This set mixes the five skills from biological mechanisms in mixed tasks: membrane explanations, plant exchange rates, energy tracing, food-web conclusions and inheritance scope. All data is invented for practice.
Write your answer first, then open the answer. Questions get harder as you go. Log each slip in the mistake log and retest queue.
Questions
Q1. A tissue sample changes mass from 3.00 g to 3.27 g in a solution. Calculate the percentage change in mass.
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Change = 3.27 − 3.00 = 0.27 g. Percentage change = 0.27 ÷ 3.00 × 100 = +9.0%. Check: 9% of 3.00 is 0.27.
Q2. A plant tissue is placed in a solution more concentrated than its cell contents. State the direction of water movement and the change in mass.
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Water moves out of the cells by osmosis through the partially permeable membrane, so the mass of the tissue decreases.
Q3. A leafy shoot takes up 1.8 cm³ of water in 12 minutes. Calculate the rate in cm³/min.
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1.8 ÷ 12 = 0.15 cm³/min. Check: 0.15 × 12 = 1.8.
Q4. The rate in still air is 0.20 cm³/min and in wind it is 0.30 cm³/min. Calculate the percentage increase and explain it.
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Increase = 0.10, and 0.10 ÷ 0.20 × 100 = 50%. Wind removes the humid air next to the leaf, so the water vapour concentration gradient stays steep and more vapour diffuses out through the stomata.
Q5. A food sample heats 40 g of water by 8.0 °C. Calculate the energy gained by the water. Use 4.2 J/g°C.
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40 × 4.2 = 168, and 168 × 8.0 = 1344 J. Check: 4.2 × 8.0 = 33.6, and 33.6 × 40 = 1344.
Q6. The food sample in Q5 had a mass of 0.60 g. Calculate the energy per gram in J/g and kJ/g, and say why the true value is likely to be higher.
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1344 ÷ 0.60 = 2240 J/g, which is 2.24 kJ/g. Check: 2240 × 0.60 = 1344. The true value is likely higher because some heat escapes to the air and equipment instead of reaching the water.
Q7. In the food chain grass → grasshopper → frog → snake, name the secondary consumer and the producer.
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The producer is grass. The grasshopper is the primary consumer, so the secondary consumer is the frog.
Q8. A population falls from 50 to 35. Calculate the percentage change, then write one cautious sentence about a possible cause in a food web.
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(35 − 50) ÷ 50 × 100 = −15 ÷ 50 × 100 = −30%. A cautious sentence: “The fall may be linked to a reduced food supply, but one survey cannot prove this, so more data is needed.”
Q9. Cross Tt × tt, where T (tall) is dominant. State the ratio of offspring genotypes and how many tall plants you expect among 60 offspring.
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Offspring are 1 Tt : 1 tt. Tall plants are Tt, with probability 1 ÷ 2, so 60 × 1/2 = 30 tall plants expected. The actual count may differ by chance.
Q10. Piece A goes from 4.0 g to 4.4 g. Piece B goes from 1.0 g to 1.15 g. Which piece gained more by percentage, and why is the raw gain misleading?
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A: 0.4 ÷ 4.0 × 100 = 10.0%. B: 0.15 ÷ 1.0 × 100 = 15.0%. Piece B gained more by percentage. The raw gain of A (0.4 g) is larger only because A started heavier, so percentage change is the fair comparison.
If you got these wrong
Match your error to the lesson that fixes it.
- Missing the “reason” or naming sugar instead of water (Q1, Q2, Q10): see connect a cell observation with a membrane explanation.
- Rate or unit errors, or explaining wind and humidity (Q3, Q4): see link plant exchange with a supplied environmental dataset.
- Using the wrong mass or mixing J and kJ (Q5, Q6): see trace energy between a biological process and a physical measurement.
- Treating a food-web link as proven or percentage errors (Q7, Q8): see use a food-web change to test an evidence-based conclusion.
- Punnett square or probability errors (Q9): see check the depth of an inherited-trait explanation against 0653 scope.
The scientific investigation critic lets you practise judging a method, and the module overview shows the study order.
If one error type keeps returning, online one-to-one Combined Science tuition can target that pattern directly.