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. Concentrations are in mmol L⁻¹ and areas in µm². A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data. All data sets are fictional.
Question 1 (multiple choice · foundation)
Which process requires a direct input of metabolic energy from the cell?
- (A) Oxygen moving from the blood into a respiring muscle cell
- (B) Carbon dioxide moving out of a respiring muscle cell
- (C) A gland cell releasing digestive enzymes by exocytosis
- (D) Water moving into a cell through protein channels, down its gradient
Answer and explanation
Answer: (C). Exocytosis needs ATP to move vesicles to the plasma membrane and to fuse them with it. It is bulk transport, and all bulk transport uses energy.
- (A) is simple diffusion. The muscle cell uses oxygen, so the concentration inside stays low and oxygen moves in down its gradient.
- (B) is also simple diffusion, in the other direction: CO₂ is made inside the cell, so it is more concentrated there.
- (D) involves a protein, but movement down a gradient through a channel is passive. A protein does not by itself mean energy is used.
Question 2 (multiple choice · foundation)
A membrane separates two solutions of glucose. Glucose can cross the membrane passively. After several hours, the glucose concentration is 5 mmol L⁻¹ on both sides. Which statement describes what is happening now?
- (A) Glucose molecules have stopped crossing the membrane.
- (B) Glucose molecules cross in both directions at equal rates, so there is no net movement.
- (C) Glucose molecules cross only from the side that originally had the higher concentration.
- (D) Glucose molecules now cross only if the cell supplies ATP.
Answer and explanation
Answer: (B). Molecules keep moving randomly. With equal concentrations, equal numbers hit the membrane from each side, so they cross at equal rates in both directions. This is a dynamic equilibrium.
- (A) mistakes “no net movement” for “no movement”.
- (C) Random motion has no preferred direction.
- (D) is wrong because passive movement continues at equilibrium; it is just balanced.
Question 3 (multiple choice · core)
A cell is placed in a solution of an uncharged sugar at 1.0 mmol L⁻¹. After 30 minutes, the sugar inside the cell is at 6.0 mmol L⁻¹. In a second test, a drug that stops ATP production is added first; after 30 minutes the sugar inside is at 0.1 mmol L⁻¹. Which conclusion is best supported?
- (A) The sugar enters by simple diffusion through the bilayer.
- (B) The sugar enters by facilitated diffusion, which needs ATP.
- (C) The sugar enters by active transport, which needs ATP and can move it from low to high concentration.
- (D) The drug damaged the membrane, so the sugar leaked out.
Answer and explanation
Answer: (C). The sugar reaches six times its outside concentration (6.0 ÷ 1.0 = 6), which passive transport cannot do. When ATP production is blocked, almost no sugar accumulates. Both observations point to active transport.
- (A) Simple diffusion could at most bring the inside up to 1.0 mmol L⁻¹, and a polar sugar crosses the bilayer poorly anyway.
- (B) Facilitated diffusion is passive. It uses a protein but no ATP, and it cannot build up a substance above its outside concentration.
- (D) A leaky membrane would let sugar in towards 1.0 mmol L⁻¹, not keep it at 0.1.
Question 4 (data analysis · core)
Discs cut from a carrot root were kept in a solution of an uncharged nutrient, compound X. The air bubbled through the solution had different percentages of oxygen. The rate of uptake of X was measured.
| Oxygen in air / % | 0 | 5 | 10 | 21 |
|---|---|---|---|---|
| Rate of uptake of X / µmol g⁻¹ h⁻¹ | 0.6 | 2.9 | 4.6 | 4.8 |
(a) Describe the effect of oxygen on the rate of uptake, using numbers from the table. (b) Explain why oxygen affects the rate of uptake of X. (c) Calculate the rate at 0% oxygen as a percentage of the rate at 21% oxygen, and suggest why uptake does not fall to zero. (d) Suggest why the rate barely changes between 10% and 21% oxygen.
Worked solution and suggested Marlbridge rubric
(a) As oxygen rises from 0% to 10%, the rate rises steeply, from 0.6 to 4.6 µmol g⁻¹ h⁻¹ (an increase of 4.0). From 10% to 21% it rises only from 4.6 to 4.8, so the rate levels off.
(b) The carrot cells use oxygen for aerobic respiration, which makes ATP. More oxygen means more ATP. The steep dependence on oxygen suggests that most uptake of X is active transport, which needs a direct input of energy from ATP.
(c) 0.6 ÷ 4.8 × 100 = 12.5% (about 13%). Uptake continues without oxygen because some X may enter passively down its gradient, and the cells can make a small amount of ATP without oxygen (anaerobic respiration).
(d) Above about 10% oxygen, ATP supply is no longer the limiting factor. Something else now limits the rate, most likely the number of transport proteins in the membranes, which are all working as fast as they can.
| Point | What earns it |
|---|---|
| 1 | Trend described as a rise then a levelling off, with at least two quoted values |
| 1 | Oxygen linked to aerobic respiration and ATP and active transport |
| 1 | 12.5% (accept 12% or 13%) with a valid reason for uptake at 0% oxygen |
| 1 | Another factor named as limiting, such as the number of transport proteins |
Do not award (b) for “oxygen gives the cells energy”: oxygen is not an energy source; it allows respiration to make ATP.
Question 5 (experimental design · core)
A student adds a radioactively labelled sugar to a culture of yeast. After 10 minutes, she washes the yeast and finds radioactivity inside the cells. She concludes: “The sugar enters yeast cells by active transport.”
(a) Explain why her evidence does not show active transport. (b) Propose an investigation that would test her conclusion. Include the independent variable, the dependent variable, two controlled variables and a control condition. (c) State the results that would support active transport.
Model answer and suggested Marlbridge rubric
(a) Finding the sugar inside only shows that it entered. It could have entered passively, down its gradient. To show active transport, she needs evidence that the sugar builds up to a higher concentration inside than outside, or that its entry depends on ATP. She measured neither.
(b) One good design:
- Measure the sugar concentration inside the cells and in the medium at several times (for example every 5 minutes for 60 minutes), so the two can be compared.
- Independent variable: presence or absence of a metabolic inhibitor that blocks ATP production.
- Dependent variable: concentration of labelled sugar inside the cells.
- Controlled variables (any two): temperature, starting sugar concentration in the medium, number or mass of yeast cells, pH, time of incubation, age of the culture.
- Control condition: yeast without the inhibitor (with the inhibitor’s solvent only). Use at least three replicate cultures in each group.
(c) Active transport is supported if, without the inhibitor, the inside concentration rises above the outside concentration, and if, with the inhibitor, uptake slows and does not rise above the outside concentration.
| Point | What earns it |
|---|---|
| 1 | Explains that entry alone could be passive; needs a comparison with the outside concentration or ATP dependence |
| 1 | Valid independent and dependent variables, with internal concentration measured |
| 1 | Two controlled variables and a suitable control condition (and replicates) |
| 1 | Predicted results: build-up above the outside concentration that is lost when ATP is blocked |
Accept other ways of lowering ATP if the answer explains how ATP is lowered. Removing oxygen alone is weak for yeast, because yeast can still make some ATP by fermentation.
Question 6 (calculation · stretch)
A secretory cell has a plasma membrane area of 800 µm². It releases its product by exocytosis. Each vesicle adds 0.050 µm² of membrane when it fuses, and 1200 vesicles fuse every minute. Over several hours, the area of the plasma membrane stays the same.
(a) Calculate the area of membrane added by exocytosis in one hour, and express it as a multiple of the cell’s surface area. (b) Explain how the cell’s surface area can stay constant. (c) A drug blocks endocytosis but not exocytosis. Predict the surface area after 10 minutes, and give one possible consequence for the cell. (d) Explain why exocytosis needs energy even though the product is not moved against a gradient through a protein.
Worked solution and suggested Marlbridge rubric
(a) Per minute: 1200 × 0.050 µm² = 60 µm². Per hour: 60 × 60 = 3600 µm². As a multiple of the surface: 3600 ÷ 800 = 4.5 times the whole plasma membrane every hour.
(b) The cell removes membrane by endocytosis at the same rate as exocytosis adds it, and recycles it into new vesicles. The total area stays roughly constant.
(c) Added in 10 minutes: 60 × 10 = 600 µm². New area: 800 + 600 = 1400 µm², an increase of 75%. Possible consequences (any one): the cell runs short of membrane to make new vesicles, so secretion slows; the cell surface folds or the cell enlarges; membrane proteins pile up at the surface.
(d) The cell uses ATP to build the vesicle, move it along the cytoskeleton and drive fusion. The energy moves membrane and cargo; it does not push cargo against a gradient.
| Point | What earns it |
|---|---|
| 1 | 3600 µm² per hour |
| 1 | 4.5 times the surface area |
| 1 | Endocytosis named as removing membrane at an equal rate (recycling) |
| 1 | 1400 µm² (or +600 µm², 75%) with a sensible consequence |
| 1 | Energy linked to vesicle formation, movement or fusion |
Common error: forgetting to multiply by 60 gives only 60 µm², which is the amount per minute.
Question 7 (constructed response · stretch)
A single-celled freshwater alga is analysed after being kept in the dark, so it is not photosynthesising. The table shows three substances (fictional data).
| Substance | Outside the cell | Inside the cell | Other observation |
|---|---|---|---|
| Oxygen (O₂) | 0.26 mmol L⁻¹ | 0.20 mmol L⁻¹ | the cell respires aerobically |
| Sugar S (uncharged) | 0.05 mmol L⁻¹ | 1.5 mmol L⁻¹ | uptake continues while inside is higher |
| Protein P (large) | found in the water | found inside vesicles | vesicles seen fusing with the plasma membrane |
(a) For each substance, identify the most likely way it crosses the plasma membrane and justify each choice from the data. (b) Calculate the inside : outside ratio for sugar S. (c) A toxin stops ATP production. Predict what happens to the movement of each substance, with a reason.
Model answer and suggested Marlbridge rubric
(a)
- Oxygen: simple diffusion into the cell. It is small and nonpolar, so it crosses the bilayer, and it is more concentrated outside (0.26) than inside (0.20), because respiration uses it up.
- Sugar S: active transport into the cell. It is far more concentrated inside, and uptake continues even so, which means it is moving from low to high concentration.
- Protein P: exocytosis out of the cell. It is too large for channels or pumps, it is packaged in vesicles and the vesicles are seen fusing with the plasma membrane.
(b) 1.5 ÷ 0.05 = 30. The inside is 30 times as concentrated as the outside.
(c)
- Oxygen: still crosses by simple diffusion, as long as there is a gradient. (If respiration stops, the cell stops using oxygen, so the gradient will disappear over time and net movement will stop.)
- Sugar S: uptake against the gradient stops, because the pump needs ATP. The inside concentration can no longer be raised, and it may fall if sugar leaks out down its gradient.
- Protein P: secretion stops, because vesicle movement and fusion need ATP.
| Point | What earns it |
|---|---|
| 1 | Oxygen: simple diffusion, justified by the gradient and small nonpolar size |
| 1 | Sugar S: active transport, justified by uptake against the gradient |
| 1 | Protein P: exocytosis, justified by vesicles or size |
| 1 | Ratio of 30 |
| 1 | Correct predictions for all three, each with a reason linked to ATP or the gradient |
Do not award the oxygen point for “facilitated diffusion”: O₂ needs no protein.
How did you do?
- Q1 or Q2 wrong: re-read “Concentration gradients and diffusion” and “Passive transport” in the study guide.
- Q3, Q4 or Q7 wrong: rework Worked example 1; check that you used both lines of evidence (build-up above the outside concentration and ATP dependence).
- Q5 incomplete: re-read Worked example 2, part (c).
- Q6 wrong: re-read “Membrane bookkeeping” in the bulk transport section.
Then tick off the topic checklist.