2025 OCR AS Level Biology A - H020 Practice Paper with Answers

The hierarchy runs domain, kingdom, phylum, class, order, family, genus, species. Option A swaps class and phylum, B swaps order/class and genus/family, D places kingdom above domain.

C = 1 mark (AO1). Any other letter = 0.

Meiosis I halves the chromosome number, so each cell has 4 chromosomes. Chromatids have not yet separated at metaphase II, so each chromosome has 2 chromatids, giving 8 chromatids.

A = 1 mark (AO2). Any other letter = 0.

Surface area = \(6 \times 4^2 = 96\) mm\(^2\). Volume = \(4^3 = 64\) mm\(^3\). Ratio = \(96/64 = 1.5\).

D = 1 mark (AO2). Method: surface area / volume with 96 and 64. Common errors: 0.67 (inverted), 6.0 (6 faces only).

Blood and water flow in opposite directions, so water always meets blood of lower oxygen concentration and diffusion continues along the whole lamella.

B = 1 mark (AO1). Other options: A confuses with lamellae structure; C and D are incorrect as equilibrium would reduce uptake.

\(-200\) kPa is higher than \(-400\) kPa, so water moves by osmosis down the water potential gradient into the cell.

C = 1 mark (AO2). Osmosis is passive, so D is rejected.

In respiring tissues raised carbon dioxide lowers haemoglobin's affinity for oxygen, so the curve shifts right and oxygen is unloaded more readily.

A = 1 mark (AO1). Other options have wrong direction or wrong gas relationship.

\(Q_{10} = R_2 / R_1 = 2.0 / 0.8 = 2.5\), as the temperature interval is 10 °C.

D = 1 mark (AO2). 0.4 results from inverting the ratio.

Acid hydrolysis splits sucrose into glucose and fructose, which are reducing sugars. The acid must be neutralised because Benedict's reagent needs alkaline conditions; heating then gives a brick-red precipitate.

B = 1 mark (AO1). A gives no change as sucrose is non-reducing; iodine tests starch and biuret tests protein.

The three-domain system separates Archaea from Bacteria using evidence such as rRNA base sequences and membrane lipid differences, even though both are prokaryotic. A is wrong because Bacteria, plants and fungi all have cell walls. B is wrong because they are in different domains. D is wrong because domains are not based on cell number.

C (1 mark; AO1). Only one correct option. Distractors: A confuses cell wall presence with domain; B ignores the separation of the two prokaryotic domains; D uses an irrelevant criterion.

After meiosis I the cells are haploid, so each has 6 chromosomes. Chromatids have not yet separated at metaphase II, so each chromosome has two chromatids, giving 12 chromatids.

B (1 mark; AO2). Distractor A ignores that chromosomes are still duplicated; C and D use the diploid number, which applies only before meiosis I ends.

During activity, anaerobic respiration products lower the water potential of muscle cells, so fluid is drawn out of the tracheole ends by osmosis. This leaves more air-filled tracheole surface, so oxygen diffuses faster to the cells. Insect haemolymph does not transport oxygen, closed spiracles reduce gas exchange, and the heart does not ventilate tracheae.

A (1 mark; AO1). Distractors: B incorrectly gives haemolymph a respiratory pigment role; C reverses the function of spiracles; D confuses circulation with tracheal gas exchange.

The first test is negative, so the sugar is not a reducing sugar. Acid hydrolysis of the glycosidic bond releases reducing monosaccharides (glucose and fructose), giving a positive test. Sucrose is the only non-reducing sugar listed; glucose, maltose and lactose are all reducing sugars and would have given a positive result initially.

D (1 mark; AO2). Candidates must link the negative first test to a non-reducing sugar and the positive second test to hydrolysis.

Water moves by osmosis from the higher (less negative) water potential, \(-250\) kPa, to the lower, \(-400\) kPa. Water therefore enters the cell, increasing turgor. Active transport is not involved in osmosis.

B (1 mark; AO2). Distractor A reverses the gradient; C ignores the gradient; D misidentifies the mechanism.

\(Q_{10} = R_2 / R_1 = 2.0 / 0.8 = 2.5\), since the temperatures differ by \(10\,^{\circ}\mathrm{C}\).

C (1 mark; AO2). Method: rate at higher temperature divided by rate at lower temperature. A is the inverted ratio; B is the difference; D is a power-of-ten error.

In respiring tissues, high carbon dioxide levels lower the affinity of haemoglobin for oxygen. The curve shifts to the right, so more oxygen is unloaded at a given partial pressure of oxygen.

D (1 mark; AO1). A and B state the wrong direction or the wrong cause; C describes a different process.

A competitive inhibitor competes with substrate for the active site, so a higher substrate concentration makes substrate binding more likely and reduces inhibition. Non-competitive inhibitors bind elsewhere and alter the shape of the active site, so extra substrate cannot overcome them.

A (1 mark; AO1). B and C swap the properties of the two inhibitor types; D is incorrect because non-competitive inhibition is not relieved by extra substrate.

Variation (from random mutation and meiosis) comes first. Selection pressure then favours individuals with advantageous alleles, so they survive and reproduce more. They pass the alleles to offspring, so the allele frequency rises over generations. Only B has this order.

B = 1 mark. A, C and D are incorrect because each places selective survival, passing on of alleles or the rise in allele frequency before the stage that causes it. No partial marks.

Surface area = \(2(4\times2 + 4\times1 + 2\times1) = 2(8+4+2) = 28\ \text{cm}^2\). Volume = \(4\times2\times1 = 8\ \text{cm}^3\). Ratio = \(28/8 = 3.5:1\).

C = 1 mark. A is volume/surface area (8/28). B (16/8) omits faces from the surface area. D (28/4) uses an incorrect volume. No partial marks.

Homologous chromosomes separate in anaphase I, so each daughter cell receives one chromosome from each homologous pair: 14/2 = 7 chromosomes (each still made of two chromatids).

A = 1 mark. B is the chromosome number before meiosis I. C has no basis. D is the number of chromatids in the cell at prophase I, not chromosomes at the end of meiosis I.

Sucrose is a non-reducing sugar, so Benedict's stays blue. Boiling with dilute hydrochloric acid hydrolyses the glycosidic bond, giving glucose and fructose, which are reducing sugars. After neutralisation, Benedict's gives a red/brick-red precipitate.

D = 1 mark. A: sucrose does not react before hydrolysis. B: hydrolysis releases reducing sugars. C: reverses the actual results. No partial marks.

Convert units so that both lengths match: 42 mm = 42 000 \(\mu\)m.
Magnification = 42 000 / 3.5 = 12 000.
The magnification is \(\times 12\,000\).

Answer: 12 000 (ALLOW \(\times 12\,000\), 1.2 \(\times 10^4\)) = 2 marks (check the final answer first).
If the answer is incorrect, award 1 mark (method) for converting units correctly (42 000 \(\mu\)m or 0.0035 mm) OR for the correct rearranged calculation using consistent units.
DO NOT ALLOW 12 (unit conversion not done) for any marks beyond the method mark if units are inconsistent.
AO2 (quantitative data) x2.

Rate at 15 °C = 9 / 60 = 0.15 cm³ s⁻¹.
Rate at 25 °C = 27 / 90 = 0.30 cm³ s⁻¹.
The temperatures differ by 10 °C, so \(Q_{10} = 0.30 / 0.15 = 2.0\).

Answer: 2.0 = 2 marks (ALLOW 2; check the final answer first).
If the answer is incorrect, award 1 mark (method) for both rates correctly calculated (0.15 and 0.30 cm³ s⁻¹, or 9 and 18 cm³ min⁻¹ etc.) OR for the correct use of \(R_2 / R_1\) with ECF from one incorrect rate.
DO NOT ALLOW 0.5 (inverted ratio) for the accuracy mark; ALLOW 1 method mark for the rates.
AO2 (quantitative data) x2.

Convert cardiac output: 15.2 dm³ min⁻¹ = 15 200 cm³ min⁻¹.
Heart rate = cardiac output / stroke volume = 15 200 / 95 = 160 beats per minute.

Answer: 160 (beats per minute) = 2 marks (check the final answer first).
If the answer is incorrect, award 1 mark (method) for converting to 15 200 cm³ min⁻¹ OR for the correct rearrangement heart rate = cardiac output / stroke volume with consistent units.
DO NOT ALLOW 0.16 (units not converted) for the accuracy mark.
AO2 (quantitative data) x2.

Statement 1: water enters when the buccal cavity floor is lowered, so its volume increases and its pressure falls below that of the water outside; water then flows in down the pressure gradient. Statement 2: water and blood flow in opposite directions (counter-current), which maintains a concentration gradient for oxygen along the whole length of the lamella. Statement 3: a thin, one-cell-thick epithelium gives a short diffusion distance; a thick one would lengthen it.

One mark for each statement, awarded only when the error is identified and the correction given (the correction is the creditworthy element).

Statement 1 [1 mark, AO1]: volume (of buccal cavity) increases / pressure decreases (AW). ALLOW 'mouth' for buccal cavity. DO NOT ALLOW correction of pressure alone if volume is stated as still decreasing.

Statement 2 [1 mark, AO1]: water flows in the opposite direction to the blood / counter-current (flow). IGNORE 'different direction' without 'opposite' or 'counter-current'.

Statement 3 [1 mark, AO1]: epithelium / lamellae walls are thin / one cell thick (AW). ALLOW 'thin walls / thin surface'. DO NOT ALLOW 'thin lamellae' with no reference to the epithelium/wall/diffusion distance being short.

Maximum 3 marks.

The insecticide acts as a selection pressure. Beetles with a resistance allele survive, reproduce and pass the allele on, so its frequency rises in the next generations.

Question 1 | 1 mark | AO1 | Resistant beetles survive/reproduce more (than non-resistant) and pass on the allele(s) to offspring. ALLOW selective advantage / natural selection. IGNORE 'beetles adapt' / 'beetles mutate in response'. DO NOT ALLOW resistance 'developing because of' the insecticide.

Counter-current flow means blood always meets water with a higher oxygen concentration, so diffusion occurs along the whole length of the lamella.

Question 2 | 1 mark | AO1 | Counter-current flow / blood and water flow in opposite directions OR dense capillary network / good blood supply removes oxygen (maintaining gradient). IGNORE 'large surface area' alone. DO NOT ALLOW 'blood and water flow in the same direction'.

Cardiac output = heart rate \(\times\) stroke volume = \(72 \times 85 = 6120\) cm\(^3\) min\(^{-1}\).

Question 3 | 1 mark | AO2 | 6120 (cm\(^3\) min\(^{-1}\)). ALLOW 6.12 dm\(^3\) min\(^{-1}\) or 6.12 litres per minute if the unit is given. Correct answer without working scores the mark.

Beetroot cells contain a red pigment in the vacuole. High temperature damages the tonoplast and cell surface membrane, making them permeable so the pigment diffuses out.

Question 4 | 1 mark | AO2 | Membrane (proteins) denatured / phospholipids gain kinetic energy and the bilayer is disrupted so pigment leaks out / membrane more permeable. ALLOW membrane damaged AND pigment escapes (both required). IGNORE 'cells burst'.

Benedict's solution only detects reducing sugars. Non-reducing sugars such as sucrose must first be hydrolysed by acid, and the mixture neutralised before retesting.

Question 5 | 1 mark | AO3 | Non-reducing sugar could be present AND hydrolyse with acid / boil with HCl then neutralise and retest (both elements required for the mark). ALLOW sodium hydrogencarbonate to neutralise. DO NOT ALLOW 'test is faulty' without reference to reducing/non-reducing sugars.

In anaphase the centromeres split, and spindle fibres shorten, pulling the separated sister chromatids to opposite poles.

Question 6 | 1 mark | AO1 | Sister chromatids / chromosomes separate and are pulled to opposite poles (by spindle fibres/microtubules). ALLOW centromeres divide. DO NOT ALLOW 'homologous chromosomes separate'.

Convert units: 45 mm = 45 000 \(\mu\)m. Magnification = image size / real size = \(45000 \div 30 = 1500\).

Question 7 | 1 mark | AO2 | 1500 (or \(\times 1500\)). Correct answer scores the mark. ALLOW 1.5 \(\times 10^3\). If the answer is incorrect, no marks (the unit conversion is required).

Resistant aphids already existed because of a chance mutation. The insecticide acted as a selection pressure, so resistant aphids survived and reproduced, and the resistance allele became more common in later generations.

1. Resistance allele present due to (random) mutation / variation in the population, so some aphids survive the insecticide (AO1)
2. Survivors reproduce and pass on the resistance allele to offspring, so allele frequency increases over generations (AO2)
ALLOW 'gene' for 'allele'. IGNORE 'aphids adapted / became immune' without reference to alleles or inheritance. DO NOT ALLOW 'the insecticide caused the mutation'.

Fewer differences in the DNA sequence indicate less time since divergence from a common ancestor, so X is the closest relative.

1. X (AO2)
2. Fewest base differences / least difference in DNA sequence, indicating a more recent common ancestor / less time for mutations to accumulate (AO3)
Second mark requires the idea of recent divergence / accumulated mutations, not just 'fewest differences'. DO NOT ALLOW Y.

Percentage error = \( \frac{0.5}{20} \times 100 = 2.5\% \).

Correct answer 2.5 (%) = 2 marks.
If answer incorrect: 0.5 / 20 (× 100) = 1 mark (method).
ALLOW 2.5 without % sign. DO NOT ALLOW 5 (from using 1.0 total range) unless method clearly shown for 1 mark only. Answer must be given to 2 s.f. for the accuracy mark (AO2).

Meiosis halves the chromosome number, 14 to 7. The paired chromosomes that line up in metaphase I are homologous chromosomes.

(a) 7 (AO2) [1]
(b) homologous (chromosomes / pair) (AO1) [1]
ALLOW bivalent. DO NOT ALLOW 'sister chromatids' or 'homozygous'.

Ventilation rate = tidal volume \(\times\) breathing rate = \(0.50 \times 14 = 7.0\) dm\(^3\) min\(^{-1}\).

Correct answer 7 / 7.0 = 2 marks.
If incorrect: 0.50 × 14 shown = 1 mark (method).
Unit is printed so is not required. DO NOT ALLOW answers using 14 ÷ 0.50 (28) for method mark (AO2).

Sucrose must first be hydrolysed to glucose and fructose, which are reducing sugars. The acid must be neutralised because Benedict's reagent needs alkaline conditions.

1. Heat/boil with (dilute) hydrochloric acid to hydrolyse sucrose / break glycosidic bond (AO1)
2. Neutralise / add alkali (e.g. sodium hydrogencarbonate) then heat with Benedict's, giving orange/brick-red precipitate (AO2)
IGNORE 'add enzyme' unless sucrase is stated and then neutralising is not needed. ALLOW colour change to green/yellow/orange/red.

High temperature disrupts the membrane structure, increasing permeability, so pigment escapes faster.

1. Proteins (channel/carrier) in the membrane denature / phospholipids more kinetic energy and move more, disrupting the membrane (AO1)
2. Membrane more permeable / gaps form, so more pigment diffuses out (down its concentration gradient) (AO2)
IGNORE 'membrane melts'. DO NOT ALLOW 'enzymes denatured' alone.

\(Q_{10} = R_2 / R_1 = 7.4 / 3.2 = 2.3125\), which is 2.3 to 2 s.f.

Correct answer 2.3 = 2 marks.
If incorrect: 7.4 / 3.2 or 2.3125 / 2.31 shown = 1 mark (method; allow if rounding incorrect).
DO NOT ALLOW 0.43 (inverted ratio) (AO2).

Distilled water has a higher water potential than the cytoplasm of both cells, so water enters each cell by osmosis. The red blood cell has no cell wall, so the surface membrane cannot withstand the increase in volume and it bursts. The root hair cell has a cellulose cell wall, which resists expansion, so the cell becomes turgid instead.

Question: Explain why the red blood cell bursts but the root hair cell does not. Total 2 marks (AO2 x2).
1. Water enters the red blood cell by osmosis / down a water potential gradient, and the cell has no cell wall (to resist expansion) so it bursts / lyses (1 mark; both linked elements required).
2. The root hair cell has a (cellulose) cell wall which prevents bursting / the cell becomes turgid / wall exerts pressure that stops further water entry (1 mark).
ALLOW AW for 'bursts' e.g. lyses. ALLOW 'cell wall is strong / rigid / inelastic'.
IGNORE 'water diffuses' without reference to osmosis if water potential gradient is described.
DO NOT ALLOW 'cell wall is semi-permeable' or 'cell wall stops water entering'.

\(Q_{10} = \frac{R_2}{R_1}\) where the temperatures differ by 10 °C.
\(Q_{10} = \frac{1.90}{0.80} = 2.375\), which is 2.4 to 2 significant figures.

Question: Calculate Q10 to 2 s.f. Total 2 marks (AO2 x2).
Answer: 2.4 = 2 marks.
If the answer is incorrect, award 1 mark for 1.90 / 0.80 (method mark) OR for 2.375 / 2.38 shown but not rounded to 2 s.f.
ALLOW 1 mark for 2.3 (incorrect rounding of 2.375) only if working shown as 1.90 / 0.80.
IGNORE units (Q10 has no units).
DO NOT ALLOW 0.42 (inverted ratio) for the accuracy mark; award 1 method mark only if working shows 0.80 / 1.90 with an incorrect inversion noted by ECF.

Variation exists in wing size, caused by different alleles. The wind is a selection pressure: beetles with fully developed wings are removed so they die before reproducing. Beetles with small wings survive, breed and pass on their alleles to offspring, so the allele frequency for small wings increases over generations.

Max 3 from:
1. Beetles with small wings are more likely to survive / fewer are blown into the sea (the wind is the selection pressure) (AW);
2. Small-winged beetles are more likely to reproduce / breed;
3. They pass on the allele(s) for small wings to offspring;
4. The frequency of the allele for small wings increases (over generations) (AW).
IGNORE 'adapt' or 'mutate' on its own. DO NOT ALLOW 'beetles evolve' or 'beetles choose to' for the survival marks.

Surface area = \(2(6\times2 + 6\times1 + 2\times1) = 2(12+6+2) = 40\ \text{mm}^2\). Volume = \(6\times2\times1 = 12\ \text{mm}^3\). Ratio = \(40 / 12 = 3.33...\), which is 3.3 : 1 to 2 significant figures.

Correct answer 3.3 (: 1) = 3 marks.
If the answer is incorrect, award:
1. Surface area calculated as 40 (mm²);
2. Volume calculated as 12 (mm³);
3. Surface area divided by volume, given to 2 significant figures (3.3).
ALLOW ECF for the final mark from an incorrect surface area or volume. ALLOW 10 : 3. Answer given to a different number of significant figures, e.g. 3.33, loses the third mark only.

(a) Mitotic index = number of cells in mitosis / total number of cells = \(12/150 = 0.08\), which is 8%. (b) Chromosomes line up at the equator of the cell during metaphase.

(a) Correct answer 8 (%) = 2 marks. If incorrect, award 1 mark for \(12/150\) (or 0.08).
DO NOT ALLOW 12.5 (inverted calculation) for the second mark.
(b) Metaphase [1]. IGNORE 'metaphase 1' or 'metaphase 2'.

Meiosis consists of two successive nuclear divisions, so one diploid parent cell gives four haploid daughter cells. Homologous chromosomes pair up in prophase 1, line up in metaphase 1 and are pulled to opposite poles in anaphase 1; in anaphase 2 sister chromatids separate instead.

Box 1: four / 4 (1 mark, AO1). ALLOW 'four haploid cells'. DO NOT ALLOW 'two' or 'four diploid cells'.
Box 2: anaphase 1 / anaphase I / first division of meiosis (1 mark, AO1). ALLOW 'meiosis I' / 'meiosis 1'. DO NOT ALLOW 'anaphase 2' or 'anaphase' without a number, and IGNORE 'prophase 1' or 'metaphase 1'.
Total: 2 marks.

Facilitated diffusion is passive, so it needs no ATP and moves substances down their concentration gradient (P, Q). Active transport uses carrier (pump) proteins, which hydrolyse ATP to move substances against the gradient (R, S, T). Non-polar molecules dissolve in the hydrophobic core of the bilayer formed by the fatty acid tails (U).

Question: Complete table and sentence | Marks: 3 | AO1 x3

1 mark: P = No AND Q = down / along / with the concentration gradient (both required for the mark). ALLOW 'from high to low concentration'. DO NOT ALLOW 'ATP not needed' for Q.

1 mark: R = carrier protein AND S = Yes (both required). ALLOW 'pump' / 'transport protein' / 'carrier'. DO NOT ALLOW 'channel protein' for R. IGNORE 'protein' alone.

1 mark: T = against / up the concentration gradient (low to high concentration) AND U = hydrophobic (fatty acid) tails / hydrophobic core / interior (both required). DO NOT ALLOW 'hydrophilic heads' or 'phosphate heads' for U. ALLOW 'hydrocarbon tails' / 'non-polar centre'.

Convert the image length to micrometres: \(42\ \text{mm} = 42\,000\ \mu\text{m}\).
Rearrange: actual size = image size / magnification = \(42\,000 / 1400 = 30\ \mu\text{m}\).

Answer: 30 (\(\mu\text{m}\)) [1 mark; AO2 handling quantitative data].
Guidance: the unit must be correct for the mark. ALLOW 0.03 mm only if the answer line unit is not given; DO NOT ALLOW 30 without a unit equivalent to micrometres. Correct working with a unit conversion error scores 0 (single-mark item).

Cardiac output = heart rate \(\times\) stroke volume = \(64 \times 78 = 4992\ \text{cm}^3\ \text{min}^{-1}\).

Answer: 4992 (cm\(^3\) min\(^{-1}\)) [1 mark; AO2 handling quantitative data].
Guidance: ALLOW 4990 or 5000 (rounded) OR 4.992 dm\(^3\) min\(^{-1}\) if the unit is stated. IGNORE missing unit if the unit is printed on the answer line.

Convert the image size to micrometres: \(46\ \text{mm} = 46\,000\ \mu\text{m}\).
Rearrange: real size = image size / magnification = \(46\,000 / 800 = 57.5\ \mu\text{m}\).

Answer 57.5 (\(\mu\)m) = 2 marks (AO2). ALLOW 58 (\(\mu\)m) or 57.5 to any sensible rounding.
If the answer is incorrect, award 1 mark (method) for either converting 46 mm to 46 000 \(\mu\)m OR for 46 / 800 (= 0.0575 mm) OR for rearranging to real size = image size / magnification.
DO NOT ALLOW 0.0575 as the final answer on the \(\mu\)m line (no unit conversion): maximum 1 mark.

The temperatures are 10 °C apart, so the formula applies directly.
\(Q_{10} = R_2 / R_1 = 4.5 / 1.8 = 2.5\).

Answer 2.5 = 2 marks (AO2). ALLOW 2.50.
If the answer is incorrect, award 1 mark (method) for 4.5 / 1.8 OR for correct substitution of the values into the formula with the higher rate as numerator.
DO NOT ALLOW 0.4 or 0.40 (inverted ratio): 0 marks.
If 2.5 is not given to two significant figures but working is correct, award 1 mark (e.g. 2.5 is accepted; 2.500 gains method mark only).
IGNORE units (Q10 has no units).

Rate at 15 °C: \(R_1 = 0.42 \div 6.0 = 0.070\) cm\(^3\) min\(^{-1}\).
Rate at 25 °C: \(R_2 = 0.96 \div 8.0 = 0.12\) cm\(^3\) min\(^{-1}\).
\(Q_{10} = R_2 / R_1 = 0.12 \div 0.070 = 1.714...\)
To 2 significant figures, \(Q_{10} = 1.7\) (no units).

Question: Calculate Q10 | Answer: 1.7 | Marks: 3 | AO2 (quantitative data) | Guidance: check the final answer first; if it is 1.7 award 3 marks.

MP1 (method): rate at 15 °C = 0.42 / 6.0 = 0.070 cm3 min-1 (1).
MP2 (method): rate at 25 °C = 0.96 / 8.0 = 0.12 cm3 min-1 (1).
MP3 (accuracy and precision): Q10 = 0.12 / 0.070 = 1.7, given to 2 significant figures (1).

ALLOW ECF for MP3 from incorrect rates in MP1/MP2 if R2 / R1 is used and the answer is given to 2 s.f. ALLOW 1.71 or 1.714 for 2 marks (MP1 and MP2 implied) if not rounded to 2 s.f. DO NOT ALLOW the inverse ratio (R1 / R2 = 0.58). IGNORE units for Q10, but DO NOT ALLOW rates without a time component (e.g. 0.42 and 0.96 used directly gives 2.3: maximum 0).

Moving a substance against its concentration gradient using ATP is active transport, not facilitated diffusion. Facilitated diffusion is passive: it moves substances down a concentration gradient through channel or carrier proteins and needs no ATP. The student has therefore used the wrong transport process, so 'facilitated diffusion' should be replaced by 'active transport'.

Question: Identify and correct the error | Marks: 2 | AO1.1 x1, AO2.1 x1

1. Error identified: facilitated diffusion is passive / does not use ATP / does not move substances against a concentration gradient (AW) [1]
2. Correction: 'facilitated diffusion' replaced by 'active transport' OR statement changed so that facilitated diffusion is down the concentration gradient and without ATP [1]

Guidance: ALLOW 'co-transport' for active transport if linked to movement against the gradient. IGNORE comments on the shape change of the carrier protein. DO NOT ALLOW a correction that changes 'against' to 'down' while keeping ATP use. Both the error and the correction must be consistent with each other for the second mark.

Part 1 (why): the invertebrate has a large surface area to volume ratio and a very short diffusion distance, so oxygen reaches all cells quickly by simple diffusion. The fish has a much smaller surface area to volume ratio, a greater volume of respiring tissue with a higher oxygen demand, and a less permeable body surface, so diffusion across the body surface alone is too slow.
Part 2 (gills): many gill filaments covered in lamellae give a large surface area. Lamellae are thin, with a short diffusion distance between water and blood. A dense capillary network carries blood away, maintaining the gradient. Blood flows through the lamellae in the opposite direction to water (counter-current), so water always meets blood of lower oxygen concentration and equilibrium is never reached; diffusion occurs along the whole length of the lamella. The mouth and operculum pump water over the gills continuously, maintaining a supply of oxygen-rich water.

Levels of response (AO1 and AO2, 6 marks).
Level 3 (5-6 marks): Explains the SA:V ratio difference AND gives a detailed, correctly sequenced description of at least three gill features including counter-current flow with its effect on the gradient. Clear, logical, relevant throughout.
Level 2 (3-4 marks): Some explanation of the SA:V ratio difference and describes some gill features with partial explanation; counter-current may be mentioned without explaining the maintained gradient. Some structure.
Level 1 (1-2 marks): Simple statements, e.g. fish is bigger / gills have a large surface area, with little or no explanation. Limited logic.
0 marks: no creditworthy response.

Indicative content:
- small organism: large SA:V ratio / short diffusion distance so diffusion is sufficient
- fish: small SA:V ratio / high metabolic rate / large volume / less permeable body surface, so needs specialised surface
- many filaments and lamellae give large surface area
- lamellae thin / one cell thick, short diffusion distance
- rich blood supply / capillaries maintain concentration gradient
- counter-current: blood and water flow in opposite directions
- so a concentration gradient is maintained along the whole length of the lamella / more oxygen diffuses into blood
- ventilation by mouth and operculum movements maintains water flow
ALLOW AW. IGNORE 'gills have a large SA' alone as full credit for the surface area point only if linked to exchange. DO NOT ALLOW 'gills let in oxygen' or references to lungs/air.

Within the original population there was variation in tolerance, arising from random, spontaneous mutation producing different alleles. When the soil became contaminated the metal acted as a selection pressure. Plants with the tolerance allele survived, grew and reproduced more successfully than non-tolerant plants, which died or produced few seeds. Tolerant plants passed the advantageous allele to offspring, so over successive generations the frequency of the allele increased and the population became more tolerant. Evidence from modern times: bacteria evolving antibiotic resistance (e.g. in TB), where antibiotic acts as a selection pressure and resistant bacteria survive and reproduce, passing on the resistance allele; or insects evolving resistance to pesticides. These show observable changes in allele frequency over short timescales.

Levels of response (AO1 and AO2, 6 marks).
Level 3 (5-6 marks): Gives a complete, correctly sequenced explanation of natural selection applied to the plants (variation from mutation, selection pressure, differential survival/reproduction, inheritance, change in allele frequency) AND describes a named modern example with how selection acted. Clear and logical.
Level 2 (3-4 marks): Gives most steps of natural selection applied to the context, with some missing links; modern example mentioned with limited explanation.
Level 1 (1-2 marks): Limited statements, e.g. tolerant plants survive / plants adapt; example absent or vague. Little sequence.
0 marks: no creditworthy response.

Indicative content:
- variation in tolerance exists in the population due to random, spontaneous mutation (allele(s) for tolerance)
- metal in soil is a selection pressure / environmental factor
- tolerant plants more likely to survive, grow and reproduce / non-tolerant die
- tolerance allele passed on to offspring
- allele frequency increases over many generations
- modern evidence: antibiotic resistance in bacteria (e.g. MRSA, TB) OR pesticide/herbicide resistance OR other valid observed example, with selection by the antibiotic/pesticide explained
ALLOW AW. DO NOT ALLOW 'plants chose to adapt' / 'plants mutated because of the metal' (Lamarckian wording) for the variation and mutation points. IGNORE reference to Darwin or Wallace unless linked to the explanation.

Increased carbon dioxide lowers the pH of the blood (carbonic acid forms and releases hydrogen ions). This reduces the affinity of haemoglobin for oxygen (the Bohr effect), so haemoglobin releases oxygen more readily at a given partial pressure of oxygen. The dissociation curve therefore shifts to the right: at any intermediate partial pressure the percentage saturation is lower than on the original curve. Both curves start at the origin and approach 100% saturation at high partial pressures of oxygen.

Question: Sketch of curve X | Marks: 1 | AO2 | Guidance:
1 mark: curve X is S-shaped (sigmoid), labelled X, and lies to the right of / below the original curve at intermediate partial pressures of oxygen (approximately 2 to 10 kPa), starting at the origin and reaching or approaching the same plateau near 100% at high partial pressure of oxygen.
ALLOW curve X converging with the original curve at the top right.
IGNORE small differences in exact shape provided the curve is clearly shifted right.
DO NOT ALLOW a curve that lies to the left/above the original, a curve that crosses the original curve at intermediate values, a curve that does not start at or near 0, or an unlabelled curve.

The ratio simplifies to \(\frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}\). So ratio = 3 / 0.8 = 3.75, which is 3.8 to 2 significant figures (\(\mu m^{-1}\)).

Answer 3.8 (\(\mu m^{-1}\)) = 1 mark (AO2). ALLOW 3.75 only if rounded to 3.8 as required; DO NOT ALLOW 3.7. Correct unit and 2 s.f. needed for the mark. Method guidance: 3/r or separate calculation of surface area (8.04) and volume (2.14) both accepted.

Heat denatures the channel and carrier proteins and makes the phospholipid bilayer more fluid, so the membrane becomes more permeable and the red pigment diffuses out.

1 mark (AO2): membrane proteins denatured / phospholipid bilayer damaged AND membrane more permeable so pigment leaves. Both linked elements required. IGNORE 'membrane melts' without reference to permeability or leakage. ALLOW 'betalain' for pigment.

\(Q_{10} = R_2 / R_1 = 5.0 / 2.4 = 2.083\), which is 2.1 to 2 significant figures.

2.1 = 1 mark (AO2). ALLOW 2.08 only if followed by 2.1. DO NOT ALLOW 2.0 or 0.48 (inverted). No unit required; IGNORE a unit if given as none.

Resistant individuals already existed through random mutation. The insecticide is a selection pressure: resistant insects survive and reproduce, passing the allele to offspring, so the allele frequency rises in successive generations.

1 mark (AO2): resistant insects survive and reproduce AND pass on the resistance allele so its frequency increases. Linked elements required. ALLOW 'gene' for allele. DO NOT ALLOW the insects 'adapted' / 'mutated because of the insecticide'.

Any variable other than pH that alters enzyme activity must be kept constant, for example temperature, which changes kinetic energy and enzyme denaturation. Otherwise the result cannot be attributed to pH.

1 mark (AO3): a valid variable (temperature / concentration or volume of amylase / concentration or volume of starch) with a linked reason that it affects rate so the change cannot be attributed to pH. Both parts required. DO NOT ALLOW 'to make it a fair test' alone. IGNORE time or pH.

Predation selects against the pale form; dark beetles survive and breed, passing on the advantageous allele to offspring.

1. Dark beetles are less visible/better camouflaged so are more likely to survive predation (ORA) [AO2]
2. Survivors reproduce and pass on the allele for dark wing cases, so the allele frequency increases over generations (AW) [AO2]
IGNORE 'adapt' on its own. DO NOT ALLOW beetles 'chose' to change colour.

Change = 2.04 - 2.40 = -0.36 g. Percentage change = \(\frac{-0.36}{2.40} \times 100 = -15\) %.

1. (2.04 - 2.40) / 2.40 × 100 (method) [AO2]
2. -15 (%) / 15 % decrease, to 2 s.f. (accuracy) [AO2]
Correct answer gains both marks. ALLOW 1 mark for 15 % given without a sign or the word decrease only if method is shown. DO NOT ALLOW -15.0 or -14.9 for the precision mark.

In anaphase I the members of each homologous pair are pulled apart to opposite poles, halving the chromosome number.

1. Homologous chromosomes separate/are pulled to opposite poles (by spindle fibres) [AO1]
2. Each daughter cell receives one chromosome from each pair / half the chromosome number, so haploid [AO1]
DO NOT ALLOW chromatids separating in anaphase I. IGNORE 'half the DNA' without reference to chromosome number.

Blood always meets water with a higher oxygen concentration, so a favourable gradient persists along the gill.

1. Blood always meets water of higher oxygen concentration / concentration gradient maintained along the whole length of the capillary (ORA) [AO1]
2. So oxygen continues to diffuse into the blood along the whole length, giving a higher percentage of oxygen absorbed / blood reaches equilibrium later (AW) [AO2]
IGNORE 'faster diffusion' without reference to gradient.

Sucrose is a non-reducing sugar. Hydrolysis by acid gives reducing monosaccharides that reduce copper(II) ions to give a coloured precipitate.

1. Colour change to orange/brick-red/red precipitate (ALLOW green/yellow if described as positive) [AO1]
2. Acid hydrolyses the glycosidic bond in sucrose, releasing reducing sugars glucose and fructose [AO1]
DO NOT ALLOW 'sucrose is broken down' without naming reducing sugars / hydrolysis.

-200 kPa is higher than -600 kPa, so water moves from the solution into the cell by osmosis across the partially permeable membrane.

1. Into the cell / from the solution into the cell by osmosis [AO2]
2. The solution has a higher (less negative) water potential than the cell, so water moves down the water potential gradient (across a partially permeable membrane) [AO2]
DO NOT ALLOW 'higher concentration of water' as the only explanation without link to gradient. Direction must be correct to credit mark 2.

The temperatures differ by 10 °C. \(Q_{10} = \frac{R_2}{R_1} = \frac{1.70}{0.80} = 2.125 \approx 2.1\).

1. 1.70 / 0.80 (method) [AO2]
2. 2.1 (accuracy, 2 s.f.) [AO2]
Correct answer gains both marks. ALLOW 2.125 for the method mark only. Inverted ratio (0.47) gains 0 marks.

Cardiac output = heart rate × stroke volume = \(72 \times 70 = 5040\) cm\(^3\) min\(^{-1}\).

1. 72 × 70 (method) [AO2]
2. 5040 cm^3 min^-1 (accuracy, unit required) [AO2]
Correct answer with no working gains both marks. ALLOW 5.04 dm^3 min^-1 if unit is correct. Missing or incorrect unit loses the second mark.

Both strands act as templates. Each daughter molecule retains one parental strand. DNA polymerase catalyses joining of free nucleotides to the growing strand.

1. Each new DNA molecule consists of one original strand and one new strand [AO1]
2. DNA polymerase joins (complementary) nucleotides / catalyses formation of phosphodiester bonds in the new strand [AO1]
IGNORE 'half the DNA' without reference to strands. DO NOT ALLOW DNA polymerase unwinding the helix.

Rates are proportional to the volume released in equal times (60 s). \(Q_{10} = R_2 / R_1\), where the temperature interval is 10 °C. \(Q_{10} = 7.5 / 3.0 = 2.5\).

Question: Calculate Q10 | Answer: 2.5 | Marks: 2 | AO2 (quantitative data)

Guidance: Correct answer of 2.5 = 2 marks. If the answer is incorrect or missing, award 1 mark (method) for 7.5 / 3.0 (rate at 30 °C divided by rate at 20 °C). ALLOW 1 mark for 7.5/60 divided by 3.0/60 (working with rates). DO NOT ALLOW 0.4 (inverted ratio) for the accuracy mark; this may earn the method mark only if the working is shown as 3.0/7.5. ALLOW 2.50. IGNORE a missing unit as Q10 has no unit.

Dark moths are camouflaged on dark trunks and are less likely to be predated. They therefore survive longer, reproduce more and pass the advantageous allele to their offspring. Over generations the frequency of the allele, and so the proportion of dark moths, increases (natural selection).

Question: Explain how the proportion of dark moths can increase | Marks: 2 | AO2 (theoretical context)

Marking points (max 2):
1. dark moths are better camouflaged / less likely to be eaten (by birds) / have a selective advantage, so more survive AW;
2. (more dark moths) reproduce / breed and pass on the (advantageous) allele to offspring;
3. allele frequency / proportion of dark moths increases over generations AW.

Guidance: ALLOW 'survival of the fittest' only if linked to survival and reproduction. IGNORE 'moths adapt / change colour / need to be dark'. DO NOT ALLOW 'the mutation is caused by pollution'.

(a) Rate at 15 °C = 3.9 / 10 = 0.39 cm\(^3\) min\(^{-1}\); rate at 25 °C = 8.8 / 10 = 0.88 cm\(^3\) min\(^{-1}\). The temperature interval is 10 °C, so \(Q_{10} = R_2 / R_1 = 0.88 / 0.39 = 2.256... = 2.3\) (2 s.f.). (b) At temperatures near 55 °C, enzymes involved in respiration begin to denature (hydrogen bonds break and active sites change shape), so rate increases less, or falls, giving a lower Q10.

(a) 2 marks. Correct final answer 2.3 = 2 marks (ALLOW 2.26 or 2.256 for 1 mark only if not rounded to 2 s.f.). Method mark: \(R_2/R_1\) = 8.8/3.9 OR 0.88/0.39 (1 mark). DO NOT ALLOW inverted ratio. Units not required (Q10 has no unit); IGNORE a unit if given. AO2 (quantitative data). (b) 1 mark: enzymes / proteins (of respiration) denature / lose shape of active site at higher temperature AW, so rate does not increase as much / decreases (the linking of denaturation to a lower rate increase is needed). IGNORE 'enzymes die'. DO NOT ALLOW 'enzymes are killed'. AO2 (theoretical context).

(a) \(Q_{10} = R_2 / R_1 = 1.90 / 0.80 = 2.375\), which is 2.4 to 2 significant figures.
(b) Between 40 °C and 50 °C the enzyme is denatured: hydrogen and ionic bonds break, the tertiary structure and active site change shape, and substrate can no longer bind. Fewer enzyme-substrate complexes form, so the rate at 50 °C is lower than at 40 °C and \(R_2 / R_1 < 1\).

(a) 2 marks (AO2, quantitative).
Correct answer 2.4 = 2 marks.
If the answer is wrong or not to 2 s.f.: 1 mark for 1.90 / 0.80 (method) OR 2.375 / 2.38 seen (unrounded). ALLOW 2.4 with no working for 2 marks.
(b) 2 marks (AO2, theoretical).
1. Enzyme denatured / active site changes shape / tertiary structure altered (by bonds breaking) at high temperature;
2. Fewer enzyme-substrate complexes formed / substrate no longer fits active site, so rate decreases (AW);
DO NOT ALLOW 'enzyme killed'. IGNORE 'enzyme dies'. ALLOW a mark for 'hydrogen bonds break' for point 1.

(a) Small worm: surface area = \(6 \times 0.5^2 = 1.5\) mm²; volume = \(0.5^3 = 0.125\) mm³; ratio = \(1.5 / 0.125 = 12\) (12:1).
Large worm: surface area = \(6 \times 4.0^2 = 96\) mm²; volume = \(4.0^3 = 64\) mm³; ratio = \(96 / 64 = 1.5\) (1.5:1).
(b) The larger worm has a much lower surface area to volume ratio. Its surface area is too small relative to the oxygen demand of its large volume of respiring cells, and diffusion distances to inner cells are long, so diffusion alone is too slow. A specialised exchange surface (large surface area, thin, with a concentration gradient maintained) is needed.

(a) 2 marks (AO2, quantitative).
1. Surface area and volume both calculated correctly for at least one worm (1.5 and 0.125, or 96 and 64) (method);
2. Both ratios correct: 12 (:1) and 1.5 (:1) (accuracy).
ALLOW ECF for one ratio if an arithmetic slip is made in the other. Units mm⁻¹ allowed but not required.
(b) 2 marks (AO2, theoretical).
1. Larger worm has smaller surface area to volume ratio (ORA) / ratio is 8 times smaller;
2. Diffusion distance is greater / surface area too small to supply enough oxygen for the volume (of respiring cells) / diffusion too slow to meet demand (AW);
IGNORE 'bigger so needs more oxygen' without reference to ratio or diffusion.

The three domain system comprises Archaea, Bacteria and Eukarya. Bacteria are their own domain, named Bacteria. Eukaryotic cells have larger 80S ribosomes in the cytoplasm, whereas prokaryotes have 70S ribosomes.

Q1 | Gap 1: Bacteria (1 mark, AO1). ALLOW Eubacteria. DO NOT ALLOW Prokaryotae / Prokaryota / Monera.
Gap 2: 80 (1 mark, AO1). IGNORE 70. Both gaps are independent; no ECF is needed.

In a bony fish, water enters through the mouth and leaves under the operculum. Blood in the gill lamellae flows in the opposite direction to the water, which is a counter-current system. This maintains a concentration gradient for oxygen along the whole length of the lamella.

Q2 | Box 1: operculum / gill slit / gill opening (1 mark, AO1). IGNORE gill alone; DO NOT ALLOW spiracle.
Box 2: counter-current / opposite (direction) / flow in opposite directions (1 mark, AO1). DO NOT ALLOW parallel / concurrent / co-current. Boxes are marked independently.

Movement against a concentration gradient using ATP and carriers is active transport. Small non-polar molecules such as oxygen pass freely down a concentration gradient through the bilayer by simple diffusion. Movement of water from a higher (less negative) water potential to a lower (more negative) water potential across a partially permeable membrane is osmosis.

Question: 1 | Marks: 3 | AO1 x2, AO2 x1
1 mark each for:
- active transport (ALLOW primary active transport; DO NOT ALLOW facilitated diffusion)
- (simple) diffusion (ALLOW passive diffusion; DO NOT ALLOW facilitated diffusion)
- osmosis (IGNORE diffusion of water unless osmosis stated)
Guidance: Mark each gap independently. Spelling must be unambiguous.

Insects move air through the tracheal system partly by diffusion and partly by abdominal pumping that changes pressure in the tracheae. Bony fish exchange gases across the thin gill lamellae on gill filaments, using counter-current flow. In humans, contraction of the diaphragm and external intercostal muscles increases thoracic volume, lowering pressure so air is drawn in.

Question: 2 | Marks: 3 | AO1 x3
1 mark each for:
(i) abdominal pumping / contraction and relaxation of abdominal muscles (AW); ALLOW air moves through spiracles by diffusion; DO NOT ALLOW lungs or breathing through the skin
(ii) (gill) lamellae / secondary lamellae / gill filaments; IGNORE gills alone; DO NOT ALLOW alveoli
(iii) diaphragm and (external) intercostal muscles contract/move, changing volume and pressure in the thorax (AW); ALLOW ventilation / breathing in and out by pressure changes; IGNORE just 'breathing' or 'inhaling'
Guidance: Mark each box independently.