Question

A particular gene has two alleles, a dominant allele A and a recessive allele a. The frequency of allele A is 0.55 . If the population is in Hardy-Weinberg equilibrium with respect to the gene, then what is the expected frequency of genotype Aa?

Answer 1

The expected frequency of the genotype Aa in a population that is in Hardy-Weinberg equilibrium, with a dominant allele A frequency of 0.55, is calculated as 2(0.55)(0.45), resulting in a genotype frequency of 49.5%.

Step-by-step explanation:

In a population that is in Hardy-Weinberg equilibrium, the expected frequency of the genotype Aa can be found using the formula 2pq, where p is the frequency of the dominant allele (A), and q is the frequency of the recessive allele (a). Given that the frequency of allele A is 0.55, we first need to calculate the frequency of allele a, which is q = 1 - p = 1 - 0.55 = 0.45. Consequently, the expected frequency of genotype Aa would be 2(0.55)(0.45), which equals 0.495, or 49.5%.

Answer 2

0.495

Step-by-step explanation:

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