This set mixes the five skills from integrated biological reasoning: linking transport and gas exchange, enzyme patterns, inheritance, ecology numbers and evaluating conclusions. All data is invented for practice.
Work in order, easy first. Write your own answer before opening the working. Record slips in the mistake log, and use the scientific investigation critic on the last two questions.
Questions
1. (Transport) A person at rest takes 18 breaths per minute, each of 0.40 dm³. Calculate the ventilation rate.
Show answer
18 × 0.40 = 7.2 dm³ per minute.
2. (Transport) A heart beats 75 times per minute with a stroke volume of 64 cm³. Calculate the cardiac output in dm³ per minute.
Show answer
75 × 64 = 4800 cm³ per minute. 4800 ÷ 1000 = 4.8 dm³ per minute.
3. (Transport) During a run, a student’s breathing rate and heart rate both rise. Explain how these two changes are linked to respiration in the leg muscles.
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The muscle cells respire faster, so they use more oxygen and release more carbon dioxide. Faster, deeper breathing increases gas exchange in the alveoli. A faster heart rate moves blood more quickly, so oxygen reaches the muscles sooner and carbon dioxide is carried to the lungs sooner.
4. (Enzymes) An enzyme (invented data) breaks down 5 units of substrate per minute at 20 °C, 11 units at 30 °C and 2 units at 40 °C. Calculate the percentage increase from 20 °C to 30 °C and the percentage decrease from 30 °C to 40 °C.
Show answer
Increase: (11 − 5) ÷ 5 = 1.2, so 120%.
Decrease: (11 − 2) ÷ 11 = 9 ÷ 11 = 0.818, so 82% (2 significant figures).
5. (Enzymes) Use the data in question 4 to explain what happens at 40 °C, and say what this would mean for a process such as digestion if the body were at that temperature.
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At 40 °C the active site is denatured, so the substrate no longer fits and the rate falls to 2 units. In digestion, large molecules would be broken down much more slowly, so fewer small molecules would be absorbed into the blood.
6. (Inheritance) In a plant, blue flowers (B) are dominant to white (b). A Bb plant is crossed with a bb plant. State the expected ratio and the expected numbers in 60 offspring. The student observes 34 blue and 26 white. Comment.
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Offspring are Bb and bb in equal numbers, so the ratio is 1:1, which is 30 blue and 30 white in 60.
Observed values differ by 4 from expected in each group. The result is close to 1:1, so it is consistent with the cross, and with only 60 offspring, some difference is normal. It does not prove the genotypes.
7. (Inheritance) A Tt × Tt cross gives 200 offspring, with 141 tall. Calculate the expected number of tall plants and comment.
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P(tall) = 3/4, so expected = 0.75 × 200 = 150. Observed 141 is 9 below.
That is a small gap for 200, so the result is consistent with a 3:1 ratio, but it suggests rather than proves both parents are Tt.
8. (Ecology) A field is 300 m². Eight quadrats of 0.5 m² contain 1, 3, 2, 0, 4, 2, 3 and 1 plants. Estimate the population.
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Total = 1 + 3 + 2 + 0 + 4 + 2 + 3 + 1 = 16. Mean = 16 ÷ 8 = 2.0 per quadrat.
Quadrat areas in the field: 300 ÷ 0.5 = 600. Estimate = 2.0 × 600 = 1200 plants.
Check: 2.0 per 0.5 m² is 4 per m², and 4 × 300 = 1200.
9. (Ecology) Twenty-five animals are marked and released. A later sample of 60 contains 10 marked animals. Estimate the population, and name one assumption.
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25 × 60 ÷ 10 = 1500 ÷ 10 = 150 animals.
Assumption: marked animals mix evenly with the rest of the population (or no animals enter or leave between samples).
10. (Ecology) A producer holds 12 000 kJ. The primary consumer receives 1200 kJ and the secondary consumer receives 120 kJ. Calculate the percentage transferred at each step.
Show answer
Producer to primary: 1200 ÷ 12 000 × 100 = 10%.
Primary to secondary: 120 ÷ 1200 × 100 = 10%.
11. (Evaluation) A student claims that a new feed increases the mass of tomatoes. Feed A tomatoes (invented data): 40, 44, 42 g. Feed B tomatoes: 41, 45, 43 g. The balance reads to 1 g. Evaluate the claim.
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Mean A = (40 + 44 + 42) ÷ 3 = 126 ÷ 3 = 42.0 g. Mean B = (41 + 45 + 43) ÷ 3 = 129 ÷ 3 = 43.0 g.
The difference is 1.0 g, equal to the balance resolution. The ranges (40 to 44 and 41 to 45) overlap strongly, and only three tomatoes were used per group.
So the data does not support the claim. A fair conclusion is that any difference is within the measurement limit. Using more tomatoes per group and a balance reading to 0.1 g would help.
If you got these wrong
| Where the slip happened | Go back to |
|---|---|
| Questions 1 to 3: unit conversion, or confusing breathing with respiration | connect transport and gas exchange |
| Questions 4 and 5: percentage change, or saying enzymes are killed | enzyme patterns and larger processes |
| Questions 6 and 7: expecting exact ratios, or stating “proves” | inheritance and variation without overclaiming |
| Questions 8 to 10: forgetting to scale by quadrat area | ecology evidence and numerical reasoning |
| Question 11: vague evaluation, ignoring resolution or overlap | evaluating a conclusion |
When a pattern of errors repeats, a one-to-one teacher can find the root cause faster than extra practice can. That is part of online one-to-one Co-ordinated Sciences tuition.