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

Answer:

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.

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  -

$A_{1}A_{1} = 77\\p^{2} = \frac{77}{140} \\= 0.55$

$A_{2}A_{2} = 18\\q^{2} = \frac{18}{140} \\= 0.128$

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} } \\= \sqrt{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.