Hemophilia is a sex-linked recessive disorder commonly carried on the X chromosome. In this case, we have:
- The male with hemophilia has the genotype \( X^hY \) (where \( X^h \) represents the X chromosome carrying the hemophilia allele and \( Y \) represents the Y chromosome).
- The homozygous normal female has the genotype \( X^X \) (where both X chromosomes are normal).
To find the probability that their female child will be a carrier, we can construct a Punnett square. The possible gametes contributed by each parent are:
- Male with hemophilia: \( X^h \) or \( Y \)
- Homozygous normal female: \( X \) or \( X \) (both gametes are the same)
Now, create a Punnett square:
\[ \begin{array}{c|c|c} & X & X \ \hline X^h & X^hX & X^hX \ \hline Y & XY & XY \ \end{array} \]
The potential offspring from this cross would be:
- \( X^hX \) (female, carrier)
- \( X^hX \) (female, carrier)
- \( XY \) (male, normal)
- \( XY \) (male, normal)
Out of the four possible combinations, the two female offspring (\( X^hX \)) are carriers of hemophilia.
Thus, the probability that the female child will be a carrier for hemophilia is 100%.
The correct answer is 100%.