

Award 1.5 marks for C. A and B claim causation; D contradicts the data.

Cross Hh × Hh gives HH, Hh, Hh, hh in a 1:2:1 ratio. Only hh is affected, so 3 out of 4 offspring are unaffected = 75%.

Award 1.5 marks for B. Accept 3 in 4 in the working. A is the probability of an affected offspring.

Meiosis halves the chromosome number, so gametes have 9. Mitosis produces genetically identical daughter cells with the full number, 18.

Award 1.5 marks for D. Both numbers must be correct.

Movement from low to high concentration is against the gradient, so it needs active transport, which uses energy released by respiration.

Award 1.5 marks for C. A is wrong because diffusion is down a gradient; B is wrong because osmosis is movement of water.

Surface area = 6 × 3 × 3 = 54 cm². Volume = 3 × 3 × 3 = 27 cm³. Ratio = 54 : 27 = 2 : 1.

Award 1.5 marks for B. Method: surface area 54 (0.5), volume 27 (0.5), ratio simplified (0.5).

Percentage = (900 ÷ 8000) × 100 = 11.25%, which is 11% to 2 significant figures.

Award 1.5 marks for A. Method: 900 ÷ 8000 × 100 (1 mark), rounding (0.5 mark). B is the inverted ratio.

Genes are lengths of DNA that code for particular proteins. The genome is the entire genetic material of an organism, found in nearly all its cells.

Award 1.5 marks for D. A, B and C each contain a factual error.

Cross Ff x Ff. Gametes F or f from each parent. Offspring: FF, Ff, Ff, ff in a 1 : 2 : 1 ratio. Two of the four offspring genotypes are Ff (carriers without the disorder), so the probability is 2/4 = 50%.

B (50%) = 1.5 marks (AO2 1.5). Award 1 mark for the correct Punnett square / offspring genotypes FF, Ff, Ff, ff if shown in working. ALLOW 1 in 2, 0.5 or 2/4. 25% is the chance of an affected child (ff) only; 75% is the chance of not being affected; 100% is incorrect.

Nitrate ions are normally absorbed from a lower concentration in the soil into root hair cells against a concentration gradient by active transport. Active transport needs energy from aerobic respiration. With little oxygen, aerobic respiration falls, less ATP/energy is available and uptake falls.

C = 1.5 marks (AO2 1.5). Distractors: A is wrong because uptake is by active transport against a concentration gradient and dilution would not stop it entirely; B is wrong because waterlogged soil does not draw water out of root cells; D is not a consequence of low oxygen and is not described in the specification.

Mean height rises from 12 cm to 18 cm to 25 cm between 0 and 20 mg per dm\(^3\), then stays at 25 cm at 30 mg per dm\(^3\). Only option A describes this accurately. B claims too much (other factors were not tested), C extrapolates beyond the data, and D is false: doubling from 10 to 20 raises height from 18 to 25 cm, not 36 cm.

A = 1.5 marks (AO3 1.5). ALLOW reasoning that the plateau suggests another factor has become limiting. Each distractor goes beyond or contradicts the data: B (only factor), C (extrapolation), D (not proportional).

Actual size = image size ÷ magnification = \(42 \div 12\,000 = 0.0035\) mm. Convert to micrometres: \(0.0035 \times 1000 = 3.5\ \mu\text{m}\).

First check the answer on the answer line. If answer = 3.5 (\(\mu\)m) award 2 marks.
If answer is incorrect:
1 mark for \(42 \div 12\,000\) (= 0.0035 mm) (method).
1 mark for converting mm to \(\mu\)m by multiplying by 1000 (accuracy).
ALLOW ECF for one conversion error. ALLOW 0.0035 mm with 1 mark only. DO NOT ALLOW answers without a correct unit conversion for the second mark.

Total area sampled = \(12 \times 0.25 = 3\) m\(^2\). Density = \(30 \div 3 = 10\) snails per m\(^2\). Population = \(10 \times 800 = 8000\).

First check the answer on the answer line. If answer = 8000 award 2 marks.
If answer is incorrect:
1 mark for calculating density = 10 per m\(^2\) (or mean 2.5 per quadrat).
1 mark for multiplying density by 800 m\(^2\) (or 2.5 × 3200 quadrat areas).
ALLOW ECF from an incorrect density.

Change in mass = \(2.64 - 2.40 = 0.24\) g. Percentage change = \(\frac{0.24}{2.40} \times 100 = 10\%\) increase.

First check the answer on the answer line. If answer = 10 (%) or +10 (%) award 2 marks.
If answer is incorrect:
1 mark for change in mass = 0.24 g.
1 mark for \(0.24 \div 2.40 \times 100\).
ALLOW 1 mark for \(-10\) (sign error). ALLOW ECF from an incorrect change in mass.

Cross Cc × Cc gives CC, Cc, Cc, cc, so probability a child is unaffected = \(\frac{3}{4}\). For two children: \(\frac{3}{4} \times \frac{3}{4} = \frac{9}{16} = 0.5625 \approx 0.56\).

First check the answer on the answer line. If answer = 0.56 award 2 marks.
If answer is incorrect:
1 mark for probability of one unaffected child = 3/4 (0.75 or 75%).
1 mark for \(0.75 \times 0.75\) (= 0.5625) correctly rounded to 2 s.f.
ALLOW 9/16 or 0.5625 for 1 mark only if not rounded. ALLOW ECF.

Rate = volume ÷ time = \(4.5 \div 15 = 0.30\) cm\(^3\) per minute.

First check the answer on the answer line. If answer = 0.3 or 0.30 award 2 marks.
If answer is incorrect:
1 mark for \(4.5 \div 15\) (method).
1 mark for correct evaluation with unit cm\(^3\)/min (accuracy).
ALLOW 1 mark for 3.3 (15 ÷ 4.5).

Three bases code for one amino acid, so number of amino acids = \(1260 \div 3 = 420\).

First check the answer on the answer line. If answer = 420 award 2 marks.
If answer is incorrect:
1 mark for idea that three bases code for one amino acid / dividing by 3.
1 mark for correct calculation \(1260 \div 3\).
ALLOW 1 mark for 3780 (multiplying by 3).

Number of divisions = \(24 \div 4 = 6\). Each division doubles the cells, so cells = \(2^6 = 64\).

First check the answer on the answer line. If answer = 64 award 2 marks.
If answer is incorrect:
1 mark for 6 divisions (24 ÷ 4).
1 mark for \(2^6\).
ALLOW 1 mark for 12 (6 × 2) or 128 (counting the original as dividing seven times). ALLOW ECF from an incorrect number of divisions.

The 19.8 cm value is far from the others, so it is anomalous. Mean of the other four = (12.4 + 12.9 + 12.1 + 12.6) / 4 = 50.0 / 4 = 12.5 cm.

First check the answer on the answer line. If answer = 12.5 award 2 marks.
1 mark: identifies 19.8 (cm) / seedling 3 as anomalous and excludes it.
1 mark: correct mean of remaining values = 12.5.
DO NOT ALLOW 14.0 (mean including the anomaly) for the second mark. ALLOW 12.50. AO3 (3.1b).

(a) Uptake rises steeply from 2 to 33 units between 0 and 15% oxygen, then plateaus at about 34 units. (b) Active transport needs energy from respiration; without oxygen, only anaerobic respiration supplies a little ATP, or a small amount of passive diffusion occurs.

(a) 1 mark: idea of uptake increasing with oxygen concentration up to (about) 15-20%, then no further increase/levels off. ALLOW positive correlation then plateau. Quoting data only is not enough unless the levelling off is clear. AO3 3.1a.
(b) 1 mark: any one from: anaerobic respiration (still) releases some energy / ATP; some ions move in by diffusion; some dissolved oxygen was still present. DO NOT ALLOW 'no respiration'. AO2 2.1.

(a) Oxygen rises by 1.5 cm\(^3\) per unit up to light intensity 3, then the increase slows, so the straight proportional section ends at 3. (b) Once more light does not increase the rate, another factor is limiting: carbon dioxide concentration or temperature.

(a) 1 mark: 3 (ALLOW between 3 and 4). AO3 3.1a.
(b) 1 mark: carbon dioxide concentration OR temperature. DO NOT ALLOW water, oxygen, 'light'. AO1 1.1.

The line crosses 0% change at 0.4 mol/dm\(^3\). At this concentration there is no net osmosis, so mass stays the same, meaning the solutions inside and outside are equal in concentration.

(a) 1 mark: 0.4 (mol/dm\(^3\)). AO3 3.1a.
(b) 1 mark: no net movement / gain / loss of water (by osmosis) OR no change in mass. ALLOW water enters and leaves at the same rate. DO NOT ALLOW 'no water moves'. AO2 2.1.

Total counted = 4+2+0+5+3+1+4+2+3+1 = 25. Mean per quadrat = 25 / 10 = 2.5 plants per m\(^2\). Estimated population = 2.5 × 400 = 1000.

First check the answer on the answer line. If answer = 1000 award 2 marks.
1 mark: mean = 2.5 per m\(^2\) (or total 25 / 10).
1 mark: mean × 400 = 1000.
ALLOW ECF for an incorrect mean multiplied by 400. ALLOW 1 mark for 25 × 400 / 10 shown. AO2 2.2.

The mean number of visits rises from 12 to 22 as width rises from 1 m to 8 m, but only by 1 between 4 m and 8 m. A correlation does not show cause because other factors were not controlled.

Award 2 marks (scaled to 2.4) for two of:
1. Positive correlation: more insects as width increases, with figures quoted (AO3.1a)
2. Increase levels off / smaller increase between 4 m and 8 m (AO3.1a)
3. Correlation does not show causation, or other variables (flower type, weather, location) not controlled (AO3.3b)
ALLOW AW. IGNORE 'results are unreliable' with no reason.

A long-eared rabbit must be ee, so each parent carried and passed on one e. Parents show the dominant phenotype so must be Ee. Ee x Ee gives a 25% chance of ee.

Award 2 marks (scaled to 2.4):
1. Both parents Ee / heterozygous / carriers of the recessive allele (AO2.1)
2. Both pass on the e allele (in gametes) so offspring is ee / homozygous recessive (AO2.1)
ALLOW a correct Punnett square for both marks. DO NOT ALLOW genotypes written as EE for either parent.

Movement against a concentration gradient is active transport and requires energy from respiration, which occurs in mitochondria.

Award 2 marks (scaled to 2.4):
1. Active transport (AO1.1)
2. Mitochondria release energy / ATP (by respiration) needed to move ions against the concentration gradient (AO2.1)
DO NOT ALLOW 'produce energy'. IGNORE diffusion/osmosis unless named as the process.

Thin or small organisms have a high surface area to volume ratio and short distances for diffusion. Larger organisms do not, so a transport system is needed.

Award 2 marks (scaled to 2.4):
1. Flatworm: large surface area to volume ratio / short diffusion distance so diffusion is sufficient (AO2.1)
2. Earthworm: smaller surface area to volume ratio / longer diffusion distance so diffusion too slow, blood needed to carry oxygen to cells (AO2.1)
ORA. IGNORE 'bigger so needs more oxygen' without link to diffusion.

Decomposers digest material extracellularly using enzymes. Minerals are released into the soil and taken up by plant roots, completing the nutrient cycle.

Award 2 marks (scaled to 2.4):
1. Decomposers/microorganisms (fungi, bacteria) secrete enzymes that break down the wood (AO1.1)
2. Mineral ions/nutrients released into the soil and absorbed by plants (AO2.1)
IGNORE 'worms eat it' alone.

Several triplets can code for the same amino acid, some DNA does not code for proteins, and a changed amino acid may not alter the protein's shape.

Any two from (1 mark each, scaled to 2.4):
- new triplet codes for the same amino acid (AO2.1)
- mutation in non-coding DNA (AO2.1)
- amino acid change does not alter the protein's folding/shape/function (AO2.1)
ALLOW AW. IGNORE 'mutation is harmless' alone.

Root tips contain meristem cells that divide for growth. DNA replicates in the cell cycle before mitosis.

Award 2 marks (scaled to 2.4):
1. Root tip is where growth occurs / contains cells that divide rapidly (AO2.1)
2. Chromosomes/DNA are copied / replicated before division (AO1.1)
DO NOT ALLOW 'chromosomes divide' without copying.

Reliability improves by repeating and averaging, and by reducing variation between samples and in the method.

Any two from (scaled to 2.4):
- repeat and calculate a mean (AO3.3b)
- use several cylinders per solution (AO3.3b)
- control size of cylinders / same potato (AO3.3b)
- consistent drying before weighing (AO3.3b)
IGNORE 'use more concentrations' and 'be more careful'.

Electron beams have a much shorter wavelength than light, giving greater resolution and magnification, so small organelle details are visible.

Award 2 marks (scaled to 2.4):
1. Electron microscope has higher resolution / resolving power (AO1.1)
2. Mitochondria / their inner membranes are too small to be seen in detail by a light microscope, or electron microscope gives higher magnification (AO2.1)
DO NOT ALLOW 'clearer' alone.

Mean mass rises from 2.1 kg to 4.6 kg between 0 and 40 mg/dm³. Above 40 mg/dm³ the change is only 0.1 kg or less (4.7 then 4.6), so something else is limiting growth. Using 80 mg/dm³ gives no benefit over 40 or 60 mg/dm³ but costs more.

1 mark: mass increases from 0 to 40 (mg/dm³) (ALLOW 2.1 to 4.6 kg; quoting data).
1 mark: levels off / no significant further increase above 40 (mg/dm³) (ALLOW 4.6 to 4.7; IGNORE 'decreases' on its own).
1 mark: 80 gives no extra yield so wastes money / fertiliser (AW; may run off and cause pollution).
AO 3.1a / 3.2a.

Each parent is Bb and produces gametes B and b. Combining gives BB, Bb, Bb and bb. Only bb is brown, so 1 in 4.

1 mark: parental gametes shown as B and b for both parents.
1 mark: offspring genotypes BB, Bb, Bb, bb correctly combined (ECF from gametes).
1 mark: probability of brown = 1 in 4 / 0.25 / 25% / 1:3 ratio (ALLOW ECF from the diagram).
DO NOT ALLOW 3:1 as a probability for brown. AO 2.1 / 2.2.

Nitrate is at a lower concentration in the soil than in the root cells, so it must be taken up by active transport, which needs energy from aerobic respiration. Little oxygen means less respiration, so less energy and less uptake.

1 mark: less (aerobic) respiration / less energy released because of little oxygen.
1 mark: less active transport (of nitrate ions) into root (hair) cells.
IGNORE 'no respiration'. DO NOT ALLOW 'less diffusion' for the second mark. AO 2.1.

A wall one cell thick means substances have only a short distance to diffuse. A rich blood supply removes absorbed molecules quickly, keeping the concentration gradient steep so diffusion continues.

1 mark: wall one cell thick so short diffusion distance (AW, faster diffusion).
1 mark: many capillaries / blood flow removes absorbed molecules, keeping a steep concentration gradient.
IGNORE 'large surface area' (not one of the features given). AO 2.1.

Decay is carried out by microorganisms. Cold slows enzyme activity and respiration, and they need water to be active, so fewer and less active decomposers mean slower breakdown.

1 mark: decomposers / microorganisms (bacteria, fungi) less active or fewer in number.
1 mark: reason linked to condition: low temperature slows enzyme action / respiration OR water needed for microbial activity.
IGNORE 'microbes die' unless linked; AO 2.1.

Bases are read in groups of three, each group coding for an amino acid. A different base can change an amino acid, altering the chain and so how it folds, giving a different shape and function.

1 mark: (base) sequence determines the order / type of amino acids in the protein (a changed base can change an amino acid).
1 mark: protein folds into a different shape so cannot function properly (e.g. enzyme active site no longer fits).
ALLOW 'triplet / three bases code for one amino acid'. AO 1.1 / 2.1.

Anything other than sugar concentration that could change the mass change must be kept constant, for example the volume of solution and the temperature.

Any two from:
volume of solution (1)
temperature (1)
time in solution (1)
size / surface area / length of cylinders (1)
same potato / same variety (1)
IGNORE 'amount of potato' unless mass or size is specified; DO NOT ALLOW sugar concentration. AO 2.2 / 3.3a.

Cells at the tip divide by mitosis, increasing cell number and allowing growth. As cells leave the dividing region they differentiate: different genes are switched on, so they develop the structures needed for specific functions.

1 mark: mitosis / cell division increases number of cells so root grows (in length).
1 mark: new cells differentiate / become specialised.
1 mark: because different genes are expressed (switched on or off) (ALLOW to carry out a particular function, e.g. absorbing water).
DO NOT ALLOW 'cells change into different genes'. AO 1.1 / 2.1.

The grey cub must be ww, so it inherited one w allele from each parent. The parents are white, so they carry W as well, so both are Ww. Gametes W or w combine at random; the cub received w from each parent. Ratio expected: 3 white : 1 grey.

Any three from:
1. Both parents are heterozygous / Ww / carriers of the recessive allele (1)
2. Each parent produces gametes with W or w (1)
3. The cub inherited a recessive allele (w) from each parent / fertilisation of w gamete with w gamete (1)
4. Cub is homozygous recessive / ww so has grey fur as no dominant allele is present (1)
ALLOW a correct Punnett square for marks 2-4. DO NOT ALLOW 'gene' for 'allele' if it makes the statement incorrect. AO2 1.2

Flat leaves have a large surface area to volume ratio, capturing more light. Thinness shortens the diffusion path for carbon dioxide to reach the chloroplasts. Movement of ions against a concentration gradient needs energy from respiration: active transport.

1. Large surface area (to absorb more light) (1)
2. Short diffusion distance / thin so carbon dioxide reaches photosynthesising cells quickly (1)
3. Active transport (1)
ALLOW 'more light absorbed' for 1. DO NOT ALLOW diffusion or osmosis for 3. AO1 1.1, AO2 2.1

Removing a consumer lowers predation on its prey, so algae numbers rise. Newts lose their food source, so starvation and competition reduce their numbers.

1. Algae increase (1)
2. because fewer snails to eat / graze them (1)
3. Newts decrease because less food / more competition for the remaining snails (1)
IGNORE references to herons. ALLOW 'population' for 'number'. AO2 2.1

The sugar solution has a lower water potential than the potato cells, so water moves out by osmosis across the partially permeable membranes, reducing the mass.

1. Water moves out of the potato / cells by osmosis (1)
2. from higher water potential (dilute) to lower water potential (more concentrated solution) through a partially permeable membrane (1)
ALLOW 'water concentration' for water potential. DO NOT ALLOW 'sugar moves in' as the explanation. AO2 2.1

Only temperature should change. Other factors that affect yeast activity must be kept constant. Repeats and means improve reliability.

Any two control variables from: mass/amount of yeast; volume of sugar solution; concentration of sugar; pH; time (ALLOW 1 mark for one variable if no reliability point; total max 2 across the question).
Plus repeat and calculate a mean (1)
Max 2 marks overall: 1 mark for two correct variables, 1 mark for repeats/mean. DO NOT ALLOW temperature as a control variable. AO3 3.3b

The values fall steadily from 82 to 60 as steps rise from 2 to 12 thousand: negative correlation. A correlation does not show causation; other variables may differ between adults and only six people were sampled.

(i) Negative correlation / as steps increase heart rate decreases (1)
(ii) Any two from: correlation does not mean causation (1); other factors such as age / fitness / diet / medication not controlled (1); small sample of only six adults (1)
DO NOT ALLOW 'no relationship'. AO3 3.1a

The points slope upwards, so the correlation is positive. A correlation alone does not prove causation because other factors (for example light or temperature) could differ between ponds. A overstates the evidence, B has the wrong direction, D ignores the data.

C = 2 marks. Any other answer = 0 marks. (Award 2 marks for the correct letter only; no partial marks for a multiple-choice item.) AO3 - interpreting data and correlation versus causation.

Let N = normal allele and n = recessive allele. Cross Nn x Nn gives NN, Nn, Nn, nn. Two of the four offspring are Nn (carriers), so the probability is 2/4 = 50%.

C = 2 marks. Any other answer = 0 marks. Working in the solution (Punnett square NN, Nn, Nn, nn) is for guidance only. AO2 - applying knowledge of monohybrid inheritance.

Nitrate ions are at a lower concentration in the soil than inside root hair cells, so they are taken up by active transport. This needs energy from aerobic respiration, which is reduced when the soil has little oxygen.

D = 2 marks. Any other answer = 0 marks. AO2 - applying knowledge of mineral uptake and active transport.

Change in mass = 3.4 - 4.0 = -0.6 g. Percentage change = \(\frac{-0.6}{4.0} \times 100 = -15\%\). The cylinder lost water by osmosis, so the change is a decrease.

D = 2 marks. Any other answer = 0 marks. Common errors: dividing by the final mass (-17.6%), omitting the x100 (-0.6), wrong sign (+15%). AO2 - percentage change calculation.

Percentage transferred = \(\frac{600}{5000} \times 100 = 12\%\). The rest is lost, for example as heat from respiration, in waste, or in parts not eaten.

B = 2 marks. Any other answer = 0 marks. Common errors: dividing 5000 by 600 (8.3), percentage lost (88). AO2 - efficiency of energy transfer calculation.

The genetic code is such that several different triplets code for the same amino acid. A single base substitution can therefore leave the amino acid sequence, and so the protein, unchanged.

B = 2 marks. Any other answer = 0 marks. AO2 - applying knowledge of the genetic code and mutations.

The distance of the lamp is the independent variable and the bubble count is the dependent variable. Light intensity changes as a result of the distance. Water temperature must be kept the same so it does not affect the rate of photosynthesis, so it is a control variable.

A = 2 marks. Any other answer = 0 marks. AO1/AO2 - identifying variables in an investigation.

Fraction of cells in mitosis = 30/200 = 0.15. Time in mitosis = 0.15 x 24 hours = 3.6 hours.

B = 2 marks. Any other answer = 0 marks. Common error: 200/30 x 24 = 160. AO2 - using sample data to calculate the duration of a stage of the cell cycle.

Brown offspring have genotype bb, so each must have inherited one b allele from each parent. Both green parents must therefore carry b, so both are Bb. A Bb x Bb cross gives BB : Bb : Bb : bb, so 3 green : 1 brown, which matches 48 : 16 (3 : 1). The probability of a brown (bb) offspring is 1 in 4. Option A is wrong because BB x BB cannot give bb offspring without mutation, which is unlikely to produce a quarter of the offspring. Option B would give no brown offspring. Option D is wrong because bb parents would be brown, not green.

Answer: C [2 marks].
Award 2 marks for C.
Award 1 mark (partial credit, only if the candidate shows working or annotation) for correctly identifying that both parents must be heterozygous (Bb) or that the ratio is 3 : 1.
Do not award any marks if more than one box is ticked.
AO2 (2.1): applying knowledge of monohybrid inheritance to an unfamiliar context.
Guidance: ALLOW working such as a Punnett square showing Bb x Bb giving BB, Bb, Bb, bb.

\(\text{real size} = \frac{\text{image size}}{\text{magnification}} = \frac{48}{800} = 0.06\) mm. Convert to micrometres: \(0.06 \times 1000 = 60\ \mu\)m.

First check the answer on the answer line. If answer = 60 (\(\mu\)m) award 2 marks.
If not, award 1 mark for \(48 \div 800\) (= 0.06 mm) OR for correct conversion of their value from mm to \(\mu\)m (\(\times 1000\)).
ALLOW ECF for one error. DO NOT ALLOW 0.06 without conversion for the second mark. AO2.1

Tt \(\times\) Tt gives offspring TT, Tt, Tt, tt, so 1 in 4 is tt (dwarf). Probability = 0.25. Expected dwarf seedlings = \(160 \times 0.25 = 40\).

(i) 0.25 / 1/4 / 1 in 4 / 25% (1 mark). DO NOT ALLOW a 1:3 ratio.
(ii) 40 (1 mark). ALLOW ECF from (i), i.e. 160 \(\times\) their probability. AO2.1

Change in mass = \(2.07 - 2.40 = -0.33\) g. Percentage change = \(\frac{-0.33}{2.40} \times 100 = -13.75\), which is -13.8 % to 3 significant figures.

First check the answer on the answer line. If answer = -13.8 (%) or 13.8 (% decrease) award 2 marks.
If not, award 1 mark for \(\frac{0.33}{2.40} \times 100\) or an unrounded answer of 13.75. ALLOW 1 mark for correct rounding to 3 s.f. of an incorrect value (ECF).
IGNORE missing sign if the word decrease is stated. AO2.1

Mean per quadrat = \(\frac{4+2+5+3}{4} = 3.5\). Mean per m² = \(3.5 \div 0.25 = 14\). Population = \(14 \times 2400 = 33\,600\).

First check the answer on the answer line. If answer = 33 600 award 2 marks.
If not, award 1 mark for mean of 3.5 per quadrat OR 14 per m².
ALLOW ECF for one error. ALLOW 3.36 \(\times 10^4\). AO2.1

Mean for high light = (148 + 152 + 139 + 155 + 146) / 5 = 740 / 5 = 148.
Increase = 148 - 113 = 35.
Percentage increase = \( \frac{35}{113} \times 100 = 30.97 \) % = 31 % (2 s.f.).

First check the answer on the answer line. If answer = 31 (%) award 1.5 marks.
If answer is incorrect, award:
0.5 mark: mean for high light = 148.
0.5 mark: (148 - 113) / 113 × 100 (ECF from the mean).
0.5 mark: correct rounding to 2 significant figures.
ALLOW 30.97 or 31.0 for 1 mark if not rounded to 2 s.f. DO NOT ALLOW 24 (calculated using 148 as the denominator) except for 0.5 mark for the mean. AO2.2

Two black parents produced brown offspring, so the brown allele must be hidden in the parents: black is dominant and both parents are carriers, Bb.
Bb × Bb gives BB : Bb : Bb : bb, so 1 in 4 are brown (bb).
Expected brown = 40 ÷ 4 = 10.

(i) 0.5 mark: black allele (B) is dominant (idea that brown appears from two black parents).
0.5 mark: both parents Bb (heterozygous).
(ii) 0.5 mark: 10.
ALLOW ECF from (i) only if the genotype given is consistent with the answer. ALLOW 'brown is recessive'. DO NOT ALLOW parents BB or bb. AO2.1, AO3.1a

(i) Mean heart rate falls from 78 to 63 bpm as exercise rises, so there is a negative correlation.
(ii) A correlation does not show cause: other factors that affect heart rate may also differ between groups, e.g. fitness-related health, age, smoking.
(iii) Age, sex, smoking status or body mass could be matched across groups.

(i) 0.5 mark: as exercise increases, resting heart rate decreases / negative correlation.
(ii) 0.5 mark: idea that another (named or unnamed) factor could explain the difference / only a correlation shown.
(iii) 0.5 mark: any one from: age; sex; smoking; body mass; existing health conditions; diet.
IGNORE 'number of people' for (iii). AO3.1a, AO3.2b

(i) The change in mass goes from +5 % at 0.2 to -4 % at 0.4, a total of 9 % over 0.2 mol/dm³. The zero point is 5/9 of the way: \( 0.2 + 0.2 \times \frac{5}{9} = 0.31 \) mol/dm³ (or read where a line of best fit crosses 0 %).
(ii) At this concentration the solution and the potato cells have the same water potential, so no net osmosis occurs.

(i) 0.5 mark: method shown, e.g. interpolation between 0.2 and 0.4, or line of best fit crossing 0 %.
0.5 mark: answer in the range 0.30 to 0.32 (mol/dm³).
(ii) 0.5 mark: same water potential / concentration of water inside and outside the cells so no net movement of water (by osmosis).
ALLOW 'isotonic' for (ii). DO NOT ALLOW 'no movement of water'. AO2.2, AO3.1b

1. Nitrate concentration is higher inside the root cell than outside, so diffusion cannot move it in; it must be active transport using carrier proteins in the cell membrane.
2. Active transport needs energy released by aerobic respiration, which needs oxygen.
3. Table 1: with no oxygen uptake is only 1.1; it rises steadily to 7.5 at 6 units, so more oxygen means more energy (ATP) for uptake. A small uptake at 0 may be from anaerobic respiration.
4. Above 6 units uptake stays at about 7.6-7.7, so oxygen is no longer limiting; the number of carrier proteins (or another factor) limits the rate.
5. Waterlogged soil has little oxygen (air spaces filled with water), so less aerobic respiration, less active transport, less nitrate absorbed, less protein/amino acid synthesis, so poor growth.

Level 3 (5-6 marks): Clear, linked explanation covering active transport against the concentration gradient AND the need for energy from aerobic respiration AND a description of the trend with data quoted (at least one pair of values) AND an explanation of the plateau AND the waterlogging link. Logical and well structured.
Level 2 (3-4 marks): Explains most of the above, e.g. active transport and energy needed, with some use of data, but the plateau or waterlogging is missing or only described. Some structure.
Level 1 (1-2 marks): Simple statements, e.g. oxygen increases uptake or roots need oxygen, with little linking or no data.
Level 0 (0 marks): No creditable response.

Indicative content (AO1 / AO2 / AO3):
AO1: active transport moves ions from low to high concentration against the gradient; uses carrier proteins; needs energy from respiration; oxygen needed for aerobic respiration.
AO2: nitrate is higher inside the root so diffusion cannot be responsible; waterlogged soil contains little oxygen so respiration and active transport decrease; less nitrate means less protein synthesis so poor growth.
AO3: uptake increases from 1.1 to 7.5 as oxygen increases from 0 to 6; levels off at about 7.6 from 6 to 10 units, so oxygen no longer limiting and the number of carrier proteins (or another factor) limits rate; some uptake at 0 oxygen may be from anaerobic respiration.

ALLOW ATP for energy. IGNORE 'produced' energy / 'creates' energy. DO NOT ALLOW diffusion or osmosis as the mechanism for nitrate uptake. Level is set by science content; mark within a level is set by quality of communication.

1. The puppy is hh, so each parent gave one h allele. Parents are healthy so each also has H: both are Hh.
2. Gametes: H or h from each parent.
3. Punnett square: HH, Hh, Hh, hh.
4. Probability of hh = 1/4 = 25% (0.25). Probability a puppy is a carrier (Hh) = 1/2.
5. Option A: no extra cost but 25% of puppies affected and 50% carriers, so the allele stays in the population. Option B: only HH dogs bred gives no affected puppies and allele h is gradually removed; disadvantages are testing costs, fewer breeding dogs, reduced genetic variation (inbreeding risk), and the ethical question of not breeding healthy carriers.

Level 3 (5-6 marks): Correct genotypes of both parents (Hh) with reasoning, a correct genetic cross with the 1 in 4 / 25% probability, AND a balanced evaluation with at least two advantages/disadvantages for both options, ending in a judgement or recommendation. Clear and logically structured.
Level 2 (3-4 marks): Correct parent genotypes and probability (may have a minor error) AND some advantages or disadvantages of at least one option, with limited balance.
Level 1 (1-2 marks): Some correct statements, e.g. parents are carriers or the test shows whether dogs carry the allele, with little reasoning or no cross.
Level 0 (0 marks): No creditable response.

Indicative content (AO1 / AO2 / AO3):
AO1: recessive allele only shows when two copies are present; heterozygous dogs are carriers; gametes carry one allele; genetic testing identifies carriers.
AO2: both parents are Hh because the puppy hh received one h from each; cross gives HH : Hh : Hh : hh; probability of hh = 1/4 or 25% or 0.25; 50% of puppies are carriers.
AO3: Option A advantages: no cost, no testing, keeps the pair that otherwise may be good dogs; disadvantage: 25% risk of affected puppies each time, suffering, carrier puppies pass on allele. Option B advantages: no affected puppies if only HH (non-carrier) dogs are used, allele removed from the breed over time; disadvantages: cost of testing, fewer dogs available to breed from, reduced genetic diversity so other harmful alleles may become more common, may lose otherwise good dogs. Valid judgement supported by the evidence.

ALLOW 1:3 ratio of affected to unaffected, 'one in four'. ALLOW other letters if defined. DO NOT ALLOW 'the puppy will definitely/not be ill' for a probability statement. Level is set by science content; mark within a level is set by quality of communication.

The value 25 mm is far from the others, so it is anomalous. A possible cause is a misread ruler or a seedling with different conditions (e.g. nearer the lamp). Mean without it: \(\frac{12+14+13+15}{4}=\frac{54}{4}=13.5\) mm.

25 mm identified as anomalous (1)
Any one cause from: measurement/reading error; recording error; seedling not identical / different position or conditions (1)
First check the answer line. If answer = 13.5 (mm) award 1 mark. Otherwise award method mark for (12 + 14 + 13 + 15) / 4 (ECF if wrong value excluded) - this mark is awarded within the final mark.
IGNORE 'human error' without detail.
Total 3 marks (AO3 / AO2).

Both parents are Tt. Gametes: T or t from each parent. Combining gives TT, Tt, Tt and tt. Only tt has a short tail, so the probability is \(\frac{1}{4}\).

Both parents' gametes shown as T and t (1)
Offspring genotypes TT, Tt, Tt, tt shown correctly (1)
Probability 1/4, 0.25 or 25% (1)
ALLOW ECF from incorrect offspring genotypes.
DO NOT ALLOW 1:3 or 3:1 as the probability.
Total 3 marks (AO2).

Diffusion cannot move ions from low to high concentration. Active transport does this, using energy released by respiration.

Active transport (1)
Against the concentration gradient / from low to high concentration, using energy from respiration (1)
DO NOT ALLOW diffusion or osmosis.
IGNORE 'uses energy' alone.
Total 2 marks (AO1/AO2).

(i) Change = 2.85 - 2.50 = 0.35 g. \(\frac{0.35}{2.50}\times100 = 14\%\) increase. (ii) The solution was more dilute than the cell contents, so water entered the cells by osmosis, increasing mass.

(i) If answer = 14 (%) award 2 marks. Otherwise 0.35 / 2.50 x 100 or 0.35/2.50 (1). ALLOW ECF. Increase need not be stated.
(ii) Water moved into the cells by osmosis from a higher water potential (dilute solution) to a lower water potential (1). DO NOT ALLOW sugar moving out.
Total 3 marks (AO2).

Decomposers need oxygen for aerobic respiration. Waterlogged mud has little oxygen, so fewer decomposers and slower breakdown.

Any two from:
less oxygen available (1)
idea that decomposers / microorganisms respire less / are fewer or less active (1)
cooler temperature in water (1)
IGNORE 'no oxygen' being the only point twice.
Total 2 marks (AO2).

Base order determines amino acid order. A different amino acid sequence changes folding and so the active site shape; the substrate no longer binds.

Changes the order / sequence of amino acids in the protein (1)
Changes the shape of the protein / active site so substrate no longer fits (1)
ALLOW enzyme denatured only if linked to shape change.
Total 2 marks (AO1/AO2).

Other factors that affect the rate of photosynthesis must be controlled so only distance changes.

Any two from:
temperature of the water (1)
carbon dioxide / hydrogencarbonate concentration (1)
size / mass / species of pondweed (1)
lamp power / type of lamp (1)
IGNORE time, unless 'time bubbles counted for'.
DO NOT ALLOW distance from lamp.
Total 2 marks (AO3).

In mitosis the DNA replicates, the copies separate and the cell divides into two identical daughter cells. Becoming specialised is differentiation.

DNA/chromosomes copied, then divided to form two genetically identical cells / nuclei (1)
Differentiation (1)
IGNORE 'cells split' without detail.
Total 2 marks (AO1).

Green is dominant, so a brown beetle must be gg. Each parent must therefore carry one g allele, so both are Gg. Gametes G or g from each parent. Punnett square: GG, Gg, Gg, gg. One in four offspring (25%) is gg and brown.

1 mark: both parents are heterozygous / Gg (carry a recessive allele).
1 mark: gametes G and g from each parent shown (correct cross diagram).
1 mark: brown offspring inherit g from both parents (gg) / 1 in 4 or 25% chance.
ALLOW any letters if clearly defined. DO NOT ALLOW parents shown as GG or homozygous.

Movement against a concentration gradient is active transport. It needs energy, which is released by respiration.

1 mark: active transport.
1 mark: (respiration releases) energy / ATP to move ions against the concentration gradient.
DO NOT ALLOW diffusion or osmosis. IGNORE 'makes energy'.

(a) Change = 0.56 - 0.50 = 0.06 g. \(\frac{0.06}{0.50}\times100 = 12\%\). (b) The sucrose solution had a higher water potential than the cell cytoplasm, so water entered by osmosis, increasing the mass.

(a) If answer = 12 (%) award 2 marks. Otherwise 1 mark for 0.06 / 0.50 x 100 (method).
(b) 1 mark: water moves into the cells by osmosis (from a more dilute solution / higher water potential to lower).
ALLOW ECF for (b) only if consistent with a decrease in (a). DO NOT ALLOW sucrose moves out/in as the explanation.

Bases code for amino acids. A changed base may alter an amino acid, changing how the chain folds, so the shape of the protein is different and it cannot function.

1 mark: change in amino acid sequence (a different amino acid in the chain).
1 mark: protein folds to a different shape / active site or binding site not the correct shape.
IGNORE 'protein is damaged' alone.

Fertiliser provides minerals so algae multiply and cover the surface. Light cannot reach submerged plants, which die. Decomposers and bacteria multiply on dead material and use up oxygen in respiration, so the minnows lack oxygen.

Any three from:
- minerals / nitrates cause (rapid) growth of algae;
- algae block light so plants / algae below die;
- more dead material / more decomposers (bacteria);
- decomposers use up oxygen (in respiration);
- minnows cannot get enough oxygen / suffocate.
ALLOW eutrophication described. IGNORE 'poisoned' unless linked to a mechanism.

Each feature is linked to its effect on the rate of diffusion or absorption.

Any two from:
- large surface area so more (digested food) can be absorbed at once / faster diffusion;
- thin wall so short diffusion distance;
- rich blood supply removes absorbed substances so a steep concentration gradient is maintained.
Each feature must be linked to its function. IGNORE 'absorbs quicker' without the feature.

Resolution is limited by wavelength. Electron beams have shorter wavelengths so very close points can be seen as separate.

1 mark: higher resolution / resolving power (not just magnification).
1 mark: electrons have a shorter wavelength than light.
ALLOW 'can see smaller objects / more detail' for the first mark only if resolution is implied. DO NOT ALLOW 'stronger magnification' alone.

Any variable other than light intensity that affects photosynthesis must be kept constant. Reliability is improved by repeating and averaging, and by using a heat shield or water bath so temperature is constant.

1 mark: two control variables from temperature, carbon dioxide concentration (e.g. sodium hydrogencarbonate), same pondweed / length, time counted.
1 mark: repeat readings at each distance and calculate a mean / time for longer / collect gas in a syringe to measure volume.
DO NOT ALLOW light intensity as a control variable.

Growth uses mitosis. DNA is replicated before division, then the copies are separated into two identical daughter nuclei.

1 mark: mitosis.
1 mark: chromosomes / DNA copied (replicated), then separated so each daughter cell gets a complete identical set.
DO NOT ALLOW meiosis. ALLOW 'chromosomes duplicate'.

41 mm is far from the other values so it is anomalous. Mean of the rest: \( (24+26+25+27) \div 4 = 102 \div 4 = 25.5 \) mm.

41 (mm) identified as anomalous [1]
First check the answer line. If answer = 25.5 award 2 marks.
Otherwise: adds 24, 26, 25 and 27 = 102 [1]; divides by 4 [1] (ECF if a different anomaly is chosen and the mean is calculated correctly from the remaining four values).
ALLOW 25.5 mm only with unit; IGNORE a mean that includes 41 (31.6) for the method marks except the first.

Brown fur needs two copies of the recessive allele (bb). The black parents must each carry one b allele hidden by the dominant B, so both are Bb. Each parent passes on b in a gamete, and these combine at fertilisation.

Both parents are Bb / heterozygous [1]
Each parent passes on the b allele (in a gamete) [1]
Offspring inherits two recessive alleles / is bb so shows the brown fur [1]
ALLOW a correct Punnett square for marks 1 and 2 if the offspring bb is identified.
DO NOT ALLOW 'gene' for 'allele' in the final mark if the idea is otherwise unclear.

Active transport needs energy from respiration. With little oxygen, aerobic respiration in root cells slows, so less energy is available to move nitrate ions into the cells against the concentration gradient.

Little oxygen so less aerobic respiration [1]
Less energy / ATP released [1]
Active transport (of nitrate) needs energy / works against the concentration gradient so slows [1]
IGNORE 'no respiration'; ALLOW 'less energy made' AW.

Surface area = \( 6 \times 6 \times 6 = 216 \) mm². Volume = \( 6^3 = 216 \) mm³. Ratio = 216 : 216 = 1 : 1.

First check the answer line. If answer = 1 : 1 award 2 marks.
Surface area = 216 and volume = 216 [1 for both]; ratio simplified to 1 : 1 [1].
ALLOW 1 for correct surface area or volume only; ECF on an arithmetic slip in one value.

Fewer otters means fewer urchins are eaten, so the urchin population grows and eats more kelp, so the amount of kelp falls.

Kelp decreases / falls [1]
Because more urchins survive / urchin population increases as fewer are eaten, so more kelp is eaten [1]
IGNORE 'otters die so kelp dies'.

The order of bases determines the order of amino acids. A changed base can change an amino acid, altering how the chain folds. The active site shape changes so the substrate no longer fits.

Base sequence / triplet code changed so a different amino acid is (coded for) in the protein [1]
Protein folds differently / enzyme has a different shape [1]
Active site no longer complementary to / does not fit the substrate [1]
ALLOW 'denatured' only if linked to shape change.

Reliability improves by repeating the readings and averaging them, and by controlling other variables or counting over a longer time to reduce counting error.

Any two from:
repeat the readings at each distance and calculate a mean [1]
identify and ignore anomalous results [1]
count bubbles for a longer time / collect gas and measure its volume [1]
keep other variables constant, e.g. temperature using a water bath / heat shield [1]
IGNORE 'do it again' without mention of a mean; IGNORE 'more accurate' alone.