Homework Answers
According to the Hardy-Weinberg equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1.
So p+q =1
p2+ 2pq+q2 =1
here p is the frequency of A allele
q is the frequency of B allele.
A - Frequency of AA genotype = AA population/ total population = 68240 / 100000 = 0.6824
Frequency of AB genotype = AA population/ total population = 28735/ 100000 = 0.28735
Frequency of BB genotype = AA population/ total population = 3025 / 100000 = 0.03025
AA= p2
AB = 2pq
BB = q2
B- Frequency of AA genotype is = 0.6824
Frequency of A allele = 
=
0.82
C - Frequency of BB genotype is = 0.03025
Frequency of B allele - 
=
0.18
D- Freqency of BB genotype is 0.03025
So number of individual which has BB genotype will be 250000 * 0.03025 = 7563 individual.
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