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Old 07-12-2012
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Genetics 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
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Old 07-12-2012
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Default 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???
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Old 07-12-2012
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B............
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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)
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Old 07-12-2012
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I do not know =)
and I am happy to forget about it
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Old 07-12-2012
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Quote:
Originally Posted by DocSikorski View Post
I do not know =)
and I am happy to forget about it
lol
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Old 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


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Old 07-12-2012
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Correct Answer Correct answer

Quote:
Originally Posted by BritneySpears View Post
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 ..
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Quote:
Originally Posted by INCOGNITO View Post
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.
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Old 07-13-2012
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why is the probability of mother and father 1/2..??? i dont get this????
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yeah i dnt get it too.... how is the probability of a carrier in AR disease can be 1/2??
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Old 07-13-2012
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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
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Old 07-15-2012
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Default 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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Old 07-16-2012
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Quote:
Originally Posted by bendect View Post
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
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