Question

Hemophilia is a sex-linked, recessive genetic disorder. If a male with hemophilia and a homozygous normal female have a female child, what is the probability that the girl will be a carrier for hemophilia?
Responses

0%
0%

25%
25%

50%
50%

100%

Answer 1

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:

1. \( X^hX \) (female, carrier)
2. \( X^hX \) (female, carrier)
3. \( XY \) (male, normal)
4. \( XY \) (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%**.