Measures of Biodiversity

This lesson covers: 

  1. Species richness and species evenness
  2. The Simpson’s diversity index

Species richness and species evenness

Species richness and species evenness are measures of biodiversity.


Species richness:

  • The total number of different species in a habitat.
  • It is quantified by taking random samples and counting the species present.
  • A higher species richness indicates greater diversity.


Species evenness:

  • A comparison of the numbers of individuals of each species in a community.
  • It is measured by taking samples and counting individuals of each species.
  • More even abundances mean higher species evenness and diversity.

The Simpson’s diversity index

The Simpson's index of diversity calculates biodiversity using the number of species and their relative abundances. That is, it takes both species richness and species evenness into account.


It is calculated as:

D=1−(∑(Nn​)2)


Where:

  • n = number of individuals of each species
  • N = total number of all individuals


D ranges from 0 to 1, with 1 representing maximum diversity.

Worked example - Calculating the Simpson’s diversity index

A biologist is assessing the biodiversity of two fields. field A contains 150 individual plants, consisting of 85 daisies, 35 buttercups, and 30 clovers. field B contains 150 individual plants, consisting of 55 daisies, 45 buttercups, and 50 clovers.


Calculate and compare the diversity indices for each habitat.


Step 1: Equation

D=1−(∑(Nn​)2)


Step 2: Construct a table to calculate ∑(Nn​)2 for field A

SpeciesNumber of individuals (n)(nN)2\left(\frac{n}{N}\right)^2
Daisies85(85150)2=0.3211...\left(\frac{85}{150}\right)^2 = 0.3211\text{...}
Buttercups35(35150)2=0.0544...\left(\frac{35}{150}\right)^2 = 0.0544\text{...}
Clovers30(30150)2=0.04\left(\frac{30}{150}\right)^2 = 0.04
SumN=∑n=150N = \sum{n} = 150∑(nN)2=0.4156...\sum\left(\frac{n}{N}\right)^2 = 0.4156\text{...}


Step 3: Substitution and correct evaluation for field A

diversity of field A =1−(∑(Nn​)2)

diversity of field A =1−0.4156...

diversity of field A =0.584 (to 3 s.f.)  


Step 4: Construct a table to calculate ∑(Nn​)2 for field B

SpeciesNumber of individuals (n)(nN)2\left(\frac{n}{N}\right)^2
Daisies55(55150)2=0.1344...\left(\frac{55}{150}\right)^2 = 0.1344\text{...}
Buttercups45(45150)2=0.09\left(\frac{45}{150}\right)^2 = 0.09
Clovers50(50150)2=0.1111...\left(\frac{50}{150}\right)^2 = 0.1111\text{...}
SumN=∑n=150N = \sum{n} = 150∑(nN)2=0.3356...\sum\left(\frac{n}{N}\right)^2 = 0.3356\text{...}


Step 5: Substitution and correct evaluation for field B

diversity of field B =1−(∑(Nn​)2)

diversity of field B =1−0.3356...

diversity of field B =0.664 (to 3 s.f.)


Step 6: Interpretation of result

the Simpson’s diversity index for field A is 0.584 and for field B it is 0.664, which indicates that field B has a higher species diversity compared to field A