AQA GCSE Biology 8461: Transport in cells – Revision Notes

These notes condense section 4.1.3 Transport in cells of the AQA GCSE Biology (8461) specification, for first teaching from September 2016 and exams from 2018 onwards: 4.1.3.1 Diffusion, 4.1.3.2 Osmosis (with required practical activity 3) and 4.1.3.3 Active transport. The section is examined on Paper 1, at Foundation and Higher Tier, and none of it is marked (HT only), so all of it applies to both tiers.

For full explanations and worked examples, use the Transport in cells study guide. Test yourself afterwards with the Transport in cells practice questions. The Cell biology revision notes cover the rest of Topic 1. Course hub: AQA GCSE Biology. Tick off each statement on the printable checklist.

Definitions to learn exactly

Term	Definition (specification wording)
Diffusion	The spreading out of the particles of any substance in solution, or particles of a gas, resulting in a net movement from an area of higher concentration to an area of lower concentration
Osmosis	The diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane
Active transport	Moves substances from a more dilute solution to a more concentrated solution (against a concentration gradient); requires energy from respiration

Key words that carry marks: net, concentration (diffusion), partially permeable membrane (osmosis), and respiration (active transport).

4.1.3.1 Diffusion

Named examples in the specification:

• oxygen and carbon dioxide in gas exchange
• urea from cells into the blood plasma, for excretion in the kidney.

Rate of diffusion goes up when:

Factor	Increase rate by	Reason
Concentration gradient	Making it steeper	Bigger net movement per second
Temperature	Raising it	Particles move faster
Surface area of membrane	Increasing it	More particles cross at once

Surface area to volume ratio (SA:V)

cube:   SA = 6 × L²        V = L³
cuboid: SA = 2 × (lw + lh + wh)    V = l × w × h
ratio = SA ÷ V, written as  x : 1

Method in steps

1. Work out total surface area (all six faces).
2. Work out volume.
3. Divide SA by V. Give it as x : 1.
4. Compare: the larger the ratio, the more surface there is for each unit of volume.

Small reminder: a cube of side 0.5 cm has SA = 6 × 0.25 = 1.5 cm² and V = 0.125 cm³, so SA:V = 12 : 1. Halving the side doubles the ratio.

Why multicellular organisms need more than their outer surface:

• Single-celled organism: relatively large SA:V, so diffusion through its surface meets its needs.
• Multicellular organism: small SA:V, so the outer surface is too small and the inner cells are too far away.
• So it needs exchange surfaces (large area) and a transport system (to carry substances to and from every cell).

Exchange surfaces

Four features that increase effectiveness (from the specification):

1. large surface area
2. thin membrane, so a short diffusion path
3. efficient blood supply (animals)
4. ventilation (animals, gas exchange).

Features 3 and 4 work by keeping the concentration gradient steep.

Surface	Area	Short path	Gradient kept by
Small intestine	Villi	Wall one cell thick	Capillary network
Lungs	Alveoli	Thin alveolar and capillary walls	Blood flow and breathing
Gills	Filaments with thin plates	Thin layer between water and blood	Blood flow and water flowing over
Roots	Root hair cells	Thin cell wall	Water and ions moving on into the root
Leaves	Flat, thin; internal air spaces	Short distance for gases	Stomata open to the air

4.1.3.2 Osmosis

What happens to plant tissue:

Solution compared with cell contents	Net water movement	Mass
More dilute	Into the cells	Increases
Same concentration	None	No change
More concentrated	Out of the cells	Decreases

Calculations you must be able to do:

percentage change in mass = (final − initial) ÷ initial × 100
rate of water uptake      = mass of water gained ÷ time taken   (e.g. g/min, g/h)
mean                      = sum of repeats ÷ number of repeats

• Gains are positive, losses negative. Keep the sign.
• Always divide by the initial mass.
• Give units for every rate.

Small reminder: 5.00 g → 4.60 g is (4.60 − 5.00) ÷ 5.00 × 100 = −8.0 %.

Required practical activity 3

Aim: investigate the effect of a range of concentrations of salt or sugar solutions on the mass of plant tissue.

Method in steps

1. Cut equal-sized cylinders of plant tissue (for example potato) with a cork borer.
2. Blot, then measure the initial mass of each.
3. Put each in a different concentration of solution, same volume, same time, same temperature.
4. Remove, blot the same way, measure the final mass.
5. Calculate percentage change; repeat; find the mean.
6. Plot concentration (x-axis) against percentage change in mass (y-axis); draw a line of best fit.

Variable	In this practical
Independent	Concentration of salt or sugar solution
Dependent	Percentage change in mass
Control	Tissue type, cylinder size, volume of solution, time, temperature, blotting method

Reading the graph: where the line crosses 0 %, the solution has the same concentration as the cell contents. Points above zero mean water entered; points below zero mean water left.

Main source of error: uneven blotting. Extra surface water raises the final mass.

4.1.3.3 Active transport

• Moves substances against the concentration gradient (dilute → concentrated).
• Needs energy from respiration.

Example	From	To	Why it matters
Mineral ions into root hair cells	Very dilute soil solution	Root hair cell (higher concentration)	Plants need ions for healthy growth
Sugar from the gut into the blood	Lower concentration in the gut	Blood (higher sugar concentration)	Sugar is used for cell respiration

Less respiration (for example, little oxygen in waterlogged soil) means less active transport.

Must-know distinctions

	Diffusion	Osmosis	Active transport
Substance	Any, in solution or a gas	Water only	Solutes such as ions, sugar
Gradient	Down	Down (for water)	Against
Energy from respiration	No	No	Yes
Membrane	Any it can cross	Partially permeable	Cell membrane

• Osmosis vs diffusion: osmosis is a special case of diffusion. Only water moves, and a partially permeable membrane is needed.
• Osmosis vs active transport: both can be described as “dilute to concentrated solution”. In osmosis water moves down its own gradient with no energy. In active transport the solute moves up its gradient using energy.
• Surface area vs SA:V: a bigger organism has more surface area in total but a smaller ratio.

Quick self-test

1. State the waste product that diffuses from cells into the blood plasma.
2. Give three factors that affect the rate of diffusion.
3. Calculate the SA:V of a cube of side 3 cm.
4. Which has the larger SA:V, a cube of side 3 cm or a cube of side 0.5 cm?
5. State two features of all effective exchange surfaces.
6. Define osmosis.
7. A piece of tissue goes from 1.60 g to 1.84 g. Calculate the percentage change in mass.
8. A cylinder gains 0.60 g of water in 40 minutes. Calculate the rate of water uptake in g/min.
9. In required practical 3, why is percentage change used rather than change in mass?
10. Why does active transport stop if respiration stops?
11. Give one example of active transport in plants and one in animals.
12. On a graph of percentage change in mass against concentration, what does the point where the line crosses zero tell you?

Answers

1. Urea.
2. Concentration gradient, temperature, surface area of the membrane.
3. SA = 6 × 3² = 54 cm²; V = 3³ = 27 cm³; SA:V = 54 ÷ 27 = 2 : 1.
4. The 0.5 cm cube (12 : 1 compared with 2 : 1).
5. Any two of: large surface area; thin membrane (short diffusion path); efficient blood supply (animals); ventilated (animals, gas exchange).
6. The diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
7. (1.84 − 1.60) ÷ 1.60 × 100 = 0.24 ÷ 1.60 × 100 = +15 %.
8. 0.60 ÷ 40 = 0.015 g/min.
9. The pieces start at different masses, so percentage change allows a fair comparison.
10. Active transport needs energy, and that energy comes from respiration.
11. Plants: mineral ions absorbed into root hair cells from the soil. Animals: sugar absorbed from the gut into the blood.
12. The concentration of solution that is the same as the concentration of the cell contents (no net movement of water).

Where marks are usually lost

• Leaving out “concentration” in “from a higher to a lower concentration”.
• Saying osmosis is the movement of “particles” or “solute” instead of water.
• Forgetting “partially permeable membrane” in the osmosis definition.
• Writing “active transport uses energy” without “from respiration”.
• Using L² instead of 6 × L² for the surface area of a cube.
• Dividing by the final mass in a percentage change calculation, or losing the minus sign for a loss.
• Giving a rate of water uptake with no unit.
• Naming an exchange-surface feature without saying how it speeds up diffusion.
• Saying a larger organism has a “smaller surface area” when the point is a smaller surface area to volume ratio.

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.