Quantifying biodiversity requires metrics that capture different aspects of community structure. VCE Environmental Science requires proficiency with three key measures: species richness, endemism and Simpson’s Index of Diversity (SID).
Definition: The total number of species present in a defined area or community.
When to use: Comparing diversity between regions; tracking colonisation after disturbance; rapid assessments.
Limitations:
- Sensitive to sampling effort — more sampling almost always finds more species
- Cannot distinguish between communities with very different structures
- Treats a community with 100 common species equally to one with 100 species where 99 are extremely rare
Definition: A species is endemic to a region if it naturally occurs there and nowhere else in the world.
Significance:
- Endemic species have restricted ranges → local extinction = global extinction
- Regions with high endemism are conservation priorities
- Endemism reflects evolutionary history: isolation, stable refugia and adaptive radiation
Measuring endemism:
- Number of endemic species in a region
- Percentage endemism = (endemic species ÷ total species) × 100%
- Weighted endemism accounts for geographic range size
Australian endemism examples:
- ~80% of Australia’s flowering plants are endemic
- ~80% of freshwater fish are endemic
- All monotremes (platypus, echidnas) are endemic to the Australia-New Guinea region
SID is a combined measure of species richness and evenness, providing a more complete picture of community diversity.
$$SID = 1 - \frac{\sum n_i(n_i - 1)}{N(N-1)}$$
Where:
- $n_i$ = number of individuals of species $i$
- $N$ = total number of individuals
- The summation ($\sum$) is over all species present
Site A has three species:
- Eucalyptus obliqua: 15 individuals
- Acacia melanoxylon: 8 individuals
- Pomaderris aspera: 2 individuals
Step 1: Calculate $n_i(n_i - 1)$ for each species:
- \$15 \times 14 = 210$
- \$8 \times 7 = 56$
- \$2 \times 1 = 2$
Step 2: Sum: $\sum n_i(n_i-1) = 210 + 56 + 2 = 268$
Step 3: $N = 15 + 8 + 2 = 25$; $N(N-1) = 25 \times 24 = 600$
Step 4: $SID = 1 - \frac{268}{600} = 1 - 0.447 = 0.553$
Interpretation: Moderate diversity (0.55); dominated by E. obliqua.
| Community | Species | Counts | SID | Interpretation |
|---|---|---|---|---|
| Forest A | 3 | 15, 8, 2 | 0.55 | Moderate, one dominant |
| Forest B | 3 | 8, 8, 9 | 0.67 | Higher, more even |
| Monoculture | 1 | 25 | 0.00 | No diversity |
Forest B has the same number of species as Forest A but greater evenness, resulting in a higher SID.
| Measure | Captures Richness? | Captures Evenness? | Captures Endemism? |
|---|---|---|---|
| Species richness | Yes | No | No |
| Endemism | Partially | No | Yes |
| SID | Yes | Yes | No |
EXAM TIP: VCAA frequently provides species count data in a table and asks you to calculate SID, compare values between sites, and draw conclusions. Practice the formula multiple times. A common error is forgetting to subtract 1 (i.e. calculating $\sum n_i^2$ instead of $\sum n_i(n_i-1)$).