Edexcel International GCSE Biology 4BI1: Movement of substances into and out of cells – Revision Notes

For full explanations and worked examples, read the study guide first.

These notes cover 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, and no point in this sub-topic has a B reference, so everything here can come up on Paper 1 or Paper 2. They apply to the June and November series examined under Issue 3.

Other pages for this topic: practice questions, cell structure revision notes and cell structure practice. Course hub: Edexcel IGCSE Biology. Checklist: printable 4BI1 checklist.

2.15 Definitions to learn word for word

Term	Definition
Diffusion	Net movement of particles from a region of higher concentration to a region of lower concentration (down a concentration gradient). Passive.
Osmosis	Net movement of water molecules from a dilute solution to a more concentrated solution through a partially permeable membrane. Passive.
Active transport	Movement of substances from a region of lower concentration to a region of higher concentration (against a concentration gradient), using energy from respiration.
Partially permeable membrane	A membrane that lets small molecules such as water through but not larger ones such as sucrose.
Turgid	A plant cell that has taken in water; the vacuole presses the cytoplasm against the cell wall.
Flaccid	A plant cell that has lost water and is no longer firm.
Plasmolysis	Loss of so much water that the cytoplasm and membrane pull away from the cell wall.

Key examples

• Diffusion: oxygen into respiring cells; carbon dioxide out of respiring cells; gases in and out of leaf cells.
• Osmosis: water into root hair cells from soil water; water into or out of potato tissue in sucrose solution.
• Active transport: mineral ions (such as nitrate) into root hair cells; glucose from the gut into intestinal cells when gut glucose is low.

Must-know distinctions

• Diffusion vs osmosis. Osmosis is diffusion of water only, across a partially permeable membrane. Diffusion can be any particle and needs no membrane.
• Passive vs active. Diffusion and osmosis use no energy from respiration. Active transport does, and needs carrier proteins in the membrane.
• Direction. Diffusion: high to low concentration. Osmosis: dilute to concentrated solution. Active transport: low to high concentration.
• Animal vs plant cell in water. Animal cell swells and may burst (no wall). Plant cell becomes turgid but does not burst (wall resists).
• Flaccid vs plasmolysed. Flaccid = lost some water, soft. Plasmolysed = membrane pulled away from the wall.

What a cell does in each solution

Outside solution compared with cell contents	Net water movement	Animal cell	Plant cell
More dilute	In	Swells, may burst	Turgid
Equal	None	No change	No change
More concentrated	Out	Shrinks, shrivels	Flaccid, then plasmolysed

2.16 Factors affecting rate

Factor	Change that speeds movement up	Reason to give
Concentration gradient	Steeper (bigger difference)	Larger net movement down the gradient
Temperature	Higher	Particles have more kinetic energy, move faster
Distance	Shorter	Less distance for particles to travel
Surface area to volume ratio	Larger	More membrane for each unit of volume to be supplied

Active transport also depends on respiration, so a lack of oxygen or a respiratory poison slows it. Diffusion and osmosis are not directly affected by respiration.

Method: surface area to volume ratio

1. Surface area = add the areas of all the faces
   (cube of side s: 6 × s²;  cuboid a × b × c: 2(ab + bc + ac))
2. Volume = length × width × height
3. Ratio = surface area ÷ volume, written as n : 1

Reminders, checked:

• Cuboid 2 cm × 2 cm × 4 cm: surface area = 2(4 + 8 + 8) = 40 cm²; volume = 16 cm³; ratio 2.5 : 1.
• Cube of side 6 cm: surface area = 216 cm²; volume = 216 cm³; ratio 1 : 1.
• Cube of side 0.5 cm: surface area = 1.5 cm²; volume = 0.125 cm³; ratio 12 : 1.

Smaller objects have larger ratios. As an object grows, volume rises faster than surface area.

2.17 Practical checklist

You need one non-living and one living system for diffusion and osmosis.

Non-living

• Agar cubes with indicator in acid: time for complete colour change. Smaller cubes change faster (larger surface area to volume ratio). Warmer acid also speeds it up.
• Visking tubing: acts as the partially permeable membrane. Sucrose solution inside, water outside: tubing gains mass by osmosis. Starch and glucose inside: only glucose is found in the water outside.

Living

• Potato (or other plant tissue) cylinders in a range of sucrose solutions.

Method in steps: potato cylinders

1. Cut equal cylinders with a cork borer; remove skin
2. Blot dry, weigh (starting mass)
3. One cylinder per sucrose concentration, same time, same temperature
4. Remove, blot the same way, reweigh (final mass)
5. % change = (final − start) ÷ start × 100
6. Plot % change against concentration; read where the line crosses 0

Reminders, checked:

• 5.00 g → 4.60 g: (4.60 − 5.00) ÷ 5.00 × 100 = −8.0% (water left; solution more concentrated than cell contents).
• 3.0 g → 3.3 g: +10% (water entered; solution more dilute).
• Rate from a time: if a cube takes 250 s to change colour, rate = 1000 ÷ 250 = 4.0 s⁻¹ (using 1000/time).

Variable type	Potato practical
Independent	Concentration of sucrose solution
Dependent	Percentage change in mass
Control	Cylinder size, potato source, time, temperature, volume of solution, blotting method

Why percentage change? Starting masses differ, so raw change in mass is not a fair comparison.

Where the line crosses zero there is no net water movement: an estimate of the concentration of the cell contents.

Improvements: repeats and a mean; more concentrations near the crossing point; consistent blotting.

Method in steps: explaining an osmosis result

Most “explain” marks on this practical come from a chain of three linked points. Write all three.

1. Compare concentrations: the solution was more dilute / more
   concentrated than the cell contents (cell sap)
2. State the water movement: water moved into / out of the cells
   by osmosis, through the partially permeable cell membrane
3. Link to the measurement: so the mass increased / decreased

For a plant cell that lost water, add the state: flaccid, or plasmolysed if the membrane pulled away from the wall. For no change in mass, say the concentrations were equal, so there was no net movement of water.

Quick self-test

1. Define diffusion.
2. Give two ways osmosis differs from diffusion in general.
3. Which of the three processes needs energy from respiration?
4. A plant cell is placed in pure water. Name the state it ends in and say why it does not burst.
5. Calculate the surface area to volume ratio of a cube of side 5 cm.
6. A 2.00 g potato cylinder has a mass of 2.30 g after an hour in a sucrose solution. Calculate the percentage change in mass.
7. Is the solution in question 6 more or less concentrated than the cell contents?
8. Why do root hair cells need active transport to take up nitrate ions?
9. Give the reason a temperature rise speeds up diffusion.
10. In the potato practical, name the dependent variable.
11. Calculate the surface area to volume ratio of a cuboid 1 cm × 1 cm × 3 cm, to 2 significant figures.
12. Visking tubing containing sucrose solution is placed in water. State and explain what happens to its mass.

Answers

1. Net movement of particles from a region of higher concentration to a region of lower concentration.
2. Osmosis involves water only; it needs a partially permeable membrane.
3. Active transport.
4. Turgid; the cell wall is strong and resists further expansion.
5. Surface area = 6 × 25 = 150 cm²; volume = 125 cm³; ratio = 1.2 : 1.
6. (2.30 − 2.00) ÷ 2.00 × 100 = +15%.
7. Less concentrated (more dilute), because water entered the cells.
8. The concentration of nitrate is often lower in the soil water than in the cell, so ions must move against the gradient.
9. Particles gain kinetic energy and move faster.
10. Percentage change in mass.
11. Surface area = 2(1 + 3 + 3) = 14 cm²; volume = 3 cm³; ratio = 4.7 : 1.
12. It increases: water moves into the tubing by osmosis from the dilute water to the concentrated sucrose solution; sucrose molecules are too large to leave.

Where marks are usually lost

• Leaving out net or concentration from the diffusion definition.
• Saying osmosis moves water from concentrated to dilute. It is dilute to concentrated.
• Describing osmosis as movement of sugar or “solution”. Only water molecules move.
• Missing partially permeable membrane from the osmosis definition.
• Saying active transport needs energy but not saying the energy comes from respiration.
• Writing that a plant cell bursts in water, or that an animal cell becomes turgid.
• Inverting the ratio (volume ÷ surface area) or giving it without “: 1”.
• Explaining the temperature effect without the words kinetic energy.
• Comparing raw mass changes instead of percentage change when starting masses differ.
• Naming “concentration” as a control variable in the potato practical; it is the independent variable.

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.