Hardy-Weinberg Equilibrium

Hardy-Weinberg Equilibrium

Question:

The frequency of homozygotes in a diploid population is 0.68. Assuming that the population is in Hardy-Weinberg equilibrium, the frequencies of the two alleles are

  1. 1 and 0.9
  2. 2 and 0.8
  3. 4 and 0.6
  4. 5 and 0.5

Rephrasing the Question:

What is the equation for Hardy-Weinberg equilibrium?

Answer:

This is the option where the product of the two allele frequencies is 0.16, which is 0.2 and 0.8 (option 2).

Explanation:

Hardy-Weinberg Equilibrium

Recall that in a population in Hardy-Weinberg equilibrium, the frequencies of two alleles p and q follow the equation:

p2 + 2pq + q2 = 1

Here, p2 represents homozygotes of the p allele, q2 represents the homozygotes of the q allele, and pq represents heterozygotes.

The question tells us that the frequency of homozygotes (so both p homozygotes and q homozygotes combined) is 0.68. We can use this to find the value of pq.

p2 + 2pq + q2 = 1

(p2 + q2) + 2pq = 1

0.68 + 2pq = 1

2pq = 1 – 0.68

2pq = 0.32

pq = 0.16

So we simply choose the option where the product of p and q is 0.16, which is option 2 (0.2 x 0.8 = 0.16).

Sources:

3 thoughts on “Hardy-Weinberg Equilibrium”

Leave a Comment

Your email address will not be published.