Question

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

a. The frequencies of the genotypes "AA" and "Aa"

b. The frequencies of the two possible phenotypes if "A" is completely dominant over "a"

Answer 1

hardyweinberg formula is p2 + 2pq +q2=1 : where p= frequency of A ; q= frequency of a 2pq = frequency of Aa

given that homozygous recessive (aa) genotype=36% i.e. 36/100= = 0.36 =q2.

thus, q= (0.36) square root =0.6

thus, p= 1-0.6= 0.4

a) thus, frequency of AA i.e. p2= (0.4)2= 0.16=16%

frequency of Aa= 2pq= 2 X 0.4 X 0.6 =0.48= 48%

b) two possible phenotypefrequency if A is dominant over a completely= frequency of AA + frequency of Aa= 16+48= 64%

Answer 2

According to Hardy-Weinberg law:-

p² + q²+2pq= 1 and p + q =1

p = Frequency of dominant allele in a population (A)
q = Frequency of recessive allele in a population (a)
2pq = percentage of heterozygous individuals (Aa)

a. Given in the question the frequency of aa= 36% so q²= 0.36 and q= 0.6

From the formula: p+q=1; p= 0.4

So, p²=0 .16 and frequency of genotype AA is 16%.

Similarly, 2pq or 2 x .4 x .6 = 0.48 and frequency of genotype Aa is 48%.

b. If "A" is completeley dominant over "a" both the genotypes "AA" and "Aa" will show dominant phenontypes and the frequency of dominant phenotype will be the sum of frequencies of "AA" genotype and "Aa' genotypes

i.e; 16% + 48% = 64%

And since the recessive phenotype will occur in only "aa" genotype the frequency of recessive phenotype will be 36%