OCR A Level Biology A: Measuring biodiversity (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 4.2.1(a) to (e) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: levels of biodiversity, sampling, richness and evenness, Simpson’s Index of Diversity and genetic diversity. Revise first with the study guide and the revision notes. See also the OCR A Level Biology hub, the checklist and the free diagnostics. Simpson’s Index is D = 1 − Σ(n/N)².

Questions

1. State what is meant by habitat biodiversity, species biodiversity and genetic biodiversity, and give an example of each. [3]

2. Name the most suitable sampling technique for each of the following. [3]

(a) Ground-living beetles in a woodland. (b) Small insects on the leaves of a bush, collected alive without damage. (c) Flying insects in long meadow grass.

3. Describe how you would use random sampling with quadrats to estimate the abundance of a plant species in a field measuring 20 m by 10 m. [4]

4. A student wants to find how the plants change from the strandline to the back of a sand dune. Explain why systematic sampling is more suitable than random sampling, and state one limitation of systematic sampling. [4]

5. Ten random quadrats, each 0.5 m by 0.5 m, were placed in a lawn of area 600 m². The numbers of plantain plants were 4, 7, 2, 6, 5, 3, 8, 5, 4 and 6. Estimate the number of plantain plants in the lawn. [3]

6. A sample from a stream contained five invertebrate species, with 34, 21, 12, 8 and 5 individuals. Calculate Simpson’s Index of Diversity (D). Give your answer to 3 significant figures. [5]

7. Two grassland sites were sampled in the same way. Each sample had 80 invertebrates in six species.

Species	1	2	3	4	5	6
Site A	22	18	15	12	8	5
Site B	60	6	5	4	3	2

(a) The value of D for site A is 0.80. Calculate D for site B to 2 decimal places. [3] (b) Explain the difference between the two values of D, although both sites have the same species richness. [3]

8. In a captive population of an endangered bird, 55 gene loci were studied. Twenty-two loci were polymorphic. Ten years later, after a fall in the number of breeding birds, 14 of the same loci were polymorphic.

(a) Calculate the percentage of polymorphic loci in the first study. [2] (b) Calculate the percentage decrease in the number of polymorphic loci. Give your answer to 3 significant figures. [2]

9. A rare breed of pig is kept in three herds. The table shows the loci examined and the number that were polymorphic.

Herd	Loci examined	Polymorphic loci
A	40	14
B	36	9
C	40	23

(a) Calculate the proportion of polymorphic loci in each herd, as a percentage. [3] (b) Which herd is most at risk from a new disease? Explain your answer. [3] (c) Suggest two ways in which the herds could be managed to maintain genetic diversity. [2]

10. A student sampled a woodland and a meadow once each. The woodland sample had six species, with 12, 9, 7, 6, 4 and 2 individuals. The meadow sample had four species, with 40, 25, 10 and 5 individuals.

(a) Calculate D for each habitat to 3 significant figures. [4] (b) The student concludes that woodland is always more diverse than meadow. Evaluate this conclusion. [3]

11. Explain why Simpson’s Index of Diversity gives more information about a habitat than a count of the species present. [3]

12. Plan an investigation to compare the biodiversity of ground-living invertebrates in a woodland edge and an open meadow. [9]

Answers

1. Habitat biodiversity is the range of different habitats in an area, for example sand dunes, woodland, meadows or streams [1]. Species biodiversity is the variety of species in a habitat, made up of species richness and evenness [1]. Genetic biodiversity is the variety of alleles within a species, for example the different breeds of sheep [1]. [3] Examiner insight: Naming a type of each level, not just defining it, is the safest way to show you can tell them apart; “genetic” must refer to alleles or genes within one species.

2. (a) Pitfall traps [1]. (b) A pooter [1]. (c) A sweep net [1]. [3] Examiner insight: Match the technique to where the animal lives and how it moves. Answers such as “net” without saying sweep, or “quadrat” for moving insects, are too vague or unsuitable.

3. Lay two tape measures at right angles along two sides of the field [1]. Take pairs of random digits or generator output as coordinates, within 20 m along one side and 10 m along the other [1]. Place the quadrat at each coordinate and count the plants of the species inside it [1]. Take many quadrats, calculate the mean per quadrat, and scale up to the field area [1]. [4] Examiner insight: Say how the positions are chosen at random (random numbers as coordinates). Throwing a quadrat is not a reliable random method.

4. The conditions change along the dune, so a belt or line transect with samples at regular intervals shows the gradient [1]. Random samples might cluster in one zone and miss others [1]. The same effort at each distance allows comparison between zones [1]. A limitation is that samples are not random, so a pattern in the habitat at the interval chosen could bias the results (or only one line is sampled, which may not represent the whole dune) [1]. [4] Examiner insight: A good answer links systematic sampling to a gradient. “It is easier” alone gains no credit.

5. Total = 50, so mean = 50 ÷ 10 = 5.0 plants per quadrat [1]. Quadrat area = 0.5 × 0.5 = 0.25 m², so density = 5.0 ÷ 0.25 = 20 plants per m² [1]. Population = 20 × 600 = 12 000 plants [1]. [3] Examiner insight: Work out the quadrat area from its sides before dividing; using 0.5 m² instead of 0.25 m² halves the answer. State that this is an estimate.

6. N = 34 + 21 + 12 + 8 + 5 = 80 [1]. The fractions are 0.425, 0.2625, 0.15, 0.1 and 0.0625 [1]. The squares are 0.1806, 0.0689, 0.0225, 0.01 and 0.0039 [1]. Σ(n/N)² = 1830 ÷ 6400 = 0.2859 [1]. D = 1 − 0.2859 = 0.714 [1]. [5] Examiner insight: Show each stage, since rounding the squares early can shift the final digit. Subtracting from 1 is a separate step that is easy to omit.

7. (a) N = 80 and Σ(n/N)² = (3600 + 36 + 25 + 16 + 9 + 4) ÷ 6400 = 3690 ÷ 6400 = 0.5766 [2]. D = 1 − 0.5766 = 0.42 [1]. (b) The two sites have the same richness, so the difference is due to evenness [1]. In site B, species 1 makes up 75% of individuals, so the numbers are very uneven [1]. In site A the individuals are spread more evenly, so D is higher (0.80) and site A is more diverse [1]. [6] Examiner insight: For an explanation, state which of richness and evenness is the same, which differs, and use figures from the table.

8. (a) 22 ÷ 55 = 0.40 [1], so 40% [1]. (b) Decrease = 22 − 14 = 8 loci, and 8 ÷ 22 × 100 [1] = 36.4% [1]. [4] Examiner insight: A percentage decrease is calculated relative to the original number of polymorphic loci. Dividing by 55 gives percentage points, not the percentage decrease.

9. (a) A: 14 ÷ 40 = 35.0% [1]. B: 9 ÷ 36 = 25.0% [1]. C: 23 ÷ 40 = 57.5% [1]. (b) Herd B [1] has the lowest proportion of polymorphic loci, so the least genetic variation [1]. A new disease is less likely to find some individuals with resistance alleles, so the herd is more likely to be badly affected [1]. (c) Any two: exchange breeding animals between herds [1]; use pedigree records to avoid mating between close relatives [1]; increase the number of breeding animals (no further credit). (Maximum 2.) [8] Examiner insight: In (b) name the herd and use its figure; link low variation to the chance that some individuals can survive the change.

10. (a) Woodland: N = 40, Σ(n/N)² = 330 ÷ 1600 = 0.20625 [1], D = 0.794 [1]. Meadow: N = 80, Σ(n/N)² = 2350 ÷ 6400 = 0.3672 [1], D = 0.633 [1]. (b) The woodland has the higher D and more species in this sample, which supports the conclusion for these two samples [1]. “Always” is not justified: one sample of each, on one occasion, with different sample sizes (40 and 80 individuals), may not represent the habitats [1]. More random samples at different times, using the same effort for each, would give means and allow a statistical test [1]. [7] Examiner insight: Give a verdict on the claim, then say what the data support and what limits the conclusion.

11. Species richness only counts the species, so a habitat with one dominant species and a habitat with evenly spread species can have the same count [1]. D uses the number of individuals of each species, so it takes both richness and evenness into account [1]. A high or low D can then be interpreted, for example to compare habitats or detect a disturbance [1]. [3] Examiner insight: Mention evenness explicitly. “D is more accurate” is too vague.

12. Choose several sites in each habitat using random coordinates (or stratified sampling if the habitat has distinct zones) [1]. Use pitfall traps sunk level with the soil, with covers against rain [1]. Use the same number of traps and the same trapping time in both habitats [1]. Control other variables, such as trap type, season, weather and time of day [1]. Identify the animals with a key and count the individuals of each species [1]. Release the animals unharmed, and handle them with care [1]. Calculate richness and D for each habitat [1]. Compare the means, with a Student’s t-test (or the standard deviation to show spread) [1]. Repeat on several occasions, because a single trapping period may not be representative [1]. [9] Examiner insight: Plans earn credit for specific, testable detail, such as number of traps, trap duration and the calculation made. A list of general phrases such as “keep it fair” does not.

Where marks are usually lost

• Describing a method as random without saying how the positions are chosen.
• Using one technique and claiming it samples the whole range of organisms.
• Failing to square n/N, or forgetting to subtract from 1.
• Using the number of species as N instead of the total number of individuals.
• Rounding intermediate values and then giving the answer to too many significant figures.
• Saying equal richness means equal diversity.
• Discussing a high D without mentioning evenness.
• Dividing by the wrong total when calculating a proportion of polymorphic loci.
• Confusing percentage points with percentage change.
• Treating one sample on one day as proof about a whole habitat.

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 4.2.1(a) to (e), Biodiversity.