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(B) $1$. The pedigree shows an $X$-linked recessive trait because the affected mother $(A)$ passes the trait to her sons.$2$. Since $A$ is affected,he

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$XX^c$ and $XY$

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(B) $1$. The pedigree shows an $X$-linked recessive trait because the affected mother $(A)$ passes the trait to her sons.$2$. Since $A$ is affected,her genotype is $X^cX^c$. Her husband is normal,so his genotype is $XY$.$3$. Individual $C$ is a normal female daughter of $A$. She must have inherited one $X^c$ from her mother and one $X$ from her father. Thus,the genotype of $C$ is $XX^c$.$4$. Individual $D$ is a normal male son of $C$. Since he is normal,he must have inherited the $X$ chromosome from his mother $(C)$ and the $Y$ chromosome from his father. Thus,the genotype of $D$ is $XY$.

In the provided pedigree,assume that no outsider marrying into the family carries the disease. Determine the genotypes of $C$ and $D$.

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