Transport in CellsAQA GCSE Biology: Revision notes
Section 1
What is diffusion and what factors affect its rate?
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. It occurs because particles move randomly, but there is a net movement down the concentration gradient until equilibrium is reached.
Factors affecting the rate of diffusion:
| Factor | Effect | Explanation |
|---|---|---|
| Concentration gradient | Steeper gradient = faster diffusion | Greater difference between high and low concentration regions increases the net movement |
| Temperature | Higher temperature = faster diffusion | Particles have more kinetic energy and move more rapidly |
| Surface area | Larger surface area = faster diffusion | More space available for particles to cross the boundary |
Diffusion requires no energy from the cell and occurs in liquids and gases. The rate of diffusion increases when any of these three factors increases.
Examiners want to see that you understand diffusion is a net movement down a concentration gradient. Use the word 'net' in your answer to show particles move in both directions, but more move down the gradient.
Think of diffusion like perfume spreading in a room – particles spread out randomly from where there is lots of perfume (high concentration) to where there is little (low concentration) until the smell is even everywhere.
Section 2
What is osmosis and how does it affect plant and animal cells?
Osmosis is the diffusion of water molecules through a partially permeable membrane from a dilute solution (higher water potential) to a more concentrated solution (lower water potential).
Key differences between osmosis and diffusion:
- Osmosis specifically involves water molecules only
- Osmosis requires a partially permeable membrane
- Osmosis occurs between solutions of different concentrations
Effects of osmosis on animal cells:
- In hypotonic solution (dilute, lower solute concentration): water enters the cell, causing it to swell and potentially burst (lysis)
- In hypertonic solution (concentrated, higher solute concentration): water leaves the cell, causing it to shrivel (crenation)
- In isotonic solution (equal concentration): no net movement of water; cell remains unchanged
Effects of osmosis on plant cells:
- In hypotonic solution: water enters by osmosis, cell becomes turgid (firm and swollen). The cell wall prevents bursting, maintaining cell rigidity
- In hypertonic solution: water leaves by osmosis, vacuole shrinks and cell becomes plasmolysed (cytoplasm separates from cell wall). The cell becomes limp and wilts
- In isotonic solution: the cell remains normal with no change
Plant cells are less likely to burst because the cell wall provides structural support, unlike animal cells which only have a cell membrane.
When describing osmosis effects, always mention the direction of water movement and explain what happens to the cell. For plant cells, state whether it becomes turgid or plasmolysed; for animal cells, state whether it undergoes lysis or crenation.
Students often forget that osmosis is only water movement through a partially permeable membrane. Saying 'osmosis is diffusion of water' without mentioning the membrane is incomplete and will lose marks.
A plant cell is placed in concentrated salt solution. Water leaves the cell by osmosis because the solution has lower water potential. The vacuole shrinks and the cytoplasm separates from the cell wall – the cell becomes plasmolysed and wilts.
Section 3
How do you investigate osmosis in plant tissue and calculate percentage change?
Practical method for investigating osmosis:
- Cut uniform strips of plant tissue (e.g. potato or rhubarb) of equal length
- Measure and record the initial mass of each strip using a balance
- Place strips in solutions of different concentrations (distilled water, dilute salt solution, concentrated salt solution)
- Leave for a set time (e.g. 15–20 minutes)
- Remove strips, blot gently with paper towel to remove excess moisture
- Measure the final mass of each strip
- Calculate the change in mass and percentage change
Calculating percentage change in mass:
Use this formula:
Percentage change = (Final mass − Initial mass) / Initial mass × 100
Key observations:
- In distilled water (hypotonic): tissue gains mass (positive percentage change) as water enters by osmosis
- In concentrated solution (hypertonic): tissue loses mass (negative percentage change) as water leaves by osmosis
- The more concentrated the solution, the greater the water loss
Variables to control:
- Length and surface area of tissue strips
- Time left in solution
- Temperature
- Volume of solution
- Type of blotting (consistent pressure)
Plotting results: Graph percentage change (y-axis) against solution concentration (x-axis). The curve shows that as solution concentration increases, percentage change becomes more negative.
Always show your working when calculating percentage change. Examiners award marks for method, so write out the formula and substitute numbers. Include the × 100 to convert to a percentage.
A potato strip has initial mass 4.5 g and final mass 5.4 g. Percentage change = (5.4 − 4.5) / 4.5 × 100 = 0.9 / 4.5 × 100 = 20%. This positive change shows water entered the tissue by osmosis.
Students often forget to blot the tissue strips with a paper towel after removing them from the solution. Excess water on the surface artificially increases the final mass, leading to incorrect results.
Section 4
What is active transport and why is it important?
Active transport is the movement of substances against the concentration gradient (from lower to higher concentration) using energy from respiration in the form of ATP.
Key features:
- Requires energy (ATP) from respiration – makes it different from diffusion and osmosis
- Moves particles against the concentration gradient (uphill)
- Occurs at the cell membrane using protein carriers
- Slower than diffusion but allows cells to control what enters and leaves
Importance of active transport:
1. Absorption of mineral ions by root hair cells:
- Mineral ions (e.g. nitrate, phosphate, potassium) are needed for plant growth
- These ions are often in low concentration in the soil
- Root hair cells use active transport to absorb mineral ions against the concentration gradient
- This allows plants to accumulate essential minerals even when soil concentration is low
2. Absorption of glucose by intestinal epithelial cells:
- Glucose is produced from carbohydrate digestion in the small intestine
- Intestinal epithelial cells use active transport to absorb glucose against the concentration gradient
- This ensures efficient absorption of all glucose, even when concentration is low
- Glucose is then used for respiration to provide energy for the cell
Active transport vs. diffusion:
| Feature | Active Transport | Diffusion |
|---|---|---|
| Direction | Against concentration gradient | Down concentration gradient |
| Energy required | Yes (ATP from respiration) | No |
| Speed | Slower | Faster |
| Selectivity | Selective (specific carriers) | Non-selective |
| Examples | Mineral ion absorption, glucose absorption | Gas exchange, osmosis |
When explaining active transport importance, always state: (1) what is being transported, (2) that it moves against the concentration gradient, and (3) why this is important (e.g. to absorb minerals when soil concentration is low). Three-part answers score full marks.
Active transport is like swimming upstream against a river current – you need energy (muscle power) to move in the opposite direction to the natural flow. Diffusion is like floating downstream – no energy needed.
Root hair cells absorb nitrate ions from soil using active transport. Although nitrate concentration is low in soil, the cells transport nitrate ions into the cell against the concentration gradient using energy from respiration. This allows the plant to accumulate nitrates for protein synthesis.
Section 5
How are exchange surfaces adapted for efficient transport?
Exchange surfaces are specialized structures that allow efficient transport of substances into and out of cells. They are adapted by having specific features that maximise the rate of transport.
Three key adaptations of exchange surfaces:
1. Large surface area to volume ratio:
- Exchange surfaces have large surface areas relative to their volume
- Examples: root hair cells have hair-like projections; intestinal epithelial cells have finger-like villi; alveoli in the lungs are numerous and clustered
- Large surface area increases the number of particles that can cross the membrane at once, speeding up transport
2. Thin membranes:
- Exchange surfaces have thin cell membranes and are often only one cell thick
- This reduces the distance particles must travel
- Shorter distance = faster transport rates
- Example: alveolar walls are thin, allowing rapid gas exchange
3. Steep concentration gradient:
- Exchange surfaces maintain a steep concentration gradient across the membrane
- This is achieved by:
- Removing absorbed substances quickly (e.g. glucose is transported away in blood after absorption in the intestine)
- Providing a constant supply of fresh solution (e.g. blood flow maintains low glucose concentration inside intestinal cells)
- Continuous active transport which removes accumulated substances
- Steep gradient = faster diffusion/active transport
Example – intestinal epithelial cells: These cells are adapted for glucose absorption with:
- Large surface area: microvilli on the cell membrane
- Thin membrane: single layer of cells
- Steep gradient: glucose transported away by blood; fresh glucose arrives from digestion
- Result: rapid and efficient glucose absorption
Example – root hair cells: These cells are adapted for mineral ion absorption with:
- Large surface area: hair-like extension of the cell
- Thin walls: single cell thickness
- Steep gradient: mineral ions transported to xylem; fresh ions arrive from soil
- Result: efficient mineral ion absorption
Examiners want you to link structure to function. Don't just list features – always explain how each adaptation increases the rate of transport. Write: 'Large surface area increases the rate of transport because more particles can cross the membrane at once.'
Students often say 'thin membranes speed up transport' without explaining why. The correct explanation is: thin membranes reduce the distance particles must travel, so transport is faster.
Must Know
- Diffusion is net movement of particles from high to low concentration; it is affected by concentration gradient, temperature, and surface area
- Osmosis is the diffusion of water through a partially permeable membrane; in hypotonic solutions, plant cells become turgid and animal cells may lyse; in hypertonic solutions, plant cells become plasmolysed and animal cells crenate
- Percentage change in mass = (Final − Initial) / Initial × 100; use this formula in osmosis investigations
- Active transport moves substances against the concentration gradient using energy from respiration; it is essential for absorbing mineral ions in roots and glucose in the intestine
- Exchange surfaces are adapted with large surface area (more transport sites), thin membranes (shorter distance), and steep concentration gradients (faster rate) to maximise efficient transport
That's the notes covered.
Carry on to the next subtopic.