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 waardenburg  



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Author12 Posts
  #1

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






  #2

A?



  #3

1/2*1/2*1/2 - for any of the three
i think A also


  #4

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!


  #5

Fox, why do you add independent events?


  #6

Hmm..Ur right they are independant events . I think I was misleading all the way.


  #7

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?



  #8

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



  #9

this is the trick genetics guy killed him self explainingraised eyebrow



  #10

Without having reviewed genetics I would suppose that the answer is E.

1/2*1+1/2*(1-1/2)+1/2*[1/2*(1-1/2)]=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)


  #11

confused


  #12

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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