OCR A Level Biology A: Variation, adaptation and natural selection (H420) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.2.2(e) to (i) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: evidence for evolution, variation, adaptation, natural selection, and pesticide and drug resistance. Revise first with the study guide and the revision notes. See also the OCR A Level Biology hub, the checklist and the free diagnostics.

Questions

1. Describe the contributions of Darwin and Wallace to the theory of evolution by natural selection. [3]

2. The same gene, 1,500 bases long, was sequenced in four species. Species K, L and M differ from species J in 15, 45 and 120 bases respectively.

(a) Calculate the percentage difference from species J for each of K, L and M. [3] (b) State which species is most closely related to J, and explain your choice. [1]

3. Give two reasons why the fossil record gives incomplete evidence for evolution. [2]

4. State whether each characteristic shows continuous or discontinuous variation. [3]

(a) Mass of tomatoes picked from one plant. (b) ABO blood group in humans. (c) Diameter of bacterial colonies on an agar plate.

5. Eight leaves were taken from one tree and their lengths (mm) measured: 41, 44, 46, 43, 45, 47, 42, 44.

(a) Calculate the mean length. [1] (b) Calculate the standard deviation, using n − 1. [3] (c) Calculate the values of the mean plus and minus two standard deviations, and state what proportion of a normally distributed population lies between them. [2]

6. Classify each adaptation as anatomical, physiological or behavioural. [3]

(a) A beetle in a fog desert stands head down on a dune crest. (b) Some Antarctic fish make proteins that stop ice crystals growing in their blood. (c) A mole has broad, spade-like forelimbs.

7. The sugar glider (a marsupial) and the flying squirrel (a placental mammal) both have a membrane between the fore and hind limbs. Explain why these two animals have similar features although they are not closely related. [3]

8. A small island has dark basalt rock. Pale lizards and dark lizards live on it. Birds catch lizards that they can see. Over many generations the proportion of dark lizards rises. Explain this change. [5]

9. A field of crop plants has 800 aphids, of which 5% carry an allele for resistance to an insecticide. After one spraying, 80% of the resistant aphids and 5% of the non-resistant aphids survive. The survivors breed and the population returns to 800, with the same proportions of the two types as the survivors.

(a) Calculate the number of aphids that survive. [2] (b) Calculate the percentage of resistant aphids in the next generation. [1] (c) Assuming the same survival rates, calculate the percentage resistant after a second spraying, to 3 significant figures. [2]

10. A culture contains 2.0 × 10⁹ bacteria. One in 10⁸ has a mutation giving resistance to an antibiotic. The bacteria divide every 20 minutes. All the non-resistant bacteria are killed by the antibiotic.

(a) Calculate how many resistant bacteria were present before treatment. [1] (b) Calculate the number of whole divisions needed for the survivors to return to 2.0 × 10⁹, and the time taken in hours. [4]

11. Suggest why doctors give a patient with tuberculosis several antibiotics together for many months, rather than one antibiotic. [4]

12. A hospital tested isolates of a bacterium for resistance to an antibiotic. In the first year, 8 of 200 isolates were resistant. Three years later, 34 of 170 isolates were resistant. During this time the hospital used the antibiotic widely.

(a) Calculate the percentage resistant in each year, and the percentage increase between them. [3] (b) Explain how the data could be the result of natural selection. [3] (c) Suggest two actions the hospital could take to slow the spread of resistance. [2]

Answers

1. Darwin collected observations on a long voyage, including from the Galapagos Islands, and published the theory in On the Origin of Species in 1859 [1]. Wallace independently reached the same explanation from his work in the Malay Archipelago, and papers from both were presented together in 1858 [1]. Both proposed that organisms vary, that more offspring are produced than survive, and that those best suited to the environment survive and reproduce, passing on inherited features [1]. [3] Examiner insight: Credit goes to the idea of selection of inherited variation and to naming both men’s roles. Saying only that Darwin “discovered evolution” earns little.

2. (a) K: 15 ÷ 1,500 × 100 = 1.0% [1]. L: 45 ÷ 1,500 × 100 = 3.0% [1]. M: 120 ÷ 1,500 × 100 = 8.0% [1]. [3] (b) Species K, because it has the fewest differences and so shared a common ancestor with J most recently [1]. [1] Examiner insight: A reason is needed as well as the name. The link between fewer differences and a more recent common ancestor is the point to state.

3. Most organisms are not fossilised, because they are eaten or decay [1]. Soft-bodied organisms rarely fossilise, and many fossils are not found or are destroyed by rock movement [1]. [2] Examiner insight: Two different reasons are needed. Two wordings of “not many fossils” count as one.

4. (a) Continuous [1]. (b) Discontinuous [1]. (c) Continuous [1]. [3] Examiner insight: Test each example by asking whether any value in a range is possible, or only separate categories.

5. (a) Sum = 352, and 352 ÷ 8 = 44 mm [1]. [1] (b) Deviations from the mean: −3, 0, 2, −1, 1, 3, −2, 0 [1]. Sum of squares = 9 + 0 + 4 + 1 + 1 + 9 + 4 + 0 = 28 [1]. s = √(28 ÷ 7) = √4 = 2.0 mm [1]. [3] (c) 44 − 2 × 2.0 = 40.0 mm and 44 + 2 × 2.0 = 48.0 mm [1]. About 95% of the population [1]. [2] Examiner insight: Show the sum of squares and the division by n − 1. Showing each step makes the method visible, and using n instead of n − 1 gives a different value.

6. (a) Behavioural [1]. (b) Physiological [1]. (c) Anatomical [1]. [3] Examiner insight: Behaviour is what the animal does, physiology is a process inside the body, and anatomy is a structure.

7. Both live in similar environments, gliding between trees, so they face similar selection pressures [1]. Individuals with a gliding membrane were more likely to survive and reproduce in each group, so the feature became common independently in each [1]. This is convergent evolution, so the similarity does not come from a recent common ancestor [1]. [3] Examiner insight: State that selection acted separately in each lineage. “They evolved from the same ancestor” is the wrong explanation.

8. Genetic variation exists, because mutation produced the dark and pale alleles [1]. Birds are the selection pressure, and they catch pale lizards more easily against dark rock [1]. Dark lizards are more likely to survive and reproduce [1]. They pass the allele for dark colour to their offspring [1]. Over generations the allele for dark colour becomes more frequent, so the proportion of dark lizards rises [1]. [5] Examiner insight: Name the selection pressure, and include reproduction and inheritance, not just survival. Saying the lizards “changed colour” describes the wrong process.

9. (a) Resistant: 800 × 0.05 = 40, and 40 × 0.80 = 32 survive [1]. Non-resistant: 760 × 0.05 = 38 survive, so total survivors = 70 [1]. [2] (b) 32 ÷ 70 × 100 = 45.7% [1]. [1] (c) Using 45.7% resistant: resistant share 0.457 × 0.80 = 0.366 and non-resistant share 0.543 × 0.05 = 0.0271 [1]. 0.366 ÷ (0.366 + 0.0271) × 100 = 93.1% [1]. [2] Examiner insight: Divide by the total survivors, not by the original 800. Carrying unrounded values through part (c) avoids a slip in the third significant figure.

10. (a) 2.0 × 10⁹ × 10⁻⁸ = 20 [1]. [1] (b) The population must grow by a factor of 10⁸ [1]. log₂(10⁸) = 26.6 [1]. Divisions are whole, so 27 [1]. Time = 27 × 20 minutes = 540 minutes = 9 hours [1]. [4] Examiner insight: The number of divisions is rounded up, because a part division does not occur. Convert minutes to hours as the question asks.

11. Resistance to an antibiotic arises by chance mutation, so a few bacteria may already be resistant to one drug [1]. With one drug, those bacteria would survive and reproduce, and resistant bacteria would increase [1]. It is much less likely that any bacterium is resistant to several drugs at once [1]. Treating for many months makes sure that all the bacteria are killed, including those that grow slowly [1]. [4] Examiner insight: Link the use of several drugs to probability. “To make the treatment stronger” is too vague.

12. (a) First year: 8 ÷ 200 × 100 = 4.0% [1]. Later: 34 ÷ 170 × 100 = 20.0% [1]. Increase = (20 − 4) ÷ 4 × 100 = 400% [1]. [3] (b) Resistant bacteria arose by chance mutation and were present in the population [1]. The antibiotic acted as the selection pressure, killing non-resistant bacteria while the resistant ones lived [1]. Resistant survivors divided and passed the allele to daughter cells, so their share of the isolates rose [1]. [3] (c) Prescribe the antibiotic only when it is needed, or use a combination of drugs [1]. Improve hygiene and screening to reduce spread [1]. Other sensible actions are also credited. [2] Examiner insight: A percentage increase is relative to the starting value. Quoting 16 percentage points gives a different quantity. In (b), the antibiotic selects, it does not cause mutation.

Where marks are usually lost

• Describing natural selection without naming the selection pressure.
• Leaving out reproduction and inheritance from an explanation.
• Writing that organisms change because they need to.
• Saying that an antibiotic or insecticide causes resistance.
• Dividing by the starting total rather than the survivors when finding a new percentage.
• Using n instead of n − 1 for the standard deviation.
• Confusing a percentage increase with a difference in percentage points.
• Rounding the number of divisions down instead of up.
• Saying that similar features in unrelated animals come from a common ancestor.
• Giving a fossil limitation as “fossils are old” rather than a reason the record is incomplete.

Next steps

• Review the revision notes and the study guide.
• Use the OCR A Level Biology hub and the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class at /trial/.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: section 4.2.2(e) to (i), Classification and evolution.