Advanced BiologyEcologyHS-LS2-2

Measuring Diversity

Simpson's diversity index D = 1 − Σ(n/N)² runs from 0 (one species) toward 1 (many even species). It rewards both richness and evenness.

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In Biodiversity and Human Impact you saw why diversity matters. To compare two places, ecologists need a number. Counting species (richness) is a start, but a meadow with 10 species where one makes up 95% of the plants is less diverse than one where all 10 are common. Simpson's diversity index captures both.

Key idea

$$D = 1 - \sum \left(\frac{n}{N}\right)^2$$ $n$ is the number of individuals of each species and $N$ the total. $D$ runs from 0 (only one species) toward 1 (many species, evenly spread). It rewards both richness (more species) and evenness (similar numbers of each).

Why it works

Each fraction $n/N$ is the chance that one randomly picked individual belongs to that species, so $(n/N)^2$ is the chance that two picks both belong to it. Adding over all species gives the chance that two random individuals are the same species. Subtracting from 1 gives the chance they are different: a natural measure of diversity. If one species dominates, two picks are usually the same, and D is low.

Worked examples

Worked example: Two meadows

Meadow A has 25 each of four plant species (100 plants). Meadow B has 85, 5, 5 and 5. Find D for each.
  1. A: each fraction is 0.25; $\sum (0.25)^2 = 4 \times 0.0625 = 0.25$; $D = 1 - 0.25 = 0.75$.
  2. B: fractions 0.85, 0.05, 0.05, 0.05; $0.7225 + 3(0.0025) = 0.73$; $D = 1 - 0.73 = 0.27$.

Answer: A: D = 0.75; B: D = 0.27. Same richness, but A is far more even, so far more diverse.

Same four species, very different evenness.

Worked example: Adding a species

A pond has 30 snails and 30 water beetles (D = 0.5). A third species, 30 water striders, moves in. What is D now?
  1. Each fraction is now 30/90 = 1/3.
  2. $\sum = 3 \times (1/3)^2 = 1/3$.
  3. $D = 1 - 1/3 \approx 0.67$.

Answer: About 0.67: more richness raised D.

In the field, ecologists count individuals in quadrats or along transects, then compute D for each site. Comparing D before and after a disturbance, or between a protected and an unprotected area, turns "it looks less diverse" into a number that can be tracked. Two cautions: D depends on how thoroughly a site was sampled, and it says nothing about which species are present, so a site full of invasive species can still score high.

Common mistake

Two slips: adding the fractions without squaring them (the plain fractions always add to 1, so D would always be 0), and forgetting to subtract from 1, which turns a low-diversity place into a high number. Check: one species only must give D = 0.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • College Board, AP Biology Course and Exam Description (science practices and the AP Biology Equations and Formulas sheet)
  • OpenStax, Biology 2e, 47.1 The Biodiversity Crisis
  • Next Generation Science Standards, HS-LS2-2: Ecosystems: Interactions, Energy, and Dynamics
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