Question

In a large isolated population, alleles p and q at a locus are at Hardy Weinberg equilibrium. The frequencies are p = 0.6 and q = 0.4. The proportion of the heterzygous genotype in the population is :

• A. 0.24
• B. 1
• C. 0.48
• D. 0.12

Answer 1

The correct option is C 0.48

Using the Hardy-Weinberg equation this problem can be solved.

$q^2$ represents the frequency of homozygous recessive organisms

As q= 0.4, $q^2$ = 0.16.

p + q = 1

1 - 0.4 = p, hence p = 0.6.

So the frequency of homozygous dominant organisms$p^2$ = 0.36.

The frequency of heterozygous plants is given by (2pq). Hence 2(0.6)(0.4) = 0.48.

Thus, the correct answer is option C.