Question

1. a woman who is a carrier marries a normal man. show the cross:

how many children will have the disease? ____
what is the sex of the child with the disease? ____

Explanation:

Step1: Define allele notation

Let $X^A$ = normal allele, $X^a$ = disease allele (X-linked recessive, as carriers are female).

Step2: Assign parental genotypes

Carrier woman: $X^A X^a$; Normal man: $X^A Y$

Step3: Set up Punnett square

$$\begin{array}{c|cc} & X^A & X^a \\ \hline X^A & X^A X^A & X^A X^a \\ Y & X^A Y & X^a Y \\ \end{array}$$

Step4: Identify affected offspring

Only $X^a Y$ has the disease.

Step5: Calculate disease probability

$\frac{1}{4}$ of offspring, or 25%.

Answer:

Punnett square:

$$\begin{array}{c|cc} & X^A & X^a \\ \hline X^A & X^A X^A & X^A X^a \\ Y & X^A Y & X^a Y \\ \end{array}$$

How many children will have the disease? $\frac{1}{4}$ (or 25%) of children
What is the sex of the child with the disease? Male