After a population of 10,000 individuals splits into two equal subpopulations of 5,000 each, the allele frequency of allele A is 0.1 in one subpopulation and 0.9 in the other after 1,000 generations. Calculate the fixation index F_ST between the two subpopulations.

Step-by-step solution

1. Compute mean allele frequency: $\bar{p}=(0.1+0.9)/2=0.5$.
2. Compute mean frequency of allele q: $\bar{q}=1-\bar{p}=0.5$.
3. Compute numerator: $(p_1-p_2{)}^2=(0.1-0.9{)}^2=(-0.8{)}^2=0.64$.
4. Compute denominator: $4\bar{p}\bar{q}=4\ast 0.5\ast 0.5=1$.
5. Divide numerator by denominator: $F_{ST}=0.64/1=0.64$.

Concept tested: Fixation index F_ST

Published 17 September 2026 · How this solution was prepared and checked