AQA GCSE Biology 8461: Transport in cells – Study Guide

This study guide teaches 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. It covers every statement in 4.1.3.1 Diffusion, 4.1.3.2 Osmosis and 4.1.3.3 Active transport, including required practical activity 3. Cell biology is assessed on Paper 1 (topics 1 to 4), which is set at Foundation and Higher Tier. The specification marks none of the statements in 4.1.3 as (HT only), so everything here is for both tiers.

For all of Topic 1 in brief, see the Cell biology study guide. The AQA GCSE Biology hub lists all topics, and the printable checklist lets you tick off each statement.

What this section covers

Spec	What you must be able to do
4.1.3.1	Define diffusion; name oxygen, carbon dioxide and urea as substances that diffuse
4.1.3.1	Explain how concentration gradient, temperature and membrane surface area affect the rate of diffusion
4.1.3.1	Calculate and compare surface area to volume ratios
4.1.3.1	Explain why multicellular organisms need exchange surfaces and a transport system
4.1.3.1	Explain how the small intestine, lungs, gills, roots and leaves are adapted for exchange
4.1.3.2	Define osmosis; calculate percentage gain and loss of mass; use rates of water uptake; plot and interpret graphs
4.1.3.2	Required practical activity 3: effect of salt or sugar concentration on the mass of plant tissue
4.1.3.3	Describe active transport, why it needs energy, and its role in root hairs and the gut
4.1.3.3	Describe and explain the differences between diffusion, osmosis and active transport

Diffusion (4.1.3.1)

Substances move into and out of cells across the cell membrane by diffusion. Learn the specification’s definition:

Diffusion is 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.

Particles move randomly, so more leave a crowded area than enter it: the net movement is down the concentration gradient. Diffusion needs no energy from the cell.

The specification names three substances that move by diffusion:

• oxygen and carbon dioxide in gas exchange (for example, oxygen diffuses from the air in the lungs into the blood)
• urea, a waste product, which diffuses from cells into the blood plasma so that it can be removed by the kidney.

Factors that affect the rate of diffusion

Factor	Change that increases the rate	Why
Concentration gradient	A bigger difference in concentration	More particles move down the gradient than move back each second
Temperature	A higher temperature	Particles have more kinetic energy, so they move faster
Surface area of the membrane	A larger surface area	More particles can cross at the same time

When you explain a factor, give the change and the reason, not just “temperature affects diffusion”.

Surface area to volume ratio

A single-celled organism has a relatively large surface area to volume ratio, so diffusion across its membrane can supply enough oxygen and remove enough waste to meet its needs.

Model organisms as cubes or cuboids:

surface area of a cube = 6 × (side length)²
volume of a cube       = (side length)³
surface area to volume ratio = surface area ÷ volume, written as  x : 1

Worked example 1. Compare cubes with sides of 1 cm, 2 cm and 4 cm.

Side 1 cm: SA = 6 × 1² = 6 cm²     V = 1³ = 1 cm³     ratio 6 ÷ 1  = 6 : 1
Side 2 cm: SA = 6 × 2² = 24 cm²    V = 2³ = 8 cm³     ratio 24 ÷ 8 = 3 : 1
Side 4 cm: SA = 6 × 4² = 96 cm²    V = 4³ = 64 cm³    ratio 96 ÷ 64 = 1.5 : 1

Each time the side doubles, the ratio halves.

Worked example 2. Shape matters as well as size. Compare a cube of side 4 mm with a flat cuboid 1 mm × 8 mm × 8 mm. Both have a volume of 64 mm³.

Cuboid SA = 2 × (1×8 + 8×8 + 1×8) = 2 × 80 = 160 mm²
Cuboid ratio = 160 ÷ 64 = 2.5 : 1
Cube ratio   = 96 ÷ 64  = 1.5 : 1

The flat shape has the larger ratio, which is why thin, flat structures make good exchange surfaces.

Why multicellular organisms need exchange surfaces and transport systems

A large multicellular organism has a small surface area to volume ratio. Its outer surface is too small to supply all its cells by diffusion, and inner cells are too far from the surface. So it needs:

1. Specialised exchange surfaces (organs such as the lungs or the small intestine) that give a large area for substances to move into and out of the body.
2. A transport system (such as the blood in animals) that carries substances between the exchange surface and every cell, so each cell is only a short distance from a supply.

Adaptations of exchange surfaces

The specification lists four features that increase the effectiveness of an exchange surface:

• a large surface area
• a thin membrane, giving a short diffusion path
• (in animals) an efficient blood supply
• (in animals, for gaseous exchange) being ventilated.

Blood supply and ventilation keep the concentration gradient steep by removing absorbed substances and bringing in fresh air or water.

Exchange surface	Large surface area	Short diffusion path	Other features
Small intestine	Millions of villi lining the inner wall	Villus wall is one cell thick	Good capillary network carries absorbed food away
Lungs	Millions of alveoli	Alveolar and capillary walls are thin	Rich blood supply; breathing ventilates the alveoli
Fish gills	Many gill filaments, each with many thin plates	Thin layer of cells between water and blood	Rich blood supply; water flows over the gills continuously
Roots	Root hair cells with long projections	Thin cell walls	Ions and water are taken from the soil
Leaves	Flat, thin shape; air spaces inside the leaf	Gases only travel a short distance	Stomata let carbon dioxide in and oxygen and water vapour out

Link each feature to its effect: “villi give a large surface area, so more glucose is absorbed at once”.

Osmosis (4.1.3.2)

Water moves across cell membranes by osmosis:

Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.

A partially permeable membrane lets water through but not larger solute molecules such as sucrose. A dilute solution has more water and less solute than a concentrated one. Osmosis needs no energy.

Applied to plant tissue:

• In a solution more dilute than the cell contents, the cells gain water and the tissue gains mass.
• In a solution more concentrated than the cell contents, the cells lose water and the tissue loses mass.
• In a solution at the same concentration as the cell contents, there is no net movement of water and the mass does not change.

Percentage change in mass

Pieces of tissue start at different masses, so compare percentage change:

percentage change in mass = (final mass − initial mass) ÷ initial mass × 100

A gain gives a positive answer and a loss gives a negative answer. Keep the sign.

Worked example 3. A potato cylinder goes from 2.50 g to 2.80 g. A second goes from 2.40 g to 2.04 g.

Cylinder 1: (2.80 − 2.50) ÷ 2.50 × 100 = 0.30 ÷ 2.50 × 100 = +12.0 %
Cylinder 2: (2.04 − 2.40) ÷ 2.40 × 100 = −0.36 ÷ 2.40 × 100 = −15.0 %

If three repeats at one concentration give +11.5 %, +12.5 % and +12.0 %, the mean is (11.5 + 12.5 + 12.0) ÷ 3 = +12.0 %.

Rate of water uptake

A rate is a compound measure: an amount divided by a time.

Worked example 4. A potato cylinder gains 0.36 g of water in 30 minutes. Calculate the rate of water uptake.

rate of water uptake = 0.36 g ÷ 30 min = 0.012 g/min
in grams per hour: 0.36 g ÷ 0.5 h = 0.72 g/h

Always give the unit.

Required practical activity 3

The specification asks you to investigate the effect of a range of concentrations of salt or sugar solutions on the mass of plant tissue. A typical method:

1. Cut potato cylinders with a cork borer and trim them to the same length.
2. Blot each dry and record its mass.
3. Place one in each of a range of sugar or salt solutions (for example 0.0 to 0.8 mol/dm³), same volume in each.
4. Leave for the same time at the same temperature.
5. Remove, blot dry the same way, and record the final mass.
6. Calculate percentage change in mass, repeat, and find the mean for each concentration.

Variables. Independent: concentration of the solution. Dependent: percentage change in mass. Control: type of tissue, size and shape of the cylinders, volume of solution, time, temperature, and how the cylinders are blotted.

Graph. Plot concentration on the x-axis and percentage change in mass on the y-axis, with a zero line across the middle. Draw a line or curve of best fit. Where the line crosses zero, the solution has the same concentration as the cell contents.

Worked example 5. At 0.2 mol/dm³ the mean change is +4.0 %; at 0.4 mol/dm³ it is −6.0 %. Assuming a straight line between these points, where is zero?

The change falls by 10.0 percentage points over 0.2 mol/dm³.
From +4.0 % to 0 is 4.0 of those 10.0 points: 4.0 ÷ 10.0 × 0.2 = 0.08
Concentration ≈ 0.2 + 0.08 = 0.28 mol/dm³

Improvements. Surface water left after blotting adds to the final mass, so blot every cylinder the same way. Use more concentrations near the crossing point, and repeat to spot anomalies.

Active transport (4.1.3.3)

Active transport moves substances from a more dilute solution to a more concentrated solution (against a concentration gradient). This requires energy from respiration.

The specification gives two examples.

• Root hair cells absorb mineral ions from very dilute solutions in the soil. The concentration of ions inside the cell is higher, so the ions must be moved against the gradient. Plants need these ions for healthy growth.
• In the gut, sugar molecules are absorbed from lower concentrations in the gut into the blood, which has a higher sugar concentration. Cells use the sugar for respiration.

Anything that reduces respiration reduces active transport. Roots in waterlogged soil get little oxygen, so aerobic respiration slows and mineral ion uptake falls.

Comparing the three processes

	Diffusion	Osmosis	Active transport
What moves	Particles of any substance in solution or a gas	Water only	Substances such as mineral ions and sugar
Direction	Higher → lower concentration	Dilute → concentrated solution	Dilute → concentrated solution (against the gradient)
Membrane	Any membrane the particle can cross	Partially permeable membrane needed	Cell membrane
Energy from respiration	Not needed	Not needed	Needed
Example	Oxygen into blood in the lungs	Water into root hair cells	Mineral ions into root hair cells

The trap: osmosis and active transport both go “dilute to concentrated” in terms of the solution, but in osmosis water moves down its own gradient, while in active transport the solute moves up its gradient.

Common errors

• Writing “from high to low” without the word concentration.
• Defining osmosis without “partially permeable membrane”, or saying solute moves in osmosis.
• Calculating surface area of a cube as side² rather than 6 × side².
• Dropping the minus sign on a percentage loss in mass.
• Dividing by the final mass instead of the initial mass.
• Saying active transport “uses energy” without saying the energy comes from respiration.
• Listing exchange-surface features with no link to how they increase the rate of diffusion.

Next, drill recall with the revision notes, try the practice questions, and find weak areas with the free diagnostics.

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.