Question

In Caucasians, the M-shaped hairline that recedes with age (sometimes to nearly complete baldness) was once thought to be due to an allele that is dominant in males but recessive in females. If this were true, and if the frequency of the baldness allele were 30 percent, what frequencies of the bald and nonbald phenotypes would be expected in males and females under random mating?

Answer 1

For Caucasians with a baldness allele frequency of 30%, the expected frequency of bald males would be 30%, and non-bald males 70%. For females, non-bald individuals would be 49%, and bald females would be 51% based on the Hardy-Weinberg principle.

Step-by-step explanation:

If the allele for baldness in Caucasians is dominant in males and recessive in females, and its frequency is 30 percent, we can use the Hardy-Weinberg principle to calculate expected phenotypic frequencies in males and females under random mating conditions. To simplify, let's represent the baldness allele as 'B' and the non-baldness allele as 'b'. The Hardy-Weinberg principle indicates that the allele frequencies in a population will remain constant if there's random mating and no other evolutionary forces are acting on the population.

For males, because they have only one copy of the X chromosome where this allele is likely located, the frequency of bald men would be equal to the frequency of the baldness allele, which is 30% (0.3). Therefore, the frequency of non-bald men will be 1 - 0.3 = 0.7 or 70%. For females, since the baldness allele is recessive, only homozygous recessive females (bb) will show non-baldness, and the frequency of these can be calculated as q2 where q represents the recessive allele frequency. If p = 0.3, then q = 1 - p = 0.7. Therefore, the frequency of non-bald women will be q2 = 0.72 = 0.49 or 49%. Consequently, the frequency of bald women (either Bb or BB genotypes) will be 1 - 0.49 = 0.51 or 51%.