Cambridge International AS & A Level Biology 9700: Energy and respiration (by subtopic) – 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 work through syllabus 9700 subtopics 12.1 (Energy) and 12.2 (Respiration), learning outcome by outcome. This is A Level content, examined in Paper 4; the practical questions also suit Paper 5. Calculators are allowed.

They do not repeat the topic-level respiration practice set; revise first with the revision notes.

Questions

1. (12.1.1–3)

(a) A root hair cell takes up nitrate ions and a root-tip cell replicates its DNA. Outline why each needs energy. [2] (b) Describe two features of ATP that make it suitable as the universal energy currency. [2] (c) Name the two ways ATP is synthesised, with one site of each in a plant cell. [2]

2. (12.1.4) C = 12, H = 1, O = 16. Glucose is C₆H₁₂O₆; palmitic acid is C₁₆H₃₂O₂.

(a) Calculate the number of moles of hydrogen atoms per gram of each substance. [2] (b) Use your answers to explain why lipids release about twice as much energy per gram as carbohydrates. [3] (c) Suggest why proteins release slightly more energy per gram than carbohydrates. [1]

3. (12.1.5–6)

(a) State what is meant by the respiratory quotient (RQ). [1] (b) The triglyceride triolein is respired aerobically: C₅₇H₁₀₄O₆ + x O₂ → 57CO₂ + 52H₂O. Calculate x and the RQ. [3] (c) The amino acid leucine is respired: 2C₆H₁₃NO₂ + 15O₂ → 12CO₂ + 10H₂O + 2NH₃. Calculate the RQ. [1]

4. (12.1.7) A respirometer holds 1.25 g of blowfly larvae; its capillary bore diameter is 0.80 mm. A control tube holds glass beads. In 10 minutes the liquid moves 33 mm towards the larvae with potassium hydroxide in the tube, and 3 mm with water instead.

(a) Explain why the liquid moves towards the larvae when potassium hydroxide is present. [2] (b) Calculate the rate of oxygen uptake in mm³ g⁻¹ min⁻¹. [3] (c) Calculate the RQ of the larvae. [2] (d) State the purpose of the control tube. [1]

5. (12.2.1–4)

(a) Outline glycolysis, naming the 6C and 3C intermediates. [4] (b) When oxygen is available, pyruvate enters the mitochondrion. Describe the link reaction, including the role of coenzyme A. [3] (c) State where each of the four stages of aerobic respiration occurs. [2]

6. (12.2.5–6) A simplified Krebs cycle has five steps.

• Step A: acetyl coenzyme A + oxaloacetate (4C) → citrate (6C) + coenzyme A
• Step B: citrate (6C) → 5C compound + CO₂; NAD is reduced
• Step C: 5C compound → 4C compound + CO₂; NAD is reduced; ATP is made
• Step D: 4C compound → different 4C compound; FAD is reduced
• Step E: 4C compound → oxaloacetate; NAD is reduced

(a) Identify the steps that involve (i) decarboxylation, (ii) dehydrogenation and (iii) substrate-linked phosphorylation. [3] (b) Calculate the CO₂ and reduced NAD made by the Krebs cycle per glucose. [2] (c) Explain why oxaloacetate is described as an acceptor. [1]

7. (12.2.7–9)

(a) Mitochondria in insect flight muscle have many closely packed cristae. Explain the advantage. [2] (b) The pH of the intermembrane space is lower than the pH of the matrix. Explain why. [2] (c) A cell contains only a small amount of NAD. Explain why respiration can continue for long periods. [2] (d) Explain the roles of ATP synthase and oxygen in oxidative phosphorylation. [3]

8. (12.2.10–11)

(a) Outline lactate fermentation in mammalian muscle. [2] (b) State two differences between lactate fermentation and ethanol fermentation. [2] (c) Two flasks held equal yeast suspensions with 5.0 g of glucose, one bubbled with air and one with nitrogen. After 24 hours yeast dry mass had increased by 2.0 g in air but only 0.40 g in nitrogen. Explain the difference. [3]

9. (12.2.12) Rice is grown in paddy fields where the roots are under water.

(a) Explain how aerenchyma helps rice roots. [2] (b) Explain how ethanol fermentation in the roots helps the plant survive. [2] (c) Explain the advantage of faster stem growth when the water level rises. [1]

10. (12.2.14) Germinating peas were kept in a respirometer with potassium hydroxide (bore radius 0.50 mm) at four temperatures.

Temperature / °C	10	20	30	40
Distance moved in 5 min / mm	6	13	25	21

(a) Calculate the rate of oxygen uptake at 30 °C in mm³ min⁻¹. [2] (b) Explain the change in rate between 10 °C and 30 °C. [2] (c) Suggest why the rate is lower at 40 °C. [1]

11. (12.2.13) Each tube held 2 cm³ of one yeast suspension, 2 cm³ of glucose solution and 1 cm³ of methylene blue at 30 °C.

Glucose concentration / %	0	1	2	4	8
Time to decolourise / s	no change after 1200 s	480	260	150	140

(a) Explain why methylene blue loses its colour. [2] (b) Calculate the rate, as 1000/t, at 2% glucose. [1] (c) Describe and explain the trend from 1% to 8%. [3] (d) Explain why the tubes must not be shaken. [1] (e) Describe how to adapt the method, using DCPIP, to find the effect of temperature. [2]

Answers

1. (a) Nitrate uptake is active transport against a concentration gradient by carrier proteins [1]. DNA replication is an anabolic reaction: energy joins nucleotides into a new strand [1]. (b) Any two: hydrolysis is one quick step; energy released in small amounts; small and soluble; quickly regenerated [2]. (c) Substrate-linked phosphorylation, in the cytoplasm (glycolysis) [1]. Chemiosmosis, on the inner mitochondrial membrane or thylakoid membranes [1]. Examiner insight: In (a), name the type of work, not just “needs energy”.

2. (a) Glucose: 12 ÷ 180 = 0.067 mol g⁻¹ [1]. Palmitic acid: 32 ÷ 256 = 0.125 mol g⁻¹ [1]. (b) Most ATP comes from oxidative phosphorylation, using hydrogen carried by reduced NAD and FAD [1]. Palmitic acid has about 1.9 times as much hydrogen per gram [1], so more hydrogen passes through the chain, giving more ATP per gram [1]. (c) Proteins contain more hydrogen atoms per gram than carbohydrates [1]. Examiner insight: “More energy in their bonds” scores nothing; link hydrogen to the electron transport chain.

3. (a) The ratio of the number of molecules of carbon dioxide produced to the number of molecules of oxygen taken in during respiration [1]. (b) Oxygen atoms on the right = 114 + 52 = 166, so x = (166 − 6) ÷ 2 = 80 [1]. RQ = 57 ÷ 80 [1] = 0.71 [1]. (c) RQ = 12 ÷ 15 = 0.80 [1]. Examiner insight: Dividing O₂ by CO₂ loses both RQ marks, though x = 80 still scores.

4. (a) The larvae take in oxygen for respiration [1]. The carbon dioxide released is absorbed by the potassium hydroxide, so the gas volume in the tube falls [1]. (b) Radius = 0.40 mm; volume = π × 0.40² × 33 = 16.6 mm³ [1]. Per minute: 16.6 ÷ 10 = 1.66 mm³ min⁻¹ [1]. Per gram: 1.66 ÷ 1.25 = 1.33 mm³ g⁻¹ min⁻¹ [1]. (c) CO₂ released = 33 − 3 = 30 mm of movement [1]. RQ = 30 ÷ 33 = 0.91 [1]. (d) To compensate for changes in temperature or atmospheric pressure [1]. Examiner insight: Using 0.80 mm as the radius gives a volume four times too big: method marks only.

5. (a) Glucose is phosphorylated using two ATP [1], forming fructose 1,6-bisphosphate (6C) [1]. This splits into two triose phosphate (3C) molecules [1]. Each is oxidised to pyruvate (3C), reducing NAD and giving a net gain of two ATP [1]. (b) Pyruvate is decarboxylated: CO₂ is removed [1]. It is dehydrogenated: hydrogen reduces NAD [1]. Coenzyme A carries the acetyl (2C) group to the Krebs cycle [1]. (c) Glycolysis: cytoplasm; oxidative phosphorylation: inner mitochondrial membrane [1]. Link reaction and Krebs cycle: mitochondrial matrix [1]. Examiner insight: Calling coenzyme A an enzyme can cost the third mark.

6. (a) (i) B and C [1]; (ii) B, C, D and E [1]; (iii) C [1]. (b) Two turns per glucose [1]: 4 CO₂ and 6 reduced NAD [1]. (c) It accepts the 2C acetyl group to form citrate and is regenerated each turn [1]. Examiner insight: (ii) needs all four steps: step D counts because FAD accepts hydrogen.

7. (a) Cristae give a large inner-membrane surface area for electron transport chains and ATP synthase [1], so more ATP is made for muscle contraction [1]. (b) Energy released by electrons in the chain transfers protons from the matrix into the intermembrane space [1]. The inner membrane is impermeable to protons except through ATP synthase, so the gradient is kept [1]. (c) NAD is reduced when it accepts hydrogen in dehydrogenation [1]. It is oxidised when it passes hydrogen to carriers in the inner membrane, so it is regenerated and reused [1]. (d) Hydrogen atoms split into protons and energetic electrons [1]. Protons return to the matrix by facilitated diffusion through ATP synthase, providing energy for ATP synthesis [1]. Oxygen is the final electron acceptor, forming water [1]. Examiner insight: In (b), the second mark needs the impermeable inner membrane, not just “protons are pumped”.

8. (a) Pyruvate is reduced to lactate by reduced NAD [1], so NAD is regenerated and glycolysis continues [1]. (b) Only ethanol fermentation releases CO₂ [1] and has two steps (via ethanal) [1]. (c) In air, reduced NAD and FAD from the link reaction and Krebs cycle drive oxidative phosphorylation [1]. In nitrogen, ATP comes only from glycolysis [1]. Much energy stays in ethanol, so less ATP is available for growth [1]. Examiner insight: No ATP total is expected; credit goes to the reasons.

9. (a) Aerenchyma has large air spaces [1], so oxygen diffuses from the shoots to the root cells for aerobic respiration [1]. (b) Root cells can carry out ethanol fermentation, regenerating NAD so glycolysis still makes ATP without oxygen [1]. They tolerate higher ethanol concentrations than most plants [1]. (c) Stems keep the leaves above the water, so gas exchange (and photosynthesis) continues [1]. Examiner insight: Other adaptations earn nothing.

10. (a) Volume = π × 0.50² × 25 = 19.6 mm³ [1]; rate = 19.6 ÷ 5 = 3.9 mm³ min⁻¹ [1]. (b) Respiratory enzymes and substrates have more kinetic energy [1], so successful collisions forming enzyme–substrate complexes are more frequent [1]. (c) Respiratory enzymes start to denature (active site changes shape) [1]. Examiner insight: “Molecules move faster” alone earns one mark at most in (b).

11. (a) Dehydrogenase enzymes remove hydrogen from glucose (respiratory substrates) [1]. The hydrogen reduces methylene blue, which is colourless when reduced [1]. (b) 1000 ÷ 260 = 3.8 s⁻¹ [1]. (c) Rate increases from 2.1 to 6.7 s⁻¹ between 1% and 4%, then levels off (7.1 s⁻¹ at 8%) [1]. More glucose means more enzyme–substrate complexes form per unit time [1]. Above 4%, enzyme concentration (number of yeast cells) becomes the limiting factor [1]. (d) Oxygen from the air would re-oxidise the methylene blue, turning it blue again [1]. (e) Keep glucose concentration and volumes constant; use water baths at a range of temperatures [1]. Time DCPIP turning from blue to colourless; rate = 1000/t [1]. Examiner insight: In (c), quote rates with units to earn the description mark.

Where marks are usually lost

• Using the bore diameter as the radius in πr².
• Saying lipids hold “more energy”, not more hydrogen per gram.
• Forgetting FAD, or the two turns per glucose.
• Calling coenzyme A an enzyme.
• Saying fermentation makes ATP; it regenerates NAD.
• Shaking methylene blue tubes.

Next steps

• Energy and respiration revision notes
• Energy and respiration study guide
• Cambridge A Level Biology hub and printable Biology checklist
• A Level Biology 9700 diagnostic
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, published by Cambridge Assessment International Education (Cambridge University Press & Assessment). Topic 12, Energy and respiration, subtopics 12.1 and 12.2.