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#1
05-15-2012
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Calculating genetic probability in cystic fibrosis

A 5-yr-old girl is diagnosed with cystic fibrosis. None of her siblings are affected. From the pedigree of the family. What is likelihood that II-3 is carrier of the trait?

Click image to enlarge

A- 25%
B- 33%
C- 50%
D- 66%
E- 100%
 The above post was thanked by: DocSikorski (05-16-2012)

#2
05-15-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,371 Times in 349 Posts Reputation: 1381

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 II-3 has trait is 50 %.

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#3
05-15-2012
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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
05-16-2012
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I think is D 66%

2/3 carrier
1/3 sick

right?
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"Disease is very old, and nothing about it has changed. It is we who change as we learn to recognize what was formerly imperceptible." JMC
#5
05-16-2012
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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
05-16-2012
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Quote:
 Originally Posted by beka-CTS 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 II-3 has trait is 50 %.
So Aa Aa parents could have 25% AA 50% Aa and 25 aa.
since the II-3 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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 The above post was thanked by: pradeeprox (05-16-2012)
#7
05-16-2012
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2/3

I always miss this question somehow.
Thank you for an extra drill!
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Ambition is a dream with a V8 engine. EP
 The above post was thanked by: drhma (05-16-2012)
#8
05-16-2012
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Quote:
 Originally Posted by modesty 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.
Quote:
 Originally Posted by Dr. Mexito I think is D 66% 2/3 carrier 1/3 sick right?
Quote:
 Originally Posted by beka-CTS So Aa Aa parents could have 25% AA 50% Aa and 25 aa. since the II-3 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.
That's conceptually incorrect. What happened to one child has no bearing on what'll happen to the next one. 25% and 50% describe the probability for each child or the overall picture you can expect to see in a very very large sample size(ideally infinity).

To give a simple and oft quoted example, if the odds of winning a mini-lotto 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
 The above post was thanked by: ..sharma (05-16-2012)
#9
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

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.
 The above post was thanked by: DocSikorski (05-16-2012)
#10
05-16-2012
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D) 2/3

Since we already know individual II-3 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
05-16-2012
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Guys, how can it be 2/3 ?

All but one (II-2) 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 II-1)- 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 II-2) 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 II-3 being a carrier. That probability would also be the same for II-1 and II-4.

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
05-16-2012
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Quote:
In AR disease, one have to eliminate the rr (homozygous) in order to make the calculation. From the possible combination:

2/3 is carrier
1/3 is non sick
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"Disease is very old, and nothing about it has changed. It is we who change as we learn to recognize what was formerly imperceptible." JMC
 The above post was thanked by: Hope2Pass (05-16-2012)
#13
05-16-2012
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Quote:
 Guys, how can it be 2/3 ? All but one (II-2) 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...
Exactly, except individual II-2 who's affected, all the other offspring can be either carriers (Aa) or normal (AA), but they cannot be (aa). So that eliminates the possibility of (aa) for all the unaffected offspring, meaning now you only have 3 choices in the punnet square, the choice (aa) can be removed. Normally the punnet square has four choices: AA, Aa, Aa, aa... So with aa removed as an option, the probability of being a carrier is 2/3.
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 The above post was thanked by: Hope2Pass (05-16-2012)
#14
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

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
05-16-2012
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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 II-3 is carrier or normal so we cancel the chance of being affected

3. II-3 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.; 05-16-2012 at 07:42 AM.

#16
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,371 Times in 349 Posts Reputation: 1381

Quote:
 Originally Posted by tarek. 3. II-3 is Dd or Dd or DD so each probability is 1/3 and thus the answer is 33%

And Dd or Dd equals to 2 Dd = 33 *2 = 66 ?

Isn't it right ?
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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
#17
05-16-2012
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Quote:
 Originally Posted by beka-CTS And Dd or Dd equals to 2 Dd = 33 *2 = 66 ? Isn't it right ?

yes thats right
#18
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,684 Threads: 213 Thanked 1,600 Times in 660 Posts Reputation: 1610

Quote:
 Originally Posted by tarek. yes thats right
#19
05-16-2012
 USMLE Forums Scout Steps History: Not yet Posts: 16 Threads: 5 Thanked 3 Times in 3 Posts Reputation: 13

Quote:
 Originally Posted by Hope2Pass hahaha nice cover up! I totally read your answer earlier but good you changed it!

shhhh!!! hahahaha :P:P I misread the question =(=(
#20
05-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 763 Threads: 77 Thanked 1,371 Times in 349 Posts Reputation: 1381

Quote:
 Originally Posted by Hope2Pass hahaha nice cover up! I totally read your answer earlier but good you changed it!

lol
__________________
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
05-16-2012
 USMLE Forums Scout Steps History: 1+CK+CS Posts: 65 Threads: 14 Thanked 23 Times in 15 Posts Reputation: 33
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
05-16-2012
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Where is this Q from ?
#23
05-16-2012
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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.
 The above post was thanked by: ..sharma (05-16-2012)
#24
05-16-2012
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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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