OCR A Level Biology A: Chemiosmosis, anaerobic respiration and respiratory quotient (H420) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover OCR A Level Biology A (H420) section 5.2.2 Respiration, learning outcomes (h) to (l), from specification Version 4.1 (April 2026). Learn the content first from the study guide and the revision notes. Course pages: the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. Explain how the electron transport chain and ATP synthase produce ATP in a mitochondrion. [5]

2. The pH of the intermembrane space of a mitochondrion is 7.0 and the pH of the matrix is 7.8.

(a) State which region has the higher concentration of hydrogen ions. [1] (b) Calculate how many times greater that concentration is. [2] (c) State what the flow of hydrogen ions through ATP synthase is used for. [1]

3. State two similarities and two differences between oxidative phosphorylation and photophosphorylation. [4]

4. A scientist adds a substance that binds to ATP synthase to isolated mitochondria supplied with a respiratory substrate and oxygen. Predict and explain the effect on:

(a) the rate of oxygen uptake [2] (b) the rate of ATP synthesis [1] (c) the concentration of hydrogen ions in the intermembrane space. [1]

5. Explain why the formation of lactate allows a muscle to keep making ATP during a sprint. [4]

6. (a) Describe how pyruvate is converted to ethanol in yeast, stating where carbon dioxide is released. [3] (b) Aerobic respiration gives about 30 ATP per glucose and anaerobic respiration in yeast gives 2. Calculate the anaerobic yield as a percentage of the aerobic yield, to 2 significant figures. [2]

7. Lipid releases 37 kJ g⁻¹ and carbohydrate 17 kJ g⁻¹ when respired.

(a) Calculate the mass of lipid that releases the same energy as 60 g of carbohydrate, to 3 significant figures. [2] (b) Explain why lipid releases more energy per gram than carbohydrate. [2]

8. (a) Lauric acid, C₁₂H₂₄O₂, is respired: C₁₂H₂₄O₂ + 17O₂ → 12CO₂ + 12H₂O. Calculate the respiratory quotient (RQ) to 2 significant figures. [2] (b) State what type of respiratory substrate this value is typical of. [1] (c) Tartaric acid, C₄H₆O₆, is respired: C₄H₆O₆ + 2.5O₂ → 4CO₂ + 3H₂O. Calculate the RQ. [2]

9. A respirometer has a capillary tube of radius 0.40 mm. Germinating seeds are placed in two tubes. In tube A, which contains potassium hydroxide solution, the liquid drop moves 52 mm. In tube B, with no absorber, it moves 14 mm in the same time.

(a) Calculate the volume of oxygen used. [2] (b) Calculate the volume of carbon dioxide produced. [2] (c) Calculate the RQ. [1] (d) Suggest the main substrate. [1]

10. Germinating peas of mass 2.0 g were placed in a respirometer in a water bath. The capillary radius is 0.50 mm.

(a) Describe how you would investigate the effect of temperature on the rate of respiration of the peas. [6] (b) At 30 °C the drop moved 49 mm in 5 min. Calculate the rate of oxygen uptake in mm³ g⁻¹ min⁻¹, to 2 significant figures. [2] (c) At 40 °C the rate was 4.8 mm³ g⁻¹ min⁻¹. Suggest why the increase from 30 °C to 40 °C is smaller than the increase from 20 °C (2.0 mm³ g⁻¹ min⁻¹) to 30 °C. [1]

11. Yeast was kept under a layer of oil in glucose solutions of different concentrations. The carbon dioxide collected in 10 min was measured.

Glucose / %	0.5	1	2	4	8	16
CO₂ / cm³	1.5	2.9	5.6	8.1	8.7	8.6

(a) Calculate the rate of carbon dioxide production at 2% glucose. [1] (b) Calculate the percentage increase in the volume collected when glucose rises from 1% to 2%. [2] (c) Describe the pattern shown. [2] (d) Explain the pattern above 8%. [2]

12. A person’s oxygen uptake and carbon dioxide output were measured at rest and during hard exercise.

	O₂ uptake / dm³ min⁻¹	CO₂ output / dm³ min⁻¹
Rest	0.25	0.205
Hard exercise	3.00	3.45

(a) Calculate the RQ at rest and during hard exercise. [2] (b) Suggest what the resting RQ shows about the substrates being respired. [1] (c) Explain why the RQ is above 1.0 during hard exercise. [2] (d) Calculate the extra carbon dioxide released per minute, over the oxygen taken in, during hard exercise. [1]

Answers

1. Reduced NAD and reduced FAD donate electrons to the electron transport chain [1]. The electrons pass along carriers, releasing energy [1]. This energy pumps H⁺ from the matrix into the intermembrane space [1], building a proton gradient across the inner membrane [1]. H⁺ flows back into the matrix through ATP synthase, which joins ADP and inorganic phosphate to make ATP [1]. [5] Examiner insight: name the side of the membrane the H⁺ is pumped to and the direction of flow through ATP synthase; “protons are pumped across” leaves the gradient unclear.

2. (a) The intermembrane space, since it has the lower pH [1]. (b) The pH difference is 0.8 [1]. 10^0.8 = 6.3 times [1]. (c) To make ATP from ADP and inorganic phosphate [1]. Examiner insight: pH is logarithmic, so a difference of 0.8 is not 0.8 times; show the power of ten.

3. Similarities, any two from: an electron transport chain pumps H⁺ across a membrane; ATP synthase makes ATP as H⁺ flows down its gradient; ATP is made from ADP and inorganic phosphate [2]. Differences, any two from: inner mitochondrial membrane versus thylakoid membrane; energy source is reduced NAD and FAD versus light; H⁺ pumped into the intermembrane space versus the thylakoid space; oxygen versus NADP as final acceptor [2]. [4] Examiner insight: a similarity must apply to both; a difference needs a statement about each process, not only one.

4. (a) Oxygen uptake decreases [1], because H⁺ cannot flow back through ATP synthase, so the gradient builds up and makes further pumping harder, so electron transport slows [1]. (b) ATP synthesis decreases [1]. (c) It increases (at first) [1]. Examiner insight: link the effect on oxygen uptake to the gradient; oxygen uptake is not changed directly by the substance.

5. In a sprint, oxygen cannot reach the muscle quickly enough for the electron transport chain to reoxidise reduced NAD [1]. Pyruvate accepts hydrogen from reduced NAD and becomes lactate [1]. This regenerates NAD [1], so glycolysis continues to make ATP [1]. [4] Examiner insight: the key idea is regeneration of NAD; “lactate is made so ATP is made” gives no mechanism.

6. (a) Pyruvate is decarboxylated to ethanal [1], and carbon dioxide is released at this step [1]. Ethanal accepts hydrogen from reduced NAD and becomes ethanol, regenerating NAD [1]. (b) 2 ÷ 30 [1] × 100 = 6.7% [1]. [5] Examiner insight: give the answer to the number of significant figures asked for, and write the unit (percent) with it.

7. (a) Energy = 60 × 17 = 1020 kJ [1]. Mass of lipid = 1020 ÷ 37 = 27.6 g [1]. (b) Lipid molecules have more hydrogen atoms for each oxygen atom [1], so more reduced NAD and FAD are produced per gram, more H⁺ is pumped and more ATP is made [1]. [4] Examiner insight: explain through hydrogen and reduced coenzymes; “lipids have more bonds” is not specific.

8. (a) RQ = CO₂ ÷ O₂ = 12 ÷ 17 [1] = 0.71 [1]. (b) Lipid [1]. (c) RQ = 4 ÷ 2.5 [1] = 1.6 [1]. [5] Examiner insight: show the division before the answer, and put carbon dioxide on top.

9. Area of the capillary = π × 0.40² = 0.503 mm². (a) Volume of oxygen = 0.503 × 52 [1] = 26 mm³ [1]. (b) Net movement in tube B is oxygen used minus carbon dioxide produced, so carbon dioxide corresponds to 52 − 14 = 38 mm [1]. Volume = 0.503 × 38 = 19 mm³ [1]. (c) RQ = 19.1 ÷ 26.1 = 38 ÷ 52 = 0.73 [1]. (d) Lipid [1]. [6] Examiner insight: the carbon dioxide volume comes from the difference between the two tubes; do not use tube B alone.

10. (a) Any six points, one mark each: place equal masses of peas in each tube above potassium hydroxide solution or soda lime, which absorbs carbon dioxide [1]; use a control tube of beads or dead peas [1]; use water baths at several temperatures, such as 10, 20, 30 and 40 °C, and equilibrate the apparatus [1]; seal and record the position of the drop at the start and at regular intervals [1]; keep other variables constant, such as mass of peas and tube volume [1]; repeat and calculate a mean, then find rate as volume ÷ time, with volume = π r² × distance [1]. (b) Volume = π × 0.50² × 49 = 38.5 mm³ [1]. Rate = 38.5 ÷ 5 ÷ 2.0 = 3.8 mm³ g⁻¹ min⁻¹ [1]. (c) Enzymes start to denature at higher temperature, so the rate rises more slowly [1]. [9] Examiner insight: in a plan, include a control and a way of calculating the rate; listing apparatus alone does not show how the results are used.

11. (a) 5.6 ÷ 10 = 0.56 cm³ min⁻¹ [1]. (b) (5.6 − 2.9) ÷ 2.9 [1] × 100 = 93% [1]. (c) The volume increases as glucose concentration rises, [1] but the increase gets smaller and the volume levels off at about 8.7 cm³ above 8% [1]. (d) Glucose is no longer limiting [1]; the rate is limited by another factor, such as the number of glucose transport proteins or the enzymes of glycolysis and fermentation being saturated [1]. [7] Examiner insight: quote figures from the table in the description, and name the factor that has become limiting.

12. (a) Rest: 0.205 ÷ 0.25 = 0.82 [1]. Hard exercise: 3.45 ÷ 3.00 = 1.15 [1]. (b) A mixture of carbohydrate and lipid, since 0.82 lies between 0.7 and 1.0 [1]. (c) The muscles respire some glucose anaerobically, forming lactic acid [1]. The acid reacts with hydrogencarbonate in the blood, releasing extra carbon dioxide, so carbon dioxide output exceeds oxygen uptake [1]. (d) 3.45 − 3.00 = 0.45 dm³ min⁻¹ [1]. [6] Examiner insight: compare the output with the uptake in your explanation, rather than saying only that more carbon dioxide is made.

Where marks are usually lost

• Writing “protons are pumped” without saying from where to where.
• Saying electrons flow through ATP synthase, when protons do.
• Not linking a change in ATP synthase or in the membrane to the proton gradient.
• Forgetting that lactate and ethanol formation regenerates NAD.
• Putting oxygen on top of the RQ fraction.
• Using the diameter of the capillary in π r².
• Reading the carbon dioxide volume from the tube without absorber instead of from the difference.
• Giving a rate with no unit, or without dividing by mass or time.

Next steps

• Re-read the study guide for any question that cost you marks.
• Work through the revision notes.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Level 3 Advanced GCE in Biology A (H420), specification Version 4.1 (April 2026), first assessment 2025, published by OCR. These questions follow section 5.2.2 Respiration, learning outcomes (h) to (l).