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#1
12-17-2011
 USMLE Forums Veteran Steps History: Not yet Posts: 281 Threads: 74 Thanked 131 Times in 68 Posts Reputation: 141
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

 The above post was thanked by: drortho (01-18-2012), riya rai (12-20-2011)

#2
12-17-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,338 Times in 347 Posts Reputation: 1348

I think answer is 2/3. Uneffected Childrens genotypes maybe AA/Aa and Aa. so 2 from 3.
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Step 1 - 244 [✔] Step 2 CK - 246 [✔] Step 2 CS [✔] Step 3 - 228 [✔] Match [EMORY SOM] YOG: 2013, 4 Months of USCE University Hospital
#3
12-18-2011
 USMLE Forums Scout Steps History: 1 + CK Posts: 99 Threads: 8 Thanked 32 Times in 28 Posts Reputation: 42

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).
 The above post was thanked by: drortho (01-18-2012)
#4
12-18-2011
 USMLE Forums Veteran Steps History: Not yet Posts: 281 Threads: 74 Thanked 131 Times in 68 Posts Reputation: 141

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.

 Tags Genetics-, Step-1-Questions

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