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 Waardenburg  




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Author27 Posts
  #1

Waardenburg syndrome is an autosomal dominant condition that accounts for 1.4% of cases of congenital deafness. In addition to deafness, patients with this condition have an atypical facial appearance, including lateral displacement of the inner canthi, hypertelorism ,white forelock and white patches of skin on the ventral midline . A mother has Waardenburg syndrome, her husband 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. 3/4
D. 1/3
E. 1/2




  #2

The probability for each of 3 children will be the same.
Let's suppose that mother is heterozygose Aa, father is normal aa, so 50% children have chance to be normal aa and 50% - sick with genotype Aa,
the correct answer is E.


  #3

shocked no, is not E


  #4

A ?


  #5

I am still thinking this is Ecool


  #6

is not E,
E is only if they plan to have 1 kid and they ask probability that the child will be affected

probability that all 3 kids will be affected is 1/8, so is not A


  #7

In a question they asked "What is the probability that one of the three children will be affected? "




  #8

my first answer was 1/2 but because u said it is wrong i gave it more shoot
so if u mean only one child is affected it would be 3/8 then option D. 1/3 would be correct !!!!!!

only if i am correct read this explaination
becuse it is autosomal dominat and the risk in one child is 50%
treat is as coin flip
then u can do it in two ways
>first long way is calculate the risk of only two children by 2X2 table and u will get four option (25% each)
then add the risk of another child by 4x2 table
and you will get eight results (12.5% each)
only three of the eight results will have the content of one affected child and two normal others

> second one
imagine if the asked about the risk of the first child is affected and the next two is normal
u can calculate this as 50% x 25% = 12.5%
(50% is the chance the first child is affected
25% is the chance that the 2nd child and 3rd child are normal )
then
back to the real question : they asked about if the 1st or 2nd or 3rd
and because these risks are exclusive in big single event u should use " +" for the word " or"
so all the risks would be
12.5% + 12.5% + 12.5% = 3/8 or approx 1/3


Edited by madoo on Sep 13, 2009 - 9:52 PM

  #9

thank you madoo, yes answer is 3/8


probabilities that one of the three children is affected are:


mad confused confused
confused mad confused
confused confused mad

so there are three different orders of birth

abnormal normal normal has a probab of 1/2 x 1/2 x 1/2 = 1/8
normal abnormal normal has a probab of 1/2 x1/2 x1/2 = 1/8
normal normal abnormal has a probab of 1/2 x 1/2 x 1/2 = 1/8

So the probability of 1 abnormal and 2 normal is the sum of the probabilities of these 3 possible orders

1/8 + 1/8 + 1/8 = 3/8






  #10

But 3/8 is not one of the options in the answers?


  #11

Yes! smiling face
Avir , where did you get this test and did it come with an answer?


  #12

I am thinking of A as the answer still.

What is the correct answer?


  #13

answer is 3/8

i also did this question but they asked probability of having 3 girls or 3 boys , so basically is the same concept (as avir and madoo said, are mutually exclusive events so you have to ADD probabilities)

(so probab of having 3 girls or 3 boys is 2/8)


  #14

usmdee wrote:
answer is 3/8

i also did this question but they asked probability of having 3 girls or 3 boys , so basically is the same concept (as avir and madoo said, are mutually exclusive events so you have to ADD probabilities)

(so probab of having 3 girls or 3 boys is 2/8)


yup
having 3 girls probability is 1/8

probability of "this or that" is calculated by adding if they are exclusive


nice question


  #15

thanks guys

Olga, is mcq from free online step1 Qbank, nothing fancy


  #16

Now I understand, I think the right answer is 1/8.
If they plan to have 3 children - each of them has 1/2 chance to be sick or healthy so probability to have one of three children affected is 1/2*1/2*1/2=1/8

Interesting question.


  #17

OlgaG wrote:
Now I understand, I think the right answer is 1/8.
If they plan to have 3 children - each of them has 1/2 chance to be sick or healthy so probability to have one of three children affected is 1/2*1/2*1/2=1/8

Interesting question.


nop answer is 3/8

1/8 is risk of 3 affected children


  #18

gringringringringrin

sorry guys, you made me smile...

so Olga, you cannot understand the concept

you have the explanation, the maths, the correct answer, is all above

ok, i will type again

correct answer is 3/8

now please answer (to yourself) what is the probability for two of three children to be affected, this way you will understand




  #19

I understand your math and concept but!
I am sorry there is no answer 3/8, and 1/3 is not really equal 3/8
1/3=0.33
3/8=0.375

They make math in this tests not so complicated and we don't need to try answer 3/8 make equal 1/3

What i understood is that your tests didn't come with a real explanation and here we trying to find it Right?




  #20

Avir never posted a Qs without explanation, and he explained very well

is free online Qbank for step1, so you can find the Qs and the explanation

it doesn't matter what the options are, u have to pick the one who approximates better 3/8, in this case is option D





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