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4.4.5Unit 4: Energy, Environment, Microbiology and Immunity

Index of Biodiversity — IAL Edexcel Linear Revision Notes & Topical Questions

5 min read 5 parts Lesson 25 of 28 Unit quiz

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Part 1 of 5The Index

4.4.5 Index of Biodiversity

An index of diversity links the number of species to the number of individuals in each, so it uses richness and abundance together.

Formula
  • D = N(N − 1) divided by the sum of n(n − 1) for every species in the community.
N
  • The total number of organisms of all species in the community.
n
  • The number of individuals of a single species.
Interpretation
  • A larger value of D means a higher level of diversity.
Use
  • Calculating D for different habitats lets their diversity be compared, or one habitat compared over time.

Part 2 of 5Calculation Steps

Work in the same order each time and show each value.

1
Count all the organisms of all species to give N, then work out N(N − 1) as value A.
2
For each species calculate n(n − 1), then add the results together to give value B.
3
Divide A by B to obtain the index of diversity D and give it to the required decimal places.
4
Compare D for different habitats: the higher value shows the more diverse habitat.

Part 3 of 5Why Use an Index

Why Use an Index
Evenness
  • If one species dominates, the denominator becomes large and D falls, correctly showing lower diversity.
Rare species
  • Rare species contribute little to the denominator, so a community with many evenly spread species gets a high D.
Minimum value
  • If all individuals are of one species, A and B are equal and D is 1, the lowest possible value.
Sampling
  • Results depend on sampling method and effort, so the same method should be used for every habitat compared.

Part 4 of 5Worked Example

Question

  • A sample of insects from a hedge contains 12, 8, 5, 3 and 2 individuals of five species.
  • Calculate the index of diversity.

  1. N = 12 + 8 + 5 + 3 + 2 = 30, so A = N(N − 1) = 30 × 29 = 870.
  2. n(n − 1): 12×11 = 132; 8×7 = 56; 5×4 = 20; 3×2 = 6; 2×1 = 2. Sum B = 216.
  3. D = 870 ÷ 216 = 4.03 (to 2 decimal places).

Answer D = 4.03

Part 5 of 5Exam Checklist & Tips

4.4.5 – Key Points for the Exam

  • Index of diversity A value that combines the number of species and their relative abundance.
  • Abundance The number of individuals of a species in an area.
  • Evenness How equally individuals are spread among the species present.
  • Sigma (Σ) The symbol meaning the sum of all the values that follow.

Common Mistakes

  • Using the number of species in place of the total number of organisms as N.
  • Forgetting to calculate n(n − 1) for every species before adding them together.
  • Dividing in the wrong order, giving Σn(n − 1) ÷ N(N − 1) instead of the other way round.
  • Concluding a habitat is diverse from a single high D value without comparing to another habitat.

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