B3: Transport in Cells
Diffusion, osmosis and active transport
B3 Transport In Cells
📋 Key Concepts
Substances move in and out of cells by three main methods: diffusion, osmosis, and active transport. Which method is used depends on the type of substance, the direction it needs to move, and whether energy is required.
| Method | What Moves | Direction | Energy Needed? |
|---|---|---|---|
| Diffusion | Gases and dissolved substances | High to low concentration | No (passive) |
| Osmosis | Water only | High to low water concentration | No (passive) |
| Active transport | Dissolved substances | Low to high concentration | Yes (from respiration) |
📝 Diffusion
Diffusion is the spreading out of particles from an area of higher concentration to an area of lower concentration. It happens in both solutions and gases because particles are free to move randomly.
Diffusion: High concentration → Low concentration
Particles move down the concentration gradient. No energy required.
Particles move down the concentration gradient. No energy required.
Examples of Diffusion
- Oxygen diffusing from the lungs into the blood (higher concentration in lungs)
- Carbon dioxide diffusing from the blood into the lungs (higher concentration in blood)
- Urea diffusing from cells into the blood for excretion by the kidneys
- Perfume spreading across a room
Factors Affecting Rate of Diffusion
- Concentration gradient - a bigger difference in concentration means faster diffusion
- Temperature - higher temperature means particles have more kinetic energy, so they move faster
- Surface area - larger surface area means more space for particles to cross
- Distance - shorter distance means particles cross faster
Key principle: Diffusion is faster when the concentration gradient is steeper, the temperature is higher, the surface area is larger, and the distance is shorter.
Example 1
Question: Explain why oxygen diffuses from the lungs into the blood.
Answer: The concentration of oxygen is higher in the air sacs (alveoli) of the lungs than in the blood. So oxygen diffuses down the concentration gradient from the alveoli into the blood.
Example 2
Question: How does increasing temperature affect the rate of diffusion?
Answer: Increasing temperature increases the kinetic energy of particles, so they move faster and more randomly. This means they spread out more quickly, increasing the rate of diffusion.
📝 Osmosis
Osmosis is the movement of water molecules from a region of higher water concentration (dilute solution) to a region of lower water concentration (concentrated solution) through a partially permeable membrane.
Osmosis: Water moves from dilute solution → concentrated solution
Through a partially permeable membrane. No energy required.
Remember: Osmosis is just diffusion of WATER across a membrane.
Through a partially permeable membrane. No energy required.
Remember: Osmosis is just diffusion of WATER across a membrane.
Osmosis in Animal Cells
- In dilute solution - water enters the cell by osmosis. The cell swells and may burst (lysis)
- In concentrated solution - water leaves the cell by osmosis. The cell shrinks (crenation)
- In isotonic solution - water enters and leaves at the same rate. The cell stays normal
Osmosis in Plant Cells
- In dilute solution - water enters the cell. The vacuole fills and pushes against the cell wall. The cell becomes turgid (firm and healthy)
- In concentrated solution - water leaves the cell. The vacuole shrinks and the cell membrane pulls away from the cell wall. The cell is plasmolysed (flaccid/wilted)
- In isotonic solution - water movement is balanced. The cell is flaccid (soft but not plasmolysed)
Important: Plant cells do NOT burst in dilute solution because the cell wall provides strong support. Animal cells have no cell wall so they CAN burst.
Example 3
Question: A plant cell is placed in a very dilute sugar solution. Explain what happens.
Answer: Water enters the cell by osmosis (from the dilute solution into the cell's more concentrated cytoplasm/cell sap). The vacuole fills with water and swells, pushing the cytoplasm against the cell wall. The cell becomes turgid. The cell wall prevents the cell from bursting.
Example 4
Question: Explain why a plant wilts when it does not receive enough water.
Answer: Without water, water leaves the plant cells by osmosis (from the cells into the more concentrated soil solution). The vacuoles shrink, the cell membranes pull away from the cell walls, and the cells become plasmolysed. The plant loses its rigidity and wilts.
📝 Osmosis Required Practical
You must know the osmosis practical: Investigating the effect of different sugar solutions on the mass of potato cylinders.
Method
- Cut equal-sized potato cylinders using a cork borer
- Measure and record the mass of each cylinder
- Place each cylinder in a different concentration of sugar solution (e.g. 0.0M, 0.2M, 0.4M, 0.6M, 0.8M, 1.0M)
- Leave for 30 minutes (controlled variable: time)
- Remove, blot dry with paper towel, and measure the mass again
- Calculate the percentage change in mass
Percentage change in mass =
((Final mass - Initial mass) ÷ Initial mass) × 100
((Final mass - Initial mass) ÷ Initial mass) × 100
Interpreting Results
- Mass increases → water entered by osmosis (solution was more dilute than potato)
- Mass decreases → water left by osmosis (solution was more concentrated than potato)
- No change in mass → the solution and potato had equal concentrations (isotonic point)
Example 5
Question: A potato cylinder has an initial mass of 5.0 g. After being placed in 0.8M sugar solution for 30 minutes, its mass is 4.2 g. Calculate the percentage change in mass.
Solution:
Change in mass = 4.2 - 5.0 = -0.8 g
Percentage change = (-0.8 ÷ 5.0) × 100 = -16%
The negative sign means the potato lost mass - water left by osmosis because the sugar solution was more concentrated than the potato cells.
📝 Active Transport
Active transport moves substances from a region of lower concentration to a region of higher concentration. This is AGAINST the concentration gradient, so it requires energy from respiration (ATP).
Active transport: Low concentration → High concentration
AGAINST the concentration gradient. Energy from respiration IS required.
Uses carrier proteins in the cell membrane.
AGAINST the concentration gradient. Energy from respiration IS required.
Uses carrier proteins in the cell membrane.
Examples of Active Transport
- Root hair cells absorb minerals from the soil. Soil has a lower concentration of minerals than the root hair cell, so active transport is needed to move minerals in against the gradient
- Small intestine absorbs glucose even when the concentration of glucose in the blood is higher than in the gut
Why active transport matters: Without it, plants couldn't absorb minerals from dilute soil solutions, and humans couldn't absorb all nutrients from food. Active transport allows cells to "pump" substances in even when they already have a higher concentration.
Example 6
Question: Explain why root hair cells use active transport to absorb minerals from the soil.
Answer: The concentration of minerals in the soil is usually lower than inside the root hair cell. Minerals cannot diffuse into the cell (they would actually diffuse out). Active transport uses energy from respiration to move minerals from the soil into the cell against the concentration gradient.
Example 7
Question: Why does active transport require energy from respiration?
Answer: Active transport moves substances against the concentration gradient (from low to high concentration). This is the opposite of the natural direction of diffusion, so energy is needed to "pump" the substances across the membrane using carrier proteins. This energy comes from respiration (which produces ATP).
📝 Surface Area to Volume Ratio
Smaller objects have a larger surface area to volume ratio. This means substances can diffuse in and out more efficiently relative to the cell's volume.
Surface area to volume ratio = surface area ÷ volume
Small cells: high SA:V ratio → efficient exchange
Large cells: low SA:V ratio → less efficient exchange
Small cells: high SA:V ratio → efficient exchange
Large cells: low SA:V ratio → less efficient exchange
This is why:
- Cells are small - to maximise the surface area for exchange
- Intestines have villi - to increase surface area for absorption
- Lungs have alveoli - to increase surface area for gas exchange
- Root hair cells have long projections - to increase surface area for water and mineral absorption
Example 8
Question: Explain why single-celled organisms do not need specialised exchange surfaces.
Answer: Single-celled organisms are very small, so they have a very large surface area to volume ratio. This means diffusion is fast enough to supply all the substances the cell needs and remove all waste products. No specialised exchange surfaces are needed.
❓ Practice Questions
Q1: Define diffusion. (2 marks)
Q2: Define osmosis. (2 marks)
Q3: Explain the difference between diffusion and active transport. (2 marks)
Q4: A potato cylinder has an initial mass of 6.0 g and a final mass of 6.9 g after being placed in a dilute sugar solution. Calculate the percentage change in mass. (2 marks)
Q5: Explain why root hair cells use active transport to absorb minerals from the soil. (3 marks)
Q6: Explain why a plant cell does not burst when placed in a very dilute solution, but an animal cell can. (3 marks)
Q7: Describe how you would investigate the effect of sugar solution concentration on osmosis in potato cylinders. (6 marks)
✅ Answers
- Diffusion is the movement of particles from an area of higher concentration to an area of lower concentration, down the concentration gradient.
- Osmosis is the movement of water molecules from a region of higher water concentration to a region of lower water concentration through a partially permeable membrane.
- Diffusion moves substances down the concentration gradient and does not require energy. Active transport moves substances against the concentration gradient and requires energy from respiration.
- Percentage change = ((6.9 - 6.0) ÷ 6.0) × 100 = (0.9 ÷ 6.0) × 100 = +15%
- Root hair cells use active transport because the concentration of minerals in the soil is lower than inside the cell. Minerals cannot diffuse in (they would diffuse out). Active transport uses energy from respiration to move minerals against the concentration gradient into the cell.
- When placed in a dilute solution, water enters both cells by osmosis. The plant cell has a rigid cellulose cell wall that provides support and prevents the cell from bursting - the cell becomes turgid instead. The animal cell has no cell wall, so as water enters, the cell swells and can burst (lysis).
- Cut equal-sized potato cylinders; measure and record initial mass; place each in a different concentration of sugar solution; leave for a set time (e.g. 30 minutes); remove and blot dry; measure final mass; calculate percentage change in mass; plot a graph of percentage change against concentration. Control variables: time, temperature, volume of solution, surface area of potato.
🎯 Exam Tips
- Always say "partially permeable membrane" when defining osmosis - you lose marks without it
- For osmosis, say "water concentration" not just "concentration" - be specific
- Remember: diffusion and osmosis are passive (no energy), active transport needs energy
- In the osmosis practical, always BLOT DRY before measuring final mass
- Percentage change can be positive OR negative - the sign tells you the direction
- SA:V ratio questions often appear - remember small = high ratio = efficient exchange
- When asked to "explain" osmosis, mention both the direction AND the membrane
🔬 Required Practical
Osmosis with Potato Cylinders
- Cut equal-sized potato cylinders using a cork borer and trim to the same length
- Measure and record the mass of each cylinder
- Place each cylinder in a different concentration of sugar solution (e.g. 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M)
- Leave for 30 minutes (control variable: time)
- Remove, blot dry with a paper towel, and re-measure the mass
- Calculate percentage change in mass
- Plot a graph of percentage change against concentration
Control variables: Volume of solution, temperature, time, surface area of potato, blotting technique.
🔢 Maths Skills
Mathematical Skills
Percentage change = ((change in mass) ÷ (original mass)) × 100
Positive value = gain in mass (water entered by osmosis)
Negative value = loss in mass (water left by osmosis)
Positive value = gain in mass (water entered by osmosis)
Negative value = loss in mass (water left by osmosis)
Maths Example
A potato cylinder has initial mass 4.8 g and final mass 5.6 g. Calculate the percentage change.
Change = 5.6 − 4.8 = +0.8 g. % change = (0.8 ÷ 4.8) × 100 = +16.7%
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Osmosis is just water diffusion Correct: Osmosis specifically requires a partially permeable membrane — without it, water movement is just diffusion
2. Wrong: Water moves to the side with more water Correct: Water moves from a dilute solution (more water, fewer solutes) to a concentrated solution (less water, more solutes) through a partially permeable membrane
✍️ 6-Mark Question
Extended Answer
6 marks: Describe the osmosis required practical and explain how to make the results valid.
Cut equal-sized potato cylinders using a cork borer and measure their initial mass. Place each cylinder in a different concentration of sugar solution (e.g. 0.0 M to 1.0 M). Leave for a fixed time such as 30 minutes, then remove, blot dry and re-measure the mass. Calculate the percentage change in mass for each cylinder. To make results valid: control variables including volume of solution, temperature, time, and surface area of potato must be kept the same; repeat the experiment to identify anomalies; blot dry to remove surface water before measuring final mass; use a sufficient range of concentrations to see the full trend.
Mark scheme: 1 mark for method (cut and measure); 1 mark for range of concentrations; 1 mark for measuring mass change; 1 mark for calculating percentage change; 1 mark for control variables; 1 mark for repeats/anomaly identification
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student obtained the following results from the osmosis practical:
| Concentration (M) | % change (Repeat 1) | % change (Repeat 2) | % change (Repeat 3) |
|---|---|---|---|
| 0.0 | +18.2 | +17.9 | +12.1 |
| 0.2 | +10.5 | +10.1 | +10.3 |
| 0.4 | +2.3 | +2.1 | +2.5 |
| 0.6 | −6.0 | −5.8 | −5.9 |
| 0.8 | −13.2 | −12.9 | −13.0 |
(a) Identify the anomaly and suggest a cause.
(b) Estimate the concentration inside the potato cells.
Answers: (a) The value +12.1 at 0.0 M is anomalous — the other two repeats are close (+18.2, +17.9). Likely cause: the cylinder was not blotted dry properly, or excess surface water was included in the mass measurement. (b) The isotonic point (where % change ≈ 0) is between 0.4 M and 0.6 M, so the concentration inside the potato cells is approximately 0.5 M.
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