The frequency of the Hemoglobin A allele is 0.9. The heterozygote with the reces

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Explanation / Answer

So we are provided with P value = 0.9 that is frequency of dominant allele A.

Now the q value or recessive allele(s) frequency= 1- P

= 0.1

We know from Hardy Weinberg law that-

P2AA + 2pq (As)+ q2ss = 1

Here 2pq = frequency of heterozygote

Therefore 2pq( As) = 2× .9 × .1

= .18

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