Question

1. a woman who is a carrier marries a normal man. show the cross. what is the probability that their children will have hemophilia? what sex will a child in the family with hemophilia be? 10. a woman who has hemophilia marries a normal man. how many of their children will have hemophilia, and what is their sex?

Explanation:

Question 9

Step1: Determine Genotypes

Hemophilia is X - linked recessive. Let \( X^H \) be normal allele, \( X^h \) be hemophilia allele. Carrier woman: \( X^H X^h \), normal man: \( X^H Y \).

Step2: Set Up Punnett Square

	\( X^H \)	\( Y \)
\( X^h \)	\( X^H X^h \)	\( X^h Y \)

Step3: Calculate Probability

Total 4 possible genotypes. Only \( X^h Y \) has hemophilia. So probability \( = \frac{1}{4}=25\% \).

Step4: Determine Sex

\( X^h Y \) is male (has Y chromosome).

Step1: Determine Genotypes

Woman with hemophilia: \( X^h X^h \), normal man: \( X^H Y \).

Step2: Set Up Punnett Square

	\( X^H \)	\( Y \)
\( X^h \)	\( X^H X^h \)	\( X^h Y \)

Step3: Analyze Offspring

Out of 4 offspring, 2 are \( X^h Y \) (hemophilic, male). So half (50%) of children will have hemophilia, and they are male.

Answer:

Probability: \( 25\% \) (or \( \frac{1}{4} \)), Sex: Male

Question 10