Pearson Edexcel International A Level Biology: Aerobic and Anaerobic Respiration (YBI11) – 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 statements 7.1 to 7.8 (aerobic and anaerobic respiration, including Core Practicals 15 and 16) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topic 7 is International A2 (IA2) and there are no tiers. Learn the content in the study guide and revision notes first. The course hub, checklist and free diagnostics show the wider course.

Questions

1. Explain what happens to the hydrogen removed from a respiratory substrate in aerobic respiration, and why it is an advantage that respiration takes place in many small steps. [4]

2. A yeast culture phosphorylates 12 hexose molecules in glycolysis. Calculate the number of molecules of ATP used, ATP made by substrate level phosphorylation, net ATP, pyruvate and reduced NAD. [5]

3. Explain why a muscle cell can still make a little ATP when oxygen is short, and why lactate is formed. [4]

4. (a) State where the link reaction and the Krebs cycle take place in a cell. [1] (b) For one molecule of glucose, state the number of carbon dioxide molecules released in the link reaction, the number released in the Krebs cycle, the number of reduced NAD made in the Krebs cycle, and the number of reduced FAD made. [4]

5. Explain why the Krebs cycle stops in a cell deprived of oxygen even though oxygen is not a reactant in it. [4]

6. Describe how ATP is made by oxidative phosphorylation. [6]

7. (a) Calculate the total ATP made from one glucose molecule in complete aerobic respiration, given 2.5 ATP per reduced NAD and 1.5 ATP per reduced FAD passing through the electron transport chain. [3] (b) Hydrolysis of one mole of ATP releases 30.5 kJ and complete oxidation of one mole of glucose releases 2870 kJ. Calculate the percentage of the energy captured as ATP. [2]

8. Isolated mitochondria are placed in a buffer containing a respiratory substrate, with an oxygen electrode. ADP is added after 1 minute and cyanide, which blocks the transfer of electrons to oxygen at the end of the chain, after 4 minutes.

Time (min)	0	1	2	3	4	5	6
Oxygen (µmol dm⁻³)	250	248	224	200	176	175	175

(a) Calculate the rate of oxygen uptake between 1 and 4 minutes. [2] (b) Compare the rate before ADP was added with the rate after, and explain the difference. [2] (c) Explain the effect of cyanide. [2]

9. A runner’s blood lactate concentration (mmol dm⁻³) was 1.1 at rest, 12.4 at the end of a sprint, 6.2 after 10 minutes of recovery, 2.0 after 30 minutes and 1.2 after 60 minutes. (a) Calculate the percentage decrease in lactate between the end of the sprint and 10 minutes. [2] (b) Calculate the mean rate of decrease between 10 and 30 minutes. [2] (c) Explain the decrease in blood lactate during recovery. [2]

10. (a) Stearic acid is respired: C₁₈H₃₆O₂ + 26O₂ → 18CO₂ + 18H₂O. Calculate its RQ. [2] (b) A tissue has an RQ of 0.85. Suggest what this indicates about the substrate. [1] (c) A yeast culture produces 28 mm³ of carbon dioxide while taking up 20 mm³ of oxygen. Calculate the RQ and explain what it suggests. [2]

11. In Core Practical 15, methylene blue was added to yeast suspensions containing different glucose concentrations at the same temperature. The time for the blue colour to disappear was recorded.

Glucose (g per 100 cm³)	0	2	4	6	8
Time (s)	540	300	195	150	148

(a) Calculate the rate (1000 ÷ time) at 4 and 8 g per 100 cm³. [2] (b) Explain the increase in rate from 0 to 6 g per 100 cm³. [2] (c) Suggest why the rate hardly changes between 6 and 8. [1] (d) State why a tube containing boiled yeast is set up. [1] (e) Explain why methylene blue loses its colour. [2]

12. Woodlice of mass 2.4 g were placed in a simple respirometer with a capillary of radius 0.40 mm. In 15 minutes the coloured drop moved 18 mm with potassium hydroxide in the tube and 3 mm without it. (a) Explain why the drop moved towards the tube when potassium hydroxide was present. [2] (b) Calculate the volume of oxygen used. [2] (c) Calculate the rate of oxygen use in mm³ g⁻¹ min⁻¹. [1] (d) Calculate the RQ of the woodlice. [3] (e) Suggest one control used in the experiment. [1]

Answers

1. Hydrogen is removed from the substrate by dehydrogenation and accepted by NAD and FAD [1]. It is passed on, and the last carrier in the electron transport chain passes it to oxygen, forming water [1]. Each step is catalysed by a specific enzyme [1]. Energy is released gradually, so more can be captured in ATP and less is lost as heat [1]. [4] Examiner insight: Name the carrier (NAD or FAD) and the final product (water). “Hydrogen is used in respiration” earns nothing without a destination.

2. ATP used = 12 × 2 = 24 [1]. ATP made = 12 × 4 = 48 [1]. Net = 48 − 24 = 24 [1]. Pyruvate = 12 × 2 = 24 [1]. Reduced NAD = 12 × 2 = 24 [1]. [5] Examiner insight: Show each product on its own line. A final answer of “24 ATP” is ambiguous unless you say whether it is net.

3. Glycolysis does not need oxygen [1]. It gives a net gain of 2 ATP per glucose by substrate level phosphorylation [1]. It needs NAD to continue [1]. Pyruvate accepts hydrogen from reduced NAD and becomes lactate, which regenerates NAD [1]. [4] Examiner insight: Lactate formation is explained by NAD regeneration, not by “getting rid of pyruvate” or “making ATP”.

4. (a) The mitochondrial matrix [1]. [1] (b) Link reaction CO₂: 2 [1]. Krebs cycle CO₂: 4 [1]. Reduced NAD from Krebs: 6 [1]. Reduced FAD: 2 [1]. [4] Examiner insight: Each of these stages happens twice for every glucose. Check you have doubled every figure that is given per turn or per pyruvate.

5. Oxygen is the final electron acceptor in the electron transport chain [1]. Without oxygen the chain stops [1]. Reduced NAD and reduced FAD are not reoxidised [1]. The Krebs cycle (and link reaction) needs NAD and FAD to accept hydrogen, so it stops [1]. [4] Examiner insight: The chain is the missing link. Answers that jump from “no oxygen” to “no Krebs cycle” leave out the step that carries marks.

6. Reduced NAD and reduced FAD give up hydrogen (protons and electrons) at the inner mitochondrial membrane [1]. Electrons pass along carriers and release energy [1]. This energy pumps H⁺ from the matrix into the intermembrane space, building a gradient [1]. H⁺ flows back into the matrix through ATP synthase, down the gradient (chemiosmosis) [1]. The flow drives ADP + Pi → ATP [1]. The final acceptor of the electrons is oxygen, which with H⁺ forms water [1]. [6] Examiner insight: Give the direction of pumping and the membrane. Writing “protons move across a membrane” without direction is vague.

7. (a) Reduced NAD = 10 and reduced FAD = 2, with 4 ATP by substrate level phosphorylation [1]. 10 × 2.5 = 25 and 2 × 1.5 = 3 [1]. 25 + 3 + 4 = 32 ATP [1]. [3] (b) Energy captured = 32 × 30.5 = 976 kJ [1]. 976 ÷ 2870 × 100 = 34% [1]. [2] Examiner insight: Do not forget the 4 ATP made without the chain. Give the percentage to a sensible precision, such as 34.0 or 34.

8. (a) (248 − 176) ÷ 3 [1] = 24 µmol dm⁻³ min⁻¹ [1]. [2] (b) Before ADP the rate is 2 µmol dm⁻³ min⁻¹ and after it is 24 [1]. ADP is needed by ATP synthase, so H⁺ can flow back through it; this uses up the gradient and lets the chain keep passing electrons to oxygen [1]. [2] (c) Cyanide blocks electron transfer to oxygen [1], so the chain stops and oxygen uptake falls to almost zero [1]. [2] Examiner insight: In a data question quote values from the table with units. Linking the ADP effect to ATP synthase and the gradient is better than “ADP speeds up respiration”.

9. (a) (12.4 − 6.2) ÷ 12.4 × 100 [1] = 50% [1]. [2] (b) (6.2 − 2.0) ÷ 20 [1] = 0.21 mmol dm⁻³ min⁻¹ [1]. [2] (c) Lactate leaves the muscle in the blood and is carried to the liver [1]. There it is converted to pyruvate, which is oxidised or converted to glycogen, using oxygen [1]. [2] Examiner insight: Percentage change is divided by the starting value. Name the organ and the conversion in (c).

10. (a) RQ = CO₂ ÷ O₂ = 18 ÷ 26 [1] = 0.69 [1]. [2] (b) A mixture of respiratory substrates such as carbohydrate and lipid (or protein being used) [1]. [1] (c) 28 ÷ 20 = 1.4 [1]. A value above 1 suggests some anaerobic respiration is happening, because carbon dioxide is released without oxygen being used [1]. [2] Examiner insight: Write RQ as a ratio with no units. Link a value above 1 to a specific process.

11. (a) 1000 ÷ 195 = 5.13 [1] and 1000 ÷ 148 = 6.76 [1]. [2] (b) Higher glucose concentration gives more substrate for glycolysis [1]. More hydrogen is released per second to reduce the indicator, so it decolourises faster [1]. [2] (c) Another factor, such as the number of enzyme molecules or glucose carriers in the yeast cells, is limiting [1]. [1] (d) To show that the colour change needs living yeast, not the glucose or the indicator alone [1]. [1] (e) Methylene blue accepts hydrogen released by dehydrogenation [1]. In its reduced form it is colourless [1]. [2] Examiner insight: A plateau explanation needs a named limiting factor; “the reaction has finished” does not describe the evidence.

12. (a) Oxygen is used by the woodlice [1]. The potassium hydroxide takes up all the carbon dioxide released, so only the oxygen loss changes the gas volume, pressure falls and the drop moves towards the tube [1]. [2] (b) π × 0.40² × 18 [1] = 9.05 mm³ [1]. [2] (c) 9.05 ÷ 15 ÷ 2.4 = 0.251 mm³ g⁻¹ min⁻¹ [1]. [1] (d) Net volume change = π × 0.40² × 3 = 1.51 mm³ [1]. Carbon dioxide = 9.05 − 1.51 = 7.54 mm³ [1]. RQ = 7.54 ÷ 9.05 = 0.83 [1]. [3] (e) A control tube with inert material (such as glass beads) of the same volume, or a water bath at constant temperature [1]. [1] Examiner insight: In (d) the no-absorber movement is oxygen used minus carbon dioxide produced. Subtracting it from the oxygen volume gives the carbon dioxide.

Where marks are usually lost

• Not saying that glycolysis makes 4 ATP and uses 2.
• Counting one turn of the link reaction and Krebs cycle for each glucose instead of two.
• Describing oxygen as used in the Krebs cycle.
• Leaving out the direction of proton pumping and the role of ATP synthase.
• Stating that lactate is made from glucose rather than pyruvate.
• Dividing oxygen by carbon dioxide for RQ.
• Using the respirometer reading without absorber as the oxygen used.
• Not giving a limiting factor to explain a plateau.

Next steps

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Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.