<Dl> <Dd> 0 = p t − 1 (p t − 1 + 2 q t − 1 + q t − 1 2 / p t − 1 − 1) = q t − 1 2 / p t − 1 − r t − 1 (\ displaystyle (\ begin (aligned) 0& = p_ (t - 1) (p_ (t - 1) + 2q_ (t - 1) + q_ (t - 1) ^ (2) / p_ (t - 1) - 1) \ \ & = q_ (t - 1) ^ (2) / p_ (t - 1) - r_ (t - 1) \ end (aligned))) </Dd> </Dl> <Dd> 0 = p t − 1 (p t − 1 + 2 q t − 1 + q t − 1 2 / p t − 1 − 1) = q t − 1 2 / p t − 1 − r t − 1 (\ displaystyle (\ begin (aligned) 0& = p_ (t - 1) (p_ (t - 1) + 2q_ (t - 1) + q_ (t - 1) ^ (2) / p_ (t - 1) - 1) \ \ & = q_ (t - 1) ^ (2) / p_ (t - 1) - r_ (t - 1) \ end (aligned))) </Dd> <P> Where the final equality holds because the allele proportions must sum to one . In both cases, q t − 1 2 = p t − 1 r t − 1 (\ displaystyle \ textstyle q_ (t - 1) ^ (2) = p_ (t - 1) r_ (t - 1)). It can be shown that the other two equilibrium conditions imply the same equation . Together, the solutions of the three equilibrium equations imply sufficiency of Hardy's condition for equilibrium . Since the condition always holds for the second generation, all succeeding generations have the same proportions . </P> <P> An example computation of the genotype distribution given by Hardy's original equations is instructive . The phenotype distribution from Table 3 above will be used to compute Hardy's initial genotype distribution . Note that the p and q values used by Hardy are not the same as those used above . </P>

How do you represent each genotype​ in hardy weinberg