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?
paint)
58
40
29
43
To calculate the number of individuals expected to have brown eyes and be heterozygous (Bb), we can use the Hardy-Weinberg equation:
p^2 + 2pq + q^2 = 1
Where:
- p^2 represents the frequency of individuals who are homozygous dominant (BB), which is the proportion of brown-eyed individuals.
- 2pq represents the frequency of individuals who are heterozygous (Bb), which is the proportion of individuals with brown eyes and are carriers of the blue eye gene.
- q^2 represents the frequency of individuals who are homozygous recessive (bb), which is the proportion of blue-eyed individuals.
- The sum of p^2, 2pq, and q^2 should equal 1, as it represents the total population.
Given that 32 percent of the population has blue eyes (bb), we know that q^2 = 0.32. Since q represents the frequency of the recessive allele, q = √0.32 ≈ 0.57.
Substituting this value into the Hardy-Weinberg equation, we can calculate the proportion of individuals who are heterozygous (Bb):
2pq = 2 * 0.57 * (1 - 0.57) = 2 * 0.57 * 0.43 ≈ 0.49
This means that approximately 49% of the population would be expected to have brown eyes and be heterozygous (Bb).
Therefore, out of the 100 babies born next year, we would expect approximately 49 of them to have brown eyes and be heterozygous.