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. Useful data: volume of a sphere = 4/3 × π × r³; a change of 1 pH unit is a tenfold change in H⁺ concentration; the cytosol is at about pH 7.2. A calculator is assumed. Keep full calculator values until the last step. All data sets and organisms in the questions are fictional.
Question 1 (multiple choice · foundation)
A student observes that the mitochondria in one cell type have many more cristae than the mitochondria in another cell type, although the mitochondria are the same size. Which statement best explains the advantage of having more cristae?
- (A) More inner membrane area fits into the same volume, so the mitochondrion can hold more ATP synthase and electron transport proteins.
- (B) The cristae enlarge the matrix, so the mitochondrion can store more glucose.
- (C) The cristae give the outer membrane more room for ATP synthase.
- (D) The cristae let the contents of the matrix and the intermembrane space mix, so the reactions meet more often.
Answer and explanation
Answer: (A). The proteins that make ATP are embedded in the inner membrane. Folding that membrane into cristae increases its area without increasing the volume of the organelle, so more of these proteins fit.
- (B) is wrong: folding adds membrane, not matrix volume, and mitochondria do not store glucose.
- (C) names the wrong membrane. The outer membrane is smooth; cristae are folds of the inner membrane.
- (D) reverses the idea of compartmentalization. The inner membrane keeps the two spaces separate, and that separation matters for making ATP.
Question 2 (multiple choice · core)
In a model organelle, a flat membrane has an area of 6.0 µm². The membrane then forms 15 folds, and each fold adds 0.80 µm² of membrane. The volume of the organelle does not change. By what factor has the total membrane area increased?
- (A) 2.0
- (B) 3.0
- (C) 15
- (D) 0.33
Answer and explanation
Answer: (B). Added area = 15 × 0.80 = 12.0 µm². New total = 6.0 + 12.0 = 18.0 µm². Factor = 18.0 ÷ 6.0 = 3.0.
- (A) compares only the added area with the original (12.0 ÷ 6.0) and forgets that the original membrane is still there.
- (C) is the number of folds, not an area ratio.
- (D) divides the old area by the new one (6.0 ÷ 18.0). That shows the original is a third of the new total, not how much the area increased.
Question 3 (multiple choice · core)
A drug makes the lysosome membrane highly permeable to H⁺ ions but does not stop the proton pumps. Which outcome is most likely?
- (A) H⁺ ions diffuse out of the lysosomes, the pH inside rises, the digestive enzymes slow down and undigested material builds up in the lysosomes.
- (B) H⁺ ions diffuse into the lysosomes from the cytosol, the pH inside falls, and the enzymes digest material faster.
- (C) The digestive enzymes escape through the openings and quickly digest the proteins of the cytosol.
- (D) There is no change, because the proton pumps are still working.
Answer and explanation
Answer: (A). The inside of a lysosome has a much higher H⁺ concentration than the cytosol. A membrane permeable to H⁺ lets them diffuse out, down their concentration gradient. The pH inside rises towards that of the cytosol, the acid-loving enzymes slow down, and material waiting to be digested accumulates.
- (B) has the direction of diffusion backwards. H⁺ moves from high concentration (inside the lysosome) to low (cytosol).
- (C) confuses a leak for small ions with a leak for large proteins. Even if enzymes did escape, they work poorly at the near-neutral pH of the cytosol.
- (D) ignores the leak. The pumps keep working, but if H⁺ leaks out as fast as it is pumped in, the pH difference cannot be kept.
Question 4 (data analysis · core)
A researcher breaks open liver cells gently and spins the mixture in a centrifuge at increasing speeds. This separates four fractions: (1) nuclei, (2) mitochondria, (3) lysosomes and other small vesicles, and (4) the liquid left at the end, which contains the cytosol. She measures the activity of three enzymes in each fraction (arbitrary units).
| Enzyme | Fraction 1 | Fraction 2 | Fraction 3 | Fraction 4 |
|---|---|---|---|---|
| P | 4 | 10 | 70 | 6 |
| Q | 3 | 88 | 5 | 4 |
| R | 2 | 3 | 2 | 93 |
Enzyme P is most active at pH 5.0.
(a) For each enzyme, calculate the percentage of its total activity found in the fraction with the most activity. (b) Identify the compartment that most likely holds each enzyme. Give one piece of evidence, other than the fraction data, that P belongs in lysosomes. (c) In a second run, the cells were broken open roughly, and 60% of the lysosomes burst before spinning. Assume enzyme released from burst lysosomes ends up in fraction 4. Predict the new fraction 3 and fraction 4 values for enzyme P. (d) Explain why the small activities of P in fractions 1 and 2 do not show that P works in the nucleus or in mitochondria.
Worked solution and suggested Marlbridge rubric
(a) Totals: P = 4 + 10 + 70 + 6 = 90; Q = 100; R = 100. P: 70 ÷ 90 × 100 = 77.8% in fraction 3. Q: 88 ÷ 100 × 100 = 88.0% in fraction 2. R: 93 ÷ 100 × 100 = 93.0% in fraction 4.
(b) P: lysosomes. Q: mitochondria. R: the cytosol (it is free in the liquid, not inside an organelle). Extra evidence for P: it works best at pH 5.0, which matches the acidic interior of a lysosome (about 4.5–5) rather than the cytosol (about 7.2).
(c) Enzyme moved = 0.60 × 70 = 42 units. Fraction 3: 70 − 42 = 28. Fraction 4: 6 + 42 = 48, which would be 48 ÷ 90 = 53.3% of the total.
(d) Fractions are never perfectly pure. Some lysosomes are large or clumped and come down with the nuclei or mitochondria. The amounts are small (14 of 90 units), so they are best explained as contamination, not as P working there.
| Point | What earns it |
|---|---|
| 1 | All three percentages correct (77.8%, 88.0%, 93.0%) |
| 1 | Correct compartment for all three enzymes and the pH 5.0 evidence linked to the lysosome’s acidic interior |
| 1 | 28 and 48 units for fraction 3 and fraction 4 |
| 1 | Contamination (impure fractions) given as the reason in (d) |
Accept (c) given as percentages of the total (31.1% and 53.3%) if the working is shown.
Question 5 (constructed response · core)
Lysosomal enzymes are made on the rough ER, modified in the Golgi complex and carried to the lysosomes in vesicles. A chemical tag added in the Golgi sends them to the lysosomes. In a mutant cell line, the tag is never added, so the enzymes are carried to the plasma membrane and released outside the cell instead.
(a) Predict what happens inside the lysosomes of the mutant cells, and explain why. (b) The fluid outside the cells is close to neutral pH. Predict whether the released enzymes will damage the outside surface of the cells, and justify your prediction. (c) Explain how making these enzymes inside the ER, Golgi and vesicles protects a normal cell before the enzymes reach the lysosome.
Model answer and suggested Marlbridge rubric
(a) The lysosomes receive few or none of their digestive enzymes. Material delivered to them (worn-out organelles, engulfed particles) is not broken down, so it builds up and the lysosomes swell. The cell also recycles fewer building blocks from that material.
(b) Little damage is expected. Lysosomal enzymes work best at about pH 4.5–5 and poorly near neutral pH, so outside the cell their activity is low. (Accept a prediction of some slow damage over a long time if it is justified by the low but non-zero activity.)
(c) From the moment they are made, the enzymes are inside membrane-bound spaces (the ER interior, the Golgi sacs, the vesicles). They never mix with the cytosol, where the cell’s own proteins, RNA and organelles would be exposed to them. The membranes keep the competing processes of making and digesting apart.
| Point | What earns it |
|---|---|
| 1 | Undigested material builds up in the lysosomes |
| 1 | Linked to the missing enzymes (the tag directs the enzymes to the lysosome) |
| 1 | Prediction about outside damage justified by low enzyme activity at near-neutral pH |
| 1 | Enzymes stay inside membrane-bound compartments and so are kept apart from the cytosol |
Question 6 (constructed response · stretch)
Two strains of a single-celled alga are grown under the same light and temperature. Strain B carries a mutation that reduces the amount of thylakoid membrane in each chloroplast. The data are means of three trials.
| Strain | Thylakoid membrane area per chloroplast / µm² | Oxygen production / µmol O₂ per 10⁶ cells per hour |
|---|---|---|
| A (normal) | 38 | 12.4 |
| B (mutant) | 17 | 5.9 |
(a) Calculate the percentage decrease from A to B in thylakoid membrane area and in oxygen production. (b) Make a claim about how thylakoid membrane area affects the rate of photosynthesis. Support it with evidence from the data and with reasoning about where the light-capturing reactions happen. (c) Calculate the oxygen production per µm² of thylakoid membrane for each strain, and explain what the result suggests. (d) Identify one other difference between the strains that could explain the result, and describe a measurement that would test it.
Model answer and suggested Marlbridge rubric
(a) Area: (38 − 17) ÷ 38 × 100 = 55.3% decrease. Oxygen production: (12.4 − 5.9) ÷ 12.4 × 100 = 52.4% decrease.
(b) Claim: less thylakoid membrane means a lower rate of photosynthesis. Evidence: strain B has 55.3% less thylakoid membrane and produces 52.4% less oxygen. Reasoning: the proteins that capture light and split water, releasing oxygen, are embedded in the thylakoid membranes. Less membrane means room for fewer of these proteins, so fewer reactions happen per second.
(c) A: 12.4 ÷ 38 = 0.33 µmol per 10⁶ cells per hour per µm². B: 5.9 ÷ 17 = 0.35. The values are similar, so the rate seems roughly proportional to the membrane area: each square micrometre of membrane does about the same amount of work in both strains.
(d) Any one difference, with a matching test: the mutation might also lower the amount of chlorophyll (measure chlorophyll per cell); strain B might have fewer chloroplasts per cell (count them under a microscope); the cells might differ in size (measure cell volume).
| Point | What earns it |
|---|---|
| 1 | Both percentage decreases correct |
| 1 | Claim supported by quoted evidence and reasoning that light-capturing proteins are in the thylakoid membranes |
| 1 | Both per-area rates correct with the conclusion that they are similar |
| 1 | A plausible alternative explanation with a measurement that would test it |
Question 7 (calculation · stretch)
A model animal cell is a sphere 20 µm in diameter. Each lysosome is a sphere 0.50 µm in diameter. One lysosome holds 1,200 molecules of a digestive enzyme.
(a) Calculate the volume of the cell and of one lysosome. (b) Calculate the concentration of the enzyme (molecules per µm³) inside the lysosome, and what it would be if the same 1,200 molecules were spread through the whole cell. How many times more concentrated is it in the lysosome? (c) The cell has 250 lysosomes. What percentage of the cell’s volume do they take up? (d) Use your answers to explain one advantage of keeping these enzymes in lysosomes.
Worked solution
(a) Cell: r = 10 µm, V = 4/3 × π × 10³ = 4,190 µm³ (4,188.8). Lysosome: r = 0.25 µm, V = 4/3 × π × 0.25³ = 0.0654 µm³.
(b) In the lysosome: 1,200 ÷ 0.0654 ≈ 18,300 molecules per µm³ (1.8 × 10⁴). Spread through the cell: 1,200 ÷ 4,190 ≈ 0.29 molecules per µm³. Ratio: 4,188.8 ÷ 0.06545 = 64,000 times. Check: the ratio of volumes of two spheres is the cube of the ratio of their diameters, (20 ÷ 0.50)³ = 40³ = 64,000.
(c) Total lysosome volume = 250 × 0.0654 = 16.4 µm³. Percentage = 16.4 ÷ 4,190 × 100 = 0.39% of the cell.
(d) The enzymes are packed at a very high concentration, so digestion is fast, yet the lysosomes fill less than 0.5% of the cell. The rest of the cell is kept free of them.
Suggested mark points (3): 1 for both volumes; 1 for the 64,000-times ratio with the two concentrations; 1 for 0.39% and an advantage linked to concentration or containment.
Common errors: using the diameter as the radius gives 33,500 µm³ for the cell, eight times too large; squaring the radius instead of cubing it gives a ratio of only 1,600 (40²).
How did you do?
- Q1 or Q2 wrong: re-read “Idea 2: folded membranes add surface area” and Worked example 1 in the study guide. Remember to keep the unfolded area in the total.
- Q3 or Q5 wrong: re-read “Idea 1: compartments keep competing reactions apart” and Worked example 2. Check the direction of diffusion and the pH where lysosomal enzymes work.
- Q4 or Q6 incomplete: practise turning data into percentages, then naming the compartment or mechanism they point to and one other explanation.
- Q7 calculations wrong: check radius versus diameter and that you cubed the radius.
Then tick off the topic checklist.