# Hardy-Weinberg question

Tags: hardyweinberg
 P: 343 It would seem to me that after monohybrid mating, the frequency of the dominant and recessive allele should change? I will donate original frequency (i.e. proportion) by f0 and that after mating by f1. f1(Dom-Dom) = f0(Dom-Dom)2 + 1/2 * f0(Dom-Dom) * f0(Dom-Rec) + 1/4 * f0(Dom-Rec)2 Am I right so far? (I hope my notation is intelligible - f0(Genotype) is the fraction of the genotype, in the original population.) Then we can replace the terms to find f1(Dom-Dom) = P(Dom)4 + 1/2 * P(Dom)2 * 2*P(Dom)*(1-P(Dom)) + 1/4 * (2*P(Dom)*(1-P(Dom)))2 = P(Dom)4 - P(Dom)3 + P(Dom)2. Where P(Dom) is the frequency of the dominant allele and P(Rec) is the frequency of the recessive allele, P(Dom)+P(Rec)=1, in the original population. Thus it would seem that the frequency of the allele has changed due to the mating?
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P: 1,378
 Quote by Big-Daddy It would seem to me that after monohybrid mating, the frequency of the dominant and recessive allele should change? I will donate original frequency (i.e. proportion) by f0 and that after mating by f1. f1(Dom-Dom) = f0(Dom-Dom)2 + 1/2 * f0(Dom-Dom) * f0(Dom-Rec) + 1/4 * f0(Dom-Rec)2
You forgot a factor of 2 in the fraction of Dom-Dom/Dom-rec matings to account for the combinatorics [i.e. the probability of a homozygous dominant individual mating with a heterozygote is 2*f(Dom-Dom)*f(Dom-rec)].
P: 343
 Quote by Ygggdrasil You forgot a factor of 2 in the fraction of Dom-Dom/Dom-rec matings to account for the combinatorics [i.e. the probability of a homozygous dominant individual mating with a heterozygote is 2*f(Dom-Dom)*f(Dom-rec)].
Ah I see. And any mating between two individuals with different genotypes would also have this multiplying factor of 2 in the calculation of its probability?

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