OCR A Level Biology A: Populations, interactions and sustainable management (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 outcomes 6.3.2(a) to 6.3.2(e), Populations and sustainability, in the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Read the study guide first. The OCR A Level Biology hub and printable checklist list the course.

Questions

1. State what is meant by a limiting factor, and name two density-dependent limiting factors for a population of woodland birds. [3]

2. A population of 1250 lizards lives on an island. In one year there were 210 births, 95 deaths, 12 immigrants and 37 emigrants.

(a) Calculate the percentage change in the population. [3] (b) Suggest why the percentage change is likely to fall as the population grows. [1]

3. A yeast culture was sampled every 2 hours. The table shows log10 of the number of cells per cm3.

Time / h	0	2	4	6	8	10	12	14	16
log10 cells per cm3	4.00	4.00	4.30	4.60	4.90	5.20	5.40	5.45	5.45

(a) Describe the growth of the culture from 0 to 16 hours. [3] (b) Calculate the number of cells per cm3 at 10 hours. Give your answer to 2 significant figures. [2] (c) Calculate the doubling time between 2 and 10 hours. The log10 of 2 is 0.30. [2] (d) Explain the change in growth after 10 hours. [2]

4. Wood mice and their predators, kestrels, were surveyed for eight years.

Year	1	2	3	4	5	6	7	8
Mice per hectare	60	150	280	190	70	110	240	170
Kestrel pairs per 10 km2	4	5	8	11	7	4	6	9

(a) Describe the relationship between the two populations. [2] (b) Calculate the percentage change in the mouse population from year 3 to year 5. [1] (c) Explain why the kestrel peak follows the mouse peak. [2] (d) Suggest why these data alone do not show that kestrels control the mouse population. [1]

5. A plant was sown at three densities. Mean dry mass per plant at harvest was 12.0 g at 5 plants per m2, 7.5 g at 20 per m2 and 3.0 g at 80 per m2.

(a) Calculate the total dry mass per m2 at each density. [2] (b) Explain why the mass per plant falls as density rises. [2] (c) Name this type of competition and state how it differs from interspecific competition. [1]

6. Distinguish between conservation and preservation, and give an example of each. [4]

7. A coastal village depends on a mangrove forest. Give one economic, one social and one ethical reason for conserving the forest. [6]

8. A 90 ha wood is coppiced on a 15-year rotation with equal areas cut each year. Each hectare yields 38 m3 of poles when cut.

(a) Calculate the area cut each year. [1] (b) Calculate the volume harvested each year. [1] (c) The owner proposes a 10-year rotation, with each hectare then yielding 24 m3. Calculate the new yearly volume and the percentage change. [2] (d) Explain how coppicing on a rotation can increase the biodiversity of a wood. [2]

9. A fish stock is 35 000 tonnes. It adds 5600 tonnes of new growth each year. A fixed quota of 7000 tonnes per year is proposed.

(a) Express the quota as a percentage of the stock. [1] (b) Show that the quota is not sustainable. [1] (c) Calculate the stock after 3 years if the quota is taken each year and growth stays at 5600 tonnes. [2] (d) State the largest percentage of the current stock that could be taken sustainably. [1] (e) Describe one measure, other than a quota, that helps maintain the stock. [1]

10. A bay of 320 km2 supports a fishing village. Annual catches over five years were 900, 860, 790, 710 and 650 tonnes. A plan would make 25% of the bay a no-take zone. In a similar bay with a no-take zone, fish biomass was 380 tonnes per km2 inside it and 160 tonnes per km2 outside.

(a) Calculate the percentage fall in catch over the five years. [1] (b) Calculate the area of the proposed no-take zone. [1] (c) Calculate how many times greater the biomass is inside the no-take zone than outside, to 2 significant figures. [1] (d) Evaluate the plan. [6]

11. Ground-nesting seabirds breed on an island. Before rats arrived there were 1200 breeding pairs and 72% of nests produced a chick. Five years after rats arrived there were 780 pairs and 31% of nests produced a chick. After trapping there were 1010 pairs.

(a) Calculate the percentage fall in breeding pairs after rats arrived. [1] (b) Calculate the percentage rise after trapping, from the 780 pairs. [1] (c) Explain how rats reduced the seabird population. [3] (d) Suggest two reasons why the population has not returned to 1200 pairs. [2] (e) State one ethical issue raised by killing rats. [1]

Answers

1. A limiting factor is a variable in short supply [1] that keeps the population below the size it could otherwise reach. Two density-dependent factors, for example competition for food or nest sites [1] and disease or predation [1]. [3] Examiner insight: Name factors whose effect grows with density; frost or flood would be density-independent.

2. (a) Net change = (210 + 12) - (95 + 37) = 90 [1]. Percentage = 90 / 1250 x 100 [1] = 7.2% increase [1]. [3] (b) Competition for food or space rises, so births fall or deaths rise [1]. [1] Examiner insight: Show the births, immigrants, deaths and emigrants in one line of working, so the logic is visible.

3. (a) No growth from 0 to 2 hours (lag phase) [1]. Steady increase on the log scale from 2 to 10 hours, which is exponential growth [1]. Growth slows after 10 hours and levels off at 5.45 from 14 hours [1]. [3] (b) 10^5.20 [1] = 1.6 x 10^5 cells per cm3 [1]. [2] (c) Rise in log10 = 5.20 - 4.00 = 1.20 over 8 hours, so 0.15 per hour [1]. Doubling time = 0.30 / 0.15 = 2.0 hours [1]. [2] (d) Limiting factors act, such as nutrients running low [1]. Births (cell division) fall to equal deaths, so numbers level off at the carrying capacity [1]. [2] Examiner insight: The axis is a logarithm. Quoting a log value as if it were a cell count loses credit in part (b).

4. (a) Both populations rise and fall in repeated cycles [1]. The kestrel peaks come one year after the mouse peaks [1]. [2] (b) (70 - 280) / 280 x 100 = -75% [1]. [1] (c) A larger mouse population feeds more kestrels, which survive and breed [1]. Kestrels need time to find prey and raise young, so their peak is delayed [1]. [2] (d) Mouse numbers may also depend on their own food supply or weather, so the data show a correlation only [1]. [1] Examiner insight: State the lag as a number of years, not just that the graphs are similar.

5. (a) 5 x 12.0 = 60 g; 20 x 7.5 = 150 g [1]; 80 x 3.0 = 240 g [1]. [2] (b) Each plant competes with more neighbours for light, water and ions [1]. Each gets a smaller share of the resources, so it grows less [1]. [2] (c) Intraspecific competition, between plants of the same species; interspecific is between different species [1]. [1] Examiner insight: Use the data in part (b): density rose 16 times while total mass rose 4 times.

6. Conservation is the maintenance of ecosystems and biodiversity through management [1], allowing sustainable human use [1]. Preservation is protection of an area or species as it is, with human interference minimised or excluded [1]. Example: coppicing a woodland (conservation) and a no-take marine zone (preservation) [1]. [4] Examiner insight: The terms differ in what human activity is allowed, so say that explicitly.

7. Economic: the forest supplies timber and fish nursery areas for income, or attracts tourists [2]. Social: it protects the village from storm waves and supports livelihoods and cultural traditions [2]. Ethical: other species have a right to exist, and people have a duty to future generations [2]. [6] Examiner insight: Link each type of reason to the mangrove context supplied; generic statements about biodiversity are less specific.

8. (a) 90 / 15 = 6 ha [1]. [1] (b) 6 x 38 = 228 m3 [1]. [1] (c) New area = 90 / 10 = 9 ha, volume = 9 x 24 = 216 m3 [1]. Change = (216 - 228) / 228 x 100 = -5.3% [1]. [2] (d) Sections are cut in turn, so parts are at different stages of regrowth [1]. Light reaches the woodland floor after cutting, so ground plants and their insects increase, giving a range of habitats [1]. [2] Examiner insight: In part (c) give the new volume and the percentage, with its sign.

9. (a) 7000 / 35 000 x 100 = 20% [1]. [1] (b) The quota of 7000 exceeds the growth of 5600 by 1400 tonnes per year, so the stock falls [1]. [1] (c) Each year the stock changes by 5600 - 7000 = -1400 [1]. After 3 years: 35 000 - 4200 = 30 800 tonnes [1]. [2] (d) 5600 / 35 000 x 100 = 16% [1]. [1] (e) Any one: larger net mesh so that young fish escape, closed season or no-take zone to protect spawning, or limits on fishing effort [1]. [1] Examiner insight: For “show that”, compare the quota with the growth and state the difference.

10. (a) (650 - 900) / 900 x 100 = -27.8% [1]. [1] (b) 0.25 x 320 = 80 km2 [1]. [1] (c) 380 / 160 = 2.375, which is 2.4 times [1]. [1] (d) Protected fish can reach breeding size and spawn [1]. Biomass is higher in a no-take zone, which suggests stocks can recover [1]. Fish may spread into fished areas, which could help catches [1]. Fishers lose 80 km2 of fishing area and income at first [1]. The catch is already falling, so doing nothing may cost more in the long term [1]. Judgement: the plan is justified because the stock is declining, though compensation may be needed, and the evidence is from a similar bay and not this one [1]. [6] Examiner insight: An evaluation gains from a judgement that weighs short-term against long-term effects.

11. (a) (780 - 1200) / 1200 x 100 = -35% [1]. [1] (b) (1010 - 780) / 780 x 100 = 29.5% [1]. [1] (c) Rats take eggs and chicks from nests on the ground [1]. Nest success fell from 72% to 31%, so fewer chicks reached breeding age [1]. Fewer young replaced the adults that died [1]. [3] (d) Some rats may remain [1]. Chicks take years to reach breeding age, so recovery is slow [1]. [2] (e) Animals are killed to protect another species, which some people think unacceptable [1]. [1] Examiner insight: In part (c), connect the nest-success figures to numbers of young and then to the population.

Where marks are usually lost

• Quoting a logarithmic value as if it were a count.
• Describing a predator-prey graph without the lag in years.
• Naming a density-independent factor when asked for density-dependent.
• Omitting units, or not saying whether a percentage change is a rise or a fall.
• Giving general reasons for conservation that ignore the context.
• Treating conservation and preservation as the same thing.
• Evaluating with no judgement.

Next steps

• Review the revision notes and the study guide.
• Go to the OCR A Level Biology hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR A Level in Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025, published by OCR. Section 6.3.2, Populations and sustainability.