AQA GCSE Biology

AQA GCSE Biology 4.1.3 Transport in Cells Exam Questions

Free AQA GCSE Biology exam questions for 4.1.3 Transport in Cells, with a mark scheme for every pack. Covers diffusion, osmosis and active transport, surface area to volume ratio, exchange surfaces and the osmosis required practical.

4.1.3 Transport in Cells exam questions

1 question pack of real past exam questions, each with its mark scheme. Do not stop after a few: answer as many as you can without notes, then mark every answer with the mark scheme. The more questions you do, the more quickly you will spot how examiners ask about each idea.

4.1.3 Transport in cells exam questions for AQA GCSE BiologyQuestionsMark scheme
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About AQA GCSE Biology 4.1.3 Transport in Cells

Transport in cells completes the AQA Cell biology section on Paper 1. It explains how substances move into and out of cells by diffusion, osmosis and active transport, and why large organisms need specialised exchange surfaces. These ideas return in gas exchange in the lungs, absorption in the small intestine, root hair cells and transpiration. Examiners test the factors affecting diffusion, surface area to volume ratio calculations, and the osmosis required practical, where you calculate percentage change in mass and plot results to estimate the concentration of the cell contents.

What you need to know

  • 4.1.3.1 Diffusion Diffusion is the spreading out of particles of a gas or solute, resulting in a net movement from an area of higher concentration to an area of lower concentration. Rate depends on the concentration gradient, temperature and surface area of the membrane. Examples: oxygen and carbon dioxide in gas exchange, urea from cells into blood plasma.
  • Surface area to volume ratio Single-celled organisms have a large surface area to volume ratio so diffusion meets their needs. Multicellular organisms need exchange surfaces and transport systems. Effective exchange surfaces have a large surface area, thin membrane, good blood supply and (in animals) ventilation.
  • 4.1.3.2 Osmosis Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane. Cells in a more concentrated solution lose water; in a more dilute solution they gain water.
  • Required practical: osmosis Place cylinders of potato of equal size in a range of sugar or salt concentrations, blot and weigh before and after. Calculate percentage change in mass and plot against concentration; where the line crosses zero gives the concentration with no net movement of water.
  • 4.1.3.3 Active transport Active transport moves substances from a more dilute solution to a more concentrated solution, against a concentration gradient, and needs energy from respiration. Examples: mineral ion uptake by root hair cells and glucose absorption from the gut into the blood when the concentration in the gut is lower.

Common mistakes in 4.1.3 Transport in Cells

  • Describing osmosis as water moving from high to low concentration of solution. Water moves from a dilute solution to a more concentrated solution through a partially permeable membrane.
  • Saying diffusion and osmosis need energy. Both are passive; only active transport needs energy from respiration.
  • Forgetting 'net movement' in definitions. Particles move randomly in both directions, so describe the overall net movement down the gradient.
  • Using change in mass instead of percentage change. Percentage change allows fair comparison because the starting masses differ.
  • Saying a larger organism has a larger surface area to volume ratio. As size increases, the surface area to volume ratio decreases.

Try three 4.1.3 Transport in Cells exam-style questions

1. A potato cylinder had a mass of 2.50 g. After 30 minutes in a sugar solution its mass was 2.20 g. Calculate the percentage change in mass. (2 marks)

Show the mark scheme answer

Change = 2.20 - 2.50 = -0.30 g; (-0.30 / 2.50) x 100 = -12%

2. Calculate the surface area to volume ratio of a cube with sides of 3 cm. (2 marks)

Show the mark scheme answer

Surface area = 6 x 3 x 3 = 54 cm2; volume = 27 cm3; ratio = 54 : 27 = 2 : 1

3. Explain why root hair cells need to use active transport to absorb mineral ions. (3 marks)

Show the mark scheme answer

Concentration of mineral ions in the soil is lower than inside the root hair cell; so ions must move against the concentration gradient; this requires energy from respiration; diffusion would move ions out of the cell

4.1.3 Transport in Cells FAQs

What does AQA GCSE Biology 4.1.3 Transport in cells cover?

It covers diffusion, osmosis and active transport, the factors that affect the rate of diffusion, surface area to volume ratio and the features of exchange surfaces.

Which paper is 4.1.3 Transport in cells examined on?

Transport in cells is part of Cell biology, assessed on Paper 1.

What do I need to know about the osmosis required practical?

Know the method, the variables to control (cylinder size, time, temperature), how to calculate percentage change in mass, and how to use a graph to find the concentration where there is no change in mass.