OxfordAQA International AS and A-level Biology: Respiration, Krebs cycle and RQ (9610) – 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 section 3.3.3, Respiration, of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, for International AS exams from May/June 2017 and International A-level exams from May/June 2018. They include required practical 8. Unit 3 is listed in the specification as International A-level only. Use the study guide and the revision notes. The course hub is OxfordAQA A-level Biology, the checklist lists every statement, and the free diagnostics give a quick check of your starting point.

Questions

1. State where in a cell each of these occurs: (a) glycolysis, (b) the Krebs cycle, (c) the electron transfer chain used in oxidative phosphorylation. [3]

2. A cell extract oxidises 0.35 mol of glucose by glycolysis. Calculate the amount of (a) ATP used, (b) ATP made, (c) net ATP gained and (d) reduced NAD formed. [4]

3. Describe the steps by which glucose is converted to pyruvate in glycolysis. [4]

4. A sprinter’s leg muscle cells are short of oxygen. (a) Explain why the cells convert pyruvate to lactate. [3] (b) State the net number of ATP molecules gained from one glucose molecule respired in this way. [1] (c) Give one difference between this pathway and the pathway in yeast without oxygen. [1]

5. A mitochondrial preparation respires 0.25 mol of glucose aerobically. Calculate the amount of (a) carbon dioxide released in the link reaction, (b) carbon dioxide released in the Krebs cycle, (c) reduced NAD made in the link reaction and Krebs cycle together, (d) reduced FAD made, and (e) ATP made directly in the Krebs cycle. [5]

6. Explain how the electron transfer chain and ATP synthase together make ATP in a mitochondrion. [5]

7. A compound that blocks the channel through ATP synthase is added to mitochondria that have been given pyruvate and oxygen. (a) Predict what happens to the concentration of protons in the intermembrane space. [1] (b) Predict what happens to the rate of oxygen uptake and explain your answer. [3]

8. Use these data. Assume 2.5 ATP for each reduced NAD made in the matrix, 1.5 ATP for each reduced NAD made in glycolysis, and 1.5 ATP for each reduced FAD. (a) Work out the ATP yield of one glucose molecule respired aerobically, counting the ATP made directly in glycolysis and in the Krebs cycle. [4] (b) Anaerobic respiration gives a net 2 ATP per glucose. Express this as a percentage of your answer to (a), to 2 significant figures. [2]

9. Cells in a fasting person respire stored triglycerides and some protein. (a) Describe how a triglyceride can be used as a respiratory substrate. [3] (b) Explain why one gram of lipid gives more ATP than one gram of carbohydrate. [2] (c) Describe how an amino acid can be used in respiration. [2]

10. Palmitic acid is respired: C16H32O2 + 23 O2 → 16 CO2 + 16 H2O. Malic acid is respired: C4H6O5 + 3 O2 → 4 CO2 + 3 H2O. (a) Calculate the RQ of palmitic acid, to 2 decimal places. [2] (b) Calculate the RQ of malic acid, to 2 decimal places. [2] (c) Explain why the RQ of malic acid is higher than that of palmitic acid. [2]

11. Germinating sunflower seeds are placed in a respirometer. The capillary tube has a diameter of 1.0 mm. In one tube, containing potassium hydroxide solution, the coloured drop moves 38 mm in 20 minutes. In a second tube with the same mass of seeds and no absorber, it moves 8 mm in the same time. (a) Explain the role of the potassium hydroxide solution. [1] (b) Calculate the volume of oxygen taken up in 20 minutes. [2] (c) Calculate the volume of carbon dioxide released in 20 minutes. [2] (d) Calculate the RQ. [1] (e) Suggest what the RQ indicates about the substrate, and comment on how firm this conclusion is. [2]

12. A student measured the volume of carbon dioxide released by yeast in a glucose solution in 10 minutes at five temperatures, using a gas syringe.

Temperature / °C	20	30	40	50	60
Volume of CO2 / cm³	3.2	7.4	12.6	6.1	0.4

(a) Calculate the rate of respiration at 40 °C in cm³ per minute. [1] (b) Calculate the percentage change in volume between 40 °C and 50 °C. [2] (c) Explain the fall in volume above 40 °C. [2] (d) State the purpose of a tube containing boiled yeast. [1] (e) Suggest two changes that would let the student locate the optimum temperature more precisely. [2]

Answers

1. (a) Cytoplasm [1]. (b) Mitochondrial matrix [1]. (c) Inner mitochondrial membrane [1]. [3] Examiner insight: Name the part of the structure. “Mitochondria” alone is less precise than “matrix”.

2. (a) ATP used = 2 × 0.35 = 0.70 mol [1]. (b) ATP made = 4 × 0.35 = 1.4 mol [1]. (c) Net = 1.4 - 0.70 = 0.70 mol [1]. (d) Reduced NAD = 2 × 0.35 = 0.70 mol [1]. [4] Examiner insight: Start from the per-glucose figures and show the multiplication on each line.

3. Glucose is phosphorylated using ATP [1]. The six-carbon sugar phosphate splits into two triose phosphate molecules [1]. Triose phosphate is oxidised to pyruvate, with NAD accepting hydrogen to become reduced NAD [1]. ATP is formed from ADP, giving a net gain of ATP [1]. [4] Examiner insight: Put the stages in order and name the molecule at each. “Glucose is broken down to pyruvate” alone gives little to credit.

4. (a) Without oxygen, reduced NAD cannot be oxidised by the electron transfer chain [1]. Reduced NAD hands its hydrogen to pyruvate, forming lactate [1]. This regenerates NAD, so glycolysis can continue and keep making ATP [1]. (b) 2 ATP [1]. (c) Yeast forms ethanol and releases carbon dioxide, but lactate formation releases no carbon dioxide [1]. [5] Examiner insight: Link the conversion to NAD. Saying lactate formation “produces energy” is weak, because no extra ATP is made at this step.

5. Per glucose, link reaction: 2 CO2, 2 reduced NAD. Krebs cycle: 4 CO2, 6 reduced NAD, 2 reduced FAD, 2 ATP. (a) 2 × 0.25 = 0.50 mol [1]. (b) 4 × 0.25 = 1.0 mol [1]. (c) (2 + 6) × 0.25 = 2.0 mol [1]. (d) 2 × 0.25 = 0.50 mol [1]. (e) 2 × 0.25 = 0.50 mol [1]. [5] Examiner insight: Multiply by the two turns per glucose. Per-acetyl CoA figures are easy to leave undoubled.

6. Electrons from reduced NAD and reduced FAD enter the electron transfer chain [1]. As electrons move between carriers in the inner membrane they release energy [1]. The energy is used to pump protons out of the matrix into the space between the membranes [1]. Protons flow back down their gradient through ATP synthase [1]. ATP synthase catalyses the formation of ATP from ADP and phosphate [1]. [5] Examiner insight: State where protons are pumped from and to. “Protons pass through the membrane” gives no direction.

7. (a) It increases [1]. (b) Oxygen uptake decreases [1]. Protons cannot flow back through ATP synthase, so the proton gradient steepens and further pumping becomes harder [1]. Electron transfer slows, so less oxygen is accepted at the end of the chain [1]. [4] Examiner insight: For “predict and explain”, give the direction of change first, then a chain of cause and effect.

8. (a) Glycolysis: 2 ATP + (2 × 1.5) = 5 [1]. Link reaction: 2 × 2.5 = 5 [1]. Krebs cycle: 2 ATP + (6 × 2.5) + (2 × 1.5) = 2 + 15 + 3 = 20 [1]. Total = 5 + 5 + 20 = 30 ATP [1]. (b) 2 ÷ 30 × 100 [1] = 6.7% [1]. [6] Examiner insight: Use the yields the question supplies, not remembered figures, and set the sum out stage by stage.

9. (a) The triglyceride is hydrolysed to glycerol and fatty acids [1]. Glycerol is converted to triose phosphate, which enters glycolysis [1]. Each fatty acid chain is cut into two-carbon pieces that become acetyl CoA and go into the Krebs cycle [1]. (b) Lipids have more hydrogen atoms per gram [1], so more reduced NAD and reduced FAD are made, more protons are pumped and more ATP is made [1]. (c) The amino group is removed (deamination) [1]. The remaining carbon skeleton enters respiration, for example as pyruvate, acetyl CoA or a Krebs cycle intermediate [1]. [7] Examiner insight: For (b), refer to hydrogen and reduced coenzymes. “Lipids have more energy” only restates the question.

10. (a) RQ = 16 ÷ 23 [1] = 0.70 [1]. (b) RQ = 4 ÷ 3 [1] = 1.33 [1]. (c) Malic acid has more oxygen atoms in the molecule than palmitic acid, so less oxygen from outside is needed to oxidise it [1]. More carbon dioxide is released for each oxygen molecule used, giving a higher ratio [1]. [6] Examiner insight: Show the fraction before the decimal. The substrate with less oxygen demand gives the larger RQ.

11. (a) It absorbs carbon dioxide, so the drop movement shows only the oxygen taken up [1]. (b) Radius = 0.50 mm, so volume = π × 0.50² × 38 [1] = 29.8 mm³ [1]. (c) Net volume without absorber = π × 0.50² × 8 = 6.28 mm³ [1]. Carbon dioxide = 29.845 - 6.283 = 23.6 mm³ [1]. (d) RQ = 23.56 ÷ 29.85 = 0.79 [1]. (e) A value of 0.79 is closer to the value for lipid (about 0.7) than carbohydrate, so the seeds may be respiring mainly lipid, perhaps with some carbohydrate or protein [1]. The conclusion is tentative: a mixture of substrates could give this value, and one run with one tube per condition gives no check on repeatability [1]. [8] Examiner insight: Convert the diameter to a radius first. In (e), give the value, a substrate and one reason for caution.

12. (a) 12.6 ÷ 10 = 1.26 cm³ per min [1]. (b) (6.1 - 12.6) ÷ 12.6 × 100 [1] = -52% [1]. (c) At higher temperatures the enzymes of respiration denature [1], so fewer substrate molecules are respired and less carbon dioxide is released [1]. (d) It shows that the gas comes from living, respiring yeast and not from the solution alone [1]. (e) Measure at more temperatures between 30 °C and 50 °C [1]. Repeat each temperature at least three times and calculate a mean [1]. [8] Examiner insight: Give a sign with a percentage change. Two suggested changes should be different from each other.

Where marks are usually lost

• Quoting 2 ATP for glycolysis without showing 4 made and 2 used.
• Confusing NAD with reduced NAD.
• Explaining anaerobic respiration as a way to make ATP, instead of regenerating NAD so glycolysis continues.
• Using per-acetyl CoA figures for a whole glucose without doubling them.
• Placing oxygen in the Krebs cycle instead of at the end of the chain.
• Describing protons as “pumped through ATP synthase” when the chain pumps them and they flow through ATP synthase.
• Dividing oxygen by carbon dioxide when working out RQ, or putting units on it.
• Using the capillary diameter as the radius.
• Naming a substrate from an RQ with no comment on how tentative that is.

Next steps

• Revise from the revision notes and re-read the study guide.
• Tick off what you know on the OxfordAQA A-level Biology checklist and visit the course hub.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, section 3.3.3 (Respiration), for International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards. Published by OxfordAQA.