Question

In an electrophoretic study of enzyme variation in a species of pelican, you find 77 A1A1, 45 A1A2, and 18 A2A2 individuals at a particular locus in a sample of 140. What are the allele frequencies for the A1 and A2 alleles? Calculate the genotype frequencies for this locus.

Answer 1

Frequencies of allele
`A_(1)` and
`A_(2)\\` are
`0.74` and
`0.26` respectively.

Frequencies of Individuals with genotype
`A_(1)A_(1), [tex]A_(2)A_(2) and [tex]A_(1)A_(2) are [tex]0.55, 0.128, 0.32` respectively.

Step-by-step explanation:

As per Hardy Weinberg's equation -

`p^(2) +q^(2) +2pq= 1` ----------Equation (A)

`p+q= 1`-----------Equation (B)

Where "p" represents the frequency of "
`A_(1)`

"q" represents the frequency of "
`A_(2)`

`p^(2)` represents frequency of individual
`A_(1)A_(1)`

`q^(2)` represents frequency of individual
`A_(2)A_(2)`

`pq` represents frequency of individual
`A_(1)A_(2)`

Here genotype frequencies are -

`A_(1)A_(1) = 77\\p^(2) = (77)/(140) \\= 0.55\\`

`A_(2)A_(2) = 18\\q^(2) = (18)/(140) \\= 0.128\\`

Substituting this values in equation A, we get

`0.55 + 0.128 + 2pq = 1\\2pq = 1-(0.128+ 0.55)pq = 0.321`

Frequencies of allele
`A_(1)` and
`A_(2)\\` are -

For
`A_(1)`
`= \sqrt{p^(2) } \\= √(0.55) \\= 0.74`

Substituting this value in equation B, we get

`p+q=1\\0.74 + q = 1\\q = 1-0.74\\q = 0.26`

Frequencies of allele
`A_(1)` and
`A_(2)\\` are
`0.74` and
`0.26` respectively.

Frequencies of Individuals with genotype
`A_(1)A_(1), A_(2)A_(2), A_(1)A_(2)`are
`0.55, 0.128, 0.32` respectively.