Biodiversity

Biodiversity depends on both species richness (number of different species) and species evenness (how equally individuals are distributed among species). Neither measure alone is sufficient to describe biodiversity fully.

Both islands have exactly 5 species and 200 total individuals, so species richness is identical. The difference in biodiversity must therefore come from species evenness.

On the first island, Species A dominates with 150 out of 200 individuals (75%), making the distribution very uneven. On the second island, each species has exactly 40 individuals — perfectly even distribution.

The second island has higher species evenness with identical species richness, making it more biodiverse. This is exactly what Simpson's Diversity Index captures — a habitat dominated by one species scores lower even if total species count is the same.

The question tests whether students understand that both richness and evenness contribute to biodiversity — not richness or individual count alone.

"Both islands have equal biodiversity because they have the same species richness and total number of individuals" is incorrect. Same richness and same total individuals does not mean equal biodiversity — the distribution among species also matters.

"The first island is more biodiverse because it has more individuals of at least one species" is incorrect. Having more individuals of one dominant species actually reduces biodiversity by lowering evenness.

"The second island is more biodiverse because it has higher species richness" is incorrect. Both islands have the same species richness (5 species each), so this reasoning is factually wrong.

The correct answer is that the second island is more biodiverse because of higher species evenness — its individuals are equally distributed, while the first island is dominated by a single species. Identical richness but better evenness = higher biodiversity.

Frogs: $n=50$, Newts: $n=30$, Water snails: $n=20$, Total $N=100$.

${\left(\frac{50}{100}\right)}^2=(0.5{)}^2=0.25$ ${\left(\frac{30}{100}\right)}^2=(0.3{)}^2=0.09$ ${\left(\frac{20}{100}\right)}^2=(0.2{)}^2=0.04$

$\sum {\left(\frac{n}{N}\right)}^2=0.25+0.09+0.04=0.38$

$D=1-0.38=0.62$

The Simpson's Diversity Index for this pond is $D=0.62$. This value closer to 1 indicates reasonably good diversity, though not perfectly even distribution.

The formula requires summing squared proportions: $\sum (n/N{)}^2$. A common error is forgetting to subtract this sum from 1, or squaring incorrectly.

"$D=0.38$" equals $\sum (n/N{)}^2$ without subtracting from 1. This is the intermediate step, not the final answer — a student who forgets the $1-$ part would choose this incorrectly.

"$D=0.67$" would result from a perfectly even distribution of 3 species (each at $1/3$). Since the pond's species are not evenly distributed, this value is too high.

"$D=0.33$" might arise from calculating $1/3$ for 3 species without using the actual data — it ignores the real $n$ values entirely.

The correct calculation gives $D=1-(0.25+0.09+0.04)=1-0.38=0.62$.