Examkin

Question

Difficulty: HardSex Determination and Sex-Linked Traits

In humans, Duchenne muscular dystrophy is inherited as an X-linked recessive disorder (XdX^d), while the normal allele is dominant (XDX^D). A phenotypically normal woman seeks genetic counseling. Her maternal grandfather had Duchenne muscular dystrophy, whereas her maternal grandmother was homozygous normal. Her father is phenotypically normal. If this woman marries a phenotypically normal man, what is the probability (expressed as a percentage) that their first male child will be affected by the disorder?

Answer: 25 %

Answer

The probability that their first male child will be affected by Duchenne muscular dystrophy is 25%.
The maternal grandfather (XdYX^d Y) passes his XdX^d chromosome to his daughter (the woman's mother), making her an obligate carrier (XDXdX^D X^d). When this carrier mother has a daughter with a normal male (XDYX^D Y), the daughter has a 50%50\% (0.50.5) chance of being a carrier (XDXdX^D X^d). If the woman is a carrier, any male child she has has a 50%50\% (0.50.5) chance of receiving the XdX^d allele and being affected. Multiplying these independent probabilities (0.5×0.50.5 \times 0.5) yields 0.250.25, or 25%25\%.

Step-by-Step Solution

1
Determine the genotype of the woman's mother from her maternal grandparents.
The woman's mother inherited XdX^d from her father (XdYX^d Y) and XDX^D from her mother (XDXDX^D X^D), making her an obligate carrier (XDXdX^D X^d).
Fathers always pass their single X chromosome to their daughters.
2
Calculate the probability that the woman inherited the recessive allele from her mother.
Probability that the woman is a carrier (XDXdX^D X^d) is 0.50.5 (or 50%50\%).
A carrier mother (XDXdX^D X^d) and normal father (XDYX^D Y) have a 50%50\% chance of producing a carrier daughter.
3
Calculate the probability that a male child of a carrier woman receives the recessive X-linked allele.
If the woman is a carrier, the probability of an affected son (XdYX^d Y) is 0.50.5 (or 50%50\%).
A male child receives his only X chromosome from his mother.
4
Multiply the independent probabilities to find the overall risk for the first male child.
P(Affected male child)=0.5 (mother is carrier)×0.5 (son inherits Xd)=0.25=25%P(\text{Affected male child}) = 0.5 \text{ (mother is carrier)} \times 0.5 \text{ (son inherits } X^d) = 0.25 = 25\%.
Both independent events (mother being a carrier and son inheriting the mutated allele) must occur.

Key Concept

Sex-Linked Recessive Inheritance and Pedigree Carrier Probability
Rate this question

Topics

Sex Determination and Sex-Linked Traits

More questions from this topic