IB DP Biology: Cell respiration: ATP, glycolysis, Krebs cycle and the electron transport chain (DP Biology) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover IB Diploma Programme Biology topic C1.2 (Cell respiration), from the IB Biology guide, first assessment 2025. Questions 1 to 6 are SL and HL. Questions 7 to 12 test additional higher level statements (C1.2.7 to C1.2.17) and are labelled “HL only”. Marks are in bold brackets.

Study first with the study guide and revision notes. Course pages: IB DP Biology hub and printable checklist.

Questions

1. State two properties of ATP that make it suitable as the energy currency of cells. [2]

2. Outline three processes in cells that are supplied with energy by ATP. [3]

3. Distinguish between cell respiration and gas exchange. [3]

4. Compare aerobic and anaerobic cell respiration in humans. [4]

5. Germinating seeds of mass 3.0 g were placed in a respirometer containing a carbon dioxide absorbent. A bead in a capillary tube of internal radius 0.60 mm moved 18 mm in 15 minutes at 15 °C, and 39 mm in 15 minutes at 25 °C.

(a) Calculate the volume of oxygen taken up at 15 °C. [2] (b) Calculate the rate of oxygen uptake at 25 °C in mm³ g⁻¹ min⁻¹, to 3 significant figures. [2] (c) Explain why the carbon dioxide absorbent was needed. [2]

6. Yeast was added to glucose solutions of different concentrations and the carbon dioxide released in 5 minutes was measured.

Glucose / %	0	1	2	4	8
CO₂ / cm³	0.5	4.0	7.5	11.0	11.5

(a) Calculate the rate of carbon dioxide release at 4% glucose, with units. [1] (b) Describe the trend shown by the data. [2] (c) Suggest two reasons for the small change between 4% and 8%. [2]

7. (HL only) (a) State what happens to NAD, and to the substrate, when hydrogen is removed from the substrate in respiration. [2] (b) Explain why removing hydrogen from a substrate counts as oxidation. [2]

8. (HL only) (a) Outline the four main events of glycolysis. [4] (b) State the net yield of ATP and reduced NAD from one glucose molecule in glycolysis. [2]

9. (HL only) Blood lactate was measured in an athlete: 1.0 mmol dm⁻³ at rest, 12.0 mmol dm⁻³ immediately after a sprint, and 4.0 mmol dm⁻³ after 10 minutes of recovery.

(a) Calculate the percentage increase in lactate from rest to immediately after the sprint. [2] (b) Explain how the conversion of pyruvate to lactate allows glycolysis to continue. [3]

10. (HL only) (a) State the two products of anaerobic respiration in yeast. [1] (b) Explain why bread dough rises when yeast is active. [2] (c) State which stage of the pathway differs between yeast and humans. [1]

11. (HL only) (a) State how many carbon dioxide molecules are released in the link reaction and Krebs cycle from one glucose molecule. [1] (b) Describe the Krebs cycle, naming the 6C and 4C intermediates. [3] (c) The oxygen supply to a tissue is cut off. Explain why the electron transport chain stops and why the Krebs cycle slows. [6]

12. (HL only) The energy released from carbohydrate is 17 kJ g⁻¹ and from lipid is 37 kJ g⁻¹ (data supplied).

(a) Calculate the mass of each needed to release 4000 kJ, to the nearest gram. [2] (b) Use the chemical composition of the two substrates to account for the difference in energy per gram. [3] (c) Explain why fatty acids cannot be used in anaerobic respiration. [3]

Answers

1. Soluble, so it moves around the cell [1]; hydrolysis releases a small, usable amount of energy [1]; quickly regenerated from ADP and phosphate (other valid properties accepted). [2] Examiner insight: Two distinct properties are needed; saying “it stores energy” is vague and gains less credit than naming a property such as solubility or rapid regeneration.

2. Active transport across membranes using pumps [1]; synthesis of macromolecules such as proteins or DNA (anabolism) [1]; movement of the cell or its parts, for example chromosomes during mitosis [1]. [3] Examiner insight: “Outline” needs three separate processes, each with a clear example; listing three types of active transport counts as one idea.

3. Respiration is a chemical process in cells [1] that releases energy from carbon compounds to make ATP [1]; gas exchange is the physical movement (diffusion) of oxygen and carbon dioxide across a surface [1]. [3] Examiner insight: A “distinguish” answer must state both sides. Writing “breathing” for respiration is a common way to lose the mark.

4. Aerobic needs oxygen, anaerobic does not [1]; aerobic can use carbohydrate and lipid, anaerobic uses carbohydrate only [1]; aerobic gives a much higher ATP yield [1]; aerobic produces carbon dioxide and water, anaerobic produces lactate (other valid points: aerobic uses mitochondria as well as the cytoplasm) [1]. [4] Examiner insight: A “compare” answer needs a point from both processes in the same sentence or row; describing each separately may leave the comparison unstated.

5. (a) Volume = π × 0.60² × 18 [1] = 20.4 mm³ [1] (b) Volume = π × 0.60² × 39 = 44.1 mm³; rate = 44.1 / 15 / 3.0 [1] = 0.980 mm³ g⁻¹ min⁻¹ [1] (c) It absorbs the carbon dioxide produced by respiration [1], so the movement of the bead shows only the oxygen taken up [1]. [6] Examiner insight: Show the formula and the substituted values; an answer with the wrong unit or without “per gram” can lose the final mark even if the working is right.

6. (a) 11.0 / 5 = 2.2 cm³ min⁻¹ [1] (b) The rate rises as glucose concentration rises from 0 to 4% (from 0.1 to 2.2 cm³ min⁻¹) [1]; above 4% it levels off, with only a small rise to 2.3 cm³ min⁻¹ at 8% [1]. (c) The number of yeast cells or enzyme molecules is limiting [1]; enzymes are saturated with substrate, or another factor such as temperature or pH is limiting [1]. [5] Examiner insight: A description of a trend scores best when it quotes values with units from the table; explanations should name the limiting factor, not just say “it stops increasing”.

7. (a) NAD is reduced [1]; the substrate is oxidised [1]. (b) Hydrogen is a proton plus an electron [1], so removing it takes an electron from the substrate, and oxidation is loss of electrons [1]. [4] Examiner insight: Define oxidation as electron loss; “gain of oxygen” is not the definition the course uses for this topic.

8. (a) Phosphorylation: two ATP are used to add phosphate groups to the sugar [1]; lysis: the 6C sugar splits into two 3C molecules [1]; oxidation: hydrogen is removed and NAD is reduced [1]; ATP formation: ATP is made (four in total) and pyruvate is formed [1]. (b) Net yield is 2 ATP [1] and 2 reduced NAD [1]. [6] Examiner insight: “Net” means the ATP used is subtracted; quoting 4 ATP without saying 2 are used is a frequent way to lose the mark.

9. (a) (12.0 − 1.0) / 1.0 × 100 [1] = 1100% [1] (b) Glycolysis needs NAD [1]; without oxygen reduced NAD cannot pass its hydrogen to the electron transport chain [1]; pyruvate takes up the hydrogen and becomes lactate, so NAD is regenerated and glycolysis continues [1]. [5] Examiner insight: For percentage change, the denominator is the starting figure, not the final one; the explanation should link NAD regeneration to the continuation of glycolysis, not just to “getting rid of pyruvate”.

10. (a) Ethanol and carbon dioxide [1]. (b) Fermentation by the yeast releases carbon dioxide [1]; bubbles of the gas are trapped and swell the dough [1]. (c) The conversion of pyruvate, which regenerates NAD [1]. [4] Examiner insight: Name both products. Naming only ethanol, or giving lactate, does not score.

11. (a) 6 [1] (b) An acetyl group (2C) joins oxaloacetate (4C) to form citrate (6C) [1]; there are four oxidations (dehydrogenations) and two decarboxylations, releasing carbon dioxide and forming ATP and reduced NAD [1]; oxaloacetate is regenerated so the cycle continues [1]. (c) Oxygen is the terminal electron acceptor, so without it electrons cannot leave the last carrier [1]; the carriers stay reduced and electron flow stops [1]; no more protons are pumped, the gradient falls and ATP synthase stops making ATP [1]; reduced NAD cannot be oxidised, so NAD is not regenerated [1]; the link reaction and Krebs cycle need NAD for their oxidations [1]; so they slow as reduced NAD builds up [1]. [10] Examiner insight: A long explanation question rewards a chain of cause and effect; name the terminal acceptor, the stalled carrier flow and the lack of NAD, rather than saying the cycle “needs oxygen”.

12. (a) Carbohydrate: 4000 / 17 = 235 g [1]; lipid: 4000 / 37 = 108 g [1] (b) Lipids contain less oxygen per gram [1] and more hydrogen and carbon that can be oxidised [1]; more oxidation gives more energy transferred to ATP production [1]. (c) Anaerobic respiration is glycolysis followed by conversion of pyruvate [1]; fatty acids are broken into 2C acetyl groups that enter as acetyl-CoA, not through glycolysis [1]; they can enter only the Krebs cycle, which depends on oxygen through the chain [1]. [8] Examiner insight: Use the correct rounding for each value stated in the question; compare “per gram” explicitly instead of saying fat is “bigger” or “denser”.

Where marks are usually lost

• Writing breathing or gas exchange when the question asks about respiration.
• Giving energy values or ATP counts without stating “net” or per glucose.
• Dividing by the final value rather than the starting value in percentage change.
• Omitting units, or “per gram”, from a rate.
• Describing a graph or table without quoting figures.
• Mixing up the anaerobic products of humans and yeast.
• Placing oxygen use in the Krebs cycle rather than at the end of the chain as terminal electron acceptor.
• Describing reduced NAD as “making” ATP directly, with no mention of the proton gradient.
• Saying fats cannot be respired anaerobically without explaining that they bypass glycolysis.

Next steps

• Revise with the revision notes and re-read the study guide.
• Open the IB DP Biology hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Topic C1.2 (Cell respiration), statements C1.2.1 to C1.2.17; C1.2.7 to C1.2.17 are additional higher level.