= One-generation random-mating equilibrium
{title2=$p=u+v/2,\quad(p^2,2p(1-p),(1-p)^2)$}
If parental <genotype> frequencies are $(u,v,w)$ and the two parental gametes are independently sampled from the same <allele> pool, <Mendelian segregation> gives allele frequency $p=u+v/2$. Their independent union has the displayed <Hardy-Weinberg proportions>, irrespective of parental genotype proportions. Equal allele frequencies in the two sexes and no transmission or viability distortion are part of this conclusion.
Back to article page
