A type of muscular dystrophy shows sex-linked recessive inheritance. Affected persons usually die by the age of 20. Suppose that a boy with the disease lives long enough to marry a woman heterozygous for the trait. If they have a son, what is the probability that he will have the disease?

Question

Quizplus · Accepted Answer

The boy with the disease must have the genotype X^mY, where X^m is the mutant allele and Y is the male sex chromosome. The woman heterozygous for the trait must have the genotype X^MX^m. 
The Punnett square for their offspring would be:

| X^M   | X^m   |
|-------|-------|
| X^mY  | X^MX^m|

Thus, there is a 50% chance that the son will inherit the X^m allele from his mother and the Y chromosome from his father, resulting in the genotype X^mY and the disease. The other 50% chance is that he will inherit the X^M allele from his mother and the Y chromosome from his father, resulting in the genotype X^MY and he will not have the disease.

A Type of Muscular Dystrophy Shows Sex-Linked Recessive Inheritance