I have solved the following question, I just don't know if I'm correct or not:
Q. A population in which the percentage of the homozygous recessive genotype (aa) is 36%. Then what is the frequency of the aa allele? Also, what are the frequencies of the AA and Aa genotypes.
i got .36+2*0.6*q+q^2=1
i used the quadratic formula after simplifying to get q=0.4, so q^2=16% thus 16% are AA and 48% are Aa. And allele frequency of aa is = .36+ 1/2(.48)= 60% ... make sense? did I go wrong somewhere?
Q. A population in which the percentage of the homozygous recessive genotype (aa) is 36%. Then what is the frequency of the aa allele? Also, what are the frequencies of the AA and Aa genotypes.
i got .36+2*0.6*q+q^2=1
i used the quadratic formula after simplifying to get q=0.4, so q^2=16% thus 16% are AA and 48% are Aa. And allele frequency of aa is = .36+ 1/2(.48)= 60% ... make sense? did I go wrong somewhere?
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aa= .36
therefore the frequency of a=.6 (square root of .36)
A= 1-0.6 = 0.04
aa= .6 x .6 = .36
AA = 0.4 x 0.4 =.16
Aa = 0.6 x 0.4= 0.24
You double check with 1= aa + 2Aa + AA
So 0.36 + 2(0.24) + 0.16 equal 1? Yes indeedy!
By using the two H-W equations, p + q = 1 and p^2 + 2pq + q^2 = 1 you can solve and check your ansers.
therefore the frequency of a=.6 (square root of .36)
A= 1-0.6 = 0.04
aa= .6 x .6 = .36
AA = 0.4 x 0.4 =.16
Aa = 0.6 x 0.4= 0.24
You double check with 1= aa + 2Aa + AA
So 0.36 + 2(0.24) + 0.16 equal 1? Yes indeedy!
By using the two H-W equations, p + q = 1 and p^2 + 2pq + q^2 = 1 you can solve and check your ansers.