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#1
07-12-2012
 USMLE Forums Master Steps History: --- Posts: 1,366 Threads: 649 Thanked 687 Times in 408 Posts Reputation: 697
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

is the answer right???
 The above post was thanked by: 5150joker (07-12-2012)
#3
07-12-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 791 Threads: 76 Thanked 698 Times in 327 Posts Reputation: 716

B............

#4
07-12-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,418 Times in 355 Posts Reputation: 1428

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
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,355 Threads: 94 Thanked 907 Times in 523 Posts Reputation: 920

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 1,418 Times in 355 Posts Reputation: 1428

Quote:
 Originally Posted by DocSikorski I do not know =) and I am happy to forget about it
lol
#7
07-12-2012
 USMLE Forums Addict Steps History: Not yet Posts: 138 Threads: 1 Thanked 53 Times in 32 Posts Reputation: 63

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 538 Times in 196 Posts Reputation: 548

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: 1,366 Threads: 649 Thanked 687 Times in 408 Posts Reputation: 697

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: 791 Threads: 76 Thanked 698 Times in 327 Posts Reputation: 716

why is the probability of mother and father 1/2..??? i dont get this????
#11
07-13-2012
 USMLE Forums Scout Steps History: Not yet Posts: 14 Threads: 0 Thanked 1 Time in 1 Post Reputation: 11

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
 USMLE Forums Guru Steps History: 1+CK+CS Posts: 395 Threads: 41 Thanked 224 Times in 103 Posts Reputation: 234
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 108 Times in 32 Posts Reputation: 118

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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