4.6Genetics, Variation and Interdependence
Biodiversity — AQA A Level Biology 7402 Revision Notes & Topical Questions
10 min read 19 parts Lesson 32 of 46 Unit quiz
Biodiversity, Index of Diversity and Farming — AQA A LEVEL BIOLOGY 7402 notes: Define biodiversity, calculate an index of diversity and evaluate the effects of farming on biodiversity.Biodiversity is the variation that exists within and between all forms of life.
- Species richness
- Worked example 1
- Worked example 2
- Worked example 3
- Monoculture
- Conservation
Section 1 of 19Biodiversity
Biodiversity is the variation that exists within and between all forms of life. It describes the range and variety of genes, species and habitats within a particular region and can be assessed on different scales: within a community, within a species (genetic diversity) and across habitats.
Section 2 of 19Species richness and index of diversity
Species richness is the number of species in a community; it is the simplest measure of species diversity, but does not take into account how many individuals of each species are present.
- Step 1: calculate N(N − 1) = 202 × 201 = 40 602 (value A).
Section 3 of 19Farming and conservation
After the Second World War there was a massive change in food production: the need to produce more food at a quicker rate meant farmers had to achieve higher yields, so modern farming practices began: removal of hedgerows to make larger fields, monoculture, use of pesticides and herbicides, artificial fertilisers, drainage of wetlands, and use of machinery. These reduce biodiversity by removing habitats and killing non-target species.
Conservation matters because it maintains biodiversity, including wild species that may have future value (for example in medicine or agriculture); supports ecosystem stability; there are ethical, aesthetic and cultural reasons; and it helps combat climate change through carbon storage (for example in woodland and peatland). The aim is to maintain or improve biodiversity while still producing enough food.
Section 4 of 19Worked examples
Calculate the index of diversity for a community with 3 species of 10, 5 and 5 individuals. [3]3 marks
N = 20; N(N − 1) = 380. Σn(n − 1) = 90 + 20 + 20 = 130. d = 380 ÷ 130 = 2.9.
Explain why species richness alone is a poor measure of biodiversity. [2]2 marks
It ignores the number of individuals of each species; a community dominated by one species may have the same richness as an even community.
Explain how removing hedgerows reduces biodiversity. [3]3 marks
It removes habitats and food sources and corridors for movement. Fewer species survive.
Section 5 of 19Balancing farming and conservation
Balancing farming and conservation is challenging, but grants (for example EU subsidies) support environmentally friendly practices to offset losses, such as leaving field margins, planting hedgerows and reducing pesticide use. Know some examples of modern farming methods and be able to explain how they affect biodiversity.
Section 6 of 19Defining species and measuring variation
A species is a group of organisms that can interbreed and produce fertile offspring; members of one species are reproductively isolated from members of another. Diversity can be investigated by comparing:
- This method is not precise enough if only one characteristic is looked at (many animals have four legs) but is useful if a species has unique characteristics.
- If two species cannot be distinguished by observable features, measurable characteristics (such as beak length) give a better understanding of similarities and differences.
- which can be compared directly; the same mRNA must be compared between samples,
- and primers that bind to specific sequences can be used to select the gene.
- Example
- step 1, align the sequences;
- step 2, determine the most similar sequences.
- step 1, align the sequences;
- If Species X and Species Z have the most similar amino acid sequences, they are the most closely related.
Section 7 of 19Random sampling
Biodiversity can be hard to measure in large or complex ecosystems such as rainforests. Accurate species lists require identifying and counting all organisms present, which is only realistic in small or simple habitats (for example woodland plots or tree species). Sampling is used to estimate the abundance and distribution of species in a representative way.
The area must be large enough to give a representative estimate for the habitat.
Section 8 of 19Worked examples
Explain why a random number generator is used to place quadrats. [2]2 marks
It avoids bias in choosing sample sites, so the sample is representative of the whole area.
Explain why observable characteristics are a limited way to identify species. [2]2 marks
Different species may look similar or one species may vary; using one characteristic is not precise enough. Measurable or molecular data are better.
Species X, Y and Z have 2, 10 and 11 differences in amino acid sequence from species W. Which pair is most likely most closely related to W? [1]1 mark
Species X, which has the fewest differences.
Section 9 of 19Quadrat sampling
Once a quadrat is placed at a sample point, the abundance of all species present can be recorded. Mark up a grid on a map or scale drawing, label with coordinates, and use a random number generator to select coordinates. The area must be large enough for a representative estimate.
Section 10 of 19Random sampling to determine distribution
Random sampling can determine whether a species grows in groups (clumped) or is evenly spread. A method should include: the sampling technique (random or systematic, such as a belt transect along an environmental gradient); data collection (quadrats, transects); what will be measured (frequency, percentage cover, physical traits); and reliability and validity (repeats, controlling variables, sufficient sample size).
Section 11 of 19Mean and standard deviation
Descriptive statistics summarise large data sets concisely.
Method (rabbit ear lengths)
- 1) calculate the mean;
- 2) find the difference between each value and the mean (x − x̄);
- 3) square each difference;
- 4) total the squared differences;
- 5) divide by (n − 1);
- 6) take the square root.
- SD = √[Σ(x − x̄)2 ÷ (n − 1)].
Section 12 of 19Worked examples
A species is found in 12 of 40 quadrats. Calculate the frequency. [1]1 mark
12 ÷ 40 × 100 = 30%.
A student counts 215 daisies in 20 quadrats of 0.25 m2. Calculate density per m2. [2]2 marks
Total area = 20 × 0.25 = 5 m2. 215 ÷ 5 = 43 per m2.
Explain what a large standard deviation indicates. [1]1 mark
Large variation in the data; the values are spread widely around the mean.
Section 13 of 19Quantitative investigations of variation
Variation can be investigated using measures of central tendency such as the mean, the average value of a data set, and the standard deviation, which describes spread.
Section 14 of 19Worked context: diet and coronary heart disease
A group of scientists investigated the effects of a diet on the risk of coronary heart disease.
- then check whether the standard deviations overlap;
- if they do not, the difference is likely significant and could suggest the diet increases lumen diameter (reducing risk), though a correlation does not prove the diet alone caused the change.
Section 15 of 19Genetic relationships between organisms
Before DNA sequencing, genetic diversity was investigated by inferring DNA differences from measurable characteristics such as phenotype. DNA sequencing now allows genetic diversity to be assessed and tracked more accurately. DNA from the nucleus, mitochondria and chloroplasts can be sequenced to show evolutionary relationships between species.
Sequence comparisons help determine evolutionary relationships by analysing DNA or protein similarities.
Section 16 of 19From observable traits to molecular methods
Older methods of assessing genetic diversity are based on observable phenotypes (for example flower colour or leaf shape), which can be affected by the environment and do not show differences in non-coding DNA or silent mutations.
Section 17 of 19Worked examples
Mean artery diameter is 3.8 ± 0.2 mm in the diet group and 3.2 ± 0.3 mm in controls. Evaluate the evidence. [3]3 marks
The means differ and the SD ranges (3.6 to 4.0 and 2.9 to 3.5) do not overlap, so the difference is likely significant. A statistical test, sample size and control of other variables are needed.
Explain why fewer sequence differences suggest a more recent common ancestor. [2]2 marks
Mutations accumulate over time after species diverge, so recently diverged species have had less time to differ.
Give a reason why DNA sequencing is more accurate than comparing phenotypes. [1]1 mark
It compares the genetic material directly, rather than traits that are affected by the environment.
Section 18 of 19Key terms
| Term | Meaning |
|---|---|
| Biodiversity | Variety of genes, species and habitats |
| Species richness | Number of different species in a community |
| Index of diversity | Measure combining richness and evenness |
| Monoculture | Growing a single crop over a large area |
| Conservation | Protecting and managing biodiversity |
| Term | Meaning |
|---|---|
| Quadrat | Frame used to sample a small area |
| Random sampling | Choosing sites by chance to avoid bias |
| Primer | Short DNA sequence that binds to a specific target sequence |
| Reproductive isolation | Inability of two species to interbreed |
| Term | Meaning |
|---|---|
| Frequency | Percentage of quadrats in which a species occurs |
| Density | Number of individuals per unit area |
| Percentage cover | Proportion of area covered by a species |
| Standard deviation | Measure of spread of data around the mean |
| Transect | Line along which samples are taken |
| Term | Meaning |
|---|---|
| Mean | Average value of a data set |
| Standard deviation | Measure of spread of data |
| Sequence alignment | Matching up of DNA or amino acid sequences to compare them |
| Divergence | Splitting of lineages from a common ancestor |
| Statistical significance | Difference unlikely to be due to chance |
Section 19 of 19Exam technique
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