### NEET Solved Paper 2014 Question 76

##### Question: In a population of 1000 individuals 360 belongs to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is [AIPMT 2014]

#### Options:

A) 0.4

B) 0.5

C) 0.6

D) 0.7

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

According to Hardy-Weinberg principle $ {{(p+q)}^{2}}=p^{2}+2pq+q^{2}=1 $

where, p = The frequency of allele ‘A’ q = The frequency of allele ‘a’ $ p^{2} $ = The frequency of individual ‘AA’ $ q^{2} $ = The frequency of individual ‘aa’ 2pq = The frequency of individual Aa (AA) $ P^{2} $ 360 out of 1000 individual. or

$ p^{2}=36 $ out of 100. $ q^{2} $ = 160 out of 1000 or $ q^{2} $ = 16 out of 100. So, $ q=\sqrt{0.16}=0.4 $ As p + q = 1 So, p is 0.6