Revision Notes
OCR A Level Biology A: Measuring biodiversity (H420) – Revision Notes
Condensed revision notes on biodiversity levels, sampling, Simpson's Index and genetic diversity for OCR A Level Biology A (H420).
- Subject
- Biology
- Level
- A LEVELS
- Topic
- Measuring biodiversity
- Author
- Marlbridge Academic Team
- Updated
Aligned to OCR A Level Biology (H420), Version 4.1 (April 2026), for first assessment in 2025. Official specification .
Syllabus page (what it covers and how it is assessed): OCR A Level Biology.
Syllabus points this page covers
H420
- 4 Biodiversity, evolution and disease (whole topic)
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These notes condense section 4.2.1(a) to (e) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025. The study guide gives the full explanations and extra worked examples. When you can recall everything below, try the practice questions, tick off topics on the printable checklist, and see the OCR A Level Biology hub. The free diagnostics help you find weak spots.
Key definitions (4.2.1(a))
| Term | Meaning |
|---|---|
| Biodiversity | The variety and complexity of life |
| Habitat biodiversity | The range of different habitats, for example sand dunes, woodland, meadows, streams |
| Species biodiversity | Species richness plus species evenness |
| Species richness | The count of distinct species found in a habitat |
| Species evenness | How similar the numbers of individuals of each species are |
| Genetic biodiversity | The variety of alleles within a species, for example different breeds |
| Sample | A small, representative part of a habitat that is studied in detail |
| Polymorphic gene locus | A gene locus with more than one allele in the population |
Sampling (4.2.1(b))
Why sample
- A whole habitat cannot be counted.
- Samples must be representative, so take enough and calculate a mean.
- Sample the range of organisms, because each technique finds only some groups.
Types of sampling
| Type | Placement | Best use | Weakness |
|---|---|---|---|
| Random | Random number coordinates | Uniform area | May miss a zone by chance |
| Systematic | Regular intervals, such as a transect | Gradient | Not random, so pattern may be missed or exaggerated |
| Stratified | Each distinct area sampled in proportion to its size | Area with clear zones | Needs prior knowledge of zones |
| Opportunistic | Convenient places | Quick survey | Biased |
Techniques
| Technique | Used for |
|---|---|
| Quadrat | Plants and slow-moving or fixed animals (count, frequency, percentage cover) |
| Sweep net | Flying and leaf-dwelling insects in vegetation |
| Pitfall trap | Ground-living invertebrates |
| Pooter | Small insects collected by suction from leaves, bark or a tray |
Method in steps: random quadrat sampling
- Lay two tape measures at right angles along two sides of the area.
- Generate pairs of random numbers within the tape lengths.
- Place a quadrat at each coordinate.
- Record the species and numbers (or percentage cover) in each.
- Take enough quadrats, then calculate the mean per quadrat.
- Scale up to the habitat area.
Quadrat measures
- Abundance: the number of individuals of a species in a quadrat.
- Frequency: the percentage of quadrats in which a species appears. Present in 12 of 40 quadrats gives 30%.
- Percentage cover: the proportion of the quadrat covered by a species. It suits plants whose separate individuals cannot be told apart, such as grass.
Field practice (PAG3)
- Use the same effort in every sample, for example the same number of sweeps or the same trap time.
- Identify organisms with a key and record unknowns for later checking.
- Handle animals with care and return them to where they were caught.
- Stratified example: a plot that is 60% grass and 40% scrub sampled with 20 quadrats uses 12 in the grass and 8 in the scrub.
Scaling up
estimated population = mean per quadrat ÷ quadrat area × habitat area
Worked reminder: mean 3.6 plants per 0.25 m² quadrat in a 250 m² plot gives 3.6 ÷ 0.25 = 14.4 per m², and 14.4 × 250 = 3600 plants.
Statistics for fieldwork
- Mean and range summarise samples. Standard deviation shows how spread out the values are around the mean.
- Student’s t-test: is the difference between two means significant?
- Spearman’s rank correlation: is there a relationship along a transect?
- Chi squared test: do observed results differ from expected results?
Richness and evenness (4.2.1(c))
- Richness = a count of species. It ignores how many individuals each has.
- Evenness needs the number of individuals per species.
- Same richness, different evenness: 8, 8, 8, 8, 8 (even) and 36, 1, 1, 1, 1 (uneven) are both five species with 40 individuals.
Simpson’s Index of Diversity (4.2.1(d))
| Item | Detail |
|---|---|
| Formula | D = 1 − Σ(n/N)² |
| n | Individuals of one species |
| N | Total individuals of all species |
| Range | 0 to 1 |
| Provided? | The specification says the formula is provided where needed and need not be recalled |
Method in steps
- Count n for each species and N in total.
- Calculate n/N for each species.
- Square each.
- Add the squares.
- Subtract from 1.
Worked reminder: pond sample
Species counts 10, 5, 3, 2 (N = 20).
Fractions: 0.5, 0.25, 0.15, 0.1. Squares: 0.25, 0.0625, 0.0225, 0.01. Sum = 0.345. D = 1 − 0.345 = 0.655.
Limiting cases
- One species only: Σ(n/N)² = 1, so D = 0.
- Five species of 8 each: Σ(n/N)² = 5 × 0.04 = 0.2, so D = 0.8.
- Two species, 30 and 10: 0.75² + 0.25² = 0.625, so D = 0.375.
Interpreting D
| High D (near 1) | Low D (near 0) |
|---|---|
| Many species, fairly even numbers | Few species, or one dominant species |
| Diverse, stable habitat | Less diverse, vulnerable to change |
| Many niches and food sources | May be disturbed, polluted, harsh or a monoculture |
Always link the number to richness and evenness. Say “greater species diversity” for a higher D, not “more individuals”.
Genetic diversity (4.2.1(e))
proportion of polymorphic gene loci = number of polymorphic gene loci ÷ total number of loci
- Multiply by 100 for a percentage.
- Used for isolated populations: zoos (captive breeding), rare breeds and pedigree animals.
- Worked reminder: 15 polymorphic loci out of 48 gives 15 ÷ 48 = 0.3125, which is 31% to 2 significant figures.
- Percentage points describe a difference between two percentages. A percentage change is relative to the starting value.
- Small, isolated populations lose alleles by chance, and inbreeding is more likely. Low diversity means less variation for selection to act on, so less ability to adapt to change such as a new disease.
Must-know distinctions
- Richness or evenness. Richness counts species. Evenness compares their numbers.
- Random or systematic. Random uses chance. Systematic uses a regular pattern, which suits a gradient.
- Stratified or opportunistic. Stratified is planned and proportional. Opportunistic is convenient and biased.
- Species diversity or genetic diversity. One is variety of species, the other variety of alleles within a species.
- n or N. Lower case is one species, capital is every individual.
Quick self-test
- Define species richness.
- Which sampling approach suits a line from the strandline to the back of a dune?
- Which technique samples ground-dwelling beetles?
- Calculate D for a sample with five species of 8 individuals each.
- Calculate D for a sample containing one species only.
- In 35 gene loci, 7 are polymorphic. Calculate the percentage.
- A mean of 3.6 plants per 0.25 m² quadrat is found in a 250 m² plot. Estimate the population.
- Two sites each have six species, but D is 0.80 for site A and 0.42 for site B. What does this show?
- Calculate D for two species with 30 and 10 individuals.
- Calculate D for a meadow sample of 14, 6, 4, 3 and 3 individuals.
- Why should more than one sampling technique be used in a habitat?
- State one reason why a captive population may have low genetic diversity.
Answers
- The count of distinct species found in a habitat.
- Systematic sampling (a transect), because conditions change along it.
- Pitfall traps.
- D = 1 − 0.2 = 0.8.
- D = 1 − 1² = 0.
- 7 ÷ 35 = 0.20, so 20%.
- 3.6 ÷ 0.25 = 14.4 per m²; 14.4 × 250 = 3600 plants.
- Both sites have the same richness. Site B has a lower D because its individuals are much less evenly spread, with one or a few species dominant. Site A is more diverse.
- 0.75² + 0.25² = 0.625; D = 0.375.
- N = 30. The squares of the fractions are 0.2178, 0.04, 0.0178, 0.01 and 0.01, which sum to 0.2956 (266 ÷ 900). D = 0.704 (3 significant figures).
- Each technique samples only some groups, so using several records more of the range of organisms in the habitat.
- The founder group was small, so few alleles were present. Mating between relatives (inbreeding) then reduces variation further.
Where marks are usually lost
- Saying richness is the number of individuals.
- Describing opportunistic sampling as random.
- Omitting the “1 −” in Simpson’s Index, so a diverse habitat gets a low answer.
- Forgetting to square n/N.
- Using the number of species as N.
- Not linking a high or low D to richness and evenness in the explanation.
- Writing “more biodiversity, so more individuals”.
- Leaving out units, or the scaling-up step, in population estimates.
- Dividing by the wrong total in a proportion of polymorphic loci.
- Suggesting a larger sample but not saying why it improves reliability.
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.
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