Suppose that on a small island off the coast of Scotland, 32 percent of the population has blue eyes, which means that these individuals must be homozygous for the blue eye color gene (bb). The only other eye color found on the island is brown, and individuals that are homozygous for the brown eye color gene (BB) or heterozygous (Bb) will have brown eyes because brown is the dominant gene. Assume this population is in Hardy-Weinberg equilibrium. If 100 babies are born next year, how many of these would you expect to have brown eyes and be heterozygous?

Question

Suppose that on a small island off the coast of Scotland, 32 percent of the population has blue eyes, which means that these individuals must be homozygous for the blue eye color gene (bb). The only other eye color found on the island is brown, and individuals that are homozygous for the brown eye color gene (BB) or heterozygous (Bb) will have brown eyes because brown is the dominant gene. Assume this population is in Hardy-Weinberg equilibrium. If 100 babies are born next year, how many of these would you expect to have brown eyes and be heterozygous?

58

49

29

43

Answer 1

The frequency of the blue eye color gene (b) in the population is 0.32. Since the population is in Hardy-Weinberg equilibrium, the frequency of the brown eye color gene (B) can be calculated as 1 - 0.32 = 0.68.

To determine the expected number of heterozygous individuals (Bb) among the 100 babies, we can use the Hardy-Weinberg equation:
2pq = number of heterozygotes,
where p is the frequency of the dominant allele (B) and q is the frequency of the recessive allele (b).

Plugging in the values:
2 * 0.68 * 0.32 = 0.4352

So, we would expect approximately 43 (rounded to the nearest whole number) of the babies to have brown eyes and be heterozygous (Bb). Therefore, the answer is 43.