Two subpopulations of a plant species, each with 4000 individuals, have allele L frequencies of 0.30 and 0.70 respectively. Calculate the fixation index $F_{ST}$ between them using $F_{ST}=\frac{H_T-H_S}{H_T}$, where $H_T=2\bar{p}\bar{q}$ and $H_S=\frac{(2p_1{q}_1+2p_2{q}_2)}{2}$.

Step-by-step solution

1. For subpopulation 1: p1 = 0.30, q1 = 0.70, H1 = 2×0.30×0.70 = 0.42.
2. For subpopulation 2: p2 = 0.70, q2 = 0.30, H2 = 2×0.70×0.30 = 0.42.
3. Compute average heterozygosity: H_S = (H1+H2)/2 = (0.42+0.42)/2 = 0.42.
4. Calculate mean allele frequency: ρ̄ = (p1+p2)/2 = (0.30+0.70)/2 = 0.50.
5. Compute total heterozygosity: H_T = 2×0.50×0.50 = 0.50.
6. Finally, F_ST = (H_T – H_S)/H_T = (0.50–0.42)/0.50 = 0.08/0.50 = 0.16.

Concept tested: Population genetics fixation index

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