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Difficulty: MediumSex Determination and Sex-Linked Traits

A man with normal blood clotting marries a phenotypically normal woman whose father had hemophilia A, an X-linked recessive disorder. What is the probability that any child born to this couple will be a carrier of the hemophilia allele?

  1. 25%Answer
  2. B
    50%
  3. C
    75%
  4. D
    0%

Answer

The probability that any child born to this couple will be a carrier of the hemophilia allele is 25%.
Since the woman's father had hemophilia (XhYX^h Y), she inherited his XhX^h allele and is a carrier (XHXhX^H X^h). When crossed with a normal male (XHYX^H Y), four total offspring genotypes are produced with equal probability: normal female (XHXHX^H X^H), carrier female (XHXhX^H X^h), normal male (XHYX^H Y), and affected male (XhYX^h Y). Only XHXhX^H X^h individuals are carriers, representing 1 out of 4 total possible outcomes, or 25%.

Step-by-Step Solution

1
Determine the parental genotypes from the pedigree description.
Father = XHYX^H Y, Mother = XHXhX^H X^h.
Because the woman's father had hemophilia (XhYX^h Y), she must have inherited his affected XhX^h chromosome, making her a heterozygous carrier (XHXhX^H X^h).
2
Construct a genetic cross between XHYX^H Y and XHXhX^H X^h.
The possible offspring genotypes are XHXHX^H X^H (25%), XHXhX^H X^h (25%), XHYX^H Y (25%), and XhYX^h Y (25%).
A Punnett square combines the maternal gametes (XH,XhX^H, X^h) and paternal gametes (XH,YX^H, Y) in equal proportions.
3
Identify the carrier genotype among total offspring possibilities.
1 out of 4 total possibilities is XHXhX^H X^h, which equals 25%.
Carrier status requires one recessive allele on an X chromosome in a phenotypically normal female (XHXhX^H X^h).

Key Concept

X-linked recessive inheritance and offspring probability calculation
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Topics

Sex Determination and Sex-Linked Traits

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