OCR GCSE Biology A: Field investigations, biodiversity and human impact on ecosystems (J247) – 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 OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), topic B6.1, B6.1a to B6.1d. Questions 11 and 12 test B6.1d, which is printed in bold in the specification and is Higher Tier only. All other questions suit both tiers.

Links: study guide, revision notes, OCR GCSE Biology hub, printable checklist and the free 10-minute diagnostics.

Questions

1. A woodland measures 60 m by 30 m. A student places ten random quadrats, each 1 m², and counts wood sorrel plants: 4, 0, 2, 7, 3, 1, 5, 2, 0, 6. Estimate the number of wood sorrel plants in the whole woodland. Show your working. [4]

2. A researcher marks 60 field voles with a small ear tag and releases them. A week later 45 voles are trapped, of which 9 carry a tag.

(a) Calculate an estimate of the population. [3] (b) State one assumption made in this method. [1]

3. A student suspects that daisies are more abundant on a mown lawn than in an unmown meadow next to it. Describe how the student could investigate this using quadrats. [6]

4. Name the equipment best suited to each job.

(a) Collecting small insects from leaf litter by suction without crushing them. [1] (b) Collecting invertebrates from the water of a pond. [1] (c) Identifying an unfamiliar plant from its features. [1]

5. A student lays a transect away from a footpath and records percentage cover of clover in a 0.5 m by 0.5 m quadrat at each point.

Distance from path (m)	0	2	4	6	8
Clover cover (%)	4	11	19	27	31

(a) Give one reason for using a transect rather than random quadrats here. [1] (b) Describe the pattern shown, quoting figures. [2] (c) Calculate the mean percentage cover. [1] (d) Suggest one factor that changes with distance from the path and could explain the pattern. [1]

6. The numbers of snails found in 20 quadrats are listed in order: 1, 2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 30.

(a) Find the 75th percentile, showing which position you used. [2] (b) State what the value in (a) tells you about the quadrats. [1] (c) Calculate the mean number of snails per quadrat. [2]

7. A hunted species of antelope fell from 1850 animals to 1110 animals over ten years.

(a) Calculate the percentage decrease. [2] (b) Explain how hunting can reduce biodiversity. [2]

8. A nature reserve receives 1200 visitors in a season. Each visitor pays £85, and 40% of the income is spent on conservation.

(a) Calculate the amount spent on conservation. [3] (b) Explain two ways in which ecotourism can help to maintain biodiversity. [2] (c) Suggest one way in which ecotourism could harm the wildlife it relies on. [1]

9. Describe two positive and two negative human interactions within ecosystems, and for each say how it affects biodiversity. [4]

10. Many countries share one migration route used by an endangered bird.

(a) Explain why it can be hard to gain agreement on a scheme to protect the whole route. [2] (b) Explain why it is hard to monitor whether such a scheme is working. [2] (c) State one benefit of maintaining high biodiversity. [1]

11. (Higher Tier only) A stream is sampled at three sites. Site 1 is upstream of a sewage outflow, Site 2 is 50 m downstream and Site 3 is 500 m downstream.

Site	Dissolved oxygen (mg/dm³)	Stonefly nymphs	Bloodworms
1	9.0	31	0
2	3.6	2	38
3	7.8	22	4

(a) Calculate the percentage fall in dissolved oxygen from Site 1 to Site 2. [2] (b) Explain the distribution of stonefly nymphs and bloodworms. [2] (c) Evaluate how well these data support the idea that the outflow caused the change in distribution. [3]

12. (Higher Tier only) A survey counts lichen species on trees along a line leading away from a city centre. Sulfur dioxide was measured at the same points.

Distance from centre (km)	1	3	5	7	9
Sulfur dioxide (arbitrary units)	52	40	24	14	9
Lichen species	2	3	6	8	9

(a) Describe the pattern in lichen species. [2] (b) Explain the pattern. [2] (c) Give two limitations of the evidence. [2] (d) State one way to make the evidence stronger. [1]

Answers

1. Total count = 4 + 0 + 2 + 7 + 3 + 1 + 5 + 2 + 0 + 6 = 30, so mean = 30 ÷ 10 = 3.0 per quadrat [1]. Quadrat is 1 m², so 3.0 per m² [1]. Area = 60 × 30 = 1800 m² [1]. Estimate = 3.0 × 1800 = 5400 plants [1]. [4] Examiner insight: Show each stage. A clear route from count to mean to area lets a slip in one step be seen, and a bare answer gives no way to give credit for method.

2. (a) Estimate = (60 × 45) ÷ 9 [1]. = 2700 ÷ 9 [1]. = 300 voles [1]. (b) For example: tagging did not harm the voles or change their chance of capture, or no animals entered or left the population [1]. [4] Examiner insight: Write the relationship in words or symbols first. The answer is an estimate, and an assumption must be something the method depends on, not a general remark about care.

3. Mark out an area of each habitat of similar size [1]. Use a measured grid and random numbers as coordinates to place quadrats, avoiding bias [1]. Use the same quadrat size in both habitats [1]. Place a suitable number of quadrats, for example 10 or more in each habitat [1]. Count the daisies in each quadrat, then calculate the mean per quadrat for each habitat [1]. Control or note other factors such as soil, shade and time of year, and compare the means, scaling up if the area is needed [1]. [6] Examiner insight: Plan answers gain credit for specific choices, such as how the quadrat positions are decided and how many are used. “Count daisies” alone is too vague to be a method.

4. (a) Pooter [1]. (b) Pond net [1]. (c) Key [1]. [3] Examiner insight: Give the piece of equipment named in the specification list.

5. (a) Conditions change gradually with distance from the path, so a transect shows how abundance changes along the gradient [1]. (b) Clover cover increases as distance from the path increases [1], from 4% at 0 m to 31% at 8 m [1]. (c) Mean = (4 + 11 + 19 + 27 + 31) ÷ 5 = 92 ÷ 5 = 18.4% [1]. (d) For example, trampling, soil compaction or light level [1]. [5] Examiner insight: Pattern questions need figures from the table. A description that only says “it goes up” shows the trend but not that you read the data.

6. (a) Position = 0.75 × 20 = 15 [1]. The 15th value is 21 [1]. (b) About 75% of the quadrats had 21 snails or fewer [1]. (c) Total = 292 [1], mean = 292 ÷ 20 = 14.6 snails per quadrat [1]. [5] Examiner insight: Say clearly how you found the position. A percentile answer is a value from the data, not the position number itself.

7. (a) Decrease = 1850 − 1110 = 740 [1]. Percentage decrease = 740 ÷ 1850 × 100 = 40.0% [1]. (b) Fewer animals of the species are left, so the species may become extinct, reducing the variety of living organisms [1]. Loss of an antelope affects the food web, for example its predators and the plants it grazes [1]. [4] Examiner insight: Divide by the starting value, not the final value. In explanations, move from the animal to biodiversity or the food web instead of stopping at “numbers fall”.

8. (a) Income = 1200 × £85 = £102 000 [1]. 40% of this = 0.40 × 102 000 [1]. = £40 800 [1]. (b) Any two: income pays for rangers and monitoring [1]; jobs for local people give a reason to protect habitats [1]. (c) Too many visitors can disturb breeding animals or damage plants and paths [1]. [6] Examiner insight: Money answers need units. An explanation of benefits should say how the benefit helps wildlife, not just that tourism is “good for the area”.

9. Positive, any two: captive breeding or reintroduction increases numbers of a threatened species [1]; nature reserves or restored habitats protect many species at once [1]. Negative, any two: clearing forest or draining wetlands destroys habitats so species decline [1]; hunting or poaching reduces a species towards extinction, or pollution such as acid rain removes sensitive species [1]. [4] Examiner insight: Each point needs an action and an effect on biodiversity. Naming only the activity is not enough.

10. (a) Countries have different priorities, such as economic development or farming, so some may not agree to restrictions [1]. Agreements across borders are hard to enforce or pay for fairly [1]. (b) Monitoring needs regular surveys over a large area, which are costly and need trained staff [1]. Remote areas make illegal hunting or habitat loss hard to detect [1]. (c) For example, more stable ecosystems, or sources of food and medicines [1]. [5] Examiner insight: Questions that ask you to explain difficulty want a reason for each. Repeating “it is difficult” scores nothing.

11. (a) Fall = 9.0 − 3.6 = 5.4 [1]. Percentage fall = 5.4 ÷ 9.0 × 100 = 60% [1]. (b) Stonefly nymphs need high oxygen, so they are few where oxygen is low and many where it is high [1]. Bloodworms tolerate low oxygen and are present at Site 2 where there is little competition [1]. (c) Oxygen, stonefly numbers and bloodworm numbers change at the outflow and recover downstream, which supports the idea [1]. However, there is only one sample at each site, so the result may not be typical [1]. Other variables, such as temperature or flow speed, were not controlled, and the data show correlation rather than proof of cause [1]. [7] Examiner insight: An evaluation scores best when it gives a strength and a limit, each tied to the data.

12. (a) Lichen species increase with distance from the centre [1], from 2 at 1 km to 9 at 9 km [1]. (b) Sulfur dioxide is highest near the centre (52 units) and falls with distance [1]. Sensitive lichens cannot survive high sulfur dioxide, so fewer species are present near the centre [1]. (c) Any two: a single line of trees only, so it may not be typical [1]; other factors such as tree type, tree age or other pollutants may differ [1]. (d) Repeat the survey along several lines from the centre, or record species and sulfur dioxide at more points [1]. [7] Examiner insight: Link the gas, the effect on the organism and the distribution. A trend alone does not show that sulfur dioxide is the cause, so keep the wording measured.

Where marks are usually lost

• Giving a mean per quadrat when the question asks for an estimate in the whole area.
• Using a quadrat of 0.25 m² as if it were 1 m².
• Calling quadrats random when they were placed by choice.
• Treating a capture-recapture answer as an exact count.
• Calculating a percentage change by dividing by the new value.
• Describing a graph or table without quoting figures.
• Claiming that a pattern proves a cause.
• Writing “do repeats” with no reason in an evaluation.

Next steps

• Review the revision notes and the study guide.
• Return to the OCR GCSE Biology hub and tick off the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source: OCR GCSE (9-1) Gateway Science Biology A (J247), Version 4.0 (August 2026), for first assessment in 2018, published by OCR. Topic B6.1 Monitoring and maintaining the environment, statements B6.1a to B6.1d.