Before you start
This review joins the five topics of Unit 3. Do the unit diagnostic first and close its gaps. Then try the questions without looking at the answers.
Big ideas of the unit
- Enzymes set the pace. An enzyme lowers activation energy when the substrate’s shape and charge fit its active site (Topic 3.1). It changes the rate, never the overall energy change.
- Shape depends on surroundings. Temperature, pH and chemicals act on the weak interactions that hold an enzyme’s shape. Inhibitors compete for the active site or bind an allosteric site (Topic 3.2).
- Life needs a constant energy input. Cells keep their order only while energy flows in; some is always lost as heat (Topic 3.3).
- Coupling pays for work. An energy-requiring reaction runs when it is linked to one that releases more energy, usually ATP hydrolysis.
- Pathways run in small, controlled steps. The product of one step is the substrate of the next, so one blocked enzyme affects the whole chain.
- Photosynthesis stores light energy. The light reactions in the thylakoids make ATP and NADPH; the Calvin cycle in the stroma uses them to fix CO₂ (Topic 3.4).
- Respiration releases it again. Glycolysis, the Krebs cycle and the ETC pass electrons from food to oxygen. Fermentation keeps glycolysis going without oxygen (Topic 3.5).
- One mechanism, three places. Electron transport pumps protons across a membrane, and ATP synthase uses the gradient. The same core machinery in all three domains is evidence for common ancestry.
Key relationships and methods
| Idea | Rule or method | Where it appears |
|---|---|---|
| Rate | change in amount ÷ time; percentage change = (new − old) ÷ old × 100 | enzyme assays, gas exchange |
| Inhibitor type | effect shrinks at high substrate → competitive; same fraction at every substrate level → noncompetitive | 3.2 data questions |
| Coupling | add the energy changes; the total must be negative; ATP = −30 kJ mol⁻¹, whole molecules only | 3.3 |
| Gas exchange | net O₂ in light = gross photosynthesis − respiration; 6 O₂ per glucose | 3.4 with 3.5 |
| ATP yield | about 30 ATP per glucose aerobic, 2 by fermentation | 3.5 |
| Proton gradient | ratio of H⁺ = 10^(pH difference); high H⁺ in thylakoid space or intermembrane space | 3.4, 3.5 |
| Blocked step | what builds up before the block, what runs out after it | any pathway |
These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not official scoring guidelines. All data sets and organisms are fictional.
Question 1 (multiple choice · mixed)
ATP synthase works the same way in chloroplasts and mitochondria. Which row correctly gives the direction in which protons flow through ATP synthase in each?
- (A) Chloroplast: stroma → thylakoid space. Mitochondrion: matrix → intermembrane space.
- (B) Chloroplast: stroma → thylakoid space. Mitochondrion: intermembrane space → matrix.
- (C) Chloroplast: thylakoid space → stroma. Mitochondrion: intermembrane space → matrix.
- (D) Chloroplast: thylakoid space → stroma. Mitochondrion: matrix → intermembrane space.
Answer and explanation
Answer: (C). In both, the ETC pumps protons into a small compartment (thylakoid space, intermembrane space). Protons flow back down the gradient through ATP synthase, into the stroma or matrix, where ATP is made.
- (A) gives the direction of pumping, not of flow through ATP synthase.
- (B) and (D) mix the two. The gradient points the same way in both organelles: from the space the ETC fills towards the fluid where ATP is used.
Question 2 (multiple choice · mixed)
A cell must run a reaction that requires +45 kJ per mole, 100 times. ATP hydrolysis releases 30 kJ per mole. How many glucose molecules must the cell use if it gets its ATP only by fermentation, and only by aerobic respiration?
- (A) 50 by fermentation; 4 by aerobic respiration
- (B) 75 by fermentation; 5 by aerobic respiration
- (C) 100 by both, because both routes start with glycolysis
- (D) 100 by fermentation; 7 by aerobic respiration
Answer and explanation
Answer: (D). One ATP is not enough (+45 − 30 = +15 kJ mol⁻¹), so each reaction needs 2 ATP (+45 − 60 = −15 kJ mol⁻¹). 100 reactions need 200 ATP. Fermentation: 200 ÷ 2 = 100 glucose. Aerobic: 200 ÷ 30 = 6.7, so 7 glucose.
- (A) uses one ATP per reaction; the coupled reaction would still require energy.
- (B) uses 1.5 ATP per reaction; a cell cannot hydrolyse half an ATP.
- (C) forgets that aerobic respiration makes far more ATP after glycolysis.
Question 3 (multiple choice · mixed)
In a fictional alga, the stroma is at pH 7.0 in the dark and pH 8.0 in bright light. A Calvin cycle enzyme works five times faster at pH 8.0 than at pH 7.0. Which statement best explains why it works faster in the light?
- (A) Light energy lowers the activation energy of the Calvin cycle reactions directly.
- (B) The light reactions pump H⁺ out of the stroma into the thylakoid space, so the stroma’s H⁺ concentration falls tenfold. The charges on R groups in the enzyme change towards those of its working shape.
- (C) The light reactions release H⁺ into the stroma, lowering its pH towards the enzyme’s optimum.
- (D) Light raises the temperature of the stroma above the enzyme’s optimum.
Answer and explanation
Answer: (B). Moving H⁺ into the thylakoids lowers H⁺ in the stroma (pH 7.0 → 8.0 is a tenfold fall). pH sets which R groups carry a charge, and the right charges hold the active site in its working shape.
- (A) is wrong: light is absorbed by chlorophyll, not by Calvin cycle enzymes.
- (C) reverses the direction of pumping, and pH 8.0 is higher, not lower.
- (D) would slow the enzyme, not speed it up.
Question 4 (constructed response · mixed)
In a fictional bacterium, a four-enzyme pathway turns substrate S into amino acid E. Enzyme 1 catalyses the first step. Its initial rate (µmol min⁻¹) is measured with two concentrations of S and three concentrations of E.
| E / mmol L⁻¹ | Rate at S = 2 mmol L⁻¹ | Rate at S = 20 mmol L⁻¹ |
|---|---|---|
| 0 | 24.0 | 60.0 |
| 1 | 16.0 | 40.0 |
| 4 | 8.0 | 20.0 |
(a) Calculate the rate with 4 mmol L⁻¹ of E as a percentage of the rate with no E, at each concentration of S. (b) Identify the type of inhibition shown by E. Justify your answer with the data. (c) E does not look like S. Explain how E reduces the activity of enzyme 1. (d) Explain why it benefits the cell that E inhibits the first enzyme of the pathway. (e) State an appropriate control for the rate measurements, and its purpose.
Worked solution and suggested Marlbridge rubric
(a) S = 2: 8.0 ÷ 24.0 × 100 = 33.3%. S = 20: 20.0 ÷ 60.0 × 100 = 33.3%.
(b) Noncompetitive. E reduces the rate by the same fraction (to 66.7% at 1 mmol L⁻¹, 33.3% at 4 mmol L⁻¹) at low and high S. Raising S tenfold does not overcome it.
(c) E binds an allosteric site away from the active site. This changes the enzyme’s shape, so the active site fits S less well and fewer enzyme–substrate complexes form.
(d) The steps run in sequence. When E builds up, the first step slows, so no S is spent making intermediates the cell does not need, and no energy is wasted. When E is used up, inhibition falls and the pathway speeds up again.
(e) A tube with S and buffer but no enzyme 1 (or with boiled enzyme), to show that S does not change without an active enzyme.
| Point | What earns it |
|---|---|
| 1 | 33.3% at both concentrations |
| 1 | Noncompetitive, with the same percentage at both S values as evidence |
| 1 | Allosteric binding changes the active site’s shape |
| 1 | First-step control saves substrate or energy when E is plentiful |
| 1 | Valid negative control with its purpose |
Total: 5 points.
Question 5 (constructed response · mixed)
A sealed flask of a fictional alga is kept at constant temperature. In bright light, the O₂ in the flask rises by 18 µmol per hour. In the dark, it falls by 4 µmol per hour. Assume respiration runs at the same rate in light and dark.
(a) Explain why O₂ falls in the dark. (b) Calculate the gross rate of O₂ production by photosynthesis. (c) The alga gets 14 hours of light and 10 hours of dark each day. Calculate the net O₂ change per day, and the net mass of glucose stored per day. (d) A student says: “In the light, the alga only photosynthesises.” Use the data to evaluate this claim. (e) Explain why the alga must keep capturing light energy to stay alive, even though it already stores glucose.
Worked solution and suggested Marlbridge rubric
(a) In the dark there are no light reactions, so no water is split and no O₂ is released. Respiration continues: oxygen accepts electrons at the end of the ETC in the mitochondria.
(b) Gross = net in light + respiration = 18 + 4 = 22 µmol O₂ per hour.
(c) Net O₂ = 14 × 18 − 10 × 4 = 252 − 40 = 212 µmol per day. Glucose = 212 ÷ 6 = 35.3 µmol. Mass = 35.3 × 10⁻⁶ mol × 180.16 g mol⁻¹ = 6.37 mg per day.
(d) The claim is not supported. Photosynthesis releases 22 µmol h⁻¹, but only 18 µmol h⁻¹ appears. The missing 4 µmol h⁻¹ matches the dark rate, so respiration continues in the light.
(e) Every transfer loses some energy as heat (second law). The alga respires its glucose to make ATP for growth, pumping and repair. Without new light energy, its stores run down, and order cannot be kept.
| Point | What earns it |
|---|---|
| 1 | Dark: no water splitting, but respiration still uses O₂ |
| 1 | 22 µmol O₂ per hour |
| 1 | 212 µmol O₂ per day |
| 1 | 35.3 µmol and 6.37 mg glucose |
| 1 | Claim rejected using 22 against 18 |
| 1 | Energy lost as heat, so a continuous input is needed |
Total: 6 points. Common error: 14 × 18 ÷ 6 = 42 µmol ignores the night.
Question 6 (constructed response · mixed)
Compound V blocks the proton channel of ATP synthase. Isolated chloroplasts are given plenty of CO₂ in bright light, with and without V.
| Measurement | Control | With V |
|---|---|---|
| CO₂ fixed / µmol h⁻¹ | 30.0 | 3.0 |
| O₂ released / µmol h⁻¹ | 30.0 | 3.0 |
| pH in thylakoid space | 5.0 | 4.6 |
| pH in stroma | 8.0 | 8.0 |
(a) Calculate the percentage fall in CO₂ fixation. (b) Use the pH values to find the ratio of H⁺ concentration (thylakoid space : stroma) in the control and with V. (c) Explain why the gradient becomes steeper with V. (d) Explain why CO₂ fixation falls. (e) V does not act on photosystem II, yet O₂ release falls. Explain why. (f) V also blocks ATP synthase in a bacterium and in yeast mitochondria. Explain how this supports common ancestry.
Worked solution and suggested Marlbridge rubric
(a) (30.0 − 3.0) ÷ 30.0 × 100 = 90%.
(b) Each pH unit is a tenfold change in H⁺. Control: 8.0 − 5.0 = 3.0 units, so the ratio is 10^3.0 = 1000 : 1. With V: 8.0 − 4.6 = 3.4 units, so 10^3.4 ≈ 2500 : 1.
(c) Water splitting and the ETC keep adding H⁺ to the thylakoid space, but H⁺ cannot flow out through ATP synthase. It accumulates.
(d) Without H⁺ flow, ATP synthase makes no ATP. The Calvin cycle in the stroma needs ATP (and NADPH) to fix CO₂, so it slows.
(e) Two effects. The very steep gradient makes further pumping hard, so electron transport slows. And with the Calvin cycle stalled, NADPH is not used, so NADP⁺ runs out and electrons from photosystem I have nowhere to go. Electrons back up, photosystem II cannot take more from water, and less O₂ is released.
(f) The same compound fits the same site in organisms from two domains and in an organelle. This suggests the ATP synthases share a structure inherited from a common ancestor, rather than arising separately.
| Point | What earns it |
|---|---|
| 1 | 90% fall |
| 1 | Ratios of 1000 : 1 and about 2500 : 1 |
| 1 | H⁺ still added but cannot leave through ATP synthase |
| 1 | No ATP, so the Calvin cycle cannot fix CO₂ |
| 1 | O₂ falls because electron flow backs up (steep gradient or no NADP⁺) |
| 1 | Shared, conserved structure across domains points to common ancestry |
Total: 6 points.
Question 7 (constructed response · mixed)
During a burst of swimming, the muscle of a fictional fish needs 60 µmol of ATP per second, but its oxygen supply cannot keep up. The activity of one of its glycolysis enzymes depends on pH:
| pH | 7.1 | 6.8 | 6.5 |
|---|---|---|---|
| Activity / % of maximum | 100 | 70 | 35 |
(a) Explain why the muscle cells switch to lactic acid fermentation when oxygen runs short. (b) Calculate the glucose needed per second to supply 60 µmol of ATP by aerobic respiration, and by fermentation. (c) During the burst, muscle pH falls from 7.1 to 6.6. Estimate the enzyme’s activity at pH 6.6. (d) Explain why the lower pH reduces the enzyme’s activity. (e) Use your answers to explain why the fish can sustain a burst for only a short time.
Worked solution and suggested Marlbridge rubric
(a) Without oxygen, the ETC has no final electron acceptor, so NADH cannot unload and NAD⁺ runs out. Glycolysis needs NAD⁺. Fermentation passes the electrons from NADH to pyruvate, forming lactic acid and regenerating NAD⁺, so glycolysis continues.
(b) Aerobic: 60 ÷ 30 = 2.0 µmol s⁻¹. Fermentation: 60 ÷ 2 = 30 µmol s⁻¹, 15 times as much.
(c) Between 6.8 and 6.5, activity falls by 35 percentage points over 0.3 pH units. pH 6.6 is 0.2 units below 6.8: 70 − 35 × (0.2 ÷ 0.3) = 46.7%, about 47%.
(d) More H⁺ changes the charges on R groups. Ionic and hydrogen bonds that hold the enzyme’s shape are disrupted, so the active site fits its substrate less well and fewer enzyme–substrate complexes form.
(e) Fermentation needs glucose 15 times faster, so stores run down quickly. As acid builds up, pH falls and a glycolysis enzyme loses about half its activity, so even fermentation slows and ATP supply falls.
| Point | What earns it |
|---|---|
| 1 | NAD⁺ regenerated so glycolysis continues |
| 1 | 2.0 and 30 µmol s⁻¹ |
| 1 | About 47% by interpolation |
| 1 | H⁺ alters R-group charges and active-site shape |
| 1 | Fast glucose use and falling enzyme activity linked to a short burst |
Total: 5 points.
How did you do?
Add up your points: 3 for the multiple-choice questions and 22 for the constructed responses. Use the total only to compare with your own next attempt; it does not predict an exam score.
- Lost points on enzyme shape or inhibitors (Questions 3, 4, 7(d)): reread Topic 3.2.
- Lost points on coupling or energy budgets (Questions 2, 5): rework the examples in Topic 3.3.
- Lost points on gradients and blocked steps (Questions 1, 6, 7(a)): trace each chain link by link in Topic 3.4 and Topic 3.5.
- Want a quick recheck? Retake the unit diagnostic.
Then tick off each topic checklist: