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#1
07-12-2012
 USMLE Forums Master Steps History: --- Posts: 1,365 Threads: 648 Thanked 584 Times in 354 Posts Reputation: 594
Calculating the genetic risk of Cystic Fibrosis?

A 27-year-old female, whose brother died of cystic fibrosis, is married to a 25-year-old whose sister (age 16) has cystic fibrosis.
What is the risk their first child will have CF?
A. 1/4
B. 1/9
C. 1/12
D. 1/16
E. 1/64
 The above post was thanked by: bendect (07-13-2012)

#2
07-12-2012
 USMLE Forums Scout Steps History: Not yet Posts: 14 Threads: 0 Thanked 1 Time in 1 Post Reputation: 11
1/9 ????

cystic fibrosis is A.R
female being a carrier -chances 2/3
male being a carrier - chances 2/3
both transmitting their mutant allel-1/2 &1/2
so, 2/3*2/3*1/2*1/2= 4/36=1/9

 The above post was thanked by: 5150joker (07-12-2012)
#3
07-12-2012
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B............
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#4
07-12-2012
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What is the risk their first child will have CF?
A. 1/4
B. 1/9
C. 1/12
D. 1/16
E. 1/64

2/3(chance of mother to be Aa) * 2/3(chance of father to be Aa) * 1/4(chance of Kid to be aa)
#5
07-12-2012
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I do not know =)
and I am happy to forget about it
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#6
07-12-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,338 Times in 347 Posts Reputation: 1348

Quote:
 Originally Posted by DocSikorski I do not know =) and I am happy to forget about it
lol
#7
07-12-2012
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d) 1/16

mother's P of being a carrier: 1/2
father's P of being a carrier: 1/2

P of both parents being carriers: 1/2 * 1/2 = 1/4

child P of getting both CF genes: 1/4

1/4 * 1/4 = 1/16

#8
07-12-2012
 USMLE Forums Guru Steps History: Step 1 Only Posts: 343 Threads: 127 Thanked 522 Times in 192 Posts Reputation: 532

Quote:
 Originally Posted by BritneySpears d) 1/16 mother's P of being a carrier: 1/2 father's P of being a carrier: 1/2 P of both parents being carriers: 1/2 * 1/2 = 1/4 child P of getting both CF genes: 1/4 1/4 * 1/4 = 1/16

YEAH THIS ONE IS CORRECT ANSWER
THIS QUESTION IS FROM UWORLD 2012 edition ..
#9
07-12-2012
 USMLE Forums Master Steps History: --- Posts: 1,365 Threads: 648 Thanked 584 Times in 354 Posts Reputation: 594

Quote:
 Originally Posted by INCOGNITO YEAH THIS ONE IS CORRECT ANSWER THIS QUESTION IS FROM UWORLD 2012 edition ..
hi , can u please post explanation from there ? my source is not uworld in this q and so i wonder how come ans be 1/16.
thnks.
#10
07-13-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 790 Threads: 76 Thanked 672 Times in 316 Posts Reputation: 690

why is the probability of mother and father 1/2..??? i dont get this????
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#11
07-13-2012
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yeah i dnt get it too.... how is the probability of a carrier in AR disease can be 1/2??
#12
07-13-2012
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 34 Threads: 1 Thanked 10 Times in 10 Posts Reputation: 20

so let try to make it clear
you know that each sibling(brother and sister ) share 1/2 of their genes and you know that to get cf you have to be homozygous aa so
chance that father of the patient has cf gene is 1/2 so its a carrier
chance that mother of the patient has cf gene is 1/2 its a carrier
probability of child to get disease from 2 carriers 1/4
 The above post was thanked by: dr.atharwa (07-13-2012)
#13
07-15-2012
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It's 1/9

P of being carrier for both parents - 2/3 considering both are phenotypically normal. Geno not known.
For every 1 affected 3 normal siblings are present, of which 2 could be carriers in AR diseases ( aa is diseased , aA,aA,AA are pheno normal and 2/3 are carrier)

P of having a affected child in AR disease two carrier parents is 1/4

= 1/4*2/3*2/3 = 1/9
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#14
07-16-2012
 USMLE Forums Addict Steps History: 1 + CS Posts: 142 Threads: 33 Thanked 97 Times in 28 Posts Reputation: 107

Quote:
 Originally Posted by bendect so let try to make it clear you know that each sibling(brother and sister ) share 1/2 of their genes and you know that to get cf you have to be homozygous aa so chance that father of the patient has cf gene is 1/2 so its a carrier chance that mother of the patient has cf gene is 1/2 its a carrier probability of child to get disease from 2 carriers 1/4

I would have to disagree with this.
You know the father is a carrier and the mother is the carrier.
But they currently ARE NOT SYMPTOMATIC FOR CF (older age and no symptoms) thus are not homozygous and suffering from CF.

With that. The carrier risk is 2/3 for mother and father. And 1/4 of the chance of the child getting CF which is an autosomal recessive disorder.

So 2/3 * 2/3 * 1/4
= 4/36
or 1/9

 Tags Genetics-, Step-1-Questions

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