OCR A Level Biology A: Ecosystems, biomass transfer and succession (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 6.3.1(a) to (e) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: ecosystems, biomass transfer, the nitrogen and carbon cycles, primary succession, and sampling with statistics. Revise first with the study guide and revision notes. See also the OCR A Level Biology hub, the checklist and the free diagnostics. Use 4.18 J per gram per °C for water.

Questions

1. A rock pool is an ecosystem.

(a) State what is meant by an ecosystem. [1] (b) Name one biotic and one abiotic factor that could change the number of periwinkles in the pool, and explain each effect. [3]

2. Describe how to find the dry biomass of grass per square metre in a meadow. [4]

3. A 0.85 g sample of dried grass seed is burned under 40 g of water. The water temperature rises from 19.0 °C to 38.5 °C.

(a) Find the energy transferred to the water, in joules. [2] (b) Calculate the energy per gram of seed. [1] (c) Explain why the true energy content is higher than your value. [1]

4. In a lake, the energy stored in the biomass of producers is 31 500 kJ m⁻² year⁻¹. Primary consumers ingest 4 800 kJ m⁻² year⁻¹ and incorporate 624 kJ m⁻² year⁻¹ into their biomass. Secondary consumers ingest 540 kJ m⁻² year⁻¹ and incorporate 54 kJ m⁻² year⁻¹.

(a) Calculate the efficiency of biomass transfer in the primary consumers. [2] (b) Calculate the efficiency in the secondary consumers. [1] (c) Calculate the energy ingested but not incorporated by the primary consumers, and give one way it is lost. [2]

5. Explain how farmers can increase the biomass available to humans by manipulating biomass transfer, and give one disadvantage. [6]

6. A farmer finds that a waterlogged field has low nitrate content and poor crop growth.

(a) Identify the overall process that produces nitrate from ammonium, and name the bacteria at each step. [3] (b) Explain why waterlogging lowers the nitrate content. [3] (c) The farmer sows clover. Explain how this can increase the nitrogen available to later crops. [2]

7. Explain how photosynthesis, respiration and decomposition cycle carbon in a woodland, and describe how the oceans can store carbon for a very long time. [5]

8. Data were collected from sand dunes of different ages.

Dune age (years)	Soil organic matter (%)	Plant species per m²
5	0.3	2
25	1.2	5
60	3.8	9
120	6.5	12

(a) Describe the trends in the data. [2] (b) Explain how pioneer species lead to an increase in the number of plant species. [3] (c) The 120-year dune has not become woodland because it is grazed by rabbits. Name this type of succession and explain why it happens. [2] (d) A sample from the 120-year dune has 7 species, with 9, 8, 7, 6, 5, 4 and 3 individuals. Calculate Simpson’s Index of Diversity, D = 1 − Σ(n/N)², to 2 decimal places. [2]

9. A student uses mark-release-recapture to estimate the number of ground beetles in a heathland. She marks 36 beetles and releases them. A second sample contains 42 beetles, of which 9 are marked.

(a) Estimate the size of the beetle population. [2] (b) Give two assumptions needed for this estimate to be reliable. [2] (c) The marked beetles are easier to trap than unmarked ones. State and explain the effect on the estimate. [2]

10. Plant species per quadrat were recorded at nine points along a transect inland from a beach. Distance from the sea (m): 2, 6, 11, 17, 25, 33, 40, 52, 61. Number of species: 3, 5, 4, 7, 9, 8, 12, 11, 15.

(a) State a null hypothesis. [1] (b) Rank both sets (1 = highest) and find Σd². [2] (c) Calculate Spearman’s rank correlation coefficient. [2] (d) The critical value for n = 9 at P = 0.05 is 0.700. State the conclusion. [2]

11. Seedling heights (mm) were measured after ten days in full light (10 seedlings: mean 28.5, standard deviation 3.03) and in shade (10 seedlings: mean 22.0, standard deviation 2.58).

(a) State a null hypothesis. [1] (b) Calculate t, using t = (mean 1 − mean 2) ÷ √(s1²/n1 + s2²/n2). [3] (c) State the degrees of freedom. At P = 0.05 the critical t value is 2.10. Give your conclusion. [3]

12. Sixty slugs were found under four types of cover. The numbers under a stone, log, plank and plastic sheet were 27, 15, 9 and 9. The null hypothesis is that slugs are spread equally.

Calculate χ², state the degrees of freedom and, given a critical value of 7.81 at P = 0.05, draw a conclusion. [5]

Answers

1. (a) All the organisms in an area, together with the non-living features and the interactions between them [1]. [1] (b) Biotic, for example predation by crabs [1], which reduces periwinkle numbers [1]. Abiotic, for example salinity rising on hot days as water evaporates, which stresses periwinkles [1]. [3] Examiner insight: A factor named with no effect on the organism, such as “temperature”, is not enough.

2. Cut the grass at ground level in a known area, for example random quadrats [1]. Weigh the sample, then dry it in an oven at about 80 °C [1]. Re-weigh at intervals until the mass is constant [1]. Divide the dry mass by the area sampled, or scale up to 1 m² [1]. [4] Examiner insight: “Dry it” without “to constant mass” is an easy mark to lose.

3. (a) Temperature rise = 38.5 − 19.0 = 19.5 °C [1]. Energy = 40 × 4.18 × 19.5 = 3 260 J (3 260.4 J) [1]. [2] (b) 3 260.4 ÷ 0.85 = 3 840 J g⁻¹ (3.84 kJ g⁻¹) [1]. [1] (c) Heat is lost to the surroundings and apparatus, so the water gains less energy than was released [1]. [1] Examiner insight: Give the unit. A bare number is weaker than a number with its unit.

4. (a) Efficiency = incorporated ÷ ingested × 100 = 624 ÷ 4 800 [1] = 13.0% [1]. [2] (b) 54 ÷ 540 × 100 = 10.0% [1]. [1] (c) 4 800 − 624 = 4 176 kJ m⁻² year⁻¹ [1]. It is lost, for example, as faeces, urine, or heat from respiration [1]. [2] Examiner insight: Divide by the intake of that trophic level, not by the producers’ energy.

5. Fertilisers or irrigation increase crop growth [1]. Herbicides remove competing weeds [1]. Pesticides or biological control reduce biomass lost to pests [1]. Keeping livestock indoors, warm and with limited movement reduces respiration, so less biomass is lost as heat and movement [1]. More of the intake is built into the animal’s biomass, which increases the transfer to humans [1]. Disadvantage, for example eutrophication, harm to non-target species or welfare concerns [1]. [6] Examiner insight: Link each method to the specific loss it reduces.

6. (a) Nitrification [1]. Nitrosomonas converts ammonium to nitrite [1]. Nitrobacter converts nitrite to nitrate [1]. [3] (b) Waterlogged soil is short of oxygen, which nitrifying bacteria need, so less nitrate is made [1]. Denitrifying bacteria are favoured in anaerobic conditions [1]. They convert nitrate to nitrogen gas, which leaves the soil [1]. [3] (c) Clover is a legume with Rhizobium bacteria in its root nodules [1]. The bacteria fix nitrogen gas into ammonium, and when the clover dies and decomposes, the nitrogen compounds are released and converted to nitrate for the next crop [1]. [2] Examiner insight: Name the organism and its step together. “Bacteria make nitrate” does not separate the two steps.

7. Photosynthesis removes carbon dioxide from the air to build organic compounds [1]. Respiration by plants, animals and decomposers returns carbon dioxide [1]. Decomposers break down dead material, releasing carbon dioxide [1]. Oceans: carbon dioxide dissolves and forms hydrogencarbonate and carbonate ions [1]. Marine organisms build calcium carbonate shells, which form limestone sediment, a long-term store [1]. [5] Examiner insight: Give the direction of each process (into or out of the atmosphere).

8. (a) Soil organic matter rises from 0.3% to 6.5% as the dunes age [1]. The number of plant species also rises, from 2 to 12 per m² [1]. [2] (b) Pioneers such as marram grass die and decompose, adding organic matter [1]. This forms soil that holds more water and nutrients, so conditions are less extreme [1]. New species can then colonise [1]. [3] (c) Deflected succession [1]. Grazing by rabbits stops shrubs and trees establishing, so the community is held at an earlier stage [1]. [2] (d) N = 42 [1]. Σ(n/N)² = 0.159, so D = 1 − 0.159 = 0.84 [1]. [2] Examiner insight: When describing a trend, quote values from the table, not only “it increases”.

9. (a) N = 36 × 42 ÷ 9 [1] = 168 [1]. [2] (b) Two from: marked animals mix evenly with the population; marking does not affect survival or capture; no births, deaths or movement in or out; the mark is not lost [1][1]. [2] (c) Marked animals make up too large a share of the second catch [1]. The calculation divides by the number of marked animals, so the population is underestimated [1]. [2] Examiner insight: Say which quantity in the formula is too high or low, and how the estimate changes.

10. (a) There is no significant correlation between distance from the sea and the number of plant species [1]. [1] (b) Distance ranks run from 9 down to 1. Species ranks: 9, 7, 8, 6, 4, 5, 2, 3, 1 [1]. d = 0, 1, −1, 0, 1, −1, 1, −1, 0, so d² sums to 6 [1]. [2] (c) rs = 1 − (6 × 6) ÷ (9 × 80) [1] = 1 − 36/720 = 0.95 [1]. [2] (d) 0.95 is greater than 0.700 [1], so the null hypothesis is rejected: more species are found further inland, and the positive correlation is significant [1]. [2] Examiner insight: A conclusion should reject or keep the null hypothesis, in context, without claiming cause.

11. (a) There is no significant difference between the mean heights in full light and in shade [1]. [1] (b) s1²/n1 = 9.18/10 = 0.918 and s2²/n2 = 6.66/10 = 0.666 [1]. √(0.918 + 0.666) = 1.258 [1]. t = 6.5 ÷ 1.258 = 5.17 [1]. [3] (c) Degrees of freedom = 10 + 10 − 2 = 18 [1]. 5.17 is greater than 2.10, so reject the null hypothesis [1]. The mean height is significantly greater in full light [1]. [3] Examiner insight: Keep unrounded values through the calculation, and state the degrees of freedom before reading the critical value.

12. Expected = 60 ÷ 4 = 15 in each [1]. χ² = (12²/15) + 0 + (6²/15) + (6²/15) = 9.6 + 0 + 2.4 + 2.4 [1] = 14.4 [1]. Degrees of freedom = 4 − 1 = 3 [1]. 14.4 is greater than 7.81, so reject the null hypothesis: the slugs are not spread equally [1]. [5] Examiner insight: Show each (O − E)² ÷ E term, and state the expected values first.

Where marks are usually lost

• Measuring fresh mass and calling it biomass.
• Dividing by the wrong total in efficiency calculations.
• Saying “energy is lost” without naming the route.
• Swapping Nitrosomonas and Nitrobacter.
• Using the wrong degrees of freedom or critical value.
• Stating that a correlation shows cause.

Next steps

• Review the revision notes and study guide.
• Use the hub and 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 6.3.1(a) to (e), Ecosystems.