A woman who is heterozygous for an allele that results in X-linked pattern baldness marries a man who is nonbald. Which of the following would be true of their offspring?
A) All would be bald
B) All of the females would be nonbald, all males would be bald
C) All of the females would be nonbald, ½ of the males would be bald
D) ½ of females would be bald, and ½ of the males would be bald
E) All would be nonbald

Question

Quizplus · Accepted Answer

Since the woman is heterozygous for an X-linked pattern baldness allele, her genotype would be XbX (b = baldness allele). The nonbald man would have the genotype XBY (Y = non-baldness allele). Therefore, their offspring could inherit an XbB allele from the mother and a Y allele from the father, resulting in a non-bald male. Or they could inherit an XbX allele from the mother AND a Y allele from the father, resulting in a bald male. Females would only be bald if they inherited two Xb alleles, which would be XbXb, which would only occur if the father is also XbXb (which is not given as an option). Therefore, all females would be non-bald since they could either inherit XBX or XBY, and only one Xb allele is needed to result in baldness. Thus, the offspring ratio would be ½ of males would be bald (XbY) and ½ would be non-bald (XBY), and all females would be non-bald (XBX or XBY).

A Woman Who Is Heterozygous for an Allele That Results