Why we need an index: Saying a habitat is “biodiverse” is too vague for science.
Two communities can have the same number of species yet feel very different: one where every species is common, and one where a single species dominates and the rest are rare.
A diversity index combines two things into one number:
Richness — how MANY species there are.
Evenness — how EQUALLY the individuals are spread between those species.
We also often need to estimate how many individuals of one species there are when we cannot count them all. Both jobs are done with a simple formula and some counting.
- Biodiversity
- The variety of living organisms in an area — it depends on both the number of species (richness) and how evenly individuals are spread among them (evenness).
- Species richness
- Simply the NUMBER of different species present in a community.
- Evenness
- How equally the individuals are distributed across the species. High evenness = no single species dominates.
- Simpson's reciprocal diversity index ()
- A single number combining richness and evenness. A LARGER D means GREATER diversity. The smallest possible value is 1 (only one species present).
- Lincoln index (capture–mark–recapture)
- A method to ESTIMATE the size of a mobile animal population that is too large to count directly, by marking some individuals and seeing what fraction turn up in a second sample.
“Reciprocal” — bigger means more diverse: We use the reciprocal form, , on purpose.
With this version a higher number = more diverse — which is the intuitive direction.
The lowest value can take is 1 (a community with only one species). There are no units.
Simpson's reciprocal diversity index: $$D = \dfrac{N(N-1)}{\sum n(n-1)}$$
where $N$ = the total number of individuals (all species added together), and $n$ = the number of individuals of each species.
Method: for every species work out , add those up to get , then divide by that total.
Pond A was sampled with a net and the catch sorted into four species. Build the column first, then add it up:
| Species | Number found, | ||
|---|---|---|---|
| Mayfly nymph | 8 | 7 | 56 |
| Freshwater shrimp | 6 | 5 | 30 |
| Pond snail | 4 | 3 | 12 |
| Water boatman | 2 | 1 | 2 |
| TOTALS | — |
IB-style question — calculate Simpson's reciprocal index for Pond A
Using the data for Pond A, calculate Simpson's reciprocal diversity index . Give your answer to 2 decimal places. [2]
Worked solution
- Write the formula. .
- Find $N$. Total individuals .
- Top line. .
- Bottom line. (the last column of the table).
- Divide. (no units). Mark 1: correct substitution. Mark 2: .
Final answer
(no units).
What the number means — compare two ponds: Pond B has the same richness (4 species) and the same $N = 20$, but one species dominates:
| Species (Pond B) | Number found, | |
|---|---|---|
| Pond snail | 17 | 272 |
| Mayfly nymph | 1 | 0 |
| Freshwater shrimp | 1 | 0 |
| Water boatman | 1 | 0 |
| TOTALS |
Same species count, very different diversity: For Pond B: .
Pond A scored 3.80; Pond B scored only 1.40 — even though both have 4 species and 20 animals.
The difference is evenness: Pond A's individuals are spread fairly evenly, while Pond B is dominated by snails. Higher $D$ = greater diversity, which is usually linked to a more stable ecosystem.
The Lincoln index (capture–mark–recapture): $$N = \dfrac{n1 \times n2}{n3}$$
$n_1$ = number caught, marked and released first; $n_2$ = number caught in the second sample; $n_3$ = number in that second sample that were already marked; $N$ = the estimated total population.
The idea: the fraction of the second catch that is marked equals the fraction of the whole population that was marked — so if few of the marked ones turn up again, the population must be large.
| Symbol | What it means | Our woodlouse study |
|---|---|---|
| Number caught, MARKED and released on day 1 | 60 | |
| Number caught in the second sample (day 2) | 50 | |
| Number in that second sample that were already MARKED | 15 | |
| ESTIMATED total population size |
IB-style question — estimate a woodlouse population with the Lincoln index
On day 1 a student catches 60 woodlice, marks them and releases them. On day 2 she catches 50 woodlice, of which 15 are marked. Estimate the total woodlouse population. [2]
Worked solution
- Write the formula and label the numbers. , with , , .
- Substitute. .
- Evaluate. woodlice. Mark 1: correct substitution. Mark 2: .
Final answer
woodlice (estimated).
Assumptions the Lincoln estimate relies on
- The marks do not harm the animals or make them easier for predators to catch.
- Marked animals mix back fully into the population before the second sample.
- There are no births, deaths, immigration or emigration between the two samples.
- Marks are not lost between the two catches.
- If any assumption fails, the estimate is less reliable — a common exam evaluation point.
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How this is tested: On Paper 1B (and in the IA) you may be given raw counts and asked to calculate Simpson's reciprocal or use the Lincoln index, then interpret the result.
The formulas are provided in the question — you are marked on the method (building the column, substituting correctly) and on a sensible interpretation, not on memorising the equation. Always show your working and keep units/decimal places as asked.
IB-style question — compare diversity before and after restoration
Before: 3 species with counts 18, 1, 1 (). After: 4 species with counts 7, 6, 4, 3 (). Using Simpson's reciprocal index , calculate for each year and state whether biodiversity increased. [4]
Worked solution
- Before — bottom line. .
- Before — $D$. .
- After — bottom line. .
- After — $D$. .
- Interpret. rose from 1.24 to 4.22, so biodiversity increased — the community is now both richer (4 vs 3 species) and far more even (no single dominant species). Marks: 1 for each correct , 1 for the comparison, 1 for the valid conclusion.
Final answer
Before ; after . D increased, so biodiversity increased — the meadow became richer and more even.
✓ Why this scores full marks: It shows the substitution for both years, gets both values, and ties the rise in to richness AND evenness before concluding.
A common slip is to compare only the species count (3 → 4) — but also captures evenness, which is why it jumps so much more than the species count alone does.
| Simpson's reciprocal | Lincoln index | |
|---|---|---|
| Question it answers | How DIVERSE is this community? | How MANY individuals are there? |
| Formula | ||
| You count | Individuals of each species () and the total () | Marked (), 2nd catch (), recaptured marked () |
| Bigger number means | GREATER diversity (min = 1) | A LARGER estimated population |