Question

You are studying a species of mouse that appears to show heterozygote advantage for coat color. In a population that lives in a habitat with brown soil, mice with brown coats (Bb) have higher fitness than mice with black coats (BB) or tan coats (bb). This fitness values for these three phenotypes are: BB = 0.8 Bb = 1.0 bb = 0.6 Given starting genotype frequencies of BB = .04, Bb = 0.32, and bb = 0.64, what is the adjusted frequency of the heterozygotes after round of selection?

Answer 1

The adjusted frequency of heterozygotes after a round of selection is approximately 40.96%.

Step-by-step explanation:

In this scenario, you are studying a species of mouse that exhibits heterozygote advantage for coat color. The fitness values for the three phenotypes, brown coat (Bb), black coat (BB), and tan coat (bb), are as follows: BB = 0.8, Bb = 1.0, and bb = 0.6. To determine the adjusted frequency of the heterozygotes after a round of selection, we need to consider the starting genotype frequencies: BB = 0.04, Bb = 0.32, and bb = 0.64.

Since each individual carries two alleles per gene, we can calculate the frequency of the heterozygote genotype Bb by multiplying the frequency of allele B (0.32) with the frequency of allele b (0.64), and then multiplying it by 2 (because there are two alleles in each genotype). This gives us a frequency of Bb of 0.4096 or 40.96% after one round of selection.