Question

You are studying a population of Eastern Blue Jays. In this population, you measure the frequencies of the two alleles at the blue&white gene using a fancy DNA sequencing protocol termed PoolSeq. Your fancy protocol tells you that the frequency of allele 1 at the blue&white gene is 0.40, while the frequency of the other allele at this locus is 0.60. What is the expected frequency of heterozygotes in this population if it is in Hardy-Weinberg Equilibrium?

A. 0.16
B. 0.24
C. 0.32
D. 0.48

Answer 1

The expected frequency of heterozygotes in the Blue Jay population in Hardy-Weinberg Equilibrium, with allele frequencies of 0.40 and 0.60, is 0.48 (option D), calculated using the formula 2pq.

Step-by-step explanation:

We are asked to find the expected frequency of heterozygotes for Eastern Blue Jays at the blue&white gene locus in a population in Hardy-Weinberg Equilibrium. Given the allele frequencies of allele 1 (p) as 0.40, and allele 2 (q) as 0.60, we can use the Hardy-Weinberg principle to calculate this.

According to the Hardy-Weinberg principle:

• The frequency of the homozygous dominant genotype (AA) is p².
• The frequency of the heterozygous genotype (Aa) is 2pq.
• The frequency of the homozygous recessive genotype (aa) is q².

Therefore, using the formula 2pq to calculate the frequency of heterozygotes, we get:

2 × 0.40 × 0.60 = 0.48

Thus, the expected frequency of heterozygotes in the population, if it is in Hardy-Weinberg Equilibrium, is 0.48 (option D).