These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Model values are given in each question where you need them; pH = −log[H⁺]. A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data. All data sets and named compounds are fictional.
Question 1 (multiple choice · foundation)
Which row correctly matches a stage of aerobic respiration in a eukaryotic cell to where it happens and one of its products?
- (A) Glycolysis — mitochondrial matrix — pyruvate
- (B) Krebs cycle — mitochondrial matrix — CO₂
- (C) Electron transport chain — outer mitochondrial membrane — water
- (D) Krebs cycle — cytosol — FADH₂
Answer and explanation
Answer: (B). The Krebs cycle runs in the matrix and releases carbon from organic molecules as CO₂ (it also makes NADH, FADH₂ and a little ATP).
- (A) has the right product but the wrong place: glycolysis is in the cytosol, which is why it works in prokaryotes and without mitochondria.
- (C) names a real product (water) but the wrong membrane: the ETC and ATP synthase are in the folded inner membrane.
- (D) names a real Krebs product but puts the cycle in the cytosol; only glycolysis happens there.
Question 2 (multiple choice · core)
Yeast cells are moved from an oxygen-rich flask into a sealed flask with no oxygen. They keep growing slowly and start producing ethanol. Which statement best explains why producing ethanol lets glycolysis continue?
- (A) Converting pyruvate to ethanol makes extra ATP, replacing the ATP lost from the ETC.
- (B) Converting pyruvate to ethanol oxidises NADH back to NAD⁺, which glycolysis needs to keep running.
- (C) Ethanol replaces oxygen as the final electron acceptor at the end of the ETC.
- (D) Ethanol production releases oxygen, which keeps the Krebs cycle running.
Answer and explanation
Answer: (B). Glycolysis reduces NAD⁺ to NADH. Without oxygen, the ETC cannot accept electrons from NADH, so NAD⁺ would run out and glycolysis would stop. Fermentation passes the electrons from NADH to a molecule made from pyruvate, forming ethanol and CO₂ and freeing NAD⁺.
- (A) is the most common error: fermentation itself makes no ATP. The cell’s only ATP is the 2 per glucose from glycolysis.
- (C) describes anaerobic respiration with an ETC. Fermentation does not use an ETC, and ethanol is a product, not an acceptor.
- (D) is wrong on two counts: fermentation releases CO₂, not O₂, and the Krebs cycle is not running in these conditions.
Question 3 (multiple choice · core)
A fictional compound, Z, blocks the proton channel of ATP synthase in isolated mitochondria that are supplied with pyruvate and oxygen. Which outcome is most likely within a few minutes?
- (A) The proton gradient disappears and oxygen consumption increases.
- (B) ATP synthesis stops, but electron transport and oxygen consumption continue at the same rate as before.
- (C) Protons build up further in the intermembrane space, electron transport slows, and oxygen consumption falls.
- (D) Protons build up in the matrix, lowering its pH, and ATP synthesis increases.
Answer and explanation
Answer: (C). With ATP synthase blocked, protons can no longer flow back into the matrix. The ETC keeps pumping until the gradient is so steep that pumping becomes too hard. Electron transport then slows, so less oxygen is reduced to water. Electron transport and ATP synthesis are coupled.
- (A) describes an uncoupler, which lets protons leak back by another route. Compound Z blocks the route instead.
- (B) ignores coupling: the ETC cannot keep pumping against an ever-steeper gradient.
- (D) puts the protons on the wrong side. They are pumped into the intermembrane space, and with no ATP synthase flow, ATP synthesis falls, not rises.
Question 4 (graph construction · core)
A student mixes yeast with sugar solution and collects the CO₂ released in 20 minutes at five temperatures. Each temperature was tested three times (fictional data):
| Temperature / °C | 15 | 25 | 35 | 45 | 55 |
|---|---|---|---|---|---|
| Trial 1 / mL | 1.6 | 3.9 | 7.2 | 4.6 | 0.4 |
| Trial 2 / mL | 1.8 | 4.1 | 7.6 | 5.2 | 0.6 |
| Trial 3 / mL | 2.0 | 4.3 | 8.0 | 5.8 | 0.8 |
(a) Calculate the mean volume of CO₂ at each temperature. (b) Construct an appropriate graph of the means, showing the spread of the trials. (c) Describe the pattern, using numbers, and explain the result at 55 °C.
Worked solution and suggested Marlbridge rubric
(a) Means: 15 °C: 1.8 mL; 25 °C: 4.1 mL; 35 °C: 7.6 mL; 45 °C: 5.2 mL; 55 °C: 0.6 mL. (Each is the sum ÷ 3, e.g. (7.2 + 7.6 + 8.0) ÷ 3 = 22.8 ÷ 3 = 7.6.)
(b) Temperature is a continuous independent variable, so plot the means as points on an x–y graph (temperature on x, mean CO₂ volume on y) and join them point to point. Do not draw a single straight best-fit line: the relationship rises then falls. Show the spread with bars from the lowest to the highest trial (range bars), and say so in a key or caption.
(c) CO₂ release rises from 1.8 mL at 15 °C to a peak of 7.6 mL at 35 °C (about 4.2 times higher), then falls to 5.2 mL at 45 °C and 0.6 mL at 55 °C, a fall of about 92% from the peak. Up to about 35 °C, faster molecular movement means more enzyme–substrate collisions in the respiration pathways. At 55 °C, the yeast’s enzymes have denatured: their active sites have changed shape, so the pathway (and CO₂ release) almost stops. The range bars at 35 °C and 45 °C do not overlap, so the drop between them is unlikely to be chance alone.
| Point | What earns it |
|---|---|
| 1 | All five means correct |
| 1 | Correct graph type and orientation: temperature on x, mean CO₂ on y, points joined (not bars, not one straight line) |
| 1 | Both axes labelled with units, even scales, points accurately plotted |
| 1 | Range (or error) bars drawn and identified in a key or caption |
| 1 | Pattern described with numbers and 55 °C result explained by enzyme denaturation |
A bar chart loses the graph-type point, because temperature is continuous. Accept standard-deviation error bars if they are calculated correctly and labelled.
Question 5 (constructed response · core)
A fictional aerobic bacterium is kept in a weakly buffered medium at pH 7.4. Its cytoplasm is at pH 7.6. When glucose and oxygen are added, the pH of the medium falls to 6.6 within minutes, while the cytoplasm stays at pH 7.6.
(a) Explain why the medium becomes more acidic. (b) Calculate how many times higher the H⁺ concentration is outside the cell than inside after glucose is added. (c) State where ATP synthase is located in this cell, and in which direction protons flow through it. (d) The same species can grow without oxygen if nitrate is present, but not if both oxygen and nitrate are absent and no fermentation pathway is available. Explain these observations.
Model answer and suggested Marlbridge rubric
(a) The bacterium respires glucose, and NADH and FADH₂ deliver electrons to the ETC in the plasma membrane. As electrons pass along the chain to oxygen, the energy released pumps protons out of the cell into the medium. More H⁺ in the medium means a lower pH.
(b) Difference = 7.6 − 6.6 = 1.0 pH unit. [H⁺]outside ÷ [H⁺]inside = 10^1.0 = 10 times higher outside.
(c) ATP synthase is in the plasma membrane (prokaryotes have no mitochondria). Protons flow into the cytoplasm through it, down their concentration gradient, driving ADP + Pᵢ → ATP by chemiosmosis.
(d) With nitrate, the bacterium can still run its ETC, using nitrate instead of oxygen as the final electron acceptor (anaerobic respiration). So protons are still pumped and ATP is still made by chemiosmosis. With neither acceptor, electrons cannot leave the chain: the ETC stops, NADH cannot be reoxidised to NAD⁺, and with no fermentation to regenerate NAD⁺, glycolysis also stops, so no ATP is made.
| Point | What earns it |
|---|---|
| 1 | Protons pumped out across the plasma membrane by electron transport |
| 1 | Ratio of 10 with working (one pH unit = tenfold) |
| 1 | ATP synthase in plasma membrane and H⁺ flowing inward through it |
| 1 | Nitrate acts as an alternative final electron acceptor so the ETC continues |
| 1 | No acceptor → ETC stops → NAD⁺ not regenerated → glycolysis stops |
Do not award the third point for “in the mitochondria”: the organism is a prokaryote.
Question 6 (constructed response · stretch)
Mitochondria are isolated from the brown fat of two groups of a fictional small mammal: one group kept at 25 °C, the other kept at 5 °C for three weeks. Both samples are given the same fuel and oxygen (rates in nmol min⁻¹ per mg protein):
| Group | O₂ consumption | ATP synthesis |
|---|---|---|
| Kept at 25 °C | 30 | 150 |
| Kept at 5 °C | 75 | 60 |
(a) Calculate the ATP made per O₂ consumed for each group, and the percentage change in each rate. (b) The cold-kept animals make far more of an uncoupling protein in their brown fat. Explain how this accounts for both changes in the table. (c) Explain how this response helps the animal survive in the cold.
Model answer and suggested Marlbridge rubric
(a) 25 °C group: 150 ÷ 30 = 5.0 ATP per O₂. 5 °C group: 60 ÷ 75 = 0.8 ATP per O₂. O₂ consumption: (75 − 30) ÷ 30 × 100 = +150%. ATP synthesis: (60 − 150) ÷ 150 × 100 = −60%.
(b) The uncoupling protein lets protons flow from the intermembrane space back into the matrix without passing through ATP synthase. So less of the gradient is used for ATP synthesis, and ATP output falls. Because the gradient is constantly drained, the ETC is no longer held back by a steep gradient; electrons flow faster, so more oxygen is reduced to water and O₂ consumption rises. Electron transport is now decoupled from ATP synthesis.
(c) The energy released by electron transport that is not captured in ATP is released as heat. Brown fat therefore acts as a heater, helping the endothermic animal keep its body temperature steady in the cold.
| Point | What earns it |
|---|---|
| 1 | Both ratios (5.0 and 0.8) and both percentage changes correct |
| 1 | Protons bypass ATP synthase → less ATP |
| 1 | Gradient drained → ETC runs faster → more O₂ used |
| 1 | Energy released as heat, linked to keeping body temperature up |
Do not award the third point for “the uncoupling protein uses oxygen”: oxygen is still used only at the end of the ETC.
Question 7 (calculation · stretch)
Use these model values: complete oxidation of 1 mol of glucose releases 2870 kJ; making 1 mol of ATP from ADP and Pᵢ stores about 30.5 kJ; aerobic respiration makes about 30 ATP per glucose; fermentation makes 2 ATP per glucose.
(a) Calculate the percentage of glucose’s energy stored in ATP by aerobic respiration and by fermentation. (b) A cell needs 300 µmol of ATP. How much glucose must it use by each route? (c) For aerobic respiration, state what happens to the energy that is not stored in ATP. For fermentation, explain why the figure in (a) is so low.
Worked solution
(a) Aerobic: 30 × 30.5 = 915 kJ stored. 915 ÷ 2870 × 100 = 31.9% (about 32%). Fermentation: 2 × 30.5 = 61 kJ stored. 61 ÷ 2870 × 100 = 2.1%.
(b) Aerobic: 300 ÷ 30 = 10 µmol glucose. Fermentation: 300 ÷ 2 = 150 µmol glucose, 15 times as much.
(c) Aerobic: the other 1955 kJ per mole (2870 − 915) is released as heat, which endotherms use to help keep warm. Fermentation: glucose is only partly broken down. Most of its energy is still held in the products, lactic acid or ethanol, which have not been oxidised to CO₂ and water.
Suggested mark points (4): 1 for 31.9%; 1 for 2.1%; 1 for 10 and 150 µmol; 1 for heat (aerobic) and energy remaining in lactic acid or ethanol (fermentation).
Common errors: using 38 ATP from an older textbook gives 40.4%, which is not what this question’s data say; dividing 2870 by 30.5 (≈ 94) answers a different question (the maximum ATP if all the energy were captured).
How did you do?
- Q1 wrong: re-read “The four stages and where they happen” and Figure 1 in the study guide.
- Q2 or Q7 wrong: re-read “Fermentation: glycolysis without oxygen”.
- Q3, Q5 or Q6 incomplete: re-read Stage 4 and Figure 2, then “Uncoupling”, and rework Worked example 2; write every link in the cause-and-effect chain.
- Q4 graph points lost: re-read Worked example 1, which lists each graph component.
Then tick off the topic checklist.