At a particular locus frequency of allele A is 0.6 and that of a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

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Arun · Answer

Dear Animesh Accordingly to Hardy-Wienberg principle, P² + 2PQ + Q² = 1 , where P and Q are the individual frequency of dominant amd recessive allele respectively at a particular locus and the term 2PQ represents the frequency of heterozygotes in a randomly mating population. Hence, the answer should be (2*0.6*0.4) = 0.48 . Regards Arun

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At a particular locus frequency of allele A is 0.6 and that of a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

At a particular locus frequency of allele A is 0.6 and that of a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

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At a particular locus frequency of allele A is 0.6 and that of a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

At a particular locus frequency of allele A is 0.6 and that of a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

1 Answers

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