Question

2. In a population of king snakes the banded pattern (B) is dominant to no banding

(b). If 12% of the population shows no banding, what percentage of the population,
to the nearest tenth, is heterozygous for banding?

Answer 1

If 12% of the population shows no banding, this means that the frequency of the recessive allele (b) is 0.12^(1/2) or 0.3464 (assuming the population is in Hardy-Weinberg equilibrium). We take the square root because the recessive phenotype occurs only in individuals who are homozygous recessive (bb), and the frequency of homozygous recessive individuals is equal to the square of the frequency of the recessive allele.

To find the frequency of the dominant allele (B), we subtract the frequency of the recessive allele from 1:

Frequency of B = 1 - Frequency of b = 1 - 0.3464 = 0.6536

Since the banded pattern (B) is dominant to no banding (b), we can use the Hardy-Weinberg equation to calculate the frequency of heterozygous individuals:

2pq = 2 x 0.6536 x 0.3464 = 0.4529

Therefore, approximately 45.3% (to the nearest tenth) of the population is heterozygous for banding.

Step-by-step explanation: