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#1
12-17-2011
 USMLE Forums Veteran Steps History: Not yet Posts: 281 Threads: 74 Thanked 134 Times in 70 Posts Reputation: 144
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 1,379 Times in 349 Posts Reputation: 1389

I think answer is 2/3. Uneffected Childrens genotypes maybe AA/Aa and Aa. so 2 from 3.
__________________
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
12-17-2011
 USMLE Forums Scout Steps History: 1 + CK Posts: 99 Threads: 8 Thanked 33 Times in 29 Posts Reputation: 43

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
12-17-2011
 USMLE Forums Veteran Steps History: Not yet Posts: 281 Threads: 74 Thanked 134 Times in 70 Posts Reputation: 144

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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