Active transport and co-transport - AQA A-Level Biology (Paper 1) Exam Questions

Practise 17 exam-style questions on Active transport and co-transport for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Explain what is meant by co-transport.

Model answer

In co-transport, a carrier protein moves two different substances across a membrane at the same time, for example sodium ions and glucose. One substance (sodium ions) moves down its concentration gradient, and this allows the other (glucose) to be moved against its concentration gradient.

Explanation

Co-transport does not use ATP directly: the energy comes from one substance diffusing down its concentration gradient. That gradient is maintained by active transport elsewhere in the cell, such as the sodium-potassium pump, which does use ATP. In the ileum, sodium ions and glucose (or amino acids) enter the epithelial cells together through co-transport proteins.

Mark scheme

1 mark
Two different substances are moved across a membrane together by the same carrier protein
Accept 'a co-transport protein moves an ion and a molecule at the same time'. Accept a correct example, e.g. sodium ions and glucose.
1 mark
One substance moves down its concentration gradient, which allows / provides the energy for the other to move against its concentration gradient
Accept 'sodium ions diffusing in down their gradient drive the uptake of glucose against its gradient'. Reject 'the co-transport protein hydrolyses ATP'.

Question 2

3 marks

The free surface of a cell lining the ileum, facing the lumen, measures 9 μm × 9 μm. It is covered by 2500 microvilli. Each microvillus is a cylinder 1.2 μm long and 0.1 μm in diameter.

Curved surface area of a cylinder = π × diameter × length. Use π = 3.14

Assume that the surface area of the membrane with microvilli = area of the flat surface + total curved surface area of the microvilli.

Calculate how many times greater the surface area of this membrane is with microvilli than it would be without microvilli. Give your answer to 2 significant figures.

Explain how microvilli increase the rate of absorption of glucose by the cell.

Model answer

Area of flat surface = 9 × 9 = 81 μm²
Curved area of one microvillus = 3.14 × 0.1 × 1.2 = 0.3768 μm²
Total curved area = 2500 × 0.3768 = 942 μm²
Surface area with microvilli = 81 + 942 = 1023 μm²
1023 ÷ 81 = 12.6 = 13 times greater (2 s.f.)

The microvilli give much more cell-surface membrane, which can hold more co-transport proteins, so more glucose can be absorbed at the same time.

Explanation

Work out the area of the flat surface (9 × 9) and of the microvilli (number × π × diameter × length), add them, and divide by the flat area. Microvilli are folds of the cell-surface membrane of a single cell; they are not the same as villi, which are folds of the whole lining, made of many cells. A larger area of membrane holds more carrier and co-transport proteins, so the rate of absorption increases.

Mark scheme

1 mark
Correct method: (81 + 2500 × 3.14 × 0.1 × 1.2) ÷ 81, i.e. 1023 ÷ 81; or a correct final answer of 13 with no working
A correct answer of 13 with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 12.6 or 12.63 (correct value not rounded to 2 significant figures) OR 942 or 1023 (a correct area in μm² but no ratio) OR 12 or 11.6 (the flat surface area not added).
1 mark
13
Must be given to 2 significant figures for this mark.
1 mark
More membrane, so more (co-transport / carrier) proteins, so more glucose can be absorbed at the same time
Insufficient: 'larger surface area' alone. Reject references to villi.

Question 3

3 marks

Some cancer cells have a carrier protein in their cell-surface membrane that uses ATP to pump anticancer drugs out of the cell. Scientists measured the relative number of these pump proteins in cells from 12 tumours, and the percentage of an anticancer drug that remained inside the cells 1 hour after it was added.

The graph shows their results.

Describe and explain the relationship shown in the graph. Suggest why adding a respiratory inhibitor would make these cancer cells more likely to be killed by the drug.

Scatter graph on a grid of small squares. The horizontal axis is "Relative number of pump proteins", from 0 to 10, numbered every 1, with 2 small squares between numbered values (each small square is 0.5). The vertical axis is "Drug remaining in cells after 1 hour (%)", from 0 to 100, numbered every 10, with 5 small squares between numbered values (each small square is 2). 12 points are plotted as crosses: (1, 90), (1.5, 84), (2, 80), (3, 70), (3.5, 62), (4, 60), (5, 48), (5.5, 46), (6, 40), (7, 30), (8, 24) and (9, 18).

Model answer

There is a negative correlation: the more pump proteins a cell has, the smaller the percentage of the drug left inside it after 1 hour (90% with 1 unit of pump protein, 18% with 9 units). More pump proteins means more drug molecules can be actively transported out of the cell at the same time. A respiratory inhibitor would reduce the production of ATP, so the pumps could not move the drug out; more drug would stay in the cells and kill them.

Explanation

The rate of active transport depends on the number of carrier proteins and on the supply of ATP. These pumps make cancer cells resistant to drugs by removing the drug before it can act, so blocking the ATP supply (or the pump itself) keeps the drug inside the cell. A correlation alone does not prove cause, but here the explanation is supported by what the pump is known to do.

Mark scheme

1 mark
As the number of pump proteins increases, the percentage of drug remaining (in the cells) decreases / negative correlation
Accept a correct description using values from the graph.
1 mark
More pump proteins, so more drug is actively transported out of the cells at the same time / per unit time
Accept 'more carriers to move the drug out against its concentration gradient'.
1 mark
A respiratory inhibitor reduces ATP production, so the pumps cannot remove the drug, so more drug stays in the cells (to kill them)
The idea of less ATP for the pump is needed. Reject 'less energy is produced'.

Question 4

5 marks

Young salmon hatch in rivers and later move into the sea. In the sea they gain sodium and chloride ions from the water they drink. Cells in their gills called chloride cells move these ions out of the blood into the sea water, against their concentration gradients.

Scientists moved young salmon from fresh water into sea water. They measured the activity of the sodium-potassium pump protein in the chloride cells, as the rate at which it released phosphate from ATP, and the number of mitochondria in these cells. The table shows their results.

Calculate the percentage increase in pump activity between day 0 and day 28. Give your answer to 3 significant figures.

Use the information to explain how these changes help the salmon to survive in sea water.

Table with columns "Time in sea water / days", "Pump activity / μmol phosphate released per mg protein per hour", "Number of mitochondria per chloride cell (relative to day 0)". Time in sea water / days: 0; Pump activity / μmol phosphate released per mg protein per hour: 2.1; Number of mitochondria per chloride cell (relative to day 0): 1.0 | Time in sea water / days: 3; Pump activity / μmol phosphate released per mg protein per hour: 4.8; Number of mitochondria per chloride cell (relative to day 0): 1.4 | Time in sea water / days: 7; Pump activity / μmol phosphate released per mg protein per hour: 8.6; Number of mitochondria per chloride cell (relative to day 0): 1.9 | Time in sea water / days: 14; Pump activity / μmol phosphate released per mg protein per hour: 11.2; Number of mitochondria per chloride cell (relative to day 0): 2.3 | Time in sea water / days: 28; Pump activity / μmol phosphate released per mg protein per hour: 11.5; Number of mitochondria per chloride cell (relative to day 0): 2.4.

Model answer

Percentage increase = (11.5 - 2.1) ÷ 2.1 × 100 = 9.4 ÷ 2.1 × 100 = 448%

After the move to sea water the pump activity rose from 2.1 to 11.5 units, so the chloride cells have more active pump proteins and can move more sodium (and chloride) ions out of the blood. Moving ions against their concentration gradient needs ATP, which the pump hydrolyses, releasing phosphate. The number of mitochondria more than doubled, so more ATP can be produced by aerobic respiration for this active transport. This stops the concentration of ions in the blood rising, so the water potential of the blood does not fall and the salmon does not lose too much water by osmosis.

Explanation

Percentage increase = (new - original) ÷ original × 100, so divide by the day 0 value. The rate of active transport depends on the number of carrier (pump) proteins and on the supply of ATP. Measuring the release of phosphate is a way of measuring ATP hydrolysis by the pump. The chloride cells adapt to sea water by having more pump proteins and more mitochondria. Remember that mitochondria produce ATP; they do not 'produce energy'.

Mark scheme

1 mark
Correct method: (11.5 - 2.1) ÷ 2.1 × 100; or a correct final answer of 448 with no working
Max 5. Points 3 to 6: max 3. A correct answer of 448% with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 4.48 (not multiplied by 100) OR 81.7 (difference divided by 11.5) OR 447.6 or 450 (correct value not given to 3 significant figures).
1 mark
448 (%)
1 mark
More pump activity / more pump proteins, so more (sodium and chloride) ions are moved out of the blood by active transport
1 mark
More mitochondria, so more (aerobic) respiration / more ATP is produced
Reject 'more energy is produced'.
1 mark
ATP is hydrolysed / needed to move the ions against their concentration gradient
Accept 'the release of phosphate shows the pump hydrolysing ATP'.
1 mark
(So) the ion concentration of the blood does not rise / the water potential of the blood does not fall, so the salmon does not lose (too much) water by osmosis
Accept 'prevents the blood becoming too concentrated'.

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