Question

In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, calculate the frequency of allele A in the population.

Answer 1

According to Hardy Weinberg equation

p^2+2pq+q^2 = 1

Also, p+q=1

where,
p- allelic frequency of A
q= allelic frequency of a

p^2 - frequency of Homozygous Dominant genotype(AA)

2pq- frequency of Heterozygous genotype (Aa)

q^2- frequency of Homozygous recessive genotype (aa)

so, p^2 or genotypic frequency of AA genotype is 360/1000*100 = 36%

Value of p (allelic frequency of A)is square root of p^2 =√36 = 6

Hence,allelic frequency of A (denoted by p)-6

Similarly we can calculate the allelic frequency of q which comes out to be 4

Answer 2

Answer:

O.6

Explanation:

Accroding to  hardy weinberg princ.

(P^2 + q^2 ) = 1 that is

P^2 + 2pq+ q^2 = 1 here

P^2 is  frequency of individual AA

q^2 is frequency of individual aa

2pq is the frequency

of individual  Aa

Also p+q =1   here

P= frequency of dominant allele(A)

q= frequency of recessive allele(a)

Now a/c to ques.

P^2 =  360 / 1000  that is  0.36

So ,

P = root sq. Of 0.36

P= 0.6 (A)