Question

In humans, male-pattern baldness is controlled by an autosomal gene that occurs in two allelic forms. Allele Hn determines nonbaldness, and allele Hb determines pattern baldness. In males, because of the presence of testosterone, allele Hb is dominant over Hn. If a man and woman both with genotype HnHb have a son, what is the chance that he will eventually be bald?

Answer 1

75%

Step-by-step explanation:

The autosomal gene for baldness has two possible alleles: Hb (baldness) and Hn (nonbaldness).

A man and a woman, both heterozygous HnHb, have a son. Each parent produces gametes Hn or Hb.

The possible genotypes of their children, resulting from the combinations of those gametes, are:

• 1/4 HnHn
• 2/4 HnHb
• 1/4 HbHb

In males, the Hb allele is dominant. Therefore, their son has a chance of 3/4 = 0.75 = 75% of having the Hb allele which determines baldness.