Imagine a gene that has 2 alleles - the dominant wild type A allele and the rece

Question

Imagine a gene that has 2 alleles - the dominant wild type A allele and the recessive disease causing mutation, called a allele. Our population is in Hardy weinberg equilibrium. There are 8 times as many heterozygous cariiers of mutation in population as there are people with affected with the disease.

1. What are the frequencies of A and a Allels in this population.

2. The population undergoes a long period od inbreeding between first cousins (F=0.0625), after population reached equilibrium, what will the frequencies homozygotes and heterozygotes

I am stuck here. Can some one help validate this for me?

1) I tried p^2+2pq+q^2 as Hardy weinberg equilibrium

I tried

2pq = 8(q^2)

p=4q

1-q = 4q

or q = 0.2, and p = 0.8

Freq of A allele is = p^2 = 0.8^2 = 0.64

Freq of a allele = q^2 = 0.2^2 = 0.04

2) F = 0.0625

   F(AA) = p^2 + F*p*q = 0.64+(0.0625)(0.8)(0.2) = 0.65

F(aa) = q^2 + F*p*q = 0.04 + (0.0625)(0.8) (0.2) = 0.05

F(Aa) = 2pq - 2*F*p*q = 2(0.8)(0.2) - 2 (0.0625)(0.8)(0.2) = 0.30

Explanation / Answer

u have done the calculation.

the frequencies are AA- 0.65,aa- 0.05 and Aa-0.30

these are in equilibrium as their summation is 1

0.65+0.05+0.30=1

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