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 and 0.9
- 2 and 0.8
- 4 and 0.6
- 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:
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).
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