Question

In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ‘$A$’ and ‘$a$’, respectively and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of '$A$' and '$a$’ in a population with $N_{AA} =90, N_{Aa}= 40$ and $N_{aa} = 70$?

• A. A = 0.55 and a = 0.45
• B. A = 0.40 and a = 0.60
• C. A = 0.35 and a = 0.65
• D. A = 0.25 and a = 0.75

Answer 1

Answer: (A)

Allele A frequency
$= \frac{2N_{AA} + A_{a}}{2N} $
Total number of individuals in population $= N$
$ N = 90 + 40 + 70 = 200 $
$ N = 200$
$ A = \frac{ 2\times90+40}{2\times200}$
$ = \frac{180+40}{400} $
$= \frac{220}{400} = 0. 55 $
Allel 'a' frequency $= \frac{ 2N_{aa} + A_{a}}{2N} $
$ a = \frac{2\times70+40}{2\times200}$
$= \frac{ 140+40}{400} $
$ = \frac{180}{400} = 0.45$
So,$ A = 0.55$
$a = 0.45$