B3: Transport in Cells
Diffusion, osmosis, active transport, and surface area to volume ratio
Key Definitions
Diffusion — The net movement of particles from an area of higher concentration to an area of lower concentration (down a concentration gradient). Passive — no energy required.
Osmosis — The net movement of water molecules across a partially permeable membrane from a region of higher water concentration (dilute solution) to a region of lower water concentration (more concentrated solution). Passive — no energy required.
Active transport — The movement of substances from a more dilute solution to a more concentrated solution (against a concentration gradient). Requires energy from respiration (ATP).
Partially permeable membrane — A membrane that allows only certain molecules (e.g. small water molecules) to pass through but not larger molecules (e.g. sucrose).
Concentration gradient — The difference in concentration between two regions.
Osmosis — The net movement of water molecules across a partially permeable membrane from a region of higher water concentration (dilute solution) to a region of lower water concentration (more concentrated solution). Passive — no energy required.
Active transport — The movement of substances from a more dilute solution to a more concentrated solution (against a concentration gradient). Requires energy from respiration (ATP).
Partially permeable membrane — A membrane that allows only certain molecules (e.g. small water molecules) to pass through but not larger molecules (e.g. sucrose).
Concentration gradient — The difference in concentration between two regions.
Diffusion, Osmosis and Active Transport Compared
| Feature | Diffusion | Osmosis | Active Transport |
|---|---|---|---|
| What moves | Any particles (gas or dissolved) | Water molecules only | Dissolved substances (ions, molecules) |
| Direction | High → low concentration | High water concentration → low water concentration | Low → high concentration (against gradient) |
| Energy required? | No (passive) | No (passive) | Yes (from respiration / ATP) |
| Membrane needed? | No (occurs in any space) | Yes (partially permeable membrane) | Yes (cell membrane with carrier proteins) |
| Examples | Oxygen into blood; perfume spreading; CO₂ into leaves | Water into plant roots; water into red blood cells | Mineral ions into root hair cells; glucose from gut into blood |
Exam tip: Osmosis is a SPECIAL CASE of diffusion — it only involves water molecules and always requires a partially permeable membrane. If the question mentions water, it is osmosis. If it mentions any other substance moving down a gradient, it is diffusion.
Factors Affecting Diffusion Rate
Diffusion is faster when:
— Concentration gradient is steeper — bigger difference in concentration between two regions.
— Temperature is higher — particles have more kinetic energy and move faster.
— Surface area is larger — more space for particles to cross.
— Diffusion distance is shorter — particles have less distance to travel.
— Concentration gradient is steeper — bigger difference in concentration between two regions.
— Temperature is higher — particles have more kinetic energy and move faster.
— Surface area is larger — more space for particles to cross.
— Diffusion distance is shorter — particles have less distance to travel.
Surface Area to Volume Ratio (SA:V)
SA:V ratio determines how efficiently substances can move in and out of cells or organisms.
— Smaller objects have a larger SA:V ratio — exchange is more efficient relative to volume.
— Larger objects have a smaller SA:V ratio — exchange is less efficient relative to volume.
Organisms adapt to increase SA:V:
— Alveoli in lungs — millions of tiny sacs give huge surface area for gas exchange.
— Villi in small intestine — finger-like projections increase surface area for absorption.
— Root hair cells — long thin projections increase surface area for water and mineral uptake.
— Smaller objects have a larger SA:V ratio — exchange is more efficient relative to volume.
— Larger objects have a smaller SA:V ratio — exchange is less efficient relative to volume.
Organisms adapt to increase SA:V:
— Alveoli in lungs — millions of tiny sacs give huge surface area for gas exchange.
— Villi in small intestine — finger-like projections increase surface area for absorption.
— Root hair cells — long thin projections increase surface area for water and mineral uptake.
Example 1: Calculating SA:V Ratio
A cube with sides of 2 cm:Surface area = 6 × 2² = 24 cm²
Volume = 2³ = 8 cm³
SA:V = 24:8 = 3:1
A cube with sides of 4 cm:
Surface area = 6 × 4² = 96 cm²
Volume = 4³ = 64 cm³
SA:V = 96:64 = 1.5:1
The smaller cube has a larger SA:V ratio, so substances diffuse in and out more efficiently relative to its volume.
Osmosis in Animal Cells
In a dilute (hypotonic) solution — Water enters the cell by osmosis. The cell swells and may burst (lysis). Red blood cells in pure water will burst.
In an isotonic solution — Water enters and leaves at the same rate. The cell stays normal — this is the ideal condition for animal cells.
In a concentrated (hypertonic) solution — Water leaves the cell by osmosis. The cell shrinks and becomes crenated (shrivelled). Red blood cells in strong salt solution will crenate.
In an isotonic solution — Water enters and leaves at the same rate. The cell stays normal — this is the ideal condition for animal cells.
In a concentrated (hypertonic) solution — Water leaves the cell by osmosis. The cell shrinks and becomes crenated (shrivelled). Red blood cells in strong salt solution will crenate.
Osmosis in Plant Cells
In a dilute solution — Water enters the cell by osmosis. The vacuole swells and pushes the cytoplasm and cell membrane against the cell wall. The cell becomes turgid (firm and swollen). This is the normal, healthy state for plant cells and provides structural support.
In an isotonic solution — Water enters and leaves at the same rate. The cell is flaccid (slightly soft).
In a concentrated solution — Water leaves the cell by osmosis. The vacuole shrinks, the cytoplasm and cell membrane pull away from the cell wall. The cell becomes plasmolysed. The cell wall does NOT collapse because it is rigid.
In an isotonic solution — Water enters and leaves at the same rate. The cell is flaccid (slightly soft).
In a concentrated solution — Water leaves the cell by osmosis. The vacuole shrinks, the cytoplasm and cell membrane pull away from the cell wall. The cell becomes plasmolysed. The cell wall does NOT collapse because it is rigid.
Example 2: Osmosis in Red Blood Cells
If red blood cells are placed in distilled water (very dilute), water enters by osmosis because the inside of the cell has a lower water concentration. The cells swell and eventually undergo lysis (burst). If placed in 2% salt solution (roughly isotonic), the cells remain normal. If placed in 5% salt solution (hypertonic), water leaves by osmosis and the cells become crenated (shrivelled).
Example 3: Osmosis in Plant Cells
If a plant cell is placed in distilled water, water enters by osmosis. The vacuole expands and pushes the cytoplasm against the cell wall. The cell becomes turgid — this is healthy and provides support. If placed in strong sugar solution, water leaves by osmosis, the cytoplasm shrinks and pulls away from the cell wall — the cell is plasmolysed. The cell wall remains intact because it is made of rigid cellulose.
Osmosis Required Practical — Potato Cylinders
Method:
1. Cut equal-sized potato cylinders using a cork borer.
2. Measure and record the mass of each cylinder.
3. 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).
4. Leave for at least 30 minutes.
5. Remove, blot dry with paper towel, and measure the final mass.
6. Calculate the percentage change in mass for each cylinder.
1. Cut equal-sized potato cylinders using a cork borer.
2. Measure and record the mass of each cylinder.
3. 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).
4. Leave for at least 30 minutes.
5. Remove, blot dry with paper towel, and measure the final mass.
6. Calculate the percentage change in mass for each cylinder.
Percentage change in mass = ((Final mass − Initial mass) ÷ Initial mass) × 100
Interpreting results:
— Positive % change — Mass increased; water entered by osmosis (solution was more dilute than the potato cell).
— Negative % change — Mass decreased; water left by osmosis (solution was more concentrated than the potato cell).
— 0% change — No net movement; the solution and potato cell had the same water concentration (isotonic).
The concentration of sugar inside the potato cells can be estimated by finding where the line of best fit crosses the x-axis on a graph of % change in mass vs sugar concentration.
— Positive % change — Mass increased; water entered by osmosis (solution was more dilute than the potato cell).
— Negative % change — Mass decreased; water left by osmosis (solution was more concentrated than the potato cell).
— 0% change — No net movement; the solution and potato cell had the same water concentration (isotonic).
The concentration of sugar inside the potato cells can be estimated by finding where the line of best fit crosses the x-axis on a graph of % change in mass vs sugar concentration.
Example 4: Calculating Percentage Change in Mass
A potato cylinder has an initial mass of 5.20 g. After being placed in 0.4 M sugar solution, its final mass is 4.65 g.Percentage change = ((4.65 − 5.20) ÷ 5.20) × 100
= (−0.55 ÷ 5.20) × 100
= −10.6%
The negative value means the potato lost mass — water left the cell by osmosis because the 0.4 M solution was more concentrated than the potato cell cytoplasm.
Example 5: Active Transport in Root Hair Cells
Root hair cells use active transport to absorb mineral ions from the soil. The concentration of minerals in the soil is often lower than inside the root hair cell, so the ions cannot move by diffusion. Instead, carrier proteins in the cell membrane use energy from respiration (ATP) to move ions against the concentration gradient into the cell. This is why root hair cells have many mitochondria — to provide the ATP needed for active transport.
Example 6: Active Transport in the Small Intestine
When the concentration of glucose in the gut is lower than in the blood, glucose cannot diffuse into the blood. Instead, active transport is used: carrier proteins in the cell membrane of the gut lining use energy from respiration to move glucose from the dilute gut into the more concentrated blood. This ensures all glucose is absorbed, even when most has already been taken up.
Exam tip: When explaining osmosis, always use the phrase "water potential" or "water concentration" — never say "water moves to where there is more sugar". Say "water moves from a region of higher water concentration to a region of lower water concentration". Also, always mention the partially permeable membrane.
Practice Questions
1. Foundation Define osmosis.
Osmosis is the net movement of water molecules across a partially permeable membrane from a region of higher water concentration (dilute solution) to a region of lower water concentration (more concentrated solution).
2. Foundation Explain the difference between diffusion and active transport.
Diffusion is the passive movement of particles from a high to low concentration (down the concentration gradient) and requires no energy. Active transport moves substances from a low to high concentration (against the concentration gradient) and requires energy from respiration (ATP).
3. Higher A potato cylinder has an initial mass of 4.80 g and a final mass of 5.52 g. Calculate the percentage change in mass.
Percentage change = ((5.52 − 4.80) ÷ 4.80) × 100 = (0.72 ÷ 4.80) × 100 = +15.0%. The positive value indicates water entered the potato by osmosis.
4. Higher Explain why a red blood cell bursts in distilled water but a plant cell does not.
In distilled water, water enters both cells by osmosis. The red blood cell has no cell wall, so it swells and bursts (lysis). The plant cell has a rigid cellulose cell wall that prevents it from bursting — the cell becomes turgid as the vacuole pushes the cytoplasm against the wall, but the wall provides structural support.
5. Higher A small cube has sides of 1 cm and a large cube has sides of 5 cm. Calculate the SA:V ratio for each and explain which exchanges substances more efficiently.
Small cube: SA = 6 × 1² = 6 cm², V = 1³ = 1 cm³, SA:V = 6:1. Large cube: SA = 6 × 5² = 150 cm², V = 5³ = 125 cm³, SA:V = 150:125 = 1.2:1. The small cube has a larger SA:V ratio, so it exchanges substances more efficiently relative to its volume.
6. Foundation Describe what happens to a plant cell when it is placed in a very concentrated sugar solution.
Water leaves the cell by osmosis (from the higher water concentration inside the cell to the lower water concentration outside). The vacuole shrinks and the cytoplasm and cell membrane pull away from the cell wall. The cell becomes plasmolysed. The cell wall remains intact because it is rigid.
🔬 Required Practical
Required Practical: Investigating Osmosis Using Potato Cylinders
Aim: To investigate the effect of sugar solution concentration on osmosis in potato tissue.
Method: 1) Use a cork borer to cut equal-sized potato cylinders and trim to the same length (e.g. 3 cm). 2) Blot each cylinder dry and measure its initial mass. 3) 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) for at least 30 minutes. 4) Remove, blot dry, and measure final mass. 5) Calculate percentage change in mass = ((final − initial) / initial) × 100. Repeat each concentration at least 3 times for reliability.
Variables: IV: concentration of sugar solution, DV: percentage change in mass of potato cylinder, Control: volume of solution, surface area of cylinder, time immersed, temperature, potato variety
🔢 Maths Skills
Mathematical Skills for this Topic
Percentage change: percentage change = (final − initial) / initial × 100. A positive value means the potato gained mass (water entered by osmosis); a negative value means it lost mass (water left by osmosis).
Calculating a mean from repeat measurements: Add all values and divide by the number of repeats. For example, percentage changes of +12%, +14%, and +13% give a mean of (12 + 14 + 13) / 3 = +13.0%. Identify anomalies before calculating the mean — any value significantly different from the others should be excluded and explained.
⚠️ Common Misconceptions
Watch Out!
Students often think osmosis is just the diffusion of water. Wrong: osmosis is simply diffusion of water molecules Correct: osmosis requires a partially permeable membrane — it is the net movement of water molecules across a partially permeable membrane from a dilute to a more concentrated solution
Students often think water always moves to the side with more water. Wrong: water moves to wherever there is more water Correct: water moves from a region of higher water potential (more dilute solution) to lower water potential (more concentrated solution) — it is about water concentration, not the amount of water
✍️ 6-Mark Question
Extended Answer Question
6 marks: Describe how you would investigate the effect of solution concentration on osmosis in potato tissue. Explain how to make your results valid and reproducible.
Cut equal-sized potato cylinders using a cork borer and trim to the same length. Blot each cylinder dry and record its initial mass. Prepare at least five different concentrations of sugar solution (e.g. 0.0 M to 1.0 M in 0.2 M increments) in separate beakers. Place one cylinder in each solution, ensuring the same volume of solution covers each cylinder. Leave for the same length of time (at least 30 minutes). Remove, blot dry, and record the final mass. Calculate percentage change in mass for each concentration. To ensure validity: control all variables except concentration (volume of solution, temperature, time, surface area of potato). To ensure reproducibility: repeat each concentration at least three times and calculate a mean, use the same potato variety and same equipment, and record all methods clearly so others can replicate the investigation.
Mark scheme: 1 mark for a clear method with at least 5 steps. 1 mark for calculating percentage change. 1 mark for controlling variables (at least 2 named). 1 mark for repeats for reliability/reproducibility. 1 mark for using a range of concentrations. 1 mark for clear explanation of how validity and reproducibility are ensured.
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigated osmosis in potato cylinders and obtained the following percentage change in mass data for 0.4 M sugar solution: −8.2%, −7.9%, +5.1%, −8.5%. Identify any anomalous results, calculate a suitable mean, and explain the effect of the anomaly on the mean if it were included.
Approach: The value +5.1% is anomalous — it differs significantly from the other three values (−8.2%, −7.9%, −8.5%) which are all close together and negative. Including the anomaly would give a mean of (−8.2 + −7.9 + 5.1 + −8.5) / 4 = −4.875%, which is much less negative than expected and misleading. Excluding the anomaly, the mean = (−8.2 + −7.9 + −8.5) / 3 = −8.2%. The anomaly may have been caused by not blotting the cylinder dry (excess surface water adding to the mass) or placing the cylinder in the wrong concentration.
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