Classification and measuring biodiversity

What you’ll learn

• Explain how molecular evidence can revise classification.
• Distinguish species richness, evenness, genetic diversity and endemism.
• Calculate the specified diversity and heterozygosity indices.

Understand the idea

Classification organises organisms using evidence about their relationships. The biological species concept groups organisms that can interbreed to produce fertile offspring, but it is difficult to apply directly to fossils and asexual organisms. Similar outward features may arise independently, so DNA, RNA and protein comparisons can reveal relationships not obvious from appearance. Molecular evidence helped establish the three domains Bacteria, Archaea and Eukarya. Scientific groups revise classifications through critical evaluation of new data; a new grouping is a conclusion from evidence, not simply a change of name.

Biodiversity includes variation in species, genes and habitats. Species richness counts the number of species in a sample, whereas evenness concerns how evenly individuals are distributed among them. An endemic species occurs naturally only in a particular geographic area, which can make local habitat loss especially important. Comparisons require similar sampling effort and methods; a larger or better-sampled area can appear richer even without a meaningful underlying difference.

The diversity index used here is $D=N(N-1)/\sum n(n-1)$, where $N$ is the total number of individuals and each $n$ is a species count. In this convention, larger values indicate greater diversity. Other sources use different Simpson-index conventions, so always check the supplied formula. At a stated genetic locus, the observed heterozygosity index is the number of heterozygous individuals divided by the total examined. Several representative loci give a broader picture than one locus alone.

Key formulae

$$D=\frac{N(N-1)}{\sum n(n-1)}$$

$$H=\frac{\text{heterozygous individuals}}{\text{total individuals}}$$

Worked example

Two samples each contain 10 individuals from two species. Sample A contains 5 of each species; B contains 9 of one and 1 of the other. Use the stated diversity index to compare them.

1. Both samples have richness 2 and numerator 10 × 9 = 90.
2. For A, the denominator is 5 × 4 + 5 × 4 = 40, so D = 90/40 = 2.25.
3. For B, the denominator is 9 × 8 + 1 × 0 = 72, so D = 90/72 = 1.25. The more even sample has the higher index.

Answer: A: 2.25; B: 1.25. A is more diverse by this index despite equal richness.