A population of 1,000 individuals suffers a severe bottleneck that reduces its size to 50 individuals for two successive generations (no migration, no mutation, random mating). The initial heterozygosity is $H_0=0.60$. Using the drift formula $H_t=H_0{\left(1-\frac{1}{2N_e}\right)}^t$, calculate the expected heterozygosity after the bottleneck and comment on the loss of genetic variation.

Step-by-step solution

1. Identify the effective population size during the bottleneck. For a diploid population $N_e\approx N=50$.
2. Write the per‑generation retention factor for heterozygosity: $\left(1-\frac{1}{2N_e}\right)=1-\frac{1}{2\times 50}=1-0.01=0.99$.
3. The bottleneck lasts $t=2$ generations, so the retention factor over two generations is $0.99^2$.
4. Apply the drift formula: $H_2=0.60\times 0.99^2=0.60\times 0.9801=0.58806$.
5. Round to three decimal places: $H_2\approx 0.588$.
6. Interpretation: Heterozygosity fell from 0.60 to about 0.588, indicating a loss of roughly 2% of genetic variation due to drift during the bottleneck.

Concept tested: Genetic drift and bottleneck effect

Published 23 September 2026 · How this solution was prepared and checked