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Difficulty: MediumSex Determination and Sex-Linked Traits

In humans, hemophilia is an X-linked recessive disorder. If a carrier woman (XHXhX^H X^h) marries a hemophilic man (XhYX^h Y), what is the probability that any daughter born to this couple will be a carrier of hemophilia?

  1. A
    25%
  2. 50%Answer
  3. C
    75%
  4. D
    100%

Answer

50%
For female offspring (XXXX), the father always contributes an XhX^h chromosome. The mother contributes either an XHX^H chromosome (50% chance) or an XhX^h chromosome (50% chance). Thus, 50% of the daughters will have the heterozygous genotype XHXhX^H X^h, making them carriers.

Step-by-Step Solution

1
Determine parental genotypes and gamete types
Mother is XHXhX^H X^h (gametes: XH,XhX^H, X^h). Father is XhYX^h Y (gametes: Xh,YX^h, Y).
Identifying parental gametes is necessary to cross all potential allele combinations.
2
Determine genotypes of female offspring (XXXX)
Female offspring inherit XhX^h from the father and either XHX^H or XhX^h from the mother, resulting in XHXhX^H X^h (carrier female) and XhXhX^h X^h (affected female).
Female children always receive one XX chromosome from each parent.
3
Calculate the probability specific to female offspring
Out of 2 possible female genotypes (XHXhX^H X^h and XhXhX^h X^h), exactly 1 is a carrier (XHXhX^H X^h), giving a probability of 12\frac{1}{2} or 50%.
The question restricts the probability domain specifically to daughters.

Key Concept

Sex-linked inheritance and gender-specific probability calculations
Estimated Time:1m 30s
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Topics

Sex Determination and Sex-Linked Traits

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