A population is made up of individuals where 188 have the A1A1 genotype, 123 have the A1A2 genotype, and 108 have the A2A2 genotype. What is the allele frequency of A1? Answer to 2 decimal places.

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

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