Answer:
Allele frequency of A1 = 0.49 (If A1 is dominant)
Allele frequency of A1 = 0.67 (If A1 is recessive)
Explanation:
Considering A1 is the dominant allele and A2 is the recessive allele
We will use the Hardy-Weinberg equations to solve this problem:
p² + 2pq + q² = 1
p + q = 1
where p² = frequency of individuals with homozygous dominant (A1A1) genotype
q² = frequency of individuals with homozygous recessive (A2A2) genotype
2pq = frequency of individuals with heterozygous (A1A2) genotype
p = frequency of dominant allele (A1) and
q = frequency of recessive allele (A2)
We are given that:
Individuals with A2A2 genotype = 108
Total number of individuals = 188 + 123 + 108 = 419 so,
frequency of Individuals with A2A2 genotype = 108/419 =0.2577
q² = 0.2577
q = √0.2577
q = 0.51
p+q = 1
so, p = 1 - q
p = 1 - 0.51
p = 0.49
Allele frequency of A1 = 0.49
In case A1 is the recessive allele,
q² = 188/419 = 0.4487
q = √0.4487
q = 0.67
Allele frequency of A1 = 0.67
