Suppose there is a population ${\displaystyle P}$ , with allele frequencies of A and a given by ${\displaystyle p}$  and ${\displaystyle q}$  respectively (${\displaystyle p+q=1}$ ). Suppose this population is split into two equally-sized subpopulations, ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ , and that all the A alleles are in subpopulation ${\displaystyle P_{1}}$  and all the a alleles are in subpopulation ${\displaystyle P_{2}}$  (this could occur due to drift). Then, there are no heterozygotes, even though the subpopulations are in a Hardy–Weinberg equilibrium.

To make a slight generalization of the above example, let ${\displaystyle p_{1}}$  and ${\displaystyle p_{2}}$  represent the allele frequencies of A in ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ , respectively (and ${\displaystyle q_{1}}$  and ${\displaystyle q_{2}}$  likewise represent a).

Let the allele frequency in each population be different, i.e. ${\displaystyle p_{1}\neq p_{2}}$ .

Suppose each population is in an internal Hardy–Weinberg equilibrium, so that the genotype frequencies AA, Aa and aa are p², 2pq, and q² respectively for each population.

Then the heterozygosity (${\displaystyle H}$ ) in the overall population is given by the mean of the two:

${\displaystyle {\begin{aligned}H&={2p_{1}q_{1}+2p_{2}q_{2} \over 2}\\[5pt]&={p_{1}q_{1}+p_{2}q_{2}}\\[5pt]&={p_{1}(1-p_{1})+p_{2}(1-p_{2})}\end{aligned}}}$ 

which is always smaller than ${\displaystyle 2p(1-p)}$  (${\displaystyle {}=2pq}$ ) unless ${\displaystyle p_{1}=p_{2}}$ 

The Wahlund effect may be generalized to different subpopulations of different sizes. The heterozygosity of the total population is then given by the mean of the heterozygosities of the subpopulations, weighted by the subpopulation size.

The reduction in heterozygosity can be measured using F-statistics.

• Cryptic relatedness