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docswapna
Forum Elite Topics: 61 Posts: 228 
Mother has waardenburg syndrome which is an autosomal dominant. Father is unaffected and they plan to have a family with three children. What is the probability that one of the three children will be affected? A.1/8 B.1/4 C.3/8 D.1/3 E.7/8  
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young_doc
Forum Guru Topics: 59 Posts: 718 
A?  
alexaE
Forum Elite Topics: 23 Posts: 328 
1/2*1/2*1/2  for any of the three i think A also  
fox
Forum Guru Topics: 70 Posts: 720 
Every child has a 50% (1/2) chance of inheriting the defective copy of the Allele. If you ask the chance for one to be affected i think it should be 50/150 (sum of all chances) = 1/3. i.e. D. Its just my assumption. Criticism welcomed!  
alexaE
Forum Elite Topics: 23 Posts: 328 
Fox, why do you add independent events?  
fox
Forum Guru Topics: 70 Posts: 720 
Hmm..Ur right they are independant events . I think I was misleading all the way.  
young_doc
Forum Guru Topics: 59 Posts: 718 
fox wrote: Every child has a 50% (1/2) chance of inheriting the defective copy of the Allele. If you ask the chance for one to be affected i think it should be 50/150 (sum of all chances) = 1/3. i.e. D. Its just my assumption. Criticism welcomed! I was under the impression that you multiply your odds of getting an affected child from each pregnancy. ALTHOUGH i do know that there are some scenarios where we do have to ADD the probabilities. Anyone here taken stats?  
yoga
Forum Guru Topics: 73 Posts: 637 
d isnt that probability 4 each preganancy?? i mean wether u r taking about 1in 3 or i in 10 it doesnt really matter right?? this is old post i hope though 1 can carify  
yoga
Forum Guru Topics: 73 Posts: 637 
this is the trick genetics guy killed him self explaining  
perfectcube
Forum Newbie Topics: 1 Posts: 10 
Without having reviewed genetics I would suppose that the answer is E. 1/2*1+1/2*(11/2)+1/2*[1/2*(11/2)]=7/8 I think that intuitively it's obvious that if the prob. of sth happening is 50% if you do the experiment 1 time the prob. is always 50% but if you repeat the experiment let's say 10 times then the prob. that at least one time this certain thing will happpen is greater than 50%. If the question was what is the prob. that all three will be affected the answer would be 1/8. If the question was what is the prob. that the second one for example would be affected the probability would be 1/2. (The same goes with the first or the third one)  
yoga
Forum Guru Topics: 73 Posts: 637 
 
littledoc
Forum Newbie Topics: 16 Posts: 950 
I think C. 3/8. 1st affected*2nd unaffected*3rd unaffected=1/2*1/2*1/2=1/8. But any of the three can be affected with remaining two unaffected so 1/8+1/8+1/8=3/8. So, C is the answer.  
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