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: percentage identity = identical positions ÷ positions compared × 100; for a cube, surface area = 6 × side² and volume = side³. A calculator is assumed. Keep full calculator values until the last step. All organisms, drugs and data sets in the questions are fictional.
Question 1 (multiple choice · foundation)
Which structure is found in both prokaryotic and eukaryotic cells?
- (A) Ribosomes
- (B) A nuclear envelope
- (C) Mitochondria
- (D) Endoplasmic reticulum
Answer and explanation
Answer: (A). All cells make proteins on ribosomes. Prokaryotic ribosomes are smaller, but they do the same job.
- (B) is wrong: prokaryotic DNA sits in a nucleoid region with no membrane around it.
- (C) is wrong: prokaryotes have no mitochondria. Aerobic prokaryotes carry the proteins for making ATP in their plasma membrane.
- (D) is wrong: the ER is part of the eukaryotic system of internal membranes. Prokaryotes typically have no membrane-bound organelles.
Question 2 (multiple choice · core)
Which observation provides the strongest support for the claim that chloroplasts evolved from free-living cyanobacteria?
- (A) Chloroplasts contain a circular DNA molecule whose gene sequences are more similar to those of cyanobacteria than to the plant’s own nuclear genes.
- (B) Chloroplasts are found in plant cells but not in animal cells.
- (C) Chloroplasts need light to carry out photosynthesis.
- (D) Chloroplasts are surrounded by cytosol and can move around inside the cell.
Answer and explanation
Answer: (A). Separate, bacteria-like DNA, with sequences closest to one bacterial group, is direct evidence of shared ancestry with that group.
- (B) is true but says nothing about where chloroplasts came from. It fits endosymbiosis (chloroplasts were acquired by one lineage), but it would also fit many other ideas.
- (C) describes what chloroplasts do, not their ancestry.
- (D) is true of many organelles, including ones that did not come from endosymbiosis.
Question 3 (multiple choice · core)
A gene from the chloroplast of a fictional alga is compared with the same gene from a cyanobacterium over 1,320 aligned positions. 1,003 positions are identical. What is the percentage identity?
- (A) 24.0%
- (B) 76.0%
- (C) 132%
- (D) 0.760%
Answer and explanation
Answer: (B). 1,003 ÷ 1,320 × 100 = 75.98…% = 76.0%.
- (A) is the percentage of positions that differ: (1,320 − 1,003) ÷ 1,320 × 100.
- (C) divides the wrong way round (1,320 ÷ 1,003 × 100 = 131.6%). A percentage identity can never exceed 100%.
- (D) gives the fraction 0.760 but labels it as a percentage, without multiplying by 100.
Question 4 (data analysis · core)
A scientist studies a structure, called S, inside the cells of a fictional single-celled protist that can photosynthesize. Her observations:
| Observation | Result |
|---|---|
| Membranes around S | two |
| DNA inside S | one small circular molecule |
| Effect of drug K (blocks bacterial ribosomes) on protein synthesis inside S | falls to 6% of control |
| Effect of drug C (blocks eukaryotic cytosolic ribosomes) on protein synthesis inside S | 97% of control |
| How new copies of S appear | an existing S splits in two |
| Identical positions in one S gene (out of 1,200) compared with a cyanobacterium | 972 |
| Identical positions in the same gene (out of 1,200) compared with the protist’s nuclear version | 624 |
(a) Calculate the two percentage identities. (b) Make a claim about the origin of S. Support it with at least three different pieces of evidence from the table, explaining how each supports the claim. (c) Describe one result that, if found instead, would weaken your claim, and explain why.
Model answer and suggested Marlbridge rubric
(a) Cyanobacterium: 972 ÷ 1,200 × 100 = 81.0%. Nuclear version: 624 ÷ 1,200 × 100 = 52.0%.
(b) Claim: S evolved from a free-living cyanobacterium (or a close relative) by endosymbiosis; it is a chloroplast. Evidence, any three:
- Its gene is 81.0% identical to the cyanobacterium’s but only 52.0% to the host’s own nuclear gene, so S shares a more recent ancestor with cyanobacteria than with the host’s nucleus.
- Two membranes fit a prokaryote that was enclosed in a host membrane when it was taken in.
- Its own small circular DNA resembles a bacterial chromosome.
- Its ribosomes are blocked by the drug that acts on bacterial ribosomes (down to 6%) and not by the drug that acts on cytosolic ribosomes (97% remains), so they behave like bacterial ribosomes.
- New S forms only by splitting, like bacterial binary fission, rather than being built by the cell.
(c) Any one, with a reason: the gene is most similar to the protist’s nuclear gene (would suggest S came from the host’s own parts); S’s protein synthesis is blocked by drug C but not drug K (its ribosomes would be like the host’s); S has no DNA of its own; S can be assembled from scratch by the cell from its membranes.
| Point | What earns it |
|---|---|
| 1 | Both percentages correct |
| 1 | Clear claim of endosymbiotic origin from a cyanobacterium-like ancestor |
| 1 | Three valid pieces of evidence from the table, each with reasoning (all three needed) |
| 1 | A result that would weaken the claim, with a reason |
Question 5 (constructed response · core)
(a) Describe two similarities between the internal organization of a prokaryotic cell and a eukaryotic cell. (b) Describe two differences. (c) A student says: “A prokaryotic cell has no compartments, so everything inside it happens in the same place.” Using one named example, explain why this statement is not accurate.
Model answer and suggested Marlbridge rubric
(a) Any two: both have a plasma membrane that controls what enters and leaves; both have ribosomes that make proteins; both store DNA in a particular region; both have specialized regions that carry out specialized jobs.
(b) Any two: eukaryotes have membrane-bound organelles and prokaryotes typically do not; eukaryotic DNA is inside a nucleus with a double envelope, prokaryotic DNA is in a nucleoid with no membrane; eukaryotes make ATP by aerobic respiration on the inner membrane of mitochondria, prokaryotes on the plasma membrane; eukaryotic cytosolic ribosomes are larger.
(c) Prokaryotes have internal regions, each with its own structures and jobs, even without membrane-bound organelles. Examples, any one: the nucleoid holds the DNA in one region; the plasma membrane carries the proteins of aerobic respiration, so ATP is made at the membrane, not throughout the cytoplasm; cyanobacteria capture light on internal thylakoid membranes; carboxysomes gather the enzyme rubisco inside a protein shell.
| Point | What earns it |
|---|---|
| 1 | Two correct similarities |
| 1 | Two correct differences, each comparing both cell types |
| 1 | States that prokaryotes have specialized internal regions |
| 1 | A correctly described named example |
Question 6 (constructed response · stretch)
In a fictional protist, the mitochondria use 1,050 different proteins. The mitochondrial DNA codes for 38 of them; the rest are coded by genes in the nucleus, made in the cytosol and imported into the mitochondria.
(a) Calculate the percentage of mitochondrial proteins that are coded by the mitochondrial DNA, and the number that must be imported. (b) A student argues: “Most mitochondrial proteins are coded in the nucleus, so mitochondria cannot have come from bacteria.” Evaluate this argument. (c) Predict what would happen to the cell’s ATP supply if a mutation stopped proteins being imported into its mitochondria. Justify your prediction.
Model answer and suggested Marlbridge rubric
(a) 38 ÷ 1,050 × 100 = 3.6% coded in the mitochondrion. Imported: 1,050 − 38 = 1,012 proteins (96.4%).
(b) The argument is not convincing. The endosymbiotic theory predicts that, over a very long time, most genes of the ancestral bacterium were transferred to the host’s nucleus or lost, because the host could supply the products. A small genome that remains is exactly what the theory expects. The other lines of evidence (two membranes, bacteria-like ribosomes, circular DNA, division by splitting, sequences closest to alphaproteobacteria) still support a bacterial origin. The data do show that modern mitochondria depend on the cell, which is a different point from where they came from.
(c) ATP production would fall sharply. Most of the proteins mitochondria need, including many of those in the inner membrane that make ATP, are coded in the nucleus and must be imported. Without import, the mitochondria could not replace or build these proteins, so aerobic respiration in the mitochondria would slow or stop, and the cell would rely on the much smaller ATP yield from reactions in the cytosol.
| Point | What earns it |
|---|---|
| 1 | 3.6% and 1,012 |
| 1 | Explains gene transfer to the nucleus as expected by the theory |
| 1 | Notes that other independent evidence still supports a bacterial origin, or distinguishes origin from present-day dependence |
| 1 | Prediction of reduced ATP production justified by loss of imported proteins needed for respiration |
Question 7 (calculation · stretch)
Model a prokaryotic cell as a cube with sides of 1 µm and a eukaryotic cell as a cube with sides of 20 µm.
(a) Calculate the surface area, volume and surface area to volume ratio of each cell. (b) In the model, the eukaryotic cell also has internal membranes with a total area 20 times its plasma membrane area. Calculate its total membrane area per µm³ of volume. (c) Use your answers to support the claim that internal membranes allow a large eukaryotic cell to carry out membrane-based reactions on a scale comparable to a small prokaryote.
Worked solution
(a) Prokaryote: surface area = 6 × 1² = 6 µm²; volume = 1³ = 1 µm³; ratio = 6.0 µm⁻¹. Eukaryote: surface area = 6 × 20² = 2,400 µm²; volume = 20³ = 8,000 µm³; ratio = 0.30 µm⁻¹, 20 times smaller.
(b) Internal membranes = 20 × 2,400 = 48,000 µm². Total membrane = 2,400 + 48,000 = 50,400 µm². Per volume: 50,400 ÷ 8,000 = 6.3 µm² per µm³.
(c) Many reactions, such as ATP synthesis, happen on membrane proteins, so the membrane area available per unit of volume limits them. With only its plasma membrane, the large cell has 20 times less membrane per µm³ than the prokaryote (0.30 against 6.0). With internal membranes it has 6.3 µm² per µm³, slightly more than the prokaryote. So internal membranes give the large cell enough membrane for its volume. To match the prokaryote exactly, the internal area would need to be 19 times the plasma membrane area (48,000 − 2,400 = 45,600 µm²).
Suggested mark points (3): 1 for both ratios; 1 for 6.3 µm² per µm³; 1 for a claim supported by comparing the two values with reasoning about membrane-bound reactions.
Common errors: using 4 × side² (only some faces) for the surface area; forgetting to add the plasma membrane to the internal membranes, which gives 48,000 ÷ 8,000 = 6.0 and hides the extra area.
How did you do?
- Q1 or Q5 wrong: re-read “Inside a prokaryotic cell” and the comparison table in the study guide.
- Q2 or Q4 incomplete: re-read “The evidence” and Worked example 1. Make sure each piece of evidence comes with a reason.
- Q3 or Q7 calculations wrong: check which number goes on the top of the fraction and that you multiplied by 100.
- Q6 incomplete: re-read “Since then: dependence” and keep the origin of mitochondria separate from their present-day dependence on the cell.
Then tick off the topic checklist.