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#1




Calculating genetic probability in cystic fibrosis
A 5yrold girl is diagnosed with cystic fibrosis. None of her siblings are affected. From the pedigree of the family. What is likelihood that II3 is carrier of the trait?
Click image to enlarge A 25% B 33% C 50% D 66% E 100% 
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DocSikorski (05162012) 
#2




So since one child is affected in family with aut. recessive disease and parents are normal it means that both of them (parents) are Aa and Aa.chance that II3 has trait is 50 %.
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#3




Becuase one child is already affected for sure you can state that the chances of a sibling to be affected is 2/3. One has to be healthy and 2 should be carriers. 50% is only true if they would have asked you to calculate the chances without knowing that anyone is sick or healthy.



#4




I think is D 66%
2/3 carrier 1/3 sick right?
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#5




I think it's A 25%.
Every child has an equal chance of getting the disease and since cystic fibrosis is autosomal recessive, it would be 1/4 = 25% 
#6




Quote:
since the II3 isn't affected (not black) it leaves only first two variants. so 2 of 3. so appr. 66% .... Didn't pay attention in the beginning.
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pradeeprox (05162012) 
#7




2/3
I always miss this question somehow. Thank you for an extra drill!
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drhma (05162012) 
#8




Quote:
Quote:
To give a simple and oft quoted example, if the odds of winning a minilotto is 1/100, and I've been buying lotto tickets without winning for the past 99 weeks, what are my chances of winning the 100th time?? That's right, 1/100! And if I've been winning the lottery for the last 99 weeks, what are the chances that I'll win the next one? Again its 1/100. Previous outcomes do not influence the next event. Sent from my Desire HD 
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..sharma (05162012) 
#9




Fcuk i misread the question ... It's asking if the offspring is a carrier ... I'd go with 50% since every time they hav a child, each one has the same risk.

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DocSikorski (05162012) 
#10




D) 2/3
Since we already know individual II3 doesn't have the disease it only leaves the possibility they they're either a heterozygous carrier or homozygous normal. The probability of being homozgous affected (1/4) is obsolete since we already know they're normal. So the probability that this individual is a carrier is 2/3 and the probability that she's homozgous normal is 1/3.
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#11




Guys, how can it be 2/3 ?
All but one (II2) of their offspring's is affected. It doesn't mean the other's have AA,Aa,Aa genotype. They can all be carriers (Aa) or normal (AA) as far as anyone knows... Think about it this way  They had their first child (lets say II1) He had 1/4 risk of being affected, 1/2 risk of being a carrier and 1/4 risk of being homozygous. Then they had their second child (let's say II2) and again, she had the same risk! Just because they had one normal child before her doesn't mean, the affected offspring had a higher chance of being affected !! IF they were quadruplets or some how the couple had all four children at one time .. then you would have 2/3 possibility of II3 being a carrier. That probability would also be the same for II1 and II4. I hope that's the right answer or I just made a fool of myself .. either way, please discuss and tell me if what I said makes sense or not? 
#12




Quote:
2/3 is carrier 1/3 is non sick
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Hope2Pass (05162012) 
#13




Quote:
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Hope2Pass (05162012) 
#14




AHHH I get it ... So when they ask the risk of being a carrier, that eliminates homozygous affected because if the genotype was aa then the person would be affected (not carrier or normal). Damn, no wonder I used to get these wrong! Well, now I won't get these wrong lol ... Good discussion fellas!

#15




read it!!
the answer is D 66% here how i get it:
1. first thing the disease is recessive because she has the disease and non of here parents are affected. 2.draw a punnett square of here parents: D d D DD Dd d Dd dd notice that 25% affected and 50% are carrier and 25% are normal we see form the pedigree that II3 is carrier or normal so we cancel the chance of being affected 3. II3 is Dd or Dd or DD so the probability of Dd is 33% so its 66% since we got 2 Dd 66% are carrier and 33% are normal Last edited by tarek.; 05162012 at 07:42 AM. 
#16




Quote:
And Dd or Dd equals to 2 Dd = 33 *2 = 66 ? Isn't it right ?
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#17




yes thats right 
#18




hahaha nice cover up! I totally read your answer earlier but good you changed it!

#19




Quote:
shhhh!!! hahahaha :P:P I misread the question =(=( 
#20




Quote:
lol
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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 
#21




correct ans is D
correct ans is D 66%
i was scratchin my head over dis question & thought of skippin it the explanation given in the source was pathetic as it was far more confusing than the question... thnx for the explanation 
#22




Where is this Q from ?

#23




If the question was asking about an unborn child, then the chance of being carrier is 50%. But since the offspring the are asking about is already not affected, then the chance of being affected should be excluded, and hence the chance for being carrier is 2 out of 3.

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..sharma (05162012) 
#24




Lets extend the question.
thanks for such a nice discussion. I got it! What you guys meant to say.
But i was thinking Just imagine if the couple had 5 children then question was asked about the II 5. Then what should be the provability. Will it be 1/4 for AA and 2/4 for Aa. May be a idiotic question for some fella's but still wanna polish my concept. Thanks
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