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Old 12-17-2011
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Genetics Cystic Fibrosis carrier state calculation

Couple with the youngest child with confirmed cystic fibrosis. Elder two children are normal. Which of the following is the likelihood that each of the unaffected child is a carrier of the disease?

a. 1 in 2
b. 2 in 2
c. 1 in 4
d. 2 in 3
e. 3 in 4

Please post answers with explanations
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Old 12-17-2011
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I think answer is 2/3. Uneffected Childrens genotypes maybe AA/Aa and Aa. so 2 from 3.
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Old 12-17-2011
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I think its . 2/3

CF is a autosomal recessive disorder.

Since, one child is affected, means parents are heterozygous for the mutation.

Possible genotypes of children by heterozygous is: cc, Cc, Cc, CC.

The probability of children being carriers is 2 of 3. (cc being homozygous for mutation).
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Old 12-17-2011
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you guys are correct.its 2/3. I got confused by the wording of the question as it mentions each of the child,so thought you have to multiply 2/3 with 2/3,but that was not a given option.
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