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Spec Mapping — OCR H420 Module 4.2.1 — Biodiversity, content statements covering the use of Simpson's Index of Diversity as a quantitative measure of biodiversity and the interpretation of its values to compare habitats (refer to the official OCR H420 specification document for exact wording). This lesson is the quantitative core of Module 4.2.1 and a routine source of AO2 calculation questions in OCR H420 papers.
Simpson's Index of Diversity is the cornerstone quantitative measure used by A-Level Biology students to compare the biodiversity of two habitats. Unlike species richness (which is just a count), Simpson's Index combines richness and evenness into a single number between 0 and 1, where higher values mean greater diversity. OCR A-Level Biology A Module 4.2.1 explicitly requires you to be able to calculate Simpson's Index and interpret the value.
The index is named after the British statistician Edward Hugh Simpson, who derived the dominance formulation in a 1949 paper for Nature. The OCR-style "Index of Diversity" form (D=1−∑(n/N)2) inverts the dominance form so that higher values correspond to higher diversity — more intuitive for biology students but mathematically equivalent. The complementary index of Shannon-Wiener (H′=−∑pilnpi), often encountered alongside Simpson, derives from information theory (Claude Shannon, 1948) and weights rare species more heavily than Simpson does.
Key Definitions:
- Simpson's Index of Diversity (D) — a measure of biodiversity combining species richness and evenness.
- n — the number of individuals of a particular species.
- N — the total number of individuals of all species.
- Richness — the number of different species present.
- Evenness — how equally abundant each species is.
D=1−∑(Nn)2
The formula works by:
The result is a number between 0 (no diversity — all individuals belong to one species) and nearly 1 (high diversity — many species, each with similar abundance).
Exam Tip: Be careful: some textbooks present Simpson's Index as D=∑(n/N)2, which runs the other way (lower = more diverse). OCR uses the Index of Diversity form (1−∑(n/N)2), where higher values mean higher diversity. Always state clearly which version you are using.
A student samples two meadows of equal area and counts the flowering plants in ten quadrats in each.
Meadow A:
| Species | n |
|---|---|
| Daisy | 40 |
| Buttercup | 35 |
| Clover | 30 |
| Dandelion | 25 |
| Plantain | 20 |
| Total (N) | 150 |
Meadow B:
| Species | n |
|---|---|
| Daisy | 130 |
| Buttercup | 10 |
| Clover | 5 |
| Dandelion | 3 |
| Plantain | 2 |
| Total (N) | 150 |
Both meadows have five species (same richness), but Meadow A has high evenness and Meadow B is dominated by daisies.
Meadow A calculation:
∑(n/N)2=(40/150)2+(35/150)2+(30/150)2+(25/150)2+(20/150)2
=0.0711+0.0544+0.0400+0.0278+0.0178=0.2111
DA=1−0.2111=0.789
Meadow B calculation:
∑(n/N)2=(130/150)2+(10/150)2+(5/150)2+(3/150)2+(2/150)2
=0.7511+0.0044+0.0011+0.0004+0.0002=0.7572
DB=1−0.7572=0.243
Interpretation: Meadow A (D = 0.789) is far more diverse than Meadow B (D = 0.243), despite both having the same species richness. This is exactly what our eyes told us — Meadow B is dominated by daisies — but now we have a single number to quantify it.
Two rocky shores are surveyed with ten 0.5 m² quadrats each; the total counts of mobile invertebrates are:
| Species | Shore 1 (n) | Shore 2 (n) |
|---|---|---|
| Common limpet | 42 | 88 |
| Dog whelk | 18 | 5 |
| Common periwinkle | 27 | 7 |
| Shore crab | 9 | 0 |
| Beadlet anemone | 14 | 0 |
| Total (N) | 110 | 100 |
Shore 1: ∑(n/N)2=(42/110)2+(18/110)2+(27/110)2+(9/110)2+(14/110)2 =0.1458+0.0268+0.0602+0.0067+0.0162=0.2557 D1=1−0.2557=0.744
Shore 2: ∑(n/N)2=(88/100)2+(5/100)2+(7/100)2=0.7744+0.0025+0.0049=0.7818 D2=1−0.7818=0.218
Shore 1 has both higher richness (5 species vs 3) and higher evenness. Shore 2 is dominated by limpets with very few other species — perhaps it is more exposed to wave action or pollution.
| D value | Interpretation |
|---|---|
| 0.0 | All individuals belong to one species; no diversity |
| 0.1–0.3 | Very low diversity; one or two species dominate |
| 0.4–0.6 | Moderate diversity |
| 0.7–0.9 | High diversity; many species present with even abundance |
| → 1.0 | Maximum diversity |
High D values are associated with:
Low D values are associated with:
Exam Tip: A common 2-mark question asks: "Suggest why D decreased from 0.75 to 0.32 between 1980 and 2020." Always link the fall in D to a cause (pollution, habitat loss, climate change) and its mechanism (sensitive species eliminated, dominant species take over).
Simpson's Index has several advantages over simply counting species:
Conservationists calculate D before and after an intervention to monitor success:
Long-term monitoring using Simpson's Index forms part of many UK conservation schemes under bodies like Natural England and the JNCC.
Two stream reaches are sampled with three-minute kick samples and the macroinvertebrates counted:
| Taxon | Clean reach (n) | Polluted reach (n) |
|---|---|---|
| Mayfly larvae (Ephemeroptera) | 28 | 0 |
| Stonefly larvae (Plecoptera) | 15 | 0 |
| Caddisfly larvae (Trichoptera) | 22 | 2 |
| Freshwater shrimp (Gammarus) | 30 | 8 |
| Bloodworm (Chironomus) — pollution-tolerant | 5 | 75 |
| Total (N) | 100 | 85 |
Clean reach: ∑(n/N)2=(28/100)2+(15/100)2+(22/100)2+(30/100)2+(5/100)2=0.0784+0.0225+0.0484+0.0900+0.0025=0.242, so Dclean=1−0.242=0.758.
Polluted reach: ∑(n/N)2=(0)2+(0)2+(2/85)2+(8/85)2+(75/85)2=0+0+0.0006+0.0089+0.7785=0.788, so Dpolluted=1−0.788=0.212.
The clean reach has more than three times the diversity of the polluted reach, driven by the loss of all sensitive taxa (mayflies, stoneflies) and the dominance of the pollution-tolerant bloodworm. The biological mechanism is organic enrichment — sewage or fertiliser raises the BOD (biological oxygen demand), suffocating sensitive taxa with high oxygen requirements while favouring those (chironomids, oligochaetes) with haemoglobin-like respiratory pigments that tolerate low O₂.
| Index | Formula | Range | Sensitive to | OCR-relevant |
|---|---|---|---|---|
| Simpson's index of diversity | D=1−∑(n/N)2 | 0 → 1 | Common species | Yes — OCR default |
| Simpson's dominance | ∑(n/N)2 | 0 → 1 | Common species (inverted) | No |
| Shannon-Wiener | H′=−∑pilnpi | 0 → ln(S) | Rare species more | No |
| Species richness (S) | count | 1 → ∞ | All species equally | Yes (qualitatively) |
| Pielou's evenness | J=H′/lnS | 0 → 1 | Evenness only | No |
A nature reserve has monitored its grassland insect community annually since 1990 using a standardised 50-quadrat sampling protocol. Simpson's Index of Diversity was 0.81 in 1990, fell to 0.42 in 2010 after intensive agriculture spread into the adjacent farmland, and recovered to 0.67 by 2024 after an agri-environment scheme reduced pesticide use and restored hedgerows.
(a) Show that a community with five species in the proportions 50 : 20 : 15 : 10 : 5 has Simpson's Index D≈0.70. Show all working. (3 marks) (b) Suggest TWO biological reasons why D fell from 0.81 to 0.42 between 1990 and 2010. (4 marks) (c) Evaluate whether the recovery from D=0.42 to D=0.67 proves the agri-environment scheme has worked. (6 marks)
| Part | AO1 | AO2 | AO3 |
|---|---|---|---|
| (a) | 0 | 3 | 0 |
| (b) | 1 | 3 | 0 |
| (c) | 1 | 2 | 3 |
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