Question

9% of Hardy Weinberg population of 800 are recessive. How many of this population are heterozygous?

• A. 336
• B. 392
• C. 372
• D. 362

Answer 1

The correct option is A 336

According to the Hardy-Weinberg principle,

p² + 2pq + q² = 1

Here, p and q represent the individual allele frequencies.

Therefore, p2= frequency of homozygous condition represented by p.

q2= frequency of homozygous alleles represented by q.

2pq = frequency of heterozygous condition.

Here, p2 = 0.09, hence p = 0.3

Using p + q = 1, q = 0.7

So, the allelic frequency of heterozygous population,

2pq = 2 × 0.3 × 0.7 = 0.42

So, the heterozygous population = 0.42 × 800 = 336.

So, the correct option is '336'.