Edexcel International GCSE Biology 4BI1: Movement of substances into and out of cells – Study Guide

This study guide teaches sub-topic 2(d), Movement of substances into and out of cells (specification points 2.15–2.17), of the Pearson Edexcel International GCSE Biology (4BI1) specification, Issue 3. The qualification is untiered. None of the three points in this sub-topic carries a B reference, so all of it can be examined on Paper 1 and on Paper 2. It applies to the June and November series examined under Issue 3.

Use it with the revision notes and the practice questions. The cell structures this topic depends on, especially the cell membrane, cell wall and vacuole, are covered in the cell structure and biological molecules study guide and the cell structure revision notes. The full course is on the Edexcel IGCSE Biology hub, and you can tick off each point on the printable checklist.

What this sub-topic covers

Spec point	What you must be able to do	Paper
2.15	Understand diffusion, osmosis and active transport as the processes that move substances into and out of cells	Papers 1 and 2
2.16	Understand how surface area to volume ratio, distance, temperature and concentration gradient affect the rate of movement	Papers 1 and 2
2.17	Practical: investigate diffusion and osmosis using living and non-living systems	Papers 1 and 2

Experimental skills are assessed through the written papers, so you need to be able to describe the practical, process its results and evaluate it on paper. A calculator may be used on both papers.

2.15 Three ways substances cross a membrane

Every cell is surrounded by a cell membrane. It is partially permeable: small molecules such as oxygen, carbon dioxide and water pass through it, but larger molecules such as sucrose and starch cannot. Three processes move substances across it.

Diffusion

Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. The particles move down a concentration gradient. It is passive: the particles move because of their own random motion, so the cell uses no energy from respiration.

The word “net” matters. Particles move randomly both ways, but more move from the crowded side. When the concentrations are equal, particles still move, but there is no net movement.

Examples you should be able to use:

• oxygen diffuses from the blood into respiring body cells, where its concentration is lower because it is being used up
• carbon dioxide diffuses out of respiring cells into the blood
• carbon dioxide diffuses into a leaf cell that is photosynthesising, and oxygen diffuses out

Osmosis

Osmosis is the net movement of water molecules from a dilute solution to a more concentrated solution through a partially permeable membrane. Put another way, water moves from where there is a higher concentration of water molecules to where there is a lower concentration of water molecules. You may also see this described as movement from a higher to a lower water potential.

Osmosis is a special case of diffusion. It is still passive, but it is about water only, and it needs a partially permeable membrane. The solute (for example sucrose) cannot cross the membrane, so only the water moves.

What happens to cells depends on the solution around them.

Solution around the cell	Animal cell (red blood cell)	Plant cell
More dilute than cell contents (e.g. pure water)	Water enters; cell swells and may burst, because there is no cell wall	Water enters; vacuole swells and pushes the cytoplasm against the cell wall; cell becomes turgid but does not burst
Same concentration as cell contents	No net movement of water; no change	No net movement; no change
More concentrated than cell contents	Water leaves; cell shrinks and becomes shrivelled	Water leaves; cell becomes flaccid; if more water is lost the cytoplasm and membrane pull away from the wall (plasmolysis)

The cell wall is the reason a plant cell does not burst in water. It is strong and resists further expansion. Turgid cells press against each other, which is what keeps non-woody plant tissue firm.

Active transport

Active transport is the movement of substances from a region of lower concentration to a region of higher concentration, against a concentration gradient. It uses energy released by respiration and needs carrier proteins in the cell membrane.

Because it depends on respiration, anything that stops respiration (lack of oxygen, a poison, a low temperature) slows or stops active transport. Cells that carry out a lot of active transport tend to contain many mitochondria.

Examples:

• root hair cells absorb mineral ions such as nitrate from the soil, even when the concentration in the soil water is lower than in the cell
• cells lining the small intestine absorb glucose from the gut when its concentration in the gut has fallen below that in the cells

Comparing the three

	Diffusion	Osmosis	Active transport
What moves	Any particles that can pass	Water only	Particular ions and molecules
Direction	High to low concentration	Dilute to concentrated solution	Low to high concentration
Energy from respiration	No	No	Yes
Membrane needed	No (can happen in air or water)	Yes, partially permeable	Yes, with carrier proteins

2.16 Factors that affect the rate of movement

The specification names four factors. For each, learn the direction of the effect and the reason.

Concentration gradient. The bigger the difference in concentration between two regions, the faster the net movement. Living systems keep gradients steep: blood flow carries oxygen away from the lungs as soon as it arrives, and respiring cells keep using oxygen up.

Temperature. Higher temperature gives particles more kinetic energy, so they move faster and diffusion and osmosis speed up. Active transport also speeds up with temperature, up to a point, because respiration is controlled by enzymes.

Distance. The shorter the distance particles must travel, the faster they arrive. This is why exchange surfaces are thin, often one cell thick.

Surface area to volume ratio. A larger surface area relative to volume means more membrane for substances to cross for each unit of volume to be supplied. Small objects have a larger surface area to volume ratio than large ones of the same shape.

To find surface area to volume ratio:

1. Surface area = total area of all the faces. For a cube of side s, that is 6 × s².
2. Volume = length × width × height. For a cube, s³.
3. Ratio = surface area ÷ volume, written as n : 1.

Worked example 1: comparing cubes

Calculate the surface area to volume ratio of a 1 cm cube and a 3 cm cube.

1 cm cube:  surface area = 6 × 1² = 6 cm²
            volume       = 1³     = 1 cm³
            ratio        = 6 ÷ 1  = 6 : 1

3 cm cube:  surface area = 6 × 3² = 54 cm²
            volume       = 3³     = 27 cm³
            ratio        = 54 ÷ 27 = 2 : 1

The 3 cm cube has 27 times the volume but only 9 times the surface area.

Worked example 2: how far does the substance get?

Agar cubes containing an indicator change colour as acid diffuses in. After 10 minutes, the acid has moved 0.5 cm in from every face of a 2 cm cube and of a 4 cm cube. What percentage of each cube has changed colour?

2 cm cube: unchanged core = (2 − 0.5 − 0.5)³ = 1³ = 1 cm³
           total volume   = 2³ = 8 cm³
           changed        = (8 − 1) ÷ 8 × 100 = 87.5%

4 cm cube: unchanged core = (4 − 1)³ = 27 cm³
           total volume   = 4³ = 64 cm³
           changed        = (64 − 27) ÷ 64 × 100 = 57.8% (3 s.f.)

The acid moves in at the same speed in both cubes, but in the larger cube a smaller fraction is reached in the same time. The ratios explain why: 3 : 1 for the 2 cm cube and 1.5 : 1 for the 4 cm cube. The same idea explains why large organisms cannot rely on diffusion across their outer surface alone.

2.17 Practical: diffusion and osmosis in living and non-living systems

The specification asks for both kinds of system. Know one example of each.

Non-living systems

• Agar cubes with indicator. Cut agar containing an indicator into cubes of different sizes, place them in acid, and time how long each takes to change colour completely. This models the effect of surface area to volume ratio (or of temperature, if you use one cube size in acid at different temperatures).
• Visking tubing. This is a partially permeable membrane. Tie off a length containing sucrose solution, record its mass, stand it in water, then blot and reweigh. The tubing gains mass by osmosis. Visking tubing containing a starch and glucose mixture in water also shows that glucose diffuses out but starch does not.

Living system: plant tissue in sucrose solutions

1. Cut cylinders of potato of equal length and diameter using a cork borer, removing the skin.
2. Blot each cylinder dry and record its starting mass.
3. Place one cylinder in each of a range of sucrose solutions (for example 0.0 to 0.8 mol/dm³) for the same time at the same temperature.
4. Remove, blot gently to remove surface liquid, and record the final mass.
5. Calculate the percentage change in mass for each.

Percentage change in mass = (final mass − starting mass) ÷ starting mass × 100.

You use percentage change because the cylinders do not start with exactly the same mass. It lets you compare them fairly.

Variables: independent, concentration of sucrose; dependent, percentage change in mass; control, cylinder size, source of potato, time in solution, temperature, volume of solution.

Worked example 3: processing the results

Sucrose / mol/dm³	Start / g	End / g	Change / g	% change
0.0	2.40	2.76	+0.36	+15.0
0.2	2.50	2.70	+0.20	+8.0
0.4	2.20	2.16	−0.04	−1.8
0.6	2.60	2.34	−0.26	−10.0
0.8	2.50	2.15	−0.35	−14.0

Check one row: for 0.4 mol/dm³, (2.16 − 2.20) ÷ 2.20 × 100 = −1.8%.

Interpreting: where mass increased, the solution was more dilute than the cell contents, so water entered the cells by osmosis. Where mass decreased, the solution was more concentrated, so water left. On a graph of percentage change against concentration, the line crosses zero between 0.2 and 0.4 mol/dm³. Reading the crossing point from a straight line between those two points gives about 0.36 mol/dm³. At that concentration there would be no net movement of water, so it is an estimate of the concentration of the potato cell contents.

Improving it: repeat each concentration and calculate a mean; use more concentrations between 0.2 and 0.4 mol/dm³ to locate the crossing point more precisely; blot every cylinder in the same way.

Common errors

• Writing “particles move from high to low” without the word concentration, or without net.
• Describing osmosis as the movement of solute, or of “water and sugar”. Only water crosses.
• Saying water moves from a concentrated to a dilute solution. It is the other way round.
• Saying plant cells burst in water. The cell wall stops them; they become turgid.
• Saying active transport “needs energy” without saying it comes from respiration.
• Giving surface area to volume ratio as a single number without “: 1”, or dividing volume by surface area.
• Explaining a temperature effect with “particles move faster” but forgetting kinetic energy.
• Comparing raw mass change when the starting masses differ.

Next steps

Condense this with the revision notes, then test yourself with the practice questions. For enzymes and cell structure, use the cell structure practice questions. To find gaps across the course, try one of the free diagnostics.

Official syllabus

Pearson Edexcel International GCSE in Biology (4BI1) specification, Issue 3, published by Pearson Education Limited: Topic 2, Structures and functions in living organisms, sub-topic (d) Movement of substances into and out of cells, points 2.15–2.17.