This set has ten original questions covering cells, enzymes, transport, inheritance and ecology, plus data and investigation skills. They are ordered from easier to harder and carry no mark labels. Check your own syllabus year on the Cambridge page before treating any topic as in scope.
Use paper, write full sentences for the explanation questions, and open each answer only when you have tried. The original mixed-practice builder can assemble longer sessions from reviewed questions.
The questions
1. Define diffusion.
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Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration, down a concentration gradient, as a result of their random movement.
Marks usually depend on “net”, “down a gradient” and “random movement”. Leaving out “net” is a common slip.
2. A drawing of a cell is 36 mm long. The real cell is 0.09 mm long. Calculate the magnification.
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Magnification = image size ÷ actual size = 36 ÷ 0.09 = ×400.
Check: 0.09 × 400 = 36. Both lengths must be in the same unit first, and here both are in millimetres.
3. Cube A has sides of 2 cm and cube B has sides of 4 cm. Calculate the surface area to volume ratio of each, and say what this shows about why large organisms need exchange surfaces.
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Cube A: surface area 6 × 2² = 24 cm², volume 2³ = 8 cm³, ratio 3 : 1.
Cube B: surface area 6 × 4² = 96 cm², volume 4³ = 64 cm³, ratio 1.5 : 1.
The larger cube has a smaller ratio. As size increases, the surface grows more slowly than the volume, so large organisms cannot rely on diffusion across the outside alone and need specialised exchange surfaces.
4. The table shows the rate of an enzyme reaction at different temperatures.
| Temperature (°C) | 20 | 30 | 40 | 50 |
|---|---|---|---|---|
| Rate (units) | 4 | 8 | 14 | 3 |
Describe the pattern, and explain the result at 50 °C.
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Describe: the rate rises from 4 units at 20 °C to 14 units at 40 °C, then falls sharply to 3 units at 50 °C.
Explain: up to 40 °C, higher temperature gives particles more kinetic energy, so substrate and active site collide more often. At 50 °C the enzyme is denatured: its shape, including the active site, changes, so the substrate no longer fits and the rate falls.
The optimum lies somewhere between 30 °C and 50 °C. The data alone do not prove it is exactly 40 °C.
5. A resting heart beats 72 times a minute and pumps 70 cm³ of blood with each beat. Calculate the volume of blood pumped in one minute.
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72 × 70 = 5040 cm³ per minute.
Check: 70 × 70 = 4900, and 2 × 70 = 140, so 4900 + 140 = 5040.
6. A potato cylinder has a mass of 5.0 g. After two hours in pure water it has a mass of 5.6 g. Calculate the percentage change in mass and explain the result.
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Change = 5.6 − 5.0 = 0.6 g. Percentage change = 0.6 ÷ 5.0 × 100 = +12%.
Explanation: the solution inside the potato cells is more concentrated than pure water, so water moves into the cells by osmosis, through the partially permeable membrane. The cells gain water, so the mass rises.
7. Two heterozygous parents (Aa) have children. The allele a is recessive. What is the probability that a child is aa? In a model of 200 children, how many would be expected to be aa?
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Gametes from each parent are A or a. The four equally likely combinations are AA, Aa, aA and aa, so the probability of aa is 1/4.
Expected number: 200 × 1/4 = 50. This is a model of chance, so a real sample of 200 may differ from 50.
8. In a food chain, the producer receives 10 000 kJ of energy and the primary consumer receives 1000 kJ of that. Calculate the percentage of energy passed on, and suggest one place where energy is lost.
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1000 ÷ 10 000 × 100 = 10%.
Energy is lost as heat from respiration, in uneaten parts of the plant, and in waste such as faeces. Any one of these answers is acceptable.
9. A student counts bubbles of oxygen from pondweed at different light intensities.
| Light intensity (units) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Bubbles per minute | 5 | 10 | 15 | 15 | 15 |
Describe the pattern and explain why it levels off.
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Describe: bubbles per minute rise by 5 for each unit of light, from 5 at 1 unit to 15 at 3 units, then stay at 15 from 3 to 5 units.
Explain: up to 3 units, light is the limiting factor, so more light gives faster photosynthesis. After 3 units another factor, such as carbon dioxide concentration or temperature, limits the rate.
10. In question 9, the student counted bubbles once at each intensity and judged the count by eye. Give one limitation and one improvement.
Show answer
Limitation: a single count may be unrepresentative, and counting bubbles is uncertain because bubbles vary in size.
Improvement: repeat each intensity at least three times and calculate a mean. Alternatively collect the gas in a syringe and measure its volume.
If you got these wrong
| What went wrong | Go to |
|---|---|
| Definition missing “net” or “gradient” (Q1) | movement across membranes |
| Magnification or ratios (Q2, Q3) | cells and microscopy |
| Enzyme shape or denaturing (Q4) | biological molecules and enzymes |
| Heart output (Q5) | circulation and blood |
| Osmosis explanation (Q6) | movement across membranes |
| Genetic crosses (Q7) | cell division and inheritance |
| Energy transfer (Q8) | ecology and energy flow |
| Limiting factors (Q9) | plant nutrition |
| Improving an investigation (Q10) | data and investigations |
If your answers list the right words but skip the reason, read the help pages on answers that are a list of keywords and on telling diffusion, osmosis and active transport apart. Log your errors using the revision guide.
Using the set well
Mark yourself on the reasoning first and the final value second. A correct answer reached by a lucky guess should be treated as a gap.
If one topic keeps appearing, return to the Biology learning guide. Where an explanation stops halfway, online one-to-one Biology tuition is the place to practise finishing it with a teacher listening.