<P> For the more general case of dioecious diploids (organisms are either male or female) that reproduce by random mating of individuals, it is necessary to calculate the genotype frequencies from the nine possible matings between each parental genotype (AA, Aa, and aa) in either sex, weighted by the expected genotype contributions of each such mating . Equivalently, one considers the six unique diploid - diploid combinations: </P> <Dl> <Dd> ((AA, AA), (AA, Aa), (AA, aa), (Aa, Aa), (Aa, aa), (aa, aa)) (\ displaystyle \ left (((\ text (AA)), (\ text (AA))), ((\ text (AA)), (\ text (Aa))), ((\ text (AA)), (\ text (aa))), ((\ text (Aa)), (\ text (Aa))), ((\ text (Aa)), (\ text (aa))), ((\ text (aa)), (\ text (aa))) \ right)) </Dd> </Dl> <Dd> ((AA, AA), (AA, Aa), (AA, aa), (Aa, Aa), (Aa, aa), (aa, aa)) (\ displaystyle \ left (((\ text (AA)), (\ text (AA))), ((\ text (AA)), (\ text (Aa))), ((\ text (AA)), (\ text (aa))), ((\ text (Aa)), (\ text (Aa))), ((\ text (Aa)), (\ text (aa))), ((\ text (aa)), (\ text (aa))) \ right)) </Dd> <P> and constructs a Punnett square for each, so as to calculate its contribution to the next generation's genotypes . These contributions are weighted according to the probability of each diploid - diploid combination, which follows a multinomial distribution with k = 3 . For example, the probability of the mating combination (AA, aa) is 2 f (AA) f (aa) and it can only result in the Aa genotype: (0, 1, 0). Overall, the resulting genotype frequencies are calculated as: </P>

How is a population in hardy weinberg equilibrium