Question

Assume that there are only two alleles at a given locus in a population, and the population is in H-W equilibrium. If the frequency of one allele is 0.6, what is the frequency of heterozygotes in the populationA. -0.6.
B. 0
C. 0.4
D. 0.6
E. 1

Answer 1

Frequency of heterozygotes in the population is 0.4.

C. 0.4

Explanation:

If a population is in Hardy-Weinberg equilibrium, the sum of the frequencies of the two alleles will be equal to 1.  The letter p denotes the frequency of the first allele and the letter q denotes the frequency of the second allele.  

p+q=1

In the problem it is given that p=0.6

p+q=1

0.6+q =1  

q= 1-0.6

=0.4

We have to calculate the frequency of heterozygotes in the population. it is given by the equation

Number of heterozygous individuals =2pq

=$2 \times 0.6 \times 0.4$

=0.48~0.4