A woman carries an X-linked recessive allele for a disorder characterized by muscle degeneration but does not express this condition. If she has multiple offspring with an unaffected man, how many of the woman's children are expected to exhibit this degenerative disorder?

Question

A woman carries an X-linked recessive allele for a disorder characterized by muscle degeneration but does not express this condition.  If she has multiple offspring with an unaffected man, how many of the woman's children are expected to exhibit this degenerative disorder?

Quizplus · Accepted Answer

11ecba2d_0fee_1b62_a3bf_c59372a39e01_MD0008_00 Sex linkage is the expression of an allele located on a sex chromosome (eg, X or Y in humans). In X-linked recessive disorders, females (XX) will express the phenotype when both X chromosomes (inherited from the mother and father) have the deleterious recessive allele. In contrast, males (XY) need only inherit a single copy of the recessive allele from their mother to exhibit the disease. The given scenario describes the genetic transmission of an X-linked recessive disorder characterized by muscle degeneration. The mother (X^MX^m) is a carrier, meaning she has the disease-causing allele but does not express the disorder, whereas the father (X^MY) is normal (unaffected by the disorder). Genetically, the woman has one copy of the wild-type dominant allele on one X chromosome (X^M) and one copy of the recessive allele on the other X chromosome (X^m). Women always pass one of their two X chromosomes to offspring; therefore, this woman would always have a 50% chance of passing the recessive allele to each child, regardless of gender. In contrast, the man would always pass on a normal X chromosome to female offspring. Accordingly, if they were to have children, their daughters would never be affected and would be either normal (X^MX^M) or carriers (X^MX^m). Their male offspring, however, would always inherit the Y chromosome from their father and inherit either the normal allele (and be unaffected) or the recessive allele (and express the disorder) from their mother. Given the parental genotypes, the overall probability (P) that a child will express an X-linked recessive trait can be calculated as follows: P(child inherits affected X-chromosome from mother) × P(child is male) = 0.5 × 0.5 = 0.25 Therefore, one-quarter (25%) of her children are expected to express this disorder and exhibit muscle degeneration (Choices A and C). (Choice B) Half (50%) of her children will inherit the abnormal allele. However, only male children will express the disorder. Educational objective: For the cross between a female carrier of an X-linked recessive trait and an unaffected man, only male children will express the trait. The probability of a child having the trait is given by P(child inherits affected X-chromosome from mother) × P(child is male) = 0.5 × 0.5 = 0.25. Therefore, 25% of all offspring will express the trait.

A Woman Carries an X-Linked Recessive Allele for a Disorder