AQA GCSE Biology 8461: Transport in cells – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.1.3 Transport in cells (4.1.3.1 Diffusion, 4.1.3.2 Osmosis and 4.1.3.3 Active transport) of the AQA GCSE Biology (8461) specification, for first teaching from September 2016 and exams from 2018 onwards. The section is assessed on Paper 1, at Foundation and Higher Tier. The specification marks none of 4.1.3 as (HT only), so every question here suits both tiers. Question 7 is based on required practical activity 3.

Learn the content first in the Transport in cells study guide and the revision notes. These questions are different from those in the Cell biology practice questions, so use both. The course hub is AQA GCSE Biology, and the printable checklist lists every statement.

Questions

1. Define diffusion. [2]

2. Urea is a waste product made in the body.

(a) Describe the movement of urea by diffusion, as given in the specification. Include where it moves from, where it moves to, and the organ that excretes it. [2] (b) Suggest what this tells you about the concentration of urea in the cells compared with the blood plasma. [1]

3. Oxygen diffuses from the air in the alveoli into the blood.

(a) Explain how a rise in body temperature would affect the rate of this diffusion. [2] (b) Explain why a continuous flow of blood through the lung capillaries increases the rate at which oxygen diffuses into the blood. [2]

4. Organism A is a cuboid measuring 1 mm × 1 mm × 4 mm. Organism B is a cube with sides of 6 mm.

(a) Calculate the surface area to volume ratio of organism A. Give your answer as a ratio x : 1. [2] (b) Calculate the surface area to volume ratio of organism B. [2] (c) Suggest which organism is more likely to get enough oxygen by diffusion through its body surface alone. Give a reason. [1]

5. Explain why a large multicellular animal needs both exchange surfaces and a transport system, in terms of surface area to volume ratio. [3]

6. Describe two ways in which the gills of a fish are adapted for gas exchange, and explain how each increases the rate of diffusion. [4]

7. A student investigated the effect of sucrose concentration on the mass of potato cylinders. The results are shown below.

Sucrose concentration (mol/dm³)	Initial mass (g)	Final mass (g)	Percentage change (%)
0.0	2.00	2.30	+15.0
0.2	2.40	2.58	+7.5
0.4	2.20	2.09
0.6	2.50	2.20
0.8	2.00	1.72	−14.0

(a) Calculate the missing percentage changes for 0.4 mol/dm³ and 0.6 mol/dm³. [3] (b) Explain why the student calculated percentage change in mass rather than just the change in mass. [2] (c) Assume the results change in a straight line between 0.2 mol/dm³ and 0.4 mol/dm³. Estimate the sucrose concentration at which the potato cylinders would show no change in mass. [2] (d) Explain the result for 0.0 mol/dm³. [3] (e) Give two variables the student should have controlled. [2]

8. A slice of carrot is placed in a concentrated salt solution for one hour. Its mass decreases. Explain why. [3]

9. Roots were kept in a dilute solution of mineral ions. The concentration of the ions inside the root hair cells was higher than in the solution. When the oxygen supply to the solution was removed, the uptake of ions fell sharply.

(a) Explain why the ions must be absorbed by active transport. [2] (b) Explain why removing oxygen reduced the uptake of ions. [2]

10. A potato cylinder of initial mass 2.10 g gained 0.42 g of water in 30 minutes.

(a) Calculate the percentage gain in mass. [2] (b) Calculate the mean rate of water uptake in g/h. [2]

11. Compare diffusion, osmosis and active transport. Refer to what moves, the direction of movement and the energy needed, and give one example of each from the specification. [6]

Answers

1. The spreading out of particles of a substance in solution or of a gas [1], resulting in a net movement from an area of higher concentration to an area of lower concentration [1]. Examiner insight: “from high to low” without the word concentration does not earn the second mark.

2. (a) Urea diffuses from cells into the blood plasma [1] and is excreted by the kidney [1]. (b) The concentration of urea is higher in the cells than in the blood plasma [1]. Examiner insight: for (b) the comparison must be the right way round; a one-sided answer such as “urea is concentrated” is not credited.

3. (a) The rate would increase [1] because the particles have more kinetic energy and move faster [1]. (b) Blood carries oxygen away from the alveoli as it is absorbed [1], so the oxygen concentration in the blood stays low and a steep concentration gradient is kept [1]. Examiner insight: the second mark in each part needs the reason, not a restatement of the effect.

4. (a) SA = 2 × (1×1 + 1×4 + 1×4) = 18 mm² and V = 1 × 1 × 4 = 4 mm³ [1]; ratio = 18 ÷ 4 = 4.5 : 1 [1]. (b) SA = 6 × 6² = 216 mm² and V = 6³ = 216 mm³ [1]; ratio = 1 : 1 [1]. (c) Organism A, because it has the larger surface area to volume ratio, so more surface for each unit of volume [1]. Examiner insight: a correct ratio with no working can still score full marks, but a wrong answer with clear SA and V working keeps the method mark.

5. A large animal has a small surface area to volume ratio [1], so its outer surface is too small for diffusion to supply all its cells, and inner cells are too far from the surface [1]. Exchange surfaces give a large area for exchange, and a transport system carries substances between these surfaces and every cell [1]. Examiner insight: “small surface area” alone is not enough; the answer must refer to the ratio.

6. First feature [1] with its linked explanation [1]; second feature [1] with its linked explanation [1]. Credit any two of these pairs:

• many gill filaments with thin plates give a large surface area, so more oxygen can diffuse at the same time
• a thin layer of cells between water and blood gives a short diffusion path
• a rich blood supply carries oxygen away and keeps the concentration gradient steep
• water flows over the gills continuously (ventilation), bringing fresh oxygen and keeping the gradient steep. Examiner insight: the explanation mark is only given if it is linked to the feature named; two features listed with no explanation score 2 out of 4.

7. (a) Percentage change = (final − initial) ÷ initial × 100 [1]. 0.4 mol/dm³: (2.09 − 2.20) ÷ 2.20 × 100 = −5.0 % [1]. 0.6 mol/dm³: (2.20 − 2.50) ÷ 2.50 × 100 = −12.0 % [1]. (b) The cylinders had different initial masses [1], so percentage change allows a fair comparison between them [1]. (c) The change falls by 12.5 percentage points over 0.2 mol/dm³; 7.5 ÷ 12.5 × 0.2 = 0.12 [1]; 0.2 + 0.12 = 0.32 mol/dm³ [1]. (d) Distilled water is more dilute than the cell contents [1], so water moves into the cells by osmosis [1] through the partially permeable cell membrane, so the mass increases [1]. (e) Any two from: size or surface area of the cylinders; volume of solution; time in the solution; temperature; type of potato; method of blotting [1] [1]. Examiner insight: in (a) the minus sign is needed for each accuracy mark, and in (c) an answer read from a correctly drawn graph is accepted within a small tolerance.

8. The salt solution is more concentrated than the cell contents [1]. Water moves out of the carrot cells by osmosis [1], from the dilute solution in the cells to the concentrated solution outside, through the partially permeable membrane, so the mass falls [1]. Examiner insight: answers that say salt moves into the carrot, or that water moves “from high to low concentration” without saying water concentration, lose marks.

9. (a) The ions are moving from a more dilute solution to a more concentrated one, against the concentration gradient [1], so diffusion cannot do this and energy is needed [1]. (b) Without oxygen, aerobic respiration in the root cells slows [1], so less energy is released for active transport [1]. Examiner insight: “no energy” alone is not enough in (b); the link to respiration is the creditworthy point.

10. (a) 0.42 ÷ 2.10 × 100 [1] = 20 % [1]. (b) 30 minutes = 0.5 h; 0.42 ÷ 0.5 [1] = 0.84 g/h [1]. Examiner insight: the unit is required for the rate; 0.014 is the rate per minute and does not earn the accuracy mark when g/h is asked for.

11. Credit any six points:

• Diffusion: net movement of particles from a higher to a lower concentration [1]; no energy from respiration needed [1]; example: oxygen or carbon dioxide in gas exchange, or urea into the blood plasma [1].
• Osmosis: movement of water only, from a dilute to a concentrated solution through a partially permeable membrane [1]; no energy needed [1].
• Active transport: moves substances against the concentration gradient using energy from respiration [1]. (Other creditworthy points: example of osmosis such as water entering plant tissue in a dilute solution; example of active transport such as mineral ions into root hair cells or sugar from the gut into the blood.) Examiner insight: AQA marks 6-mark extended responses by levels of response, so a clear comparison across all three processes scores higher than a long account of one.

Where marks are usually lost

• Leaving “concentration” or “net” out of the definition of diffusion.
• Saying solute particles move during osmosis.
• Explaining a fall in active transport as “no energy” without naming respiration.
• Giving surface area alone when the question asks about surface area to volume ratio.
• Leaving out the minus sign for a percentage loss in mass.
• Dividing by the final mass rather than the initial mass.
• Giving a rate without a unit, or in the wrong unit.
• Naming an adaptation of an exchange surface with no link to the rate of diffusion.
• In required practical questions, naming “amount of solution” or “time” without saying it should be kept the same.

Next steps

• Transport in cells revision notes
• Transport in cells study guide
• Cell biology revision notes
• AQA GCSE Biology hub
• Printable checklist for AQA GCSE Biology
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Biology (8461) specification, for first teaching 2016 and exams from 2018 onwards, published by AQA – Topic 4.1 Cell biology, section 4.1.3 Transport in cells, and required practical activity 3.