Biodiversity within a community - AQA A-Level Biology (Paper 1) Exam Questions

Practise 17 exam-style questions on Biodiversity within a community for AQA A-Level Biology (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

17 questions
1–8 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Ecologists set a light trap overnight beside a hedgerow on a farm. The next morning, they identified and counted the moths in the trap. The table shows their results.

The index of diversity, d, can be calculated using the formula

d = N(N - 1) ÷ Σn(n - 1)

where N = total number of moths of all species and n = total number of moths of each species.

Calculate the index of diversity of the moths in the trap. Give your answer to 2 decimal places.

Table with columns "Moth species", "Number of moths". Moth species: Large yellow underwing; Number of moths: 14 | Moth species: Heart and dart; Number of moths: 9 | Moth species: Common footman; Number of moths: 6 | Moth species: Dark arches; Number of moths: 11 | Moth species: Setaceous Hebrew character; Number of moths: 3 | Moth species: Brimstone moth; Number of moths: 5 | Moth species: Flame shoulder; Number of moths: 2.

Model answer

N = 14 + 9 + 6 + 11 + 3 + 5 + 2 = 50
N(N - 1) = 50 × 49 = 2450
Σn(n - 1) = (14 × 13) + (9 × 8) + (6 × 5) + (11 × 10) + (3 × 2) + (5 × 4) + (2 × 1)
= 182 + 72 + 30 + 110 + 6 + 20 + 2 = 422
d = 2450 ÷ 422 = 5.8057
d = 5.81 (to 2 decimal places)

Explanation

N is the total number of individuals of all species (50 moths), not the number of species (7) - using the number of species is the commonest error. Work out n(n - 1) for each species separately, add the values to get Σn(n - 1), then divide N(N - 1) by this total. Write down N(N - 1) and Σn(n - 1): each can earn a mark even if the final answer is wrong.

Mark scheme

1 mark
Correct method, e.g. N(N - 1) = 50 × 49 = 2450 OR Σn(n - 1) = 422; or a correct final answer of 5.81 with no working
A correct answer of 5.81 with no working scores 2 marks. Accept for 1 mark: evidence of 2450 (correct N(N - 1)) OR evidence of 422 (correct Σn(n - 1)) OR 5.8, 5.80 or 5.806 (correct value not given correctly to 2 decimal places). Do not award this mark for using N = 7 (the number of species instead of the number of moths).
1 mark
5.81
Only 5.81 scores this mark. Ignore any unit.

Question 2

3 marks

Grey partridges are birds that nest on farmland. For the first two weeks after hatching, partridge chicks feed almost entirely on insects. Many of these insects feed on weeds (wild plants) growing among cereal crops.

Explain how spraying cereal fields with herbicide can reduce the biodiversity of the farmland community and cause the number of grey partridges to fall.

Model answer

The herbicide kills the weeds growing among the crop, so there are fewer plant species in the field. Insects that feed on these weeds lose their food and habitat, so there are fewer insect species and fewer insects. Partridge chicks then have less food in their first two weeks, so fewer chicks survive and partridge numbers fall. Other animals that depend on the weeds or insects are also affected, so the biodiversity of the whole community falls.

Explanation

A herbicide is not aimed at insects or birds, but it removes the plants at the bottom of their food chains, so its effect spreads through the community. This is why herbicide use is one of the farming techniques that reduce biodiversity. Make sure you say that herbicides kill plants: confusing herbicides with insecticides is a common error.

Mark scheme

1 mark
Herbicide kills weeds / wild plants, so there are fewer plant species (in the field)
Max 3. Reject 'herbicide kills the insects' for this point (herbicides kill plants). Accept 'lower species richness of plants'.
1 mark
Fewer food plants / habitats for insects (that feed on weeds), so fewer insect species / fewer insects
1 mark
Less food for partridge chicks, so fewer chicks survive (to breed), so partridge numbers fall
Accept 'chicks starve'.
1 mark
Other species that depend on these plants or insects are also lost / the effect is passed along food chains, so the index of diversity / species richness of the community falls

Question 3

4 marks

A student identified and counted all the plants in a sample from a garden lawn. She found 120 plants belonging to 6 species. She calculated the index of diversity, d, using the formula

d = N(N - 1) ÷ Σn(n - 1)

where N = total number of plants of all species and n = total number of plants of each species.

Her value of Σn(n - 1) was 3570, giving d = 4.00.

She then found one more plant in the sample. It belonged to a seventh species that she had not recorded before.

Calculate the new value of d. Give your answer to 2 decimal places.

Use the formula to explain why d increases, even though the new species is represented by only one plant.

Model answer

New N = 120 + 1 = 121
New N(N - 1) = 121 × 120 = 14 520
For the new species, n(n - 1) = 1 × 0 = 0, so Σn(n - 1) stays at 3570
New d = 14 520 ÷ 3570 = 4.067 = 4.07

A species represented by a single individual adds nothing to Σn(n - 1), because 1 × (1 - 1) = 0, so the denominator stays the same. However, the extra plant increases N from 120 to 121, so N(N - 1) increases from 14 280 to 14 520. The numerator increases while the denominator is unchanged, so d increases.

Explanation

The formula rewards communities with many species whose individuals are spread evenly. Each species contributes n(n - 1) to the denominator, which is large for abundant species and zero for a species with one individual, while every individual adds to N. In the same way, if a community contained only one species, n would equal N, so d = N(N - 1) ÷ N(N - 1) = 1, the lowest possible value. A common error is to forget to change N.

Mark scheme

1 mark
Correct method, e.g. (121 × 120) ÷ 3570 OR 14 520 ÷ 3570; or a correct final answer of 4.07 with no working
A correct answer of 4.07 with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 14 520 (correct new N(N - 1)) OR 4.067 or 4.1 (correct value not given to 2 decimal places). Do not award this mark for 14 280 ÷ 3570 = 4.00 (N not changed).
1 mark
4.07
1 mark
For the new species n = 1, so n(n - 1) = 1 × 0 = 0, so Σn(n - 1) does not change / stays at 3570
Reject 'Σn(n - 1) increases by 1'. Accept 'the new species adds nothing to the bottom of the formula'.
1 mark
N increases (from 120 to 121), so N(N - 1) increases (from 14 280 to 14 520), so d increases
Accept 'the top of the formula increases but the bottom stays the same'. Insufficient: 'there are more species' (must use the formula).

Question 4

5 marks

Scientists compared the wheat fields on 12 organic farms with the wheat fields on 12 conventional farms in one region of England. Organic farms do not use artificial fertilisers and use very few pesticides or herbicides. The table shows some of their results.

Explain why the organic wheat fields had more plant species and a higher index of diversity of spiders than the conventional fields.

A politician said: 'Converting all farms in the UK to organic farming would increase biodiversity.' Use the data to evaluate this statement.

Table with columns "Measurement", "Organic farms", "Conventional farms". Measurement: Mean number of plant species per field; Organic farms: 21.4; Conventional farms: 9.8 | Measurement: Mean index of diversity of spiders; Organic farms: 5.6; Conventional farms: 3.9 | Measurement: Mean wheat yield / tonnes per hectare; Organic farms: 4.6; Conventional farms: 8.1.

Model answer

Organic farms use few or no herbicides, so more weeds and other wild plant species survive in the fields. These plants provide food and habitats for more species of insects, so there is more prey for spiders, and because few insecticides are used, the spiders and their prey are not killed.

The data support the statement in part: organic wheat fields had more than twice as many plant species as conventional fields (21.4 compared with 9.8) and a higher index of diversity of spiders (5.6 compared with 3.9). However, the organic farms produced much less wheat (4.6 compared with 8.1 tonnes per hectare, about 43% less), so more land would be needed to grow the same amount of food, and natural habitats could be destroyed elsewhere. Also, only wheat fields on 12 farms of each type in one region were studied, and no statistical test was done, so we do not know whether the differences are significant.

Explanation

Herbicides remove wild plants and insecticides kill non-target insects, so farms that use few of them keep more of the plants and invertebrates at the base of food chains. This is the balance between conservation and farming: methods that increase biodiversity on a farm often reduce yield. Good evaluations use the data for and against the statement and comment on how far the investigation can be generalised (sample size, crop, region, organisms measured, statistics); quote the figures from the table rather than saying 'more' or 'less'.

Mark scheme

1 mark
Few / no herbicides, so more weeds / wild plant species survive (in the organic fields)
Max 4 if the answer does not use any figures from the table. Max 5. Points 1 to 4 (the explanation): max 2. Points 5 to 10 (the evaluation): max 3. Reject 'herbicides kill insects / spiders'.
1 mark
More plant species provide more (types of) food / habitats for insects / invertebrates, so more prey (species) for spiders
Accept 'more places for spiders to build webs / more habitats for spiders'.
1 mark
Few / no insecticides / pesticides, so spiders / the insects they feed on are not killed
Accept 'non-target species / predators are not killed'.
1 mark
No artificial fertiliser, so fast-growing plants (e.g. grasses) do not outcompete other plant species
1 mark
(Supports) organic fields had more plant species (21.4 compared with 9.8 / more than twice as many) OR a higher index of diversity of spiders (5.6 compared with 3.9)
The figures must be quoted for this mark.
1 mark
The wheat yield of organic farms is lower (4.6 compared with 8.1 tonnes per hectare / about 43% lower), so more land would be needed to grow the same amount of food, so more natural habitats (in the UK or elsewhere in the world) might be converted to farmland (reducing biodiversity)
The idea of more land being needed is required for this mark. Insufficient: 'organic farms produce less wheat' on its own.
1 mark
Only wheat fields / one region / 12 farms of each type were studied, so the results may not apply to all farms in the UK
1 mark
Only plants and spiders were measured / other groups of organisms may not show the same difference
1 mark
No standard deviations / statistical test, so we do not know whether the differences are significant / could be due to chance
Reject 'the results are due to chance'.
1 mark
The organic farms may differ in other ways (e.g. more hedgerows / different soils), so the difference may not be caused by organic farming

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