AQA A-level Biology: Biodiversity, Index of Diversity and Farming (7402) – Revision Notes

These are condensed notes for AQA A-level Biology (7402) section 3.4.6, Biodiversity within a community, from the AQA AS and A-level Biology specification for first teaching 2015. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions. Course pages: AQA A-level Biology hub and the printable checklist. Not sure where your gaps are? Take a free diagnostic.

Key definitions

Term	Definition to learn
Biodiversity	The variety of living organisms, considered from a small local habitat up to the whole Earth
Community	The populations of all the different species that live together in one habitat
Species richness	A count of the different species in a community
Index of diversity	A measure relating the number of species in a community to the number of individuals in each species
Monoculture	One crop grown over a large area (an example of a farming technique)

Scales of biodiversity

• Small local habitat: a pond, a hedge base, a rotting log.
• Larger habitat: a woodland, a lake, a large field.
• Region or country: built up from many surveys.
• The Earth: only estimated, because many species are not yet described.
• Compare like with like: same scale, same sampling method, same effort.

Richness and the index of diversity

Measure	What it uses	Weakness
Species richness	Number of species only	Ignores how many individuals there are of each, so an even community and a dominated one can score the same
Index of diversity, d	Number of species and individuals in each	Needs individuals to be counted, and is unreliable for small samples

Formula

d = N(N − 1) ÷ Σn(n − 1)

N is the total number of organisms of all species. n is the number of organisms of one species. Σ means the sum over all species.

Method in steps

1. Find N by adding every count. Work out N(N − 1).
2. For each species work out n(n − 1).
3. Add those to get Σn(n − 1).
4. Divide: d = N(N − 1) ÷ Σn(n − 1).
5. Give 2 or 3 significant figures, then say what the value means.

Worked reminder. A sample has species counts 8, 4 and 2. N = 14, so N(N − 1) = 14 × 13 = 182. Σn(n − 1) = 56 + 12 + 2 = 70. d = 182 ÷ 70 = 2.6.

Second worked reminder. Site 1 has counts 16, 4 and 1 and site 2 has 9, 8 and 4. Both have N = 21, so N(N − 1) = 420. Site 1: Σn(n − 1) = 240 + 12 + 0 = 252, so d = 420 ÷ 252 = 1.67. Site 2: Σn(n − 1) = 72 + 56 + 12 = 140, so d = 420 ÷ 140 = 3.0. Both have three species, but site 2 has the more even spread and so the higher d.

Special cases

• One species only: d = 1, the lowest possible value.
• Every species has one individual: Σn(n − 1) = 0, so d cannot be calculated.
• A species with n = 0 is not in the sample, so ignore it.
• A species with n = 1 adds 0 to the denominator.
• Equal counts across species give a high d. One dominant species gives a low d.

Interpreting d

• Higher d: more species and a more even spread, so a more diverse community.
• Lower d: fewer species or one dominating, often a community more vulnerable to change.
• Compare values only when sampling effort and method are the same.
• Say whether the sample is large enough and random. Small samples may miss species.
• Describe the difference in words and quote both values, for example “d is higher at A (4.7) than at B (1.6)”.
• For percentage change: (new − old) ÷ old × 100.

Farming and biodiversity

Technique	Effect on biodiversity
Monoculture	Fewer plant species, so fewer food sources and habitats for animals
Hedgerow removal	Loss of nesting, shelter and food, and fewer routes between habitats
Herbicides	Wild plants killed, so less food for insects and animals that eat them
Pesticides	Non-target insects killed, so less food for insect-eating animals
Drainage, ploughing grassland	Habitat lost or changed
Fertiliser	Nutrient run-off can alter nearby water communities
Overgrazing	Fewer plant species survive

Chain to write: habitat or food removed, so fewer niches, so fewer species, so lower d.

Conservation and farming: the balance

• Farming gives food and income, and intensive methods give high yield from less land.
• Conservation protects species, and wild species can help farming, such as pollinators and pest predators.
• Compromises: hedgerows kept, sown field margins and buffer strips, crop rotation, fallow land, fewer pesticides, lower stocking density, protected wet areas.
• Evaluate with data: how much did d rise, and how much did yield fall?
• Give a judgement: say which factors matter most and why.

Answering the command words

Command word	What to write
Calculate	Show N(N − 1), each n(n − 1), the total and the division, then the answer to a sensible number of significant figures
Compare	Give both values of d, say which is higher, and link the difference to richness and evenness
Explain	Give the mechanism: what changed in the habitat, why that changes the species present, and what happens to d
Suggest	Offer a reasonable biological reason that fits the data supplied
Evaluate	Weigh evidence on both sides, say what the data cannot show, then reach a conclusion

Sampling points for diversity data

• Use random sampling, for example random coordinates for quadrats or trap positions, so that the sample is not biased by where you choose to look.
• Use the same method, effort and time of year in each habitat being compared.
• Take enough samples that rare species are likely to appear.
• Repeat on different days if you can, and quote values with a sensible number of significant figures.

Must-know distinctions

• Richness counts species; the index uses species and numbers.
• Species richness is not the same as the number of individuals.
• Community means many species in a habitat; population means one species.
• Higher d means more varied, not more abundant.
• The index is calculated from a sample, not from the entire community.

Quick self-test

1. Define species richness.
2. A sample has two species with 10 individuals each. Calculate d to 3 significant figures.
3. A sample has two species with 9 and 1 individuals. Calculate d to 3 significant figures.
4. A sample has counts 7, 5, 3 and 1. Calculate d to 3 significant figures.
5. A sample contains 20 individuals, all of one species. What is d?
6. Six species with 3 individuals each. Calculate d.
7. Why is d a better measure than richness for comparing two meadows?
8. Explain why a species with one individual adds nothing to Σn(n − 1).
9. Give two ways hedgerow removal reduces biodiversity.
10. Why should two values of d be compared only if the sampling effort was the same?
11. A farm adds sown field margins and d rises from 1.5 to 4.9. State one further piece of data needed before judging the margins a success.
12. Name two features of a sample that make a calculated d unreliable.

Answers

1. A count of the different species in a community.
2. N = 20, N(N − 1) = 380. Σn(n − 1) = 90 + 90 = 180. d = 380 ÷ 180 = 2.11.
3. N = 10, N(N − 1) = 90. Σn(n − 1) = 72 + 0 = 72. d = 90 ÷ 72 = 1.25.
4. N = 16, N(N − 1) = 240. Σn(n − 1) = 42 + 20 + 6 + 0 = 68. d = 240 ÷ 68 = 3.53.
5. N(N − 1) = 380 and Σn(n − 1) = 380, so d = 1 (the lowest possible value).
6. N = 18, N(N − 1) = 306. Each species gives 6, so Σn(n − 1) = 36. d = 306 ÷ 36 = 8.5.
7. It uses the number of individuals of each species as well as the number of species, so it shows whether one species dominates.
8. n(n − 1) = 1 × 0 = 0.
9. Any two from: loss of nesting sites or shelter; loss of food sources; fewer routes between habitats; loss of plant species.
10. More traps, nights or area would catch more individuals and species, so a difference in d could be due to effort instead of the habitat.
11. The effect on crop yield (or on income), so the gain in biodiversity can be weighed against the cost.
12. A small number of individuals sampled; sampling that is not random; species wrongly identified; a sample taken at only one time of year.

Where marks are usually lost

• Using n² or N² instead of n(n − 1) and N(N − 1).
• Adding the species counts for N but forgetting to subtract 1 inside N(N − 1).
• Including species with n = 0 as if they were present.
• Dividing the wrong way round, so d comes out below 1.
• Stating richness as a number of individuals.
• Describing d only as “higher” without saying it means more species or a more even spread.
• Giving a farming effect with no mechanism, such as “pesticides harm wildlife” instead of “kill insects that are food for birds”.
• Evaluating the conservation and farming balance from one side only.
• Comparing d values without commenting on sample size or method.

Official syllabus

AQA AS and A-level Biology specification (7401, 7402), for first teaching 2015, published by AQA. These notes are aligned to section 3.4.6 and maths skill MS 2.3.