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A man who is affected with hemophilia A (X-linked recessive) mates with a woman who is a heterozygous carrier of this disorder. What proportion of this couple's daughters will be affected, and what proportion of the daughters will be heterozygous carriers?

A. 0%; 50%
B. 100%; 0%
C. 0%; 100%
D. 50%; 50%
E. 2/3; 1/3

1 Answer

2 votes

Final answer:

When a hemophilic man mates with a heterozygous carrier woman, their daughters have a 50% chance of being carriers of hemophilia and a 50% chance of being affected with hemophilia. The correct answer to this inheritance pattern question is D. 50%; 50%.

Step-by-step explanation:

The question involves predicting the inheritance pattern of hemophilia A, an X-linked recessive disorder, when a hemophilic man mates with a heterozygous carrier woman. Since the man is affected by hemophilia, his X chromosome contains the gene for hemophilia versus a normal Y chromosome. The woman, being a heterozygous carrier, has one normal X chromosome and one X chromosome with the gene for hemophilia.

Daughters receive one X chromosome from each parent. Therefore, a daughter has the following possible combinations: An X chromosome with hemophilia from her father and a normal X chromosome from her mother, making her a carrier (heterozygous for the hemophilia gene), or an X chromosome with hemophilia from both parents, making her affected by hemophilia. No other scenarios are possible since the father only carries X chromosomes with the gene for hemophilia. Thus, there is a 50% chance a daughter will be a carrier, inheriting the father's X chromosome with hemophilia and the mother's normal X chromosome, and a 50% chance a daughter will be affected, inheriting the gene for hemophilia from both parents.

The correct answer is D. 50%; 50%.

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