2025 OCR A-Level Biology A - H420 模拟试题及答案详解

Carbon dioxide dissolves to form carbonic acid, which dissociates and lowers pH. This is the Bohr effect: haemoglobin's affinity for oxygen falls (curve shifts right), so more oxygen is unloaded at the same partial pressure. A is wrong (pH falls, not rises). C is wrong (CO2 does not act by displacing oxygen at the haem groups as described). D is wrong (a left shift would increase affinity).

B only (1 mark). Do not allow any other letter. Distractors: A reverses the pH change; C is an incorrect mechanism; D gives the wrong direction of shift.

In ventricular systole the ventricles contract and pressure rises above atrial pressure, closing the atrioventricular valves. When ventricular pressure exceeds that in the aorta and pulmonary artery, the semilunar valves open and blood is ejected.

C only (1 mark). A describes diastole; D describes the brief isovolumetric phase; B does not occur in the normal cardiac cycle.

The taxonomic hierarchy is domain, kingdom, phylum, class, order, family, genus, species. Taxa that share a family must also share every higher rank, including order and class. They differ at genus level, so A is wrong, and B and C are wrong because higher ranks are shared.

D only (1 mark). Do not allow other letters.

Auxin (IAA) is redistributed to the shaded side of the shoot. It stimulates cell elongation there, so the shaded side grows faster and the shoot bends towards the light (positive phototropism).

A only (1 mark). B and C give the wrong side or wrong effect; D is incorrect as auxin is redistributed, not destroyed, and the response is by elongation.

Denaturation (about 95 °C) separates the strands by breaking hydrogen bonds. Annealing (about 50–60 °C) lets primers form hydrogen bonds with complementary sequences. Extension (about 72 °C) is when Taq polymerase adds nucleotides. Only B matches correctly.

B only (1 mark). A and C swap the purposes of stages; D gives the wrong temperature and an incorrect enzyme.

N = 20. Values of n/N are 0.5, 0.3, 0.15 and 0.05. Squares: 0.25, 0.09, 0.0225 and 0.0025, which sum to 0.365. Then D = 1 − 0.365 = 0.635.

C only (1 mark). A is the sum before subtracting from 1; B and D come from using only the most abundant species.

Lactose binds to the repressor protein and changes its shape, so it can no longer bind to the operator. RNA polymerase can then transcribe the structural genes.

D only (1 mark). Other options misidentify the molecule that lactose binds to or the site to which the repressor attaches.

Each step dilutes the solution by a factor of 10 (1 cm3 in a total of 10 cm3). After three steps the dilution factor is 10^3, so the concentration is 1.0 × 10^-3 mol dm^-3.

A only (1 mark). B and C miscount the steps; D wrongly adds the dilution factors.

Respiring muscle produces carbon dioxide, which lowers the pH (Bohr effect). This reduces the affinity of haemoglobin for oxygen, so the curve shifts right and more oxygen is unloaded at a given partial pressure of oxygen. Option B is correct; A and D wrongly state that carbon dioxide concentration falls; C wrongly states that pH rises.

B only (1 mark). Do not allow other options. A and D: carbon dioxide concentration falls in exercising muscle, so incorrect. C: carbon dioxide lowers, not raises, pH.

Cardiac output = heart rate \(\times\) stroke volume, so stroke volume = \(4.8 \div 64 = 0.075\) dm\(^3\). Converting: \(0.075 \times 1000 = 75\) cm\(^3\).

C only (1 mark). Distractors: A is a power-of-ten error in unit conversion; B results from dividing heart rate by cardiac output (64 \(\div\) 4.8); D results from multiplying 4.8 by 64.

The mean value of the phenotype shifts in one direction because an extreme phenotype is favoured by a changed environmental factor. This is directional selection.

D only (1 mark). Stabilising selection would maintain the mean and reduce variation; disruptive selection would favour both extremes; genetic drift is random change in allele frequency, not driven by a selection pressure.

In shoots, auxin (IAA) is redistributed to the shaded side. Greater auxin concentration there stimulates more cell elongation, so the shaded side grows faster and the shoot bends towards the light.

A only (1 mark). B, C and D describe the wrong location or the wrong effect of auxin on shoot cell elongation.

The mixture is heated to about 95 \(^\circ\)C to break hydrogen bonds between strands, cooled to about 55 \(^\circ\)C so primers anneal by complementary base pairing, then heated to about 72 \(^\circ\)C for a thermostable DNA polymerase to extend the primers.

C only (1 mark). A: polymerase acts at about 72 \(^\circ\)C, not 95 \(^\circ\)C. B: this describes the 95 \(^\circ\)C denaturation step. D: primers are not separated at 72 \(^\circ\)C; this is the extension step.

\(N = 24\). \(\sum (n/N)^2 = 2\times(8/24)^2 + 2\times(4/24)^2 = 2\times0.1111 + 2\times0.0278 = 0.2778\). \(D = 1 - 0.2778 = 0.72\).

D only (1 mark). A is the value of \(\sum (n/N)^2\) without subtracting from 1. Other options arise from arithmetic errors.

Each stage dilutes the solution tenfold: \(1.0 \times 10^{-1}\), then \(1.0 \times 10^{-2}\), then \(1.0 \times 10^{-3}\) mol dm\(^{-3}\).

B only (1 mark). A is two dilutions only; D is four dilutions; C results from dividing by 3 rather than diluting tenfold three times.

(a) Heart rate = 60 ÷ 0.94 = 63.8 → 64 beats per minute.
(b) Cardiac output = stroke volume × heart rate = 72 × 64 = 4608 cm\(^3\) min\(^{-1}\) (using unrounded 63.83 gives 4596; both accepted).
(c) Delay lets atria finish emptying blood into ventricles before ventricles contract; the AVN delays the wave of excitation, and Purkyne tissue conducts excitation rapidly down the septum to the apex so ventricles contract from the base upward, expelling blood efficiently.
(d) Fetal haemoglobin is more saturated at every partial pressure (e.g. 80% v 60% at 4 kPa), the largest difference being at low pO2; its curve is to the left, i.e. higher affinity. The fetus can therefore load oxygen from maternal blood in the placenta, where pO2 is low, as adult haemoglobin releases oxygen there.
(e) Higher respiration in muscle produces CO2, lowering pH (carbonic acid/H+); Bohr effect reduces haemoglobin affinity so oxygen is unloaded more at the same pO2; higher heart rate and stroke volume increase cardiac output so more blood reaches muscles; vasodilation of arterioles to muscle and vasoconstriction elsewhere redirects blood; steeper concentration gradient between capillary and respiring cells; more rapid ventilation maintains loading in the lungs.

(a) 2 marks: correct answer 64 = 2 marks. If incorrect, 60 ÷ 0.94 = 1 mark. ALLOW 63.8 for 1 mark if not rounded.
(b) 2 marks: 4608 / 4600 / 4596 = 2 marks. Otherwise stroke volume × heart rate = 1 mark. Error carried forward from (a): ALLOW 72 × (their (a)) for both marks. Unit must be correct for the second mark.
(c) 2 marks, max 2: 1. AVN delays the impulse / delay allows atria to empty/ventricles to fill; 2. Purkyne tissue conducts excitation (rapidly) to the apex/base of the heart so ventricles contract from the apex upwards. IGNORE 'SAN controls' alone.
(d) 3 marks: 1. Fetal haemoglobin has a higher saturation at all partial pressures / curve to the left, (data quoted e.g. 80% v 60% at 4 kPa) ; 2. Fetal haemoglobin has a higher affinity for oxygen; 3. Fetal haemoglobin loads/takes oxygen from maternal haemoglobin at the placenta where pO2 is low. ALLOW difference is greatest at low pO2 / 4 kPa.
(e)* 6 marks, levels of response.
Indicative content: Bohr effect: more CO2 from respiration → lower pH / more H+ (carbonic acid; carbonic anhydrase); haemoglobin affinity for oxygen falls, curve shifts right; more oxygen unloaded at a given pO2 / at the muscle; still loads in lungs because pO2 is high. Increased heart rate and stroke volume → higher cardiac output / faster blood flow to muscle; vasodilation of arterioles supplying muscle, vasoconstriction in gut/skin; maintains steep diffusion gradient; more capillaries open; faster ventilation.
Level 3 (5–6 marks): thorough explanation of the Bohr effect with its cause AND at least two other circulatory changes explained; logically organised, no significant errors.
Level 2 (3–4 marks): explanation of the Bohr effect OR several circulatory changes with some explanation; some gaps in logic.
Level 1 (1–2 marks): limited, mainly descriptive points (e.g. 'heart beats faster', 'haemoglobin releases more oxygen') with little or no explanation.
0 marks: no response worthy of credit.

(a) Increase = 8.2 − 3.1 = 5.1; 5.1 ÷ 3.1 × 100 = 164.5 → 165 %.
(b) Same compost volume/type, light intensity, temperature, watering volume (control by using same batch/volume, same location).
(c)(i) Means are very similar (8.2 v 8.0) and the standard deviations (1.9 and 2.1) are large, so the ranges overlap greatly; the difference is probably due to chance, so the conclusion that 2.0 is not better is supported (but significance not proven without a test).
(c)(ii) Excess auxin is inhibitory to root growth / toxic or high concentrations inhibit root development.
(d) Null hypothesis: there is no significant difference between the mean number of roots on cuttings treated with 0 and with 1.0 mg g\(^{-1}\) auxin. Unpaired because the two groups are independent samples of different cuttings, data are continuous/numerical and approximately normally distributed, comparing two means.
(e) Water shortage: roots/leaves make abscisic acid (ABA); ABA binds to receptors on guard cells; calcium ions/ chloride ions enter cytoplasm, ion channels open, K+ leaves, water potential of guard cells rises, water leaves by osmosis, guard cells become flaccid, stomata close, reducing transpiration. Autumn: decrease in auxin from leaf; ethylene produced; ethylene stimulates cells in the abscission zone to produce enzymes that digest cell walls; a layer of cells separates; leaf falls; reduces water loss / conserves resources in winter.

(a) 2 marks: 165 = 2 marks. If incorrect, (8.2 − 3.1) ÷ 3.1 × 100 = 1 mark. ALLOW 164.5 for 1 mark if not rounded.
(b) 1 mark: variable (compost type/volume, volume of water, temperature, light intensity, same species/age of shrub, time in powder) AND how controlled (e.g. same volume daily / same greenhouse). Both needed.
(c)(i) 2 marks: 1. Means very similar / difference only 0.2; 2. standard deviations large/ overlap so difference not significant / likely due to chance. ALLOW a statement that a statistical test is needed to be certain.
(c)(ii) 1 mark: high auxin concentration inhibits root growth / toxic / supra-optimal. IGNORE 'too much auxin' alone.
(d) 3 marks: 1. null hypothesis stating no significant difference between mean numbers of roots at 0 and 1.0 mg g\(^{-1}\) (DO NOT ALLOW 'no relationship/correlation'); 2. two independent/unrelated groups (different cuttings); 3. comparing means of continuous (numerical) data / data normally distributed.
(e)* 6 marks, levels of response.
Indicative content: Stomata – abscisic acid released in water stress; binds to receptors on guard cell surface; calcium ions as second messenger / ion channels open; ions leave guard cells, water potential increases, water leaves by osmosis, guard cells lose turgor, stomata close; less transpiration. Leaf fall – fall in auxin concentration; increased ethylene; abscission zone; cell wall digesting enzymes / separation layer; leaf falls; conserves water in winter / reduces damage.
Level 3 (5–6 marks): both situations covered with clear, linked mechanisms for each, correct named hormones; logically structured.
Level 2 (3–4 marks): both situations covered, one with a full mechanism, OR one fully explained with only brief reference to the other.
Level 1 (1–2 marks): limited, mainly descriptive statements (e.g. 'ABA closes stomata', 'ethylene causes leaf fall') with little explanation.
0 marks: no response worthy of credit.

(a)(i) Cardiac output = 74 × 68 = 5032 cm\(^3\) min\(^{-1}\). Divide by 1000 to convert to dm\(^3\): 5.032, which is 5.0 dm\(^3\) min\(^{-1}\) to 2 s.f.

(a)(ii) Training enlarges and strengthens the cardiac (ventricular) muscle, so each contraction ejects more blood (higher stroke volume, 98 vs 68 cm\(^3\)). The same cardiac output (52 × 98 = 5096 cm\(^3\) min\(^{-1}\)) is then achieved with fewer beats per minute.

(b)(i) At 4 kPa: 85 − 40 = 45 percentage points.

(b)(ii) Myoglobin is almost saturated at low pO\(_2\), so it has a higher affinity for oxygen than haemoglobin. Haemoglobin unloads oxygen at the pO\(_2\) found in respiring muscle, which myoglobin picks up and stores. Myoglobin only releases its oxygen when pO\(_2\) falls very low, giving a reserve for intense exercise.

(c) The SAN (pacemaker) in the right atrial wall initiates a wave of excitation that spreads across both atria, causing atrial systole. A layer of non-conducting fibrous tissue prevents the wave passing directly to the ventricles, so it reaches the AVN, which delays transmission, allowing the atria to empty. The wave then passes down the bundle of His and Purkyne fibres in the septum to the base/apex, then up through the ventricle walls, so ventricles contract from the apex upwards. During exercise, respiring muscle produces more CO\(_2\), which lowers blood pH (forms carbonic acid, releasing H\(^+\)). This reduces haemoglobin's affinity for oxygen (Bohr effect), shifting the dissociation curve to the right, so haemoglobin unloads more oxygen at a given pO\(_2\) in the muscle capillaries.

(a)(i) [2] Correct answer 5.0 (dm\(^3\) min\(^{-1}\)) = 2 marks. If incorrect: 74 × 68 = 5032 (cm\(^3\)) = 1 mark (method). ALLOW 5 for 2 marks only if 2 s.f. shown as 5.0; 5 alone = 1 mark. DO NOT ALLOW 5.03 for the second mark.

(a)(ii) [2] Max 2:
1. Trained group has larger stroke volume / more blood pumped per beat (98 vs 68) ✓
2. (because) cardiac/ventricular muscle is larger/stronger/thicker (so contracts more forcefully) ✓
3. so fewer beats (per minute) are needed to give the same cardiac output / ORA ✓
IGNORE 'heart is fitter' / 'more efficient' alone.

(b)(i) [1] 45 ✓. ALLOW 45% / 45 percentage points. DO NOT ALLOW answers given without a number.

(b)(ii) [2] Max 2:
1. Myoglobin has a higher affinity for oxygen than haemoglobin / ORA (curve to the left) ✓
2. haemoglobin unloads oxygen to muscle / myoglobin takes up oxygen from haemoglobin at low pO\(_2\) and stores it ✓
3. myoglobin releases oxygen only when pO\(_2\) is very low (e.g. in intense exercise), providing a reserve / allowing aerobic respiration to continue ✓

(c)* [6] Level of response. Indicative content:
Heartbeat: SAN initiates wave of excitation / is the pacemaker (myogenic); wave spreads across atrial walls causing atrial systole; non-conducting tissue between atria and ventricles prevents direct spread; AVN delays the impulse (so atria empty before ventricles contract); impulse passes down bundle of His / Purkyne tissue to the apex; ventricles contract from the apex upwards.
Bohr effect: increased respiration produces more CO\(_2\); CO\(_2\) lowers pH (carbonic acid, H\(^+\)); reduces affinity of haemoglobin for oxygen; curve shifts right; haemoglobin unloads more oxygen at a given pO\(_2\) to muscle / supports aerobic respiration.

Level 3 (5–6 marks): Detailed, correct description of the coordination of the heartbeat covering the roles of SAN, AVN and Purkyne tissue in the right sequence AND a clear explanation of the Bohr effect linking CO\(_2\)/pH, haemoglobin affinity and oxygen unloading. Both parts are covered in a logical, well-organised answer with few or no errors.
Level 2 (3–4 marks): Describes the main stages of heartbeat coordination and gives some explanation of the Bohr effect, OR gives a good account of one part and a limited account of the other. Some gaps in sequence or linkage.
Level 1 (1–2 marks): Limited statements about either the heartbeat (e.g. SAN starts the beat) or the Bohr effect (e.g. more CO\(_2\) makes haemoglobin release oxygen). Mostly unlinked or lacking detail.
Level 0 (0 marks): No response or no response worthy of credit.
IGNORE nervous control of the heart (medulla) unless relevant. DO NOT ALLOW 'AVN starts the heartbeat'.

(a)(i) Cardiac output = stroke volume x heart rate = 110 x 48 = 5280 cm3 min-1 = 5.28 dm3 min-1.
(a)(ii) Cardiac output = stroke volume x heart rate, so a large stroke volume (stronger or larger-volume ventricular contraction, more complete filling) means fewer beats per minute are needed to pump the same volume of blood to the tissues.
(b)(i) At every pO2 species X haemoglobin has a higher % saturation, so its curve is to the left. The difference is greatest at low pO2: at 4 kPa X is 72% and human is 50% (22 percentage points), whereas at 12 kPa both are almost fully saturated (99% v 97%).
(b)(ii) Species X haemoglobin has a higher affinity for oxygen. At altitude the partial pressure of oxygen in the alveoli is low, so its haemoglobin still loads/associates with more oxygen (e.g. 72% at 4 kPa) than human haemoglobin would, increasing oxygen supply to tissues.
(c) Respiring muscle produces CO2, raising pCO2. Table 1.2 shows p50 increases as pCO2 increases (3.4 kPa at 0.5 kPa pCO2 to 4.9 kPa at 5.0 kPa pCO2): the oxygen dissociation curve shifts to the right (Bohr effect). Haemoglobin therefore has a lower affinity for oxygen and releases oxygen more readily, at a higher pO2, in the muscle where it is needed. Mechanism: CO2 diffuses into red blood cells and is converted by carbonic anhydrase to carbonic acid, which dissociates into H+ and hydrogencarbonate ions; H+ binds to haemoglobin (forming haemoglobinic acid), changing its shape/affinity so oxygen is released. Hydrogencarbonate ions diffuse out of the cell into plasma in exchange for chloride ions (chloride shift) and are the main way (about 85%) CO2 is carried to the lungs; a small amount dissolves in plasma or binds to haemoglobin as carbaminohaemoglobin.

(a)(i) Correct answer 5.28 (dm3 min-1) = 2 marks. ALLOW 5.3. If answer wrong, 5280 (cm3 min-1) = 1 mark (method: 110 x 48; or conversion of a wrong product to dm3). ALLOW ecf for conversion. DO NOT ALLOW answer in wrong units for 2 marks.

(a)(ii) 1. Larger/greater stroke volume (in athlete) / more blood per beat; 2. so fewer beats/lower heart rate needed for same cardiac output / CO = SV x HR (AW). ALLOW heart muscle stronger/larger ventricle for reason for larger SV. IGNORE 'heart is more efficient' alone.

(b)(i) 1. Species X has higher saturation at all (pO2) values / curve to the left (AW); 2. Quoted comparative data from Table 1.1 e.g. at 4 kPa 72% v 50% / difference greatest at low pO2 / difference smaller at high pO2 (both near 100%). Data must include units or % to credit. ALLOW p50 of X is lower.

(b)(ii) 1. (Species X haemoglobin has) higher affinity for oxygen; 2. loads/associates more oxygen at low pO2 in lungs/alveoli at altitude / becomes saturated at low pO2 (AW). DO NOT ALLOW 'more oxygen in air'. ALLOW ORA for human haemoglobin.

(c) LEVEL 3 (5-6 marks): Clear, well-structured explanation linking CO2 to the Bohr effect, using data from Table 1.2 correctly (with values and a trend), AND a correct description of CO2 transport as hydrogencarbonate including the role of carbonic anhydrase/H+ and haemoglobin. Few or no errors; logical sequence.
LEVEL 2 (3-4 marks): Explains the Bohr effect with some use of data, and gives some correct detail of CO2 transport; may lack linking or contain minor errors/omissions, OR one aspect covered well and the other only partially.
LEVEL 1 (1-2 marks): Some relevant points, e.g. states CO2 increases p50 / haemoglobin releases more oxygen, or states CO2 carried as hydrogencarbonate, with limited use of data or explanation; limited structure.
LEVEL 0 (0 marks): No response or no response worthy of credit.

Indicative scientific points (not a checklist): muscle respiration increases pCO2; p50 rises from 3.4 to 4.9 kPa (as pCO2 rises 0.5 to 5.0 kPa) / curve shifts right; lower affinity so more oxygen unloaded at a given pO2 / Bohr effect; CO2 diffuses into red blood cells; carbonic anhydrase catalyses CO2 + H2O to H2CO3; dissociates to H+ and HCO3-; H+ binds to haemoglobin / haemoglobinic acid, causing oxygen release; HCO3- diffuses out into plasma, chloride shift maintains electrical balance; most (~85%) CO2 transported as hydrogencarbonate; some as carbaminohaemoglobin or dissolved in plasma. ALLOW 'oxyhaemoglobin dissociation curve shifts to right' for shift. IGNORE named lung/alveolar CO2 removal detail unless linked to transport.

(a)(i) A higher partial pressure of carbon dioxide (lower pH) reduces the affinity of haemoglobin for oxygen, so the dissociation curve shifts to the right and oxygen is released more readily at respiring tissues.

(a)(ii) At every pO\(_2\) the sand rat haemoglobin is more saturated, e.g. 62% compared with 20% at 2.0 kPa. The sand rat curve lies to the left, showing a higher affinity for oxygen. This lets it become well saturated in the lungs even though the oxygen in the burrow air is low.

(b)(i) Cardiac output = heart rate \(\times\) stroke volume = \(310 \times 0.021 = 6.51\) cm\(^3\) min\(^{-1}\).

(b)(ii) Exercise: \(520 \times 0.018 = 9.36\) cm\(^3\) min\(^{-1}\). Percentage increase = \(\frac{9.36-6.51}{6.51}\times 100 = 43.78\) so 43.8%.

(c) The delay allows the atria to finish contracting and empty blood into the ventricles before the ventricles contract. This ensures the ventricles are full and prevents backflow. The atrioventricular valves then close under ventricular pressure.

(d) Hydrostatic pressure at the arteriolar end forces water and small solutes out through the capillary wall, and plasma proteins stay in. This gives tissue fluid. Plasma proteins lower the water potential of the blood. At the venule end the hydrostatic pressure is lower, and water returns by osmosis down the water potential gradient. Some fluid returns through lymph vessels. With low plasma protein the blood water potential is not as low, so less water returns by osmosis and excess tissue fluid accumulates as oedema.

(a)(i) [2 marks]
1. Increase in (partial pressure of) carbon dioxide / decrease in pH reduces haemoglobin's affinity for oxygen / curve shifts to the right (1)
2. so oxygen is released / dissociates more readily at respiring tissues / at lower oxygen saturation for a given pO2 (1)
IGNORE 'carbon dioxide competes with oxygen'.

(a)(ii) [3 marks]
1. Sand rat haemoglobin has a higher saturation at all pO2 values / curve is to the left / higher affinity (1)
2. Quotes comparative data with units, e.g. 62% v 20% at 2.0 kPa / 30% v 8% at 1.0 kPa (1)
3. Haemoglobin is able to load (become saturated with) oxygen in the lungs when the air in the burrow has low pO2 / oxygen is still obtained at low pO2 (1)
ALLOW AW. DO NOT ALLOW 'releases more oxygen'.

(b)(i) [1 mark]
6.51 (cm3 min-1) (1)

(b)(ii) [2 marks]
Correct answer 43.8 (%) = 2 marks.
If the answer is incorrect, award 1 mark for 9.36 (cm3 min-1) calculated for exercise, or for correct use of (new - old)/old x 100 using ecf from (b)(i).
ALLOW 1 mark for 43.78 / 44 (not to three significant figures). Answer of 143.8 (%) = 1 mark.

(c) [2 marks]
1. Allows atria to finish contracting / empty (blood into the ventricles) before ventricles contract (1)
2. So ventricles are full / maximum blood is pumped out / prevents backflow of blood to atria (1)
IGNORE 'so heart beats in the correct order' without detail.

(d)* Level of response [6 marks]
Level 3 (5-6 marks): Gives a detailed, accurate account of both the formation and return of tissue fluid and a clear explanation of how low plasma protein causes oedema, with a logical, well organised line of reasoning that links hydrostatic pressure and water potential.
Level 2 (3-4 marks): Describes both formation and return of tissue fluid and attempts an explanation of oedema, with some detail missing or one section less developed; reasoning mostly logical.
Level 1 (1-2 marks): Makes some correct statements about tissue fluid formation or return, or about oedema, but with limited detail or no links between them.
Level 0 (0 marks): No response or no response worthy of credit.

Indicative scientific points:
- high hydrostatic pressure at the arteriolar end (due to ventricular contraction / narrow lumen) forces fluid out (ultrafiltration)
- water and small molecules (glucose, oxygen, ions) leave through gaps in the capillary wall / endothelium
- plasma proteins / large molecules / cells remain in the capillary
- plasma proteins lower the water potential of the blood
- hydrostatic pressure falls along the capillary / is lower at the venule end
- water returns into the capillary by osmosis down a water potential gradient
- excess tissue fluid drains into lymph capillaries and is returned to the blood via the veins
- with low plasma protein the blood water potential is higher / less negative, so water potential gradient is smaller
- less water is reabsorbed by osmosis, so more fluid remains in tissues (oedema) / lymph drainage is overwhelmed
ALLOW AW. Indicative content supports the level judgement and is not a simple checklist.

(a)(i) Cardiac output = heart rate x stroke volume = 64 x 78 = 4992 cm3 min-1. Divide by 1000 = 4.992 dm3 min-1 = 4.99 dm3 min-1 (3 s.f.).
(a)(ii) Percentage increase = (18.5 - 4.99) / 4.99 x 100 = 270.7... = 271 % (using 4.992 gives 270.6, also 271 %).
(b)(i) 50% lies between 45% at 2 kPa and 68% at 3 kPa. Fraction = (50 - 45) / (68 - 45) = 0.217, so pO2 = 2.2 kPa.
(b)(ii) The llama curve is to the left of the human curve, so its haemoglobin has a higher affinity for oxygen. At the low pO2 in the lungs at altitude it still becomes highly saturated (e.g. 82% at 4 kPa compared with 62% in humans), so more oxygen is loaded.
(c) Respiring muscle produces CO2, which diffuses into red blood cells. Carbonic anhydrase converts CO2 and water to carbonic acid, which dissociates into H+ and hydrogencarbonate ions. Hydrogencarbonate ions diffuse out in exchange for chloride ions (chloride shift). H+ ions bind to haemoglobin forming haemoglobinic acid, which lowers its affinity for oxygen; the dissociation curve shifts right (Bohr effect). Oxygen is released at a higher rate / more oxygen is unloaded at a given pO2 in the muscle, supporting aerobic respiration at a higher rate during exercise. Haemoglobin also acts as a buffer by taking up H+.

(a)(i) Correct final answer 4.99 = 2 marks. If incorrect, 64 x 78 = 4992 (cm3) OR 4.992 (dm3) seen = 1 mark. ALLOW 5.0 for 1 mark only if 4992 not seen. DO NOT ALLOW 4.9.
(a)(ii) 271 (%) = 1 mark. ALLOW 270 / 270.6 / 270.7. ALLOW ecf from an incorrect (a)(i) answer.
(b)(i) 2.2 (kPa) = 1 mark. ALLOW 2.1 to 2.3.
(b)(ii) Max 2: (higher) affinity for oxygen / curve to the left of human (1); (so) more saturated / loads more oxygen at low pO2 (in lungs/at altitude) (1). ALLOW AW, ORA for human. IGNORE 'carries more oxygen' without reference to low pO2.
(c) Level of response, 6 marks.
Indicative content: CO2 produced by respiring tissue diffuses into red blood cells; carbonic anhydrase catalyses CO2 + H2O to H2CO3; H2CO3 dissociates to H+ and HCO3-; HCO3- moves out of the cell, chloride ions move in (chloride shift); H+ binds to haemoglobin (forming haemoglobinic acid) which acts as a buffer; haemoglobin changes shape / lower affinity for oxygen; oxygen dissociation curve shifts to the right (Bohr effect); oxygen unloaded more readily / at higher pO2 in muscle; more oxygen available for aerobic respiration during exercise; greater ATP production.
Level 3 (5-6 marks): Gives a detailed, accurate account of CO2 transport in red blood cells leading to H+ production AND links this to the effect on haemoglobin affinity and the benefit to muscle. The answer is logically organised, with correct use of terminology and no significant errors.
Level 2 (3-4 marks): Describes some stages of CO2 transport and/or explains the Bohr effect, with some link between the two parts; or covers both parts with limited detail. Mostly logical, minor omissions or errors.
Level 1 (1-2 marks): Limited statements, for example CO2 lowers pH and haemoglobin releases more oxygen, with little or no explanation of mechanism or benefit. Organisation is weak.
Level 0 (0 marks): No response or no creditworthy content.
DO NOT ALLOW CO2 binding to haemoglobin as the cause of the Bohr effect without reference to H+ / pH. DO NOT ALLOW haemoglobin 'releasing' CO2 in the muscle.

Raised carbon dioxide lowers the pH in red blood cells (the Bohr effect), which reduces the affinity of haemoglobin for oxygen. The curve shifts to the right and more oxygen is unloaded at a given partial pressure of oxygen. The answer is D.

D only (1 mark). A and B incorrect: a left shift means increased affinity. C incorrect: a right shift means more, not less, oxygen is released.

The absence of a nucleus shows the organism is prokaryotic, so it is not in Eukarya. Bacterial cell walls contain peptidoglycan, but this organism's does not, which places it in Archaea. The answer is A.

A only (1 mark). B incorrect: bacterial walls contain peptidoglycan. C and D incorrect: no nucleus means the organism is not eukaryotic.

In shoots, auxin promotes cell elongation. Extra auxin on one side makes cells on that side elongate more, so the shoot bends towards the opposite side, away from the gel. The answer is B.

B only (1 mark). A and C give the wrong direction of bending. D is wrong because auxin does stimulate elongation in shoots.

At about 95 °C the hydrogen bonds between base pairs break, producing single template strands. Annealing of primers occurs at a lower temperature and Taq polymerase extends at about 72 °C. Phosphodiester bonds remain intact. The answer is B.

B only (1 mark). A occurs at a lower temperature (about 55 °C). C occurs at about 72 °C. D is incorrect as the backbone is not broken.

\( \sum (n/N)^2 = 0.6^2 + 0.3^2 + 0.1^2 = 0.36 + 0.09 + 0.01 = 0.46 \). So \( D = 1 - 0.46 = 0.54 \). The answer is B.

B only (1 mark). A is the value of the sum without subtracting from 1. C and D are incorrect calculations.

Normally the repressor bound to the operator blocks RNA polymerase when lactose is absent. If the repressor cannot bind, RNA polymerase can bind to the promoter and the structural genes are transcribed constitutively. The answer is A.

A only (1 mark). B and C describe normal repression, which cannot occur in this mutant. D adds an irrelevant condition.

Cardiac output = heart rate × stroke volume. 5.4 dm\(^3\) = 5400 cm\(^3\). Stroke volume = 5400 ÷ 75 = 72 cm\(^3\). The answer is A.

A only (1 mark). B and D are unit or power-of-ten errors. C results from multiplying 5400 by 75 incorrectly or multiplying 5.4 by 75.

\( q^2 = 0.04 \), so \( q = 0.2 \) and \( p = 1 - 0.2 = 0.8 \). Heterozygote frequency \( 2pq = 2 \times 0.8 \times 0.2 = 0.32 \). The answer is C.

C only (1 mark). A is \(q^2\) of a wrong value (0.4²). B is q. D is \(p^2\).

This is the Bohr effect. Respiring muscle produces more CO2, which forms carbonic acid and lowers pH. The lower pH reduces the affinity of haemoglobin for oxygen, shifting the curve to the right so more oxygen is unloaded at a given partial pressure. B, C and D each contain an incorrect pH change, affinity change or loading/unloading statement.

A (1 mark). Do not award for B, C or D: each contains at least one incorrect link between CO2, pH, affinity and unloading.

Archaea and Bacteria are both prokaryotes, so 70S ribosomes, no nuclear envelope and a circular chromosome are shared. The domains are distinguished by molecular evidence such as rRNA sequences and membrane lipid composition.

C (1 mark). A, B and D are features common to both prokaryotic domains.

Unilateral light causes lateral movement of auxin to the shaded side. There it stimulates proton pumping into cell walls, lowering pH and loosening the walls so cells elongate faster than on the lit side, so the shoot bends towards the light.

A (1 mark). B places auxin on the wrong side; C gives the wrong effect; D contradicts the lateral redistribution of auxin.

Heating to about 95 °C breaks hydrogen bonds to separate the strands. Cooling to about 55 °C lets primers anneal to the single strands. At about 72 °C the thermostable polymerase adds nucleotides to extend the primers.

A (1 mark). B has the wrong order; C and D have steps matched to the wrong temperatures.

\( N = 20 \). \( \sum (n/N)^2 = 0.50^2 + 0.30^2 + 0.15^2 + 0.05^2 = 0.25 + 0.09 + 0.0225 + 0.0025 = 0.365 \). \( D = 1 - 0.365 = 0.635 \), which is 0.64 to two decimal places.

B (1 mark). A results from omitting the subtraction from 1; C results from using \( n(n-1)/N(N-1) \).

Reliability is improved by repeats, because individual Daphnia vary biologically, and a mean reduces the effect of anomalies. A finer stopwatch resolution adds little to a 10 s count. A shorter count increases the relative error. A higher-power lens does not address variation between organisms.

A (1 mark). B, C and D do not address variation between individuals.

Lactose acts as an inducer. It binds to the repressor, altering its tertiary structure so it cannot bind to the operator. RNA polymerase is then no longer blocked and transcribes the structural genes, including that for beta-galactosidase.

A (1 mark). B, C and D give incorrect binding sites for lactose.

(a)(i) Before: 62 x 88 = 5456 cm3 min-1. After: 54 x 104 = 5616 cm3 min-1. Change = (5616 - 5456) / 5456 x 100 = 2.93%, so 2.9% (2 s.f.).
(a)(ii) Each contraction ejects more blood, so the same or greater volume per minute is delivered with fewer beats.
(b)(i) Human: 50% lies between 3 kPa (40%) and 4 kPa (60%), so about 3.5 kPa. Llama: 50% occurs just below 2 kPa (52%), about 1.9 kPa.
(b)(ii) At low alveolar pO2 the llama haemoglobin is much more saturated (higher affinity, curve shifted left), so more oxygen is loaded per breath, supplying tissues despite the thin air.
(c) Haemoglobin has four polypeptide subunits each with a haem group. At the high pO2 of the lungs oxygen binds; the first oxygen binding alters the quaternary structure, making further binding easier (cooperative binding), giving the S-shaped curve with steep middle section and near-saturation in the lungs. In respiring tissues pO2 is low and CO2 is high; CO2 lowers pH (carbonic acid / H+ ions), reducing the affinity of haemoglobin for oxygen (Bohr effect), shifting the curve to the right so oxygen is unloaded more readily where it is needed, especially by the most active tissues.
(d) Add a control group of volunteers who train at sea level for the same time, so that changes due to training alone can be separated from the effect of altitude.

(a)(i) [3] Cardiac output before = 5456 and after = 5616 (cm3 min-1) (1 mark each, both required for the method marks); correct answer 2.9 (%) (1). Correct answer with no working = 3 marks. ALLOW 2.93 / 2.9% for final mark; DO NOT ALLOW 3 (not 2 s.f.) for the final mark. If one cardiac output is incorrect, ALLOW error carried forward for the final mark if the percentage change is calculated correctly from the candidate's values.

(a)(ii) [2] 1. Greater volume of blood pumped / ejected per beat / each contraction (1); 2. so fewer beats / lower heart rate needed to deliver the same volume per minute (cardiac output = HR x SV) (1). IGNORE 'heart is stronger / fitter' without link to volume.

(b)(i) [2] Human 3.4-3.6 (kPa) (1); llama 1.8-2.0 (kPa) (1). ALLOW graph-based answers within these ranges.

(b)(ii) [2] 1. Llama haemoglobin has a higher affinity for oxygen / curve shifted to the left / more saturated at low pO2 (1); 2. loads more oxygen in the alveoli (where pO2 is low at altitude) so more oxygen is transported to respiring tissues (AW) (1). DO NOT ALLOW 'absorbs more oxygen' without reference to haemoglobin saturation.

(c)* [6] Level 3 (5-6 marks): Detailed description and explanation of both loading and unloading, including cooperative binding explaining the S-shaped curve and the Bohr effect with the role of CO2; logical, well-organised answer linking each statement to the correct location (lungs / tissues). Level 2 (3-4 marks): Describes loading and unloading with some correct explanation, e.g. cooperative binding OR the Bohr effect explained; some links between pO2, CO2 and affinity, may lack detail or organisation. Level 1 (1-2 marks): Limited description, e.g. haemoglobin binds oxygen in lungs and releases it in tissues, with little or no explanation. 0 marks: no response worthy of credit.
Indicative content: haemoglobin has four polypeptide chains each with a haem / Fe2+ group; each can bind one O2 (four per molecule); high pO2 in lungs so oxyhaemoglobin forms / high affinity; first O2 binding changes quaternary structure / shape, making subsequent binding easier (cooperative); gives S-shaped / sigmoidal curve, steep section over a small pO2 change; at low pO2 in tissues O2 released; respiring tissues produce CO2, which forms carbonic acid / H+ ions, lowering pH; lower pH / high pCO2 reduces affinity of haemoglobin for O2 (Bohr effect), curve shifts right; more unloading at a given pO2 in active tissues; in lungs CO2 removed, pH rises, affinity increases so loading is favoured. Marks are for the quality of the reasoning, not a simple tally of the points.

(d) [1] Include a control group / group living and training at sea level (for four weeks) to compare with the altitude group (1) - to show that the change is due to altitude and not training / time / other factors (AW). ALLOW larger sample size with reason of reducing the effect of individual variation. Linked requirement: the improvement and the reason must both be given; ALLOW a single linked statement for the mark.

(a) Cardiac output = heart rate x stroke volume = 64 x 85 = 5440 cm3 min-1 = 5.44 dm3 min-1, which is 5.4 dm3 min-1 to 2 s.f.

(b) Saturation at 6 kPa = 85%; at 4 kPa = 62%. Oxygen unloaded = 85 - 62 = 23 percentage points. As a percentage of the oxygen carried: 23 / 85 x 100 = 27.06%, so 27%.

(c) The Bohr effect is the decrease in affinity of haemoglobin for oxygen as the partial pressure of carbon dioxide increases (the curve shifts to the right). In exercise the muscles respire faster and produce more CO2, so haemoglobin releases more oxygen to the muscles, supporting more aerobic respiration.

(d) Llama haemoglobin has a higher affinity for oxygen (curve to the left of the human curve). At the low pO2 of 6 kPa it is 85% saturated compared with 71% for human haemoglobin, and at 2 kPa 15% compared with 5%, so it loads more oxygen when little is available in the alveoli at altitude. Possible explanation: different amino acid sequence/tertiary-quaternary structure of the globin chains, hence a different shape of the oxygen-binding site. The disadvantage is that oxygen is held more tightly, so less is unloaded in the tissues: at 4 kPa the llama haemoglobin is still 62% saturated against 30% for human. Between 6 and 4 kPa the llama unloads 23 percentage points but the human 41 percentage points, so less oxygen is delivered per unit of haemoglobin to respiring tissues, which would limit aerobic respiration unless compensated, e.g. by more haemoglobin/red cells or a higher cardiac output.

(a) 2 marks. Correct answer 5.4 = 2 marks. If incorrect, 64 x 85 (= 5440 cm3) OR 5.44 seen = 1 mark. ALLOW 5.44 for 1 mark if not rounded to 2 s.f. (1 mark for method/value, 1 mark for correct rounding and unit conversion). DO NOT ALLOW 5400 or 5.5.

(b) 2 marks. Correct answer 27 = 2 marks. If incorrect: reads 85 and 62 and subtracts to give 23 (1 mark). ALLOW ECF for 1 mark for 23/85 x 100 evaluated correctly from a wrong subtraction. 23 alone = 1 mark. ALLOW 27.1.

(c) 2 marks. 1. Bohr effect: increased pCO2 / (more) carbon dioxide reduces the affinity of haemoglobin for oxygen / causes haemoglobin to release (more) oxygen / curve shifts right (1). 2. During running, muscles respire faster/more CO2 produced so more oxygen unloaded for (aerobic) respiration / ATP production (1). IGNORE 'lactic acid' alone.

(d)* 6 marks, levels of response.
Indicative content:
- Llama curve is to the left / higher affinity for oxygen than human.
- Quote data: at 6 kPa 85% v 71%; at 2 kPa 15% v 5% (comparative figures from table needed).
- Loads more oxygen in the lungs where pO2 is low at altitude / maintains saturation of blood.
- Due to different amino acid sequence / tertiary or quaternary structure of globin, giving a different shape of the oxygen binding sites.
- Disadvantage: holds oxygen more tightly so unloads less at the tissues.
- Data: at 4 kPa 62% v 30% remains saturated; unloading 6 to 4 kPa is 23 v 41 percentage points (ORA).
- Less oxygen available per unit of haemoglobin for respiring tissues / compensation needed, e.g. more red blood cells, higher cardiac output.

Level 3 (5-6 marks): Explains the adaptation with comparative data AND discusses the disadvantage using data for tissues; reasoning is logical, linked and relevant throughout.
Level 2 (3-4 marks): Explains the adaptation with some use of data, or describes both adaptation and disadvantage but with limited use of data or limited linkage.
Level 1 (1-2 marks): Some correct points, e.g. higher affinity / curve to the left, or less unloading, but with little explanation or no data; may be poorly organised.
Level 0 (0 marks): No response or no creditworthy content.

Total 10 marks.

(a)(i) Heart rate = 60 / 0.125 = 480 beats per minute.
(ii) Cardiac output = heart rate x stroke volume = 480 x 0.42 = 201.6, which is 202 cm3 min-1 (3 s.f.).
(b)(i) The high-altitude vole has the higher affinity: at any given pO2 its saturation is higher, e.g. 66% against 40% at 3 kPa (curve shifted to the left).
(ii) At low pO2 in the lungs its haemoglobin still becomes much more saturated, so more oxygen is loaded/transported to tissues despite the thin air.
(iii) High CO2 lowers blood pH (carbonic acid / hydrogen ions), so haemoglobin changes shape and releases oxygen more readily (Bohr effect); percentage saturation falls at a given pO2, so more oxygen is unloaded to the muscle.
(c) Indicative content: SAN initiates a wave of excitation (myogenic); it spreads across the atria causing atrial systole; a layer of non-conducting fibrous tissue prevents it passing directly to the ventricles; the AVN delays the impulse so atria empty before ventricles contract; impulse passes down the bundle of His/Purkyne tissue to the apex and then up the ventricle walls, so ventricles contract from the base upwards. CO2 transport: about 85% is converted in red blood cells by carbonic anhydrase to H2CO3, which dissociates to H+ and hydrogencarbonate; hydrogencarbonate leaves the cell by diffusion in exchange for chloride ions (chloride shift); H+ is buffered by haemoglobin (haemoglobinic acid); some CO2 is carried dissolved in plasma and some as carbaminohaemoglobin; in the lungs the reactions reverse and CO2 diffuses out.
(d)(i) (66 - 40) / 40 x 100 = 65%.
(ii) The data support the conclusion, because the high-altitude values are higher at every pO2 listed (by 2 to 26 percentage points); however the difference is only 2 percentage points at 10 kPa, so it is of little advantage at higher pO2; also only one temperature/pH was used and no repeats or sample size are given, so reliability is unknown, and the conclusion cannot be applied to tissue conditions where pH is lower. A good answer judges the conclusion as supported only for the pO2 range tested.

(a)(i) 480 [1]
(a)(ii) 201.6 / 202 seen [1]; correct to 3 s.f. = 202 [1]. ALLOW ecf from (a)(i) (heart rate x 0.42) with correct rounding. Final answer of 202 gains 2 marks. DO NOT ALLOW 200.
(b)(i) high-altitude vole [1]; correct comparative data quoted from the table, e.g. 66% against 40% at 3 kPa, with both values [1]. Evidence mark requires both values at the same pO2.
(b)(ii) at low pO2 / in the lungs/alveoli, haemoglobin is more saturated / loads more oxygen [1]; more oxygen delivered to tissues / respiring cells / aerobic respiration maintained [1]. IGNORE 'more oxygen is absorbed' without reference to haemoglobin.
(b)(iii) high CO2 concentration lowers pH / more H+ [1]; haemoglobin releases more oxygen / lower saturation / curve shifts right (Bohr effect) [1]. ALLOW haemoglobin shape change / lower affinity.
(c) Level 3 (5-6 marks): Clear and balanced account covering BOTH heart excitation (SAN, AVN delay and Purkyne tissue with a correct sequence) AND CO2 transport (at least two forms, with carbonic anhydrase/hydrogencarbonate/chloride shift detail). Well organised and logically sequenced, with accurate terminology. Level 2 (3-4 marks): Some coverage of both parts, or detailed coverage of one with limited reference to the other. Mostly accurate; some gaps in sequence or explanation. Level 1 (1-2 marks): Limited statements, e.g. SAN starts the heartbeat; CO2 transported in the blood; lacks detail or logical order. 0 marks: no response worthy of credit. Indicative content as in the solution: SAN initiates excitation (myogenic); atrial systole; fibrous tissue stops conduction to ventricles; AVN delay; Purkyne tissue/bundle of His carries impulse to apex; ventricular systole from the apex; CO2 diffuses into red blood cells; carbonic anhydrase; H2CO3 forms H+ and HCO3-; chloride shift; haemoglobin buffers H+; carbaminohaemoglobin; dissolved in plasma; reversed in lungs. DO NOT ALLOW 'nerves' stimulating the heart wall.
(d)(i) (66 - 40) / 40 [1]; 65 (%) [1]. Answer 65 gains 2 marks. DO NOT ALLOW 39 (wrong baseline) but ALLOW 39.4 / 39 for 1 mark for method with the high-altitude vole as baseline.
(d)(ii) any two from: data support the conclusion for all pressures tested / high-altitude saturation is higher at each value [1]; difference small (2 percentage points) at 10 kPa so not much better at high pO2 [1]; one temperature/pH / no repeats or SD / sample size unknown so reliability limited [1]; cannot extrapolate beyond the range 1-10 kPa [1]. Max 2.

(a) Light causes IAA to move/be redistributed to the shaded side of the shoot tip (by lateral transport); IAA is therefore more concentrated on the shaded side; it stimulates cell elongation (acid growth: H+ pumped into the cell wall, loosening it) so cells on the shaded side elongate more and the shoot bends towards the light.
(b) Indicative content: serial dilution of the stock to make a range of concentrations (e.g. 10, 1, 0.1 mg dm-3) with distilled water as the 0 control; cut equal-length stem sections from the same region of seedlings of the same age/variety using a sharp blade; measure the initial length; place a set number (e.g. 10) in each Petri dish with the same volume of solution; same temperature, light and time (24 h); remeasure and calculate mean increase in length per concentration; validity: only IAA concentration varies, other variables controlled, distilled water control; reliability: repeats/10 sections per concentration, mean and standard deviation, identify anomalies, same ruler/measure to same precision.
(c)(i) (6.2 - 2.1) / 2.1 x 100 = 195.238, which is 195%.
(ii) Elongation rises from 2.1 mm at 0 to a maximum of 6.2 mm at 1 mg dm-3, then falls at 10 and 100 mg dm-3, where it is below the control (1.5 mm). At very high concentration IAA is inhibitory / supraoptimal, e.g. it stimulates ethylene production or is toxic, so elongation is reduced.
(iii) Unpaired t-test: two independent samples of continuous (normally distributed) data with different sections in each group. H0: there is no significant difference between the mean increase in length of sections at 1 mg dm-3 and at 0 mg dm-3 IAA (any difference is due to chance).
(iv) The means differ by 2.4 mm and the standard deviations are 0.5 and 0.6, so the mean values are more than 2 standard deviations apart and the ranges are unlikely to overlap; the statement is supported (but a statistical test would be needed to confirm).
(v) Rooting powder/hormone to promote root growth in cuttings; or seedless fruit production; or selective weedkillers; or preventing premature fruit drop.

(a) IAA moves/is transported to the shaded side [1]; higher IAA concentration on the shaded side stimulates more cell elongation (cell wall loosening) [1]; unequal growth / shaded side grows faster so shoot bends towards light [1]. IGNORE 'auxin is made on the shaded side'. DO NOT ALLOW 'cell division' for elongation.
(b) Level 3 (5-6 marks): Clear, logical method that includes a range of IAA concentrations produced by dilution with a water control, standardised sections, measurement before and after, AND explains both validity (control variables, control) and reliability (repeats, mean/SD). Level 2 (3-4 marks): Method with some correct detail on dilutions, measurement and controls; explanation of validity or reliability is limited or only one of them explained. Level 1 (1-2 marks): Basic outline, e.g. put stem sections in IAA solutions and measure length; little or no explanation. 0 marks: no response worthy of credit. Indicative content as in the solution: serial dilution; 0 mg dm-3 control with distilled water; same region/age/length of stem; initial and final length; same volume, temperature, light, 24 h; repeats (10) and mean; SD; anomalies; same ruler. DO NOT ALLOW use of a stopwatch for 'time to grow'.
(c)(i) (6.2 - 2.1) / 2.1 x 100 or 195.2... seen [1]; 195 [1]. Answer 195 gains 2 marks. DO NOT ALLOW 195.2 or 200.
(c)(ii) increases to a maximum at 1 mg dm-3 then decreases [1]; quoted data, e.g. 6.2 mm maximum / 1.5 mm at 100 mg dm-3 below the 2.1 mm control [1]; very high IAA concentration inhibits elongation / stimulates ethylene production / is toxic [1]. ALLOW 'supra-optimal concentration'. IGNORE 'denatures'.
(c)(iii) unpaired t-test [1]; reason: two independent/different sets of sections / comparing means of two samples of continuous data [1]; null hypothesis: no significant difference between the mean increase in length at 1 mg dm-3 and at 0 mg dm-3 [1]. DO NOT ALLOW a paired t-test, chi-squared or Spearman's rank. DO NOT ALLOW null hypothesis worded as 'there is a correlation'.
(c)(iv) means are 2.4 mm apart / difference is greater than the sum of the standard deviations (0.5 + 0.6 = 1.1) so a clear difference / little overlap, statement supported [1]. ALLOW 'a statistical test is needed to confirm'.
(c)(v) any one from: rooting powder / promoting root growth in cuttings; producing seedless fruit; herbicide / selective weedkiller; delaying fruit drop; tissue culture [1]. IGNORE 'to make plants grow faster'.

(a)(i) Cardiac output = heart rate × stroke volume = 54 × 95 = 5130 cm^3 min^-1 = \(5.13 \times 10^{3}\) cm^3 min^-1.

(a)(ii) Before training: 72 × 70 = 5040 cm^3 min^-1. Change = 5130 − 5040 = 90. Percentage change = \(\frac{90}{5040} \times 100 = 1.79\%\), which is 1.8% (2 s.f.) increase.

(b) Training causes hypertrophy of the cardiac muscle of the ventricle wall, so the ventricles contract with greater force and eject more blood per beat, emptying more completely. The heart may also fill with more blood during diastole.

(c) Carbon dioxide dissolves and forms carbonic acid, which dissociates to release hydrogen ions and lowers blood pH. This changes the tertiary/quaternary structure of haemoglobin, reducing its affinity for oxygen (the Bohr effect). The oxygen dissociation curve shifts to the right, so at a given partial pressure of oxygen more oxygen is released to the respiring muscle.

(a)(i) [2]
Correct answer 5.13 × 10^3 = 2 marks (ALLOW 5.1 × 10^3 for 2 marks).
If answer is incorrect or not in standard form, 54 × 95 / 5130 seen = 1 mark (method).

(a)(ii) [2]
1.8(%) = 2 marks. ALLOW 1.79 (not to 2 s.f.) for 1 mark.
If incorrect, award 1 mark for (5130 − 5040) ÷ 5040 × 100 or for 5040 seen as the 'before' value. ALLOW error carried forward from (a)(i) for the 5130 value, with answer given to 2 s.f. DO NOT ALLOW a negative sign / 'decrease'.

(b) [max 2]
1. (Cardiac) muscle of ventricle wall hypertrophies / becomes thicker / larger / stronger ✓
2. (so) contracts with greater force / pressure, ejecting more blood per beat / more complete emptying of ventricle ✓
3. Greater filling of ventricle / larger ventricle volume ✓
IGNORE 'heart gets bigger' alone. IGNORE 'more efficient'.

(c) [max 3]
1. CO2 forms carbonic acid / releases H+ (ions) / lowers pH ✓
2. Change in shape of haemoglobin / lower affinity (of haemoglobin) for oxygen (Bohr effect) ✓
3. Oxygen dissociation curve shifts to the right / haemoglobin unloads oxygen more readily at a given partial pressure of oxygen ✓
ALLOW 'more oxygen dissociates' AW. DO NOT ALLOW 'haemoglobin is denatured'. DO NOT ALLOW 'higher affinity'.

(a) Burrowing output = 0.09 × 220 = 19.8 cm\(^3\) min\(^{-1}\). Surface-dwelling output = 0.07 × 260 = 18.2 cm\(^3\) min\(^{-1}\). Difference = 19.8 − 18.2 = 1.6. Percentage difference relative to the burrowing mammal = \( \frac{1.6}{19.8} \times 100 = 8.08\% \), so 8.1% (the surface-dwelling mammal's output is 8.1% lower).
(b) The burrowing mammal's p50 (2.4 kPa) is lower than 4.1 kPa, so its haemoglobin has a higher affinity for oxygen. It therefore becomes saturated at a lower partial pressure of oxygen, so it can load more oxygen in the lungs even though the burrow air has little oxygen.
(c) Exercising muscle produces more carbon dioxide, lowering pH. This reduces the affinity of haemoglobin for oxygen (curve shifts right), so more oxygen is unloaded (dissociates) at the muscle at a given partial pressure, supporting more aerobic respiration.
(d) The SAN (pacemaker) initiates a wave of depolarisation across the atria, causing atrial systole. A band of non-conducting fibrous tissue prevents the wave passing directly to the ventricles. The wave reaches the AVN, which delays transmission, allowing the atria to empty before ventricular contraction. The wave then passes down the bundle of His and the Purkyne tissue in the septum to the apex, so the ventricles contract from the base upwards, efficiently emptying blood. In first-degree heart block the delay at the AVN is longer, so the PR interval (P wave to QRS) is about twice as long (0.32 s against 0.16 s). The P wave and QRS complex are themselves normal in shape, and each P wave is still followed by a QRS, because the impulse is slowed, not blocked.

(a) 8.1 (%) [2]. If answer incorrect, award 1 mark for correct outputs 19.8 and 18.2 (cm3 min-1) or for (19.8 − 18.2)/19.8 × 100. Unrounded 8.08 gains 1 mark only.
(b) Max 2: 1. (lower p50 so) haemoglobin has a higher affinity for oxygen / curve shifted to the left (AW); 2. loads/associates with oxygen (more readily) at low partial pressure of oxygen / is more saturated at the low pO2 in burrow air / in the lungs (AW). IGNORE 'absorbs more oxygen' without reference to affinity or pO2.
(c) Max 2: 1. increased carbon dioxide (concentration) / lower pH (in respiring muscle) reduces (haemoglobin) affinity for oxygen / curve shifts right; 2. so more oxygen is released/unloaded (at muscle) for (more) aerobic respiration. DO NOT ALLOW 'haemoglobin gets denatured'.
(d)* Level 3 (5–6 marks): Clear, correct explanation of the roles of SAN, AVN and Purkyne tissue in the right sequence AND correct explanation of the ECG change with data (0.16 s v 0.32 s) used. Logical, well organised, relevant throughout.
Level 2 (3–4 marks): Describes most of the sequence with some roles explained, and makes a correct statement about the ECG difference, but either explanation or use of data is incomplete.
Level 1 (1–2 marks): Some correct points about the SAN/AVN/Purkyne or the ECG, mostly descriptive, with limited logical organisation.
0 marks: no creditworthy response.
Indicative content: SAN initiates (wave of) depolarisation / is the pacemaker (heart is myogenic); across atria causing atrial systole; fibrous/non-conducting tissue between atria and ventricles prevents direct spread; AVN delays impulse (~0.1 s) so atria empty/ventricles fill; impulse passes down bundle of His / Purkyne tissue in septum to apex; ventricles contract from the base/apex upwards (ventricular systole) so blood is forced out into arteries; patient: PR interval (P wave to start of QRS) longer / doubled to 0.32 s; because the AVN delay is greater/conduction slower; P wave and QRS still present/normal as the impulse still passes; heart rate/atrial-ventricular sequence preserved but with longer pause. ALLOW 'bundle of His' as part of the conducting pathway. IGNORE 'AVN passes impulse to ventricles' without the delay point for the top level.

(a)(i) Cardiac output = heart rate × stroke volume = 62 × 85 = 5270 cm\(^3\) min\(^{-1}\). Divide by 1000 to convert to dm\(^3\): 5.27 dm\(^3\) min\(^{-1}\).
(a)(ii) The Bohr effect.
(b) The llama has a lower p50 (2.4 kPa compared with 3.6 kPa), so its haemoglobin has a higher affinity for oxygen and its dissociation curve is shifted to the left. At the low partial pressure of oxygen in the alveoli at altitude, llama haemoglobin is therefore still highly saturated, so more oxygen is loaded. In running muscle, respiration produces more carbon dioxide, which forms carbonic acid and lowers pH; this changes the shape of haemoglobin, reducing its affinity so the curve shifts right (Bohr effect). Haemoglobin then unloads more oxygen at the same pO2, supplying more oxygen for aerobic respiration to make ATP. Because llama haemoglobin has a high affinity, this additional unloading prevents oxygen being held too tightly in the tissues.

(a)(i) Correct answer 5.27 (ALLOW 5.3 / 5.270) = 2 marks. If answer incorrect, 62 × 85 (= 5270) or 5270 ÷ 1000 gains 1 mark. DO NOT ALLOW 5270 without conversion (1 mark max). [2]
(a)(ii) Bohr effect / Bohr shift. [1]
(b) Level of response (6 marks).
Indicative content:
1. Llama p50 lower (2.4 vs 3.6 kPa) / quotes data;
2. Llama haemoglobin has higher affinity for oxygen / curve to the left (AW);
3. Haemoglobin loads/becomes saturated at lower pO2 in alveoli / more oxygen loaded at altitude;
4. Running muscle respires faster / produces more CO2;
5. CO2 lowers pH (carbonic acid / H+ ions);
6. Haemoglobin changes shape / affinity decreases, curve shifts right (Bohr effect);
7. More oxygen unloaded at muscle for aerobic respiration (AW);
8. Linked idea: unloading still possible in tissues despite high affinity.
Band 3 (5-6 marks): Adaptation explained using data and Bohr effect explained with correct sequence of events; logical, coherent, linked to benefit. Band 2 (3-4 marks): Some explanation of both parts, or a full explanation of one with limited reference to the other; some linkage. Band 1 (1-2 marks): Statements of affinity or Bohr effect with little or no explanation or data. 0 marks: no creditworthy response.
IGNORE references to haemoglobin carrying more oxygen per molecule. DO NOT ALLOW 'lower p50 means lower affinity'. [6]

(a) At 95 °C hydrogen bonds between complementary bases break, separating the DNA into single strands. Cooling to about 55 °C allows primers to anneal (form hydrogen bonds) to complementary sequences either side of the target. A heat-stable polymerase (e.g. Taq) is not denatured at 95 °C, so it can extend the primers at about 72 °C in every cycle without being replaced.
(b) The DNA fragments are placed in wells in an agarose gel and a potential difference is applied. DNA is negatively charged because of its phosphate groups, so fragments move towards the positive electrode. Smaller fragments move faster/further through the gel matrix, so 450 bp travels further than 600 bp. Plant 1 shows only the 450 bp band so is homozygous rr; plants 2 and 5 show only 600 bp so are RR; plants 3 and 4 show both bands so are heterozygous Rr. Plants 3 × 4 is Rr × Rr: gametes R and r from each parent, giving RR : Rr : rr in the ratio 1 : 2 : 1, so \(\frac{2}{4} = \frac{1}{2}\) (50%) of offspring are expected to be heterozygous.

(a) 1. 95 °C breaks hydrogen bonds between (complementary) bases / separates the two DNA strands;
2. 55 °C allows primers to anneal / bind (by complementary base pairing) to the single strands;
3. Taq / the polymerase is heat stable / not denatured at 95 °C so does not need to be added each cycle (AW).
IGNORE 'kills the enzyme' for 55 °C. DO NOT ALLOW 'melts' without reference to bonds/strands for mark 1. [3]
(b) Level of response (6 marks).
Indicative content:
1. DNA is negatively charged (phosphate groups);
2. Moves towards the positive electrode / anode under a potential difference;
3. Agarose gel acts as a mesh/matrix; smaller fragments move faster/further;
4. 450 bp band is further from the wells than 600 bp;
5. Plant 1 rr; plants 2 and 5 RR;
6. Plants 3 and 4 Rr (two bands);
7. Cross Rr × Rr gives genotype ratio 1 RR : 2 Rr : 1 rr;
8. 1/2 / 50% heterozygous.
Band 3 (5-6 marks): Explanation of separation with correct genotypes and the correct proportion with reasoning shown; logical order. Band 2 (3-4 marks): Some explanation of separation and some correct genotypes OR correct genotypes and proportion with limited explanation of electrophoresis. Band 1 (1-2 marks): Isolated correct statements with little linkage. 0 marks: no creditworthy response.
ALLOW error carried forward for the proportion if genotypes are wrongly assigned but a consistent cross is shown. [6]

(a)(i) Read from table: 0.5 kPa.
(ii) Myoglobin has a higher affinity, so it takes up oxygen from haemoglobin at the low pO2 found in muscle and stores it; it releases oxygen only when pO2 falls very low, e.g. during intense respiration or a dive.
(b)(i) Cardiac output = heart rate x stroke volume = \(40 \times 150 = 6000\) cm\(^3\) min\(^{-1}\).
(ii) Dive: \(12 \times 150 = 1800\) cm\(^3\) min\(^{-1}\). Decrease = \((6000-1800)/6000 \times 100 = 70\%\).
(c) Haemoglobin is 92% saturated at 8 kPa (lungs) and only 25% at 2 kPa (muscle), so it loads in lungs and unloads in tissues. The S-shaped curve arises from cooperative binding. The Bohr effect (more CO2 shifts the curve right) increases unloading in active muscle. Myoglobin is 88% saturated at 2 kPa, where haemoglobin has largely unloaded, so it accepts oxygen from haemoglobin and stores it. During the dive, with cardiac output reduced by 70%, the stored oxygen in myoglobin is released when pO2 falls below about 1 kPa, sustaining aerobic respiration.

(a)(i) 0.5 [1]
(a)(ii) max 2: myoglobin takes/accepts oxygen from haemoglobin (1); stores oxygen / releases it only at very low pO2 / when respiration is rapid or oxygen supply is low (1); ALLOW higher saturation at low pO2 (AW). IGNORE 'more oxygen'.
(b)(i) 6000 [1]. DO NOT ALLOW without correct value; ALLOW 6 dm3 min-1 only if units converted.
(b)(ii) Correct answer 70 = 2 marks. If incorrect: dive cardiac output 1800 (1); OR (6000-1800)/6000 x 100 (1). ALLOW ecf from (b)(i): percentage decrease of dive value calculated from their resting value (1 max for method).
(c) Level 3 (5-6 marks): Detailed explanation of loading/unloading of haemoglobin and the role of myoglobin, with several correct uses of data, and at least one of the Bohr effect or cooperative binding. Logical, well organised, and covers both surface and dive.
Level 2 (3-4 marks): Explains most of haemoglobin loading/unloading or myoglobin storage, with some use of data; some gaps in logic or organisation.
Level 1 (1-2 marks): Limited statements about oxygen transport or storage, little or no use of data; may be disorganised.
Level 0 (0): No response worthy of credit.
Indicative content: haemoglobin loads at high pO2 in lungs, e.g. 92% at 8 kPa; unloads at low pO2 in tissues, e.g. 25% at 2 kPa; sigmoid curve due to cooperative binding / change in shape (conformational change) making further oxygen binding easier; Bohr effect: high CO2 / lower pH in active muscle shifts curve to right, more unloading; myoglobin has higher affinity, e.g. 88% at 2 kPa when Hb only 25%; so takes oxygen from haemoglobin; myoglobin acts as an oxygen store; releases oxygen only at very low pO2, e.g. below 1 kPa, in a dive; heart rate and cardiac output fall in dive so less oxygen delivered by blood; stored oxygen supports aerobic respiration for longer.

(a) Cardiac output = heart rate x stroke volume = 62 x 78 = 4836 cm3 min-1. Divide by 1000 to convert to dm3: 4.836, which is 4.84 dm3 min-1 to 3 s.f.

(b)(i) Seal haemoglobin: at every pO2 shown it has the higher percentage saturation, e.g. 68% compared with 38% at 3 kPa.
(ii) Advantage: it becomes highly saturated even when pO2 in the lungs is low, so more oxygen is loaded during a dive. Disadvantage: it releases oxygen only at very low pO2, so less oxygen is unloaded to tissues unless pO2 falls very low.

(c) Respiring muscle produces more carbon dioxide. This dissolves to form carbonic acid (carbonic anhydrase), which dissociates to release H+ ions. H+ combine with haemoglobin (forming haemoglobinic acid), changing its shape and lowering its affinity for oxygen (Bohr effect). Oxygen is released more readily at a given pO2, supplying more oxygen for aerobic respiration.

(d) The SAN (pacemaker) initiates a wave of excitation across both atria, causing atrial systole. A layer of non-conducting fibrous tissue stops the wave passing directly to the ventricles, so it is delayed at the AVN, allowing the atria to empty. The wave then passes down the Purkyne tissue (bundle of His) to the apex and spreads upwards through the ventricle walls, so ventricles contract from the base upwards. In ventricular systole, left ventricular pressure rises above that in the left atrium, closing the atrioventricular valve and preventing backflow to the atrium. When ventricular pressure exceeds aortic pressure, the semilunar valve opens and blood is forced into the aorta. In diastole, ventricular pressure falls below aortic pressure and the semilunar valve closes, preventing backflow into the ventricle.

(e) Take blood from several seals and repeat the measurements several times at each pO2, then calculate means; keep temperature and pH constant (and the same for both species).

(a) 4836 (cm3 min-1) seen (1); 4.84 (dm3 min-1) (1). ALLOW 1 mark for 4.836 or 4.8 (wrong s.f.). Correct answer gains both marks. [2]

(b)(i) Seal (1) because higher % saturation at the same pO2 / curve is to the left / quoted paired values e.g. 68 v 38 at 3 kPa. Reason must be linked to seal for the mark. [1]
(ii) Advantage (1): loads / becomes saturated at low pO2 in the lungs / holds oxygen when pO2 falls during a dive AW. Disadvantage (1): unloads oxygen less readily / only at very low pO2 in tissues AW. DO NOT ALLOW 'carries more oxygen' without reference to low pO2. [2]

(c) Any three: more CO2 produced by (aerobic) respiration (1); carbonic acid formed (carbonic anhydrase) then dissociates giving H+ (1); H+ combine with haemoglobin / lower pH changes haemoglobin shape (1); lower affinity for oxygen (so curve shifts right / Bohr effect) (1); oxygen released more readily at the same pO2 / more oxygen available to the muscle for respiration (1). Max 3. IGNORE 'lactic acid' alone. [3]

(d)* Level of response, 6 marks.
Level 3 (5-6 marks): Describes the sequence SAN, atrial systole, delay at AVN and Purkyne conduction AND explains pressure changes and valve action in the ventricle and aorta, with a clear, logical and correctly ordered account linking excitation to contraction and to one-way flow.
Level 2 (3-4 marks): Describes most of the electrical sequence AND gives some explanation of pressure and valve action, or gives a full account of one demand with limited coverage of the other; mostly logical.
Level 1 (1-2 marks): Gives a few relevant points from either demand, with little linkage or some inaccuracy.
Level 0 (0 marks): No response worthy of credit.
Indicative content: SAN initiates excitation / myogenic; wave spreads over atria causing atrial systole; non-conducting tissue prevents direct passage to ventricles; AVN delays the wave; Purkyne tissue carries it to the apex; ventricles contract from the base up; ventricular pressure exceeds atrial pressure so atrioventricular valve closes; ventricular pressure exceeds aortic pressure so semilunar valve opens; blood forced into aorta; aortic pressure higher than ventricular pressure in diastole so semilunar valve closes. [6]

(e) Any two: repeat with blood samples from more seals / larger sample size (1); repeat measurements at each pO2 and calculate a mean (1); control temperature / pH / CO2 concentration at the same value for all samples (1); check for anomalies / calculate a measure of spread (1). IGNORE 'repeat' alone. Max 2. [2]

Total 16

(a) Cardiac output = heart rate \(\times\) stroke volume = \(52 \times 85 = 4420\) cm\(^3\) min\(^{-1}\). Dividing by 1000 gives 4.42 dm\(^3\) min\(^{-1}\), which is 4.4 to 2 s.f.

(b) Exercise: \(168 \times 130 = 21840\) cm\(^3\) min\(^{-1}\). Increase = \(21840 - 4420 = 17420\). Percentage increase = \(\frac{17420}{4420} \times 100 = 394.1\)%, so 394 %.

(c) Carbon dioxide dissolves to form carbonic acid, which dissociates to release H\(^+\) ions and lowers the pH. This reduces the affinity of haemoglobin for oxygen (the Bohr effect), shifting the oxygen dissociation curve to the right. Haemoglobin therefore unloads more oxygen at a given partial pressure in the muscle capillaries, so more oxygen is available for aerobic respiration.

(a) 4.4 (dm\(^3\) min\(^{-1}\)) [1]. DO NOT ALLOW 4420 or 4.4 without correct unit conversion.

(b) Correct answer 394 (%) = 2 marks. If the answer is incorrect, award 1 mark for either exercise cardiac output = 21840 cm\(^3\) (21.8 / 21.84 dm\(^3\)) OR \((21840 - 4420)/4420 \times 100\) shown. ALLOW 1 mark for ecf from an incorrect part (a) value used in the percentage calculation. ALLOW 394.1. DO NOT ALLOW 494 (ratio expressed as a percentage of rest).

(c) Max 2:
1. Increased CO\(_2\) lowers the pH / increases H\(^+\) concentration (of blood) / forms carbonic acid [1];
2. (Bohr effect) haemoglobin has a lower affinity for oxygen / curve shifts to the right so oxygen is released / dissociates more readily (in the muscle capillaries) [1].
IGNORE 'haemoglobin carries more oxygen'. DO NOT ALLOW 'haemoglobin has a higher affinity'. ALLOW AW for point 2 such as 'more oxygen unloaded at the same partial pressure of oxygen'.

(a)(i) Cardiac output = heart rate \(\times\) stroke volume = \(62 \times 84 = 5208\) cm\(^3\) min\(^{-1}\).

(a)(ii) Exercise cardiac output = \(170 \times 130 = 22100\) cm\(^3\) min\(^{-1}\). Increase = \(22100 - 5208 = 16892\). Percentage increase = \(16892 / 5208 \times 100 = 324.3\%\), which is 324% to 3 s.f.

(b) The AVN delays the impulse so that the atria have time to contract fully and empty their blood into the ventricles, and the ventricles are full before they contract. The atrial and ventricular walls are separated by non-conducting fibrous tissue, so the delay also means ventricular contraction starts only after atrial systole has finished.

(c) Muscles respire faster and produce more carbon dioxide (and lactic acid), which lowers blood pH. This reduces the affinity of haemoglobin for oxygen (Bohr effect), so oxygen dissociates from haemoglobin more readily at a given partial pressure of oxygen. More oxygen is therefore released to the muscle for aerobic respiration.

(a)(i) [2] Correct answer 5208 (cm\(^3\) min\(^{-1}\)) = 2 marks. ALLOW 5200 or 5.2 dm\(^3\) min\(^{-1}\) (with unit). If answer wrong, 62 \(\times\) 84 shown = 1 mark (method).

(a)(ii) [2] Correct answer 324 (%) = 2 marks. ALLOW 324.3. DO NOT ALLOW 324.35 or 3.24 \(\times\) 10\(^2\) without clear 324 value being given to 3 s.f. If answer wrong, award 1 mark for exercise cardiac output 22100 AND (22100 \(-\) 5208) / 5208 \(\times\) 100 shown (method). ALLOW ecf from an incorrect resting cardiac output from (a)(i) for the method mark and final mark if the calculation is correct.

(b) [2] Any two from:
1. allows atria to (finish) contract(ing) / atrial systole to complete;
2. so ventricles fill (completely) with blood / blood moves from atria to ventricles;
3. ventricles contract only after they are full / ensures atrial and ventricular systole do not occur at the same time / ensures maximum stroke volume.
IGNORE 'slows the heart rate'.

(c) [2]
1. (increased respiration produces) more CO\(_2\) / lactic acid, lowering pH / increasing H\(^+\) (concentration) (in the blood); (linked)
2. reduces affinity of haemoglobin for oxygen / haemoglobin releases oxygen more readily (Bohr effect), so more oxygen unloaded (for aerobic respiration);
ALLOW 'more oxygen dissociates at the same partial pressure of oxygen'. DO NOT ALLOW 'haemoglobin is denatured'. IGNORE 'temperature' alone.

(a)(i) Cardiac output = heart rate × stroke volume = 96 × 52 = 4992 cm\(^3\) min\(^{-1}\). Divide by 1000 to convert to dm\(^3\): 4.992 dm\(^3\) min\(^{-1}\) = 4.99 dm\(^3\) min\(^{-1}\) (3 s.f.).

(a)(ii) The SAN, the pacemaker, spontaneously generates a wave of depolarisation that spreads across both atria, making the atrial walls contract. A band of non-conducting fibrous tissue prevents the wave passing directly to the ventricles. The wave reaches the AVN, which delays transmission briefly so the atria finish emptying before the ventricles contract. The excitation then passes down the bundle of His and along Purkyne fibres to the apex of the heart, and the ventricles contract from the apex upwards.

(b) Human haemoglobin has four subunits. The first oxygen binds with difficulty at low pO2 (8% at 1 kPa), but binding changes the tertiary/quaternary structure, opening up the molecule so that further oxygen binds more easily (cooperative binding). This gives a steep middle portion (25% to 80% between 2 and 6 kPa). When most sites are occupied, it is hard for the last oxygen to bind, so the curve flattens at high pO2, giving an S shape. The sand vole curve is to the left: at 3 kPa it is 75% saturated against 42% in humans, so its haemoglobin has a higher affinity for oxygen. In poorly ventilated tunnels oxygen concentration is low, so the vole can still load a large amount of oxygen in the lungs at a low pO2. Increased CO2 in respiring muscle lowers pH (Bohr effect), which reduces the affinity of haemoglobin for oxygen, shifting the human curve to the right, so haemoglobin unloads more oxygen at a given pO2 to the respiring muscle.

(a)(i) [2]
- 4.99 (dm\(^3\) min\(^{-1}\)) scores 2 marks. [2]
- If the answer is incorrect: 96 × 52 (or 4992) scores 1 mark (method). [1]
- ALLOW 5.0 / 4.992 for 1 mark only if working shown; DO NOT ALLOW 4990 without conversion.

(a)(ii) [3 max]
1. SAN initiates / sends out a wave of depolarisation / excitation across the atria (walls); causing atrial systole / contraction;
2. Non-conducting tissue / fibres prevent the wave passing straight to the ventricles;
3. AVN delays the wave / impulse (so atria empty before ventricles contract);
4. Wave passes down the bundle of His / Purkyne tissue to the apex (of the heart);
5. Ventricles contract from the base / apex upwards (ventricular systole).
IGNORE references to nerves / brain controlling the heart. ALLOW 'electrical impulse' for wave of excitation.

(b)* Level of response, 6 marks. Judge the scientific coverage and the quality of the reasoning, use of data and organisation, not a simple count of points.
Level 3 (5–6 marks): Explains all three bullet points, with at least one correct and quantified use of Table 1.1 data. The explanation is detailed, logically structured, and links structure to function throughout.
Level 2 (3–4 marks): Explains at least two of the bullet points with some accurate use of data, or covers all three with limited detail. The reasoning is mostly clear.
Level 1 (1–2 marks): Makes simple statements about one or two bullet points, with little or no use of data or limited linkage. The response lacks structure.
0 marks: no response or no response worthy of credit.

Indicative scientific points:
- Haemoglobin has four polypeptide subunits, each with a haem group; first oxygen binds with difficulty at low pO2 (shallow start of the curve, e.g. 8% at 1 kPa).
- Binding of the first oxygen changes the tertiary/quaternary structure (shape), making it easier for further oxygen to bind (cooperative binding); steep gradient in the middle (e.g. 25% to 80% between 2 and 6 kPa).
- Becomes difficult for the last oxygen to bind as few sites remain (flattening at high pO2); S-shaped / sigmoid.
- Sand vole curve is to the left of the human curve / higher saturation at every pO2 (e.g. 75% against 42% at 3 kPa): higher affinity for oxygen.
- Burrow has low oxygen / high CO2 (poor ventilation) so the vole can load (more) oxygen at low pO2 and still be saturated; AW.
- Higher CO2 concentration in respiring tissue lowers pH (Bohr effect), changing the haemoglobin shape so affinity falls; human curve shifts to the right; more oxygen unloaded to the muscle at the same pO2.
ALLOW ECF for data use if a value is misread by one small square. IGNORE 'haemoglobin is more efficient' without reference to affinity. DO NOT ALLOW 'oxygen is made' / confusion between affinity and amount of haemoglobin.