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. No constants are needed. A calculator is assumed for Questions 4 and 7. Keep full calculator values until the last step. All data sets and cell types are fictional.
Question 1 (multiple choice · foundation)
Which statement about ribosomes is correct?
- (A) Ribosomes are made of rRNA and protein, have no surrounding membrane and are found in prokaryotic and eukaryotic cells.
- (B) Ribosomes are membrane-bound organelles found only in eukaryotic cells.
- (C) Ribosomes are made of DNA and protein and copy genes into mRNA.
- (D) Ribosomes are found only on the rough ER, which is why prokaryotes cannot make proteins.
Answer and explanation
Answer: (A). Ribosomes are built from ribosomal RNA and protein and are not enclosed by a membrane. Every known cell has them, which is evidence that all life shares a common ancestor.
- (B) is wrong twice: ribosomes have no membrane, and bacteria and archaea have them too.
- (C) confuses the ribosome’s job. Ribosomes read mRNA to build proteins; they do not copy DNA, and they contain rRNA, not DNA.
- (D) ignores free ribosomes in the cytosol. Prokaryotes have no ER at all, yet they make proteins on free ribosomes.
Question 2 (multiple choice · core)
A fictional drug stops vesicles from budding off the Golgi complex. Cells treated with it are given a pulse of a radioactive amino acid. What is the most likely location of a labelled secreted protein two hours later?
- (A) Mostly in the Golgi complex, with little outside the cell
- (B) Mostly in lysosomes, where it is digested
- (C) Nowhere, because ribosomes cannot make proteins while the Golgi is blocked
- (D) Mostly outside the cell, because the ER can send proteins straight to the plasma membrane
Answer and explanation
Answer: (A). The protein is made on rough ER ribosomes, enters the ER and travels to the Golgi in transport vesicles. Because no vesicles can leave the Golgi, the protein accumulates there and is not secreted.
- (B) needs vesicles to leave the Golgi for lysosomes, which the drug prevents.
- (C) is wrong because protein synthesis happens on ribosomes and does not depend on the Golgi releasing vesicles.
- (D) skips the Golgi. Secreted proteins pass through the Golgi for modification and packaging before reaching the plasma membrane.
Question 3 (multiple choice · core)
Two mitochondria have the same volume. Mitochondrion M has many deep cristae; mitochondrion N has only a few shallow ones. Which statement best explains why M can make ATP faster?
- (A) M has a larger area of inner membrane, so it can hold more of the proteins that make ATP.
- (B) M has a larger outer membrane, so more glucose can enter.
- (C) The cristae of M store ATP, so less ATP needs to be made.
- (D) M’s folds increase its volume of matrix, so it can hold more fluid.
Answer and explanation
Answer: (A). Proteins that make ATP are in the inner membrane. Deeper and more numerous folds give more membrane area in the same volume, so more ATP can be made at once.
- (B) is wrong because the outer membrane is smooth and is not where ATP is made; the stated difference between M and N is in the inner membrane. (Glucose is also broken down before its products enter the mitochondrion, as you will see in Unit 3.)
- (C) misdescribes cristae. They are folds of membrane, not storage sacs.
- (D) reverses the effect: folding adds membrane area, not matrix volume, and the volumes are stated to be equal.
Question 4 (data analysis · core)
A student measured the percentage of cell volume taken up by the central vacuole in leaf cells of a fictional plant. One group was well watered. The other group had no water for three days and was wilting. She measured three cells in each group.
| Group | Cell 1 / % | Cell 2 / % | Cell 3 / % |
|---|---|---|---|
| Well watered | 72 | 75 | 70 |
| No water for 3 days | 51 | 48 | 54 |
(a) Calculate the mean for each group to one decimal place. (b) Calculate the percentage change in mean vacuole volume after three days without water. (c) Explain how the change in the vacuole is linked to the plant wilting. (d) The ranges of the two groups do not overlap. State what this suggests, and give one reason the conclusion is still limited.
Worked solution and suggested Marlbridge rubric
(a) Well watered: (72 + 75 + 70) ÷ 3 = 217 ÷ 3 = 72.3%. No water: (51 + 48 + 54) ÷ 3 = 153 ÷ 3 = 51.0%.
(b) Percentage change = (51.0 − 72.33) ÷ 72.33 × 100 = −29.5% (a fall of nearly a third). Dividing by the new value instead (−41.8%) is a common error: always divide by the starting value.
(c) The central vacuole stores water. Without water the vacuole loses water and shrinks, so it pushes less on the cell wall. Turgor pressure falls, cells become soft (flaccid), and the non-woody parts of the plant droop.
(d) The difference between the groups is unlikely to be due to chance variation between cells alone. But only three cells per group were measured, perhaps from one plant each, so the result may not hold for the species in general.
| Point | What earns it |
|---|---|
| 1 | Both means correct |
| 1 | −29.5% (or a fall of 29.5%), using the starting mean as the denominator |
| 1 | Vacuole loses water and turgor pressure falls, linked to wilting |
| 1 | Non-overlapping ranges suggest a real difference, plus a valid limitation (small sample, few plants) |
Question 5 (constructed response · core)
Cultured cells of a fictional gland release an enzyme into the liquid around them. A researcher thinks that a chemical, compound K, damages the Golgi complex. She can measure the enzyme’s activity in the liquid.
(a) State a testable hypothesis. (b) Identify the independent variable, the dependent variable and an appropriate control group. (c) Give two variables that must be kept constant. (d) Predict the result if her idea is correct, and explain it using the route of a secreted protein.
Model answer and suggested Marlbridge rubric
(a) “If compound K damages the Golgi complex, cells treated with compound K will release less enzyme into the liquid than untreated cells.”
(b) Independent variable: presence (or concentration) of compound K. Dependent variable: enzyme activity in the liquid after a set time, as a measure of the amount secreted. Control group: identical cells given the same volume of the solvent used to dissolve compound K, but no compound K.
(c) Any two: number of cells per dish; type and age of cells; temperature; time before sampling; volume of liquid; method of measuring enzyme activity.
(d) Enzyme activity in the liquid would be lower with compound K. The enzyme is made on rough ER ribosomes and then must pass through the Golgi, where it is modified and packaged into secretory vesicles. If the Golgi is damaged, fewer vesicles carry the enzyme to the plasma membrane, so less is released.
| Point | What earns it |
|---|---|
| 1 | Testable hypothesis naming compound K and the amount of enzyme released |
| 1 | Independent and dependent variables correct and a solvent-only control |
| 1 | Two valid controlled variables |
| 1 | Prediction of less secretion, linked to the Golgi’s role in packaging for secretion |
Do not award the last point for “compound K stops the enzyme being made”: the hypothesis is about the Golgi, not the ribosomes.
Question 6 (constructed response · stretch)
Three fictional mutant cell lines each have a defect in one structure. Observations:
- Line 1: A newly made secreted protein is found in the rough ER but never reaches the Golgi or the outside of the cell.
- Line 2: Large membrane-bound sacs full of undigested proteins and lipids build up in the cytoplasm over several days.
- Line 3: Proteins are made and secreted normally, but each mitochondrion makes much less ATP. Electron micrographs show inner membranes with very few folds.
(a) For each line, make a claim about which structure or step is defective. (b) Support your claim for Line 2 with reasoning about structure and function. (c) Predict one further effect of the Line 2 defect on the cell or the organism, apart from the build-up of material.
Model answer and suggested Marlbridge rubric
(a) Line 1: transport from the rough ER to the Golgi (the transport vesicles, or their budding from the ER) is defective. Line 2: lysosomes lack one or more working hydrolytic enzymes. Line 3: the inner mitochondrial membrane has too few cristae.
(b) Lysosomes contain hydrolytic enzymes that break polymers into monomers. The sacs are membrane-bound and full of undigested polymers, which is what lysosomes would look like if material was delivered to them but not broken down. The material builds up because it cannot be recycled.
(c) Any one, with a reason: worn-out organelles are not recycled, so the cell may function poorly; monomers such as amino acids are not returned for reuse; apoptosis may not be completed normally, because lysosomes help carry out programmed cell death, so unwanted cells may not be removed properly during development.
| Point | What earns it |
|---|---|
| 1 | Line 1 claim: ER-to-Golgi transport |
| 1 | Line 3 claim: too few cristae, so less inner membrane area for ATP synthesis |
| 1 | Line 2 claim with reasoning: lysosomes, hydrolytic enzymes, undigested polymers accumulate |
| 1 | A plausible further effect with a reason, for example a link to apoptosis |
Accept for Line 1 “the protein cannot leave the ER”. Do not accept “the Golgi is defective” for Line 1 alone: the protein never arrives there.
Question 7 (calculation · stretch)
Fictional measurements were made on single mitochondria from two cell types. Each mitochondrion has an outer membrane area of 12 µm². The inner membrane area is 54 µm² in cell type X and 22 µm² in cell type Y. Assume that both inner membranes carry ATP-making enzyme complexes at the same density: 3000 complexes per µm².
(a) Calculate the ratio of inner to outer membrane area for each cell type. (b) Calculate the number of ATP-making complexes per mitochondrion in each. (c) Calculate how many percent more complexes X has than Y. (d) One cell type is from a muscle that contracts all the time. Make a claim about which one, and state one assumption behind your claim.
Worked solution
(a) X: 54 ÷ 12 = 4.5. Y: 22 ÷ 12 = 1.83, so about 1.8.
(b) X: 54 µm² × 3000 µm⁻² = 162 000 complexes. Y: 22 × 3000 = 66 000 complexes.
(c) (162 000 − 66 000) ÷ 66 000 × 100 = 145% more (X has about 2.5 times as many; 54 ÷ 22 = 2.45). Dividing by X’s value instead gives 59.3%, which answers a different question (“how many percent fewer does Y have?”).
(d) Cell type X. A constantly contracting muscle cell uses ATP quickly. X’s more folded inner membrane holds more ATP-making complexes, so each mitochondrion can make ATP faster. Assumptions (any one): the complexes work at the same rate in both cell types; both cells have similar numbers of mitochondria; the measured mitochondria are typical of their cells.
Suggested mark points (4): 1 for both ratios; 1 for both counts; 1 for 145% using Y as the base; 1 for the claim X with reasoning and an assumption.
How did you do?
- Q1 or Q2 wrong: re-read “Ribosomes” and “The endomembrane system” in the study guide, and trace Figure 1 again.
- Q3 or Q7 wrong: re-read “Mitochondria” and look at Figure 2: folds add membrane area, not volume.
- Q4 calculations wrong: keep the unrounded mean (72.33) and divide by the starting value for a percentage change; for rates and units, rework Worked example 1 (b).
- Q5 or Q6 incomplete: re-read Worked example 2 on claim, evidence and reasoning, and Worked example 1 (d) on predicting a blocked step.
Then tick off the topic checklist.