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. Data for every question: molar masses CO₂ 44.01, H₂O 18.02, O₂ 32.00 and C₆H₁₂O₆ 180.16 g mol⁻¹; each pH unit is a tenfold change in H⁺ concentration. All experiments, organisms and data sets are fictional unless a question says otherwise.
Question 1 (multiple choice · foundation)
A chloroplast is in bright light. In which region is the concentration of H⁺ ions highest?
- (A) The stroma
- (B) The thylakoid space
- (C) The cytoplasm of the leaf cell
- (D) The space between the outer and inner chloroplast membranes
Answer and explanation
Answer: (B). H⁺ released by water splitting, and H⁺ pumped by the electron transport chain, collect inside the thylakoids. This high concentration drives H⁺ back out through ATP synthase.
- (A) is where H⁺ is pumped from, so its concentration is low. It is also where NADP⁺ picks up H⁺ to form NADPH.
- (C) and (D) lie outside the thylakoid membrane, which is the membrane that the gradient forms across.
Question 2 (multiple choice · core)
In ordinary water and ordinary CO₂, about 0.20% of oxygen atoms are the heavy isotope ¹⁸O. A researcher grows a fictional alga in two sealed flasks in bright light.
- Flask 1: water enriched to 1.50% ¹⁸O; ordinary CO₂.
- Flask 2: CO₂ enriched to 1.50% ¹⁸O; ordinary water.
After 20 minutes, the O₂ released contains 1.49% ¹⁸O in flask 1 and 0.21% ¹⁸O in flask 2. Which conclusion do these data best support?
- (A) The O₂ released by photosynthesis comes from water.
- (B) The O₂ released by photosynthesis comes from CO₂.
- (C) About half of the O₂ comes from water and half from CO₂.
- (D) The O₂ is released when glucose made in the Calvin cycle is broken down.
Answer and explanation
Answer: (A). The ¹⁸O in the O₂ matches the water: almost 1.50% when the water is labelled, and only the ordinary 0.20% when the CO₂ is labelled. This fits water being split at photosystem II, with its oxygen atoms released as O₂.
- (B) If the O₂ came from CO₂, flask 2 would give O₂ with about 1.50% ¹⁸O and flask 1 would give about 0.20%. The data show the opposite.
- (C) Half from each source would give about (1.50 + 0.20) ÷ 2 = 0.85% ¹⁸O in both flasks.
- (D) The oxygen atoms in the sugar come from CO₂, so this would put the label in the O₂ of flask 2, not flask 1. Breaking down glucose also uses O₂ rather than releasing it.
Question 3 (multiple choice · core)
A fictional compound K makes the thylakoid membrane leaky to H⁺ ions, but does not stop electron transport. What is the most likely immediate effect in the light?
- (A) NADPH production stops, because electrons cannot reach photosystem I.
- (B) ATP production falls, because H⁺ leaks back to the stroma without passing through ATP synthase.
- (C) O₂ release stops, because water can no longer be split.
- (D) ATP production rises, because H⁺ moves across the membrane faster.
Answer and explanation
Answer: (B). ATP synthase makes ATP only when H⁺ flows through it. If H⁺ leaks across the membrane elsewhere, the gradient cannot build up, and ATP synthesis falls.
- (A) K does not block electron transport, so electrons still reach photosystem I and NADPH is still made.
- (C) Water splitting depends on electrons leaving photosystem II, which still happens.
- (D) The H⁺ that leaks bypasses ATP synthase, so it makes no ATP; faster leaking means less ATP, not more.
Question 4 (calculation · core)
In one hour, a fictional culture of algae releases 0.960 g of O₂.
(a) Calculate the mass of CO₂ fixed in that hour. (b) Calculate the maximum mass of glucose that could be made. (c) Give one reason why the glucose actually gained by the culture would be less.
Worked solution and suggested Marlbridge rubric
(a) Moles of O₂ = 0.960 ÷ 32.00 = 0.0300 mol. Ratio 6 O₂ : 6 CO₂, so 0.0300 mol CO₂. Mass = 0.0300 × 44.01 = 1.32 g.
(b) Ratio 6 O₂ : 1 glucose, so 0.0300 ÷ 6 = 0.00500 mol glucose. Mass = 0.00500 × 180.16 = 0.901 g.
(c) Any one: the algae also break down sugar in cellular respiration; some fixed carbon goes into other molecules (proteins, lipids, cell walls) rather than glucose; some carbohydrate is released into the water.
| Point | What earns it |
|---|---|
| 1 | Moles of O₂ (0.0300 mol) |
| 1 | 1.32 g CO₂ |
| 1 | 0.901 g glucose, using the 6 : 1 ratio |
| 1 | Valid reason |
A common error in (b) is to skip the ÷ 6, which gives 5.40 g.
Question 5 (graph · core)
The rate of O₂ production by a fictional water plant was measured at different light intensities, with temperature and CO₂ supply kept constant.
| Relative light intensity | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| O₂ production / µmol h⁻¹ | 0 | 4.2 | 8.3 | 12.1 | 14.0 | 14.4 | 14.5 |
(a) Calculate the rate of increase in O₂ production between light intensities 0 and 2, per unit of light intensity. (b) Explain the trend between intensities 0 and 3, using the light reactions. (c) Suggest why the rate levels off above intensity 4, and predict, with a reason, the effect of raising the CO₂ supply at intensity 6.
Worked solution and suggested Marlbridge rubric
(a) (8.3 − 0) ÷ (2 − 0) = 4.15 µmol h⁻¹ per unit of light intensity (about 4.2).
(b) More light means chlorophyll in photosystems II and I absorbs more energy per second, so more electrons are boosted each second. Faster electron loss from photosystem II means water is split faster, so more O₂ is released. In this range light is the factor limiting the rate.
(c) Above intensity 4, adding light has little effect (14.0 → 14.5), so something else limits the rate: for example, the CO₂ supply limits how fast the Calvin cycle uses ATP and NADPH, so ADP and NADP⁺ are not returned fast enough for the light reactions to speed up further. Raising CO₂ should increase the rate at intensity 6, because the Calvin cycle could run faster, returning NADP⁺ and ADP to the light reactions sooner. (An answer that the photosystems are saturated with light is also acceptable for the first part.)
| Point | What earns it |
|---|---|
| 1 | 4.15 (accept 4.2) with units |
| 1 | More light → more electrons excited → faster water splitting and O₂ release |
| 1 | Another factor limits above intensity 4, named with a reason |
| 1 | Prediction of a higher rate with more CO₂, linked to the Calvin cycle returning NADP⁺/ADP |
Question 6 (argumentation · core)
A student claims: “Photosynthesis first evolved in prokaryotes, and prokaryotic photosynthesis is what filled the atmosphere with oxygen.” The following statements are accepted findings.
- Rock layers show that free O₂ in the atmosphere rose sharply about 2.4 billion years ago.
- The oldest fossils that most scientists accept as eukaryotes are much younger than this rise in O₂.
- Living cyanobacteria release O₂ in photosynthesis using two photosystems, as plants do.
- Chloroplasts have a double membrane, their own circular DNA and bacteria-like ribosomes, and their DNA is most similar to that of cyanobacteria.
(a) Use statements 1–3 to support the claim that prokaryotes oxygenated the atmosphere. (b) Use statement 4 to explain how eukaryotic photosynthesis is related to prokaryotic photosynthesis.
Model answer and suggested Marlbridge rubric
(a) Statement 1 shows when O₂ appeared. Statement 2 shows that eukaryotes were not yet present (or not yet common) at that time, so they are very unlikely to have produced the O₂. Statement 3 shows that cyanobacteria have the machinery to split water and release O₂. So prokaryotes, most likely cyanobacteria, are the best explanation for the oxygen.
(b) Statement 4 shows that chloroplasts are descended from cyanobacteria that were taken into an early eukaryotic cell. So eukaryotic photosynthesis did not evolve separately: it was inherited from prokaryotes, and the prokaryotic pathway, with its two photosystems, became the foundation of photosynthesis in algae and plants.
| Point | What earns it |
|---|---|
| 1 | Uses the timing of statements 1 and 2 to rule out eukaryotes |
| 1 | Uses statement 3 to show cyanobacteria can release O₂ |
| 1 | Chloroplasts descend from cyanobacteria (endosymbiosis), with evidence from statement 4 |
| 1 | Concludes eukaryotic photosynthesis was built on the prokaryotic pathway |
Question 7 (data analysis · stretch)
Two fictional mutant strains of a single-celled alga are compared with the normal (wild-type) strain in bright light.
- Mutant A has no working photosystem I.
- Mutant B has an ATP synthase that lets H⁺ through but cannot make ATP.
| Measurement | Wild type | Mutant A | Mutant B |
|---|---|---|---|
| O₂ release / µmol h⁻¹ | 10.0 | 0.3 | 8.5 |
| pH in thylakoid space | 5.0 | 7.6 | 7.0 |
| pH in stroma | 8.0 | 7.9 | 8.0 |
| Sugar made (% of wild type) | 100 | 2 | 15 |
(a) Calculate how many times more concentrated H⁺ is inside the thylakoid than in the stroma for the wild type and for mutant B. (b) Explain why O₂ release is almost zero in mutant A, even though its photosystem II is normal. (c) Explain why mutant B still releases O₂ but makes little sugar.
Worked solution and suggested Marlbridge rubric
(a) Wild type: 8.0 − 5.0 = 3.0 pH units, so 10³ = 1000 times. Mutant B: 8.0 − 7.0 = 1.0 unit, so 10 times.
(b) Without photosystem I, electrons that reach the end of the electron transport chain have nowhere to go. The carriers fill with electrons, so photosystem II cannot pass on its excited electrons. It therefore cannot take new electrons from water, water is not split, and almost no O₂ is released (0.3, 3% of wild type). With little electron flow, little H⁺ is pumped and the gradient is small (pH 7.6 inside). No NADPH is made, so the Calvin cycle almost stops (2%).
(c) Electron flow from water through both photosystems to NADP⁺ still works, so water is still split (O₂ 8.5, 85% of wild type) and NADPH is made. But H⁺ leaks back through the faulty ATP synthase without making ATP, so the gradient is small (10 times instead of 1000) and little ATP is made. The Calvin cycle needs ATP as well as NADPH, so sugar production falls to 15%.
| Point | What earns it |
|---|---|
| 1 | 1000 times and 10 times |
| 1 | Mutant A: electrons cannot leave the chain, so PSII cannot accept electrons from water → little O₂ |
| 1 | Mutant A: no NADPH (and little ATP) → Calvin cycle stops |
| 1 | Mutant B: electron flow and NADPH continue, so O₂ is released |
| 1 | Mutant B: gradient lost through ATP synthase without ATP made; Calvin cycle needs ATP → little sugar |
Total: 5 points. Do not award the mutant A points for “photosystem I splits water”: water is split at photosystem II.
How did you do?
- Q1 or Q3 wrong: re-read “The light reactions”, step 6, in the study guide, and look again at Figure 2.
- Q2 wrong: re-read step 2 (water splitting) and the misconception about where O₂ comes from.
- Q4 wrong: rework Worked example 1; check the 6 : 1 ratio.
- Q5 wrong: link light to electron excitation, then to O₂ release; re-read “The Calvin cycle” for why the stages depend on each other.
- Q6 incomplete: re-read “Where photosynthesis came from”.
- Q7 wrong: rework Worked example 2, tracing each effect along the chain.
Then tick off the topic checklist.