In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this date, the frequency of allele A in the population is:
0.4
0.5
0.6
0.7
The Hardy-Weinberg formulas -
This formula allows detecting frequencies that change from generation thus allowing a simplified method of determining that evolution is occurring.
- wherein
In a population at equilibrium for a locus of two alleles D and d having frequencies p and q respectively, the genotype frequencies are:
DD=p2,
Dd= 2 pq,
dd= q2
and p+q =1
p2+2 pq +q2 = 1
According to Hardy - Weinberg principle
In a population at equilibrium for a locus of two alleles D and d having frequencies p and q respectively, the genotype frequencies are:
DD=p2,
Dd= 2 pq,
dd= q2
and p+q =1
p2+2 pq +q2 = 1
so,
p2= 36 out of 100 individual
q2= 16 out of 100
q= 0.161/2= 0.4
Option 1)
0.4
This is incorrect option
Option 2)
0.5
This is incorrect option
Option 3)
0.6
This is correct option
Option 4)
0.7
This is incorrect option