Simpson's Index of Diversity

By the end of this lesson you should be able to explain and apply each part of this topic — The Formula, Interpreting D Values, Why Simpson's Index Is Useful and Using Simpson's Index in Conservation — and use these ideas accurately in exam-style questions.

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 ($D = 1 - \sum (n/N)^2$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 ($H' = -\sum p_i \ln p_i$H′=−∑pi​lnpi​), 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.

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The Formula

$D = 1 - \sum\left(\frac{n}{N}\right)^2$D=1−∑(Nn​)2

The formula works by:

1. Dividing the number of individuals of each species (n) by the total number of all individuals (N), giving the proportion each species contributes.
2. Squaring that proportion.
3. Summing the squared proportions across all species.
4. Subtracting from 1.

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 $D = \sum(n/N)^2$D=∑(n/N)2, which runs the other way (lower = more diverse). OCR uses the Index of Diversity form ($1 - \sum(n/N)^2$1−∑(n/N)2), where higher values mean higher diversity. Always state clearly which version you are using.

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Worked Example 1: Two Meadows

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:

$\sum(n/N)^2 = (40/150)^2 + (35/150)^2 + (30/150)^2 + (25/150)^2 + (20/150)^2$∑(n/N)2=(40/150)2+(35/150)2+(30/150)2+(25/150)2+(20/150)2

$= 0.0711 + 0.0544 + 0.0400 + 0.0278 + 0.0178 = 0.2111$=0.0711+0.0544+0.0400+0.0278+0.0178=0.2111

$D_A = 1 - 0.2111 = 0.789$DA​=1−0.2111=0.789

Meadow B calculation:

$\sum(n/N)^2 = (130/150)^2 + (10/150)^2 + (5/150)^2 + (3/150)^2 + (2/150)^2$∑(n/N)2=(130/150)2+(10/150)2+(5/150)2+(3/150)2+(2/150)2

$= 0.7511 + 0.0044 + 0.0011 + 0.0004 + 0.0002 = 0.7572$=0.7511+0.0044+0.0011+0.0004+0.0002=0.7572

$D_B = 1 - 0.7572 = 0.243$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.

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Worked Example 2: Comparing Rocky Shores

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: $\sum(n/N)^2 = (42/110)^2 + (18/110)^2 + (27/110)^2 + (9/110)^2 + (14/110)^2$∑(n/N)2=(42/110)2+(18/110)2+(27/110)2+(9/110)2+(14/110)2 $= 0.1458 + 0.0268 + 0.0602 + 0.0067 + 0.0162 = 0.2557$=0.1458+0.0268+0.0602+0.0067+0.0162=0.2557 $D_1 = 1 - 0.2557 = 0.744$D1​=1−0.2557=0.744

Shore 2: $\sum(n/N)^2 = (88/100)^2 + (5/100)^2 + (7/100)^2 = 0.7744 + 0.0025 + 0.0049 = 0.7818$∑(n/N)2=(88/100)2+(5/100)2+(7/100)2=0.7744+0.0025+0.0049=0.7818 $D_2 = 1 - 0.7818 = 0.218$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.

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Interpreting D Values

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:

• Stable environments with long histories (tropical rainforests, ancient woodland).
• Diverse habitats with many niches.
• Absence of disturbance — pollution, agriculture and invasive species all reduce D.

Low D values are associated with:

• Harsh environments (deserts, polar regions, acidic bogs).
• Polluted habitats — only a few tolerant species survive.
• Early colonisation — pioneer communities.
• Monoculture agriculture — intentionally low diversity.

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).

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Why Simpson's Index Is Useful

Simpson's Index has several advantages over simply counting species:

1. Sensitive to evenness. Two habitats with the same number of species can have very different D values.
2. Bounded 0 to 1. Easy to compare and interpret.
3. Weights common species more. Rare species affect the value only slightly, which reduces the influence of sampling error (missing a rare species barely changes D).
4. Quick to calculate. No statistical tables needed.

Limitations

• Less sensitive to rare species. If you want to highlight rarities, other indices (e.g. Shannon-Wiener) may be better.
• Depends on accurate sampling. Biased samples give biased D values.
• Single number. Sometimes it is more useful to report richness and evenness separately.

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Using Simpson's Index in Conservation

Conservationists calculate D before and after an intervention to monitor success:

• Reintroduction projects — beaver reintroduction in Devon raised aquatic invertebrate D from 0.54 to 0.71.
• Pollution monitoring — sewage outflows lower freshwater D; successful clean-ups should raise it again.
• Habitat restoration — replanting hedgerows after intensive farming increases both richness and D.

Long-term monitoring using Simpson's Index forms part of many UK conservation schemes under bodies like Natural England and the JNCC.

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Common Exam Mistakes

• Forgetting to square (n/N). A classic arithmetic slip.
• Forgetting to subtract from 1. D is $1 - \sum(n/N)^2$1−∑(n/N)2, not $\sum(n/N)^2$∑(n/N)2.
• Comparing D with the wrong formula. Be sure your version has "1 −" at the front.
• Treating D as a percentage. It is a proportion, not a percentage; do not multiply by 100.
• Saying "high D means more species". Strictly, high D means high richness and/or high evenness. A few species evenly distributed can still give a reasonable D.
• Reporting too many decimal places. Match the precision of your input data — usually 2 or 3 significant figures.

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Quick Recap

• Simpson's Index of Diversity $D = 1 - \sum(n/N)^2$D=1−∑(n/N)2 combines species richness and evenness into one number between 0 and 1.
• Higher D = greater diversity; a habitat dominated by one species has low D even if richness is high.
• Strengths: sensitive to evenness, bounded, quick to calculate.
• Weaknesses: less sensitive to rare species; depends on unbiased sampling.
• Used by conservationists to monitor habitats over time.

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Worked Example 3: A Polluted vs Clean Stream

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: $\sum (n/N)^2 = (28/100)^2 + (15/100)^2 + (22/100)^2 + (30/100)^2 + (5/100)^2 = 0.0784 + 0.0225 + 0.0484 + 0.0900 + 0.0025 = 0.242$∑(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 $D_{\text{clean}} = 1 - 0.242 = 0.758$Dclean​=1−0.242=0.758.

Polluted reach: $\sum (n/N)^2 = (0)^2 + (0)^2 + (2/85)^2 + (8/85)^2 + (75/85)^2 = 0 + 0 + 0.0006 + 0.0089 + 0.7785 = 0.788$∑(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 $D_{\text{polluted}} = 1 - 0.788 = 0.212$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.

Comparison Table: Simpson Versus Other Indices

Index	Formula	Range	Sensitive to	OCR-relevant
Simpson's index of diversity	$D = 1 - \sum (n/N)^2$D=1−∑(n/N)2	0 → 1	Common species	Yes — OCR default
Simpson's dominance	$\sum (n/N)^2$∑(n/N)2	0 → 1	Common species (inverted)	No
Shannon-Wiener	$H' = -\sum p_i \ln p_i$H′=−∑pi​lnpi​	0 → ln(S)	Rare species more	No
Species richness (S)	count	1 → ∞	All species equally	Yes (qualitatively)
Pielou's evenness	$J = H' / \ln S$J=H′/lnS	0 → 1	Evenness only	No

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Synoptic Links

• sampling-techniques — the n and N values fed into Simpson's index come from quadrat / transect / trap data; bad sampling propagates into a misleading D.
• biodiversity-levels — Simpson's index operationalises the species level of biodiversity (it does not directly capture habitat or genetic biodiversity).
• factors-affecting-biodiversity — Simpson's D is the quantitative metric for monitoring how pollution, habitat loss, and climate change reduce biodiversity.
• conservation-in-situ-and-ex-situ — D before vs after conservation interventions (e.g. reintroductions, agri-environment schemes) is the routine evaluation tool.

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Specimen Question (modelled on OCR H420 paper format)

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 $D \approx 0.70$D≈0.70. Show all working. (3 marks) (b) Suggest TWO biological reasons why $D$D fell from 0.81 to 0.42 between 1990 and 2010. (4 marks) (c) Evaluate whether the recovery from $D = 0.42$D=0.42 to $D = 0.67$D=0.67 proves the agri-environment scheme has worked. (6 marks)

AO breakdown

Part	AO1	AO2	AO3
(a)	0	3	0
(b)	1	3	0
(c)	1	2	3

Grade-band model answers

(a) 3-mark Mid-band response: Total = 100. $\sum (n/N)^2 = (0.5)^2 + (0.2)^2 + (0.15)^2 + (0.1)^2 + (0.05)^2 = 0.25 + 0.04 + 0.0225 + 0.01 + 0.0025 = 0.325$∑(n/N)2=(0.5)2+(0.2)2+(0.15)2+(0.1)2+(0.05)2=0.25+0.04+0.0225+0.01+0.0025=0.325. $D = 1 - 0.325 = 0.675 \approx 0.70$D=1−0.325=0.675≈0.70.

Examiner-style commentary: M1 — proportions correctly squared. M2 — sum correctly accumulated. M3 — subtraction from 1 with appropriate rounding. A typical pitfall is computing Simpson's index as $\sum (n/N)^2$∑(n/N)2 and forgetting the $1 -$1− step that converts dominance into diversity.

(b) 4-mark Mid-band response: Pesticides will have killed some of the insect species, especially the more sensitive ones. Hedgerow loss will have reduced the habitat for insects so there were fewer species and one or two became dominant.

Examiner-style commentary: M1 — pesticides as a cause. M2 — link to species loss. M3 — hedgerow loss as a cause. M4 — link to habitat simplification. A solid C-grade response but could be tightened by naming a specific group (e.g. neonicotinoid impact on pollinators).

(b) 4-mark Top-band response: Two plausible reasons: (1) chemical pressure — neonicotinoid insecticides leaching from adjacent intensive farmland will have selectively eliminated pollinators and other sensitive non-target insects, reducing species richness and shifting the community toward tolerant taxa (Diptera, some Coleoptera); (2) habitat simplification — hedgerow loss and field-margin removal eliminated the small-scale habitat heterogeneity that supports specialist insects, leaving generalist species dominant. Both mechanisms reduce species evenness (M and Y dropped: one or two species now dominate, raising $\sum (n/N)^2$∑(n/N)2) and richness (some species are now absent), so D falls sharply.

Examiner-style commentary: M1, M2 — chemical mechanism with named pesticide class and target group. M3, M4 — habitat-simplification mechanism with link back to richness and evenness. The A* explicitly links the biological cause to the mathematical drivers of D (richness and evenness).

(c) 6-mark Mid-band response: D has gone up from 0.42 to 0.67 so biodiversity has increased and the scheme is working. But D is still lower than the 1990 value of 0.81, so recovery is not complete. Other factors might have changed too, like climate. So we cannot be sure the scheme alone caused the increase.

Examiner-style commentary: M1 — direction of change correctly read. M2 — recovery is partial. M3 — confounding variables noted. Basic AO3 but does not develop the experimental-design weakness.