Question

#3. 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: Determine the genotypes

Let's assume the recessive allele for the X - linked disease is \(x\) and the dominant normal allele is \(X\). The woman is a carrier, so her genotype is \(Xx\). The normal man has the genotype \(XY\).

Step2: Set up the Punnett square

Gametes of female (\(Xx\))	Gametes of male (\(XY\))
\(x\)	\(Xx\)	\(xY\)

Step3: Analyze the offspring

• The genotypes of the offspring are \(XX\) (normal female), \(Xx\) (carrier female), \(XY\) (normal male), and \(xY\) (affected male).

Answer:

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