Simpson's Diversity Index
Two hedgerows in County Kildare each hold 100 plants from five species. In one, the plants are shared out evenly. In the other, hawthorn makes up almost every plant. Both have "five species", yet you would rather walk through the first one with a notebook. How can we put a number on that difference?
What biodiversity means
Biodiversity is the variety of living things in a habitat, ecosystem or on Earth as a whole. Ecologists measure it in two ways at once.
A habitat with 4 species of 10 individuals each is both rich and even. A habitat with 4 species where one has 97 individuals and the others 1 each is just as rich, but very uneven. Most of it is one species, so it is less diverse in practice.
Richness alone cannot tell these two samples apart. We need an index that uses both richness and evenness.
Define biodiversity* and state what is meant by species richness* and species evenness*. Explain why a high value of Simpson's Diversity Index indicates a more diverse…
Define biodiversity and state what is meant by species richness and species evenness. Explain why a high value of Simpson's Diversity Index indicates a more diverse habitat when the index is calculated as $D = 1-\sum\left(\frac{n}{N}\right)^2$.
Calculating Simpson's Diversity Index
Simpson's Diversity Index ($D$) takes the number of individuals of each species into account. Let $n$ be the number of individuals of one species and $N$ the total number of individuals of all species.
Why does it work? The fraction $\frac{n}{N}$ is the share of the sample that belongs to one species. Squaring and adding the shares gives a large number when one species dominates, and a small one when the shares are even. Taking $1-$ that sum flips it, so a high $D$ means high diversity. $D$ lies between $0$ (one species only) and $1$ (never reached).
Two limiting cases help you check your answer.
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One species only (say 40 individuals): the share is $\frac{40}{40} = 1$, so $D = 1 - 1^2 = 0$. There is no diversity at all.
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$S$ species, all equally abundant: each share is $\frac{1}{S}$, so $\sum \left(\frac{n}{N}\right)^2 = S \times \frac{1}{S^2} = \frac{1}{S}$ and $D = 1 - \frac{1}{S}$. With 4 equal species, $D = 1 - \frac{1}{4} = 0.75$.
A quadrat survey of a meadow in County Kildare found: 10 daisies, 10 buttercups, 10 clover and 10 dandelions. Using $D = 1-\sum\left(\frac{n}{N}\right)^2$, calculate…
A quadrat survey of a meadow in County Kildare found: 10 daisies, 10 buttercups, 10 clover and 10 dandelions. Using $D = 1-\sum\left(\frac{n}{N}\right)^2$, calculate Simpson's Diversity Index for the meadow.
A sample from a field contains only one species, with 40 individuals. Calculate Simpson's Diversity Index using $D = 1-\sum\left(\frac{n}{N}\right)^2$ and state what the…
A sample from a field contains only one species, with 40 individuals. Calculate Simpson's Diversity Index using $D = 1-\sum\left(\frac{n}{N}\right)^2$ and state what the value tells you about the field.
Using the index to interpret a habitat
Ecologists use $D$ to compare habitats, or the same habitat at different times. Always say what the number tells you in the context given. A falling $D$ over time can signal biodiversity loss, which is why $D$ is used to monitor habitats and plan conservation.
When you compare two sites:
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Calculate $D$ for each.
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Compare richness (count the species) and evenness (look at how the numbers are spread).
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Link the result to wildlife: more species and a more even spread usually mean more food sources, more niches and more shelter, so more wildlife is supported.
A student records invertebrates in a stream in County Wicklow: | Species | Number | |---|---| | Freshwater shrimp | 30 | | Mayfly nymph | 10 | | Stonefly nymph | 5 | |…
A student records invertebrates in a stream in County Wicklow:
| Species | Number |
|---|---|
| Freshwater shrimp | 30 |
| Mayfly nymph | 10 |
| Stonefly nymph | 5 |
| Water louse | 5 |
- a) Calculate $N$.
- b) Calculate $D = 1-\sum\left(\frac{n}{N}\right)^2$.
- c) State what the result suggests about water quality if mayfly and stonefly nymphs are pollution-sensitive.
Two hedgerows are sampled. | Species | Hedgerow A | Hedgerow B | |---|---|---| | Hawthorn | 20 | 37 | | Blackthorn | 20 | 1 | | Hazel | 20 | 1 | | Ash | 20 | 1 | | Holly…
Two hedgerows are sampled.
| Species | Hedgerow A | Hedgerow B |
|---|---|---|
| Hawthorn | 20 | 37 |
| Blackthorn | 20 | 1 |
| Hazel | 20 | 1 |
| Ash | 20 | 1 |
| Holly | 20 | 0 |
- a) Calculate $D$ for each hedgerow using $D = 1-\sum\left(\frac{n}{N}\right)^2$.
- b) Compare species richness and evenness in the two hedgerows.
- c) Explain which hedgerow is likely to support more wildlife and why.
The other form of the index
The specification gives the index in a second form, where individuals are counted in a different way:
Here $n$ is the number of individuals of one species and $N$ the total of all species. This form is the probability that two individuals picked at random, without replacement, belong to the same species. So a low value means high diversity, the reverse of the first form.
A formula sometimes used is $D = \frac{\sum n(n-1)}{N(N-1)}$, where a lower value of $D$ means higher diversity. A sample of a woodland floor has 6 beetles of species X,…
A formula sometimes used is $D = \frac{\sum n(n-1)}{N(N-1)}$, where a lower value of $D$ means higher diversity. A sample of a woodland floor has 6 beetles of species X, 4 of species Y and 2 of species Z.
- a) Calculate $D$ using this formula.
- b) Calculate $1-D$.
- c) Explain why the two forms of the index are inverse in interpretation.
Evaluating and reasoning with the index
Higher Level questions ask you to go beyond the arithmetic.
Use these ideas in "discuss" and "evaluate" answers.
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A monoculture such as intensive ryegrass has very low $D$: one species fills almost all the space. It supports fewer pollinators, because there are few flowers and a short flowering season. It also gives the soil one type of root and one type of plant litter, so there is less food for a varied soil community.
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A mixed farm with hedgerows and wildflower margins has a high $D$, with more habitats, flowers and shelter.
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Evaluate a claim like "high $D$ is always better for conservation": $D$ ignores which species are present (invasive, rare, native), so a site with a low $D$ but rare specialists can matter more. Good practice is to combine $D$ with species lists and repeated sampling.
A bog in County Offaly is surveyed before and after drainage for peat extraction. | Species | Before | After | |---|---|---| | Sphagnum moss | 60 | 10 | | Heather | 25 |…
A bog in County Offaly is surveyed before and after drainage for peat extraction.
| Species | Before | After |
|---|---|---|
| Sphagnum moss | 60 | 10 |
| Heather | 25 | 45 |
| Bog cotton | 10 | 5 |
| Sundew | 5 | 0 |
| Purple moor-grass | 0 | 40 |
- a) Calculate Simpson's Diversity Index $D = 1-\sum\left(\frac{n}{N}\right)^2$ before and after drainage.
- b) Interpret the change, commenting on richness and evenness.
- c) Explain why the index alone may not show the full ecological damage to the bog, and name a limitation of the sampling.
Farm A is an intensive ryegrass monoculture with $D = 0.12$. Farm B is a mixed organic farm with hedgerows and wildflower margins, with $D = 0.78$. a) Explain why a…
Farm A is an intensive ryegrass monoculture with $D = 0.12$. Farm B is a mixed organic farm with hedgerows and wildflower margins, with $D = 0.78$.
- a) Explain why a monoculture has a low index.
- b) Discuss two ecological consequences of the difference for pollinators and soil health.
- c) A researcher claims that Farm B is always better for conservation. Evaluate this claim, considering that $D$ ignores which species are present.
In a sample, species A has $n$ individuals and species B has $20-n$ individuals, so $N = 20$. a) Write an expression for $D = 1-\sum\left(\frac{n}{N}\right)^2$ in terms…
In a sample, species A has $n$ individuals and species B has $20-n$ individuals, so $N = 20$.
- a) Write an expression for $D = 1-\sum\left(\frac{n}{N}\right)^2$ in terms of $n$.
- b) Calculate $D$ for $n = 10$, $n = 15$ and $n = 19$.
- c) Deduce the value of $n$ that gives maximum diversity and explain why.
- d) State the maximum possible value of $D$ for a community of $S$ equally abundant species, giving your reasoning.
Back to the start: both Kildare hedgerows have five species, but the one with an even spread has a high $D$ and the one dominated by hawthorn a very low $D$. Simpson's index captures richness and evenness together.
Summary
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Biodiversity is the variety of life in a habitat. Species richness is the number of species. Species evenness is how equally individuals are shared among them.
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Simpson's Diversity Index: $D = 1-\sum\left(\frac{n}{N}\right)^2$, with $n$ the individuals of one species and $N$ the total of all species. A high $D$ means high diversity.
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The form on the specification, $D = \frac{\sum n(n-1)}{N(N-1)}$, is the opposite way round: a low value means high diversity, and $1-D$ converts it.
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One species gives $D = 0$. $S$ equally abundant species give $D = 1 - \frac{1}{S}$.
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$D$ ignores which species are present, so evaluate it alongside species lists and the limits of the sampling.