[doctorF]

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

 

[gabra]

Every time they will have a child they have 50 percent chance to have an affected one.as it is an autosomal dominant pattern.
simple Mendelian inheritance
so the answer is E

 

[ath.pantelis]

The probability for EACH child, according to Mendelian inheritance, is indeed 1/2, but the question asks what is the probability that ONE out of THREE children is affected. In other words, what is the chance that an independent ratio is repeated three times, just like when a fellow tosses a coin 3 times and the result is either heads or tails (50% = 1/2 for each).

So, the correct answer is 1/2 x 1/2 x 1/2 = 1/8 (choice A).

 

[Seetal]

yes yes A

 

[doctorF]

NO WRONG!!! think again guys

anth u r on da ryt track think a bit more.... then will i post da ans

 

[ath.pantelis]

Total recall!

Ok, let's consider it another way.

If H stands for healthy children and W stands for children affected by Waardenburg syndrome, then the possible makeups are as follows:

HHH (all children are healthy)
WHH (1 child is affected)
HWH ( >> )
HHW ( >> )
WWH (2 children are affected)
WHW ( >> )
HWW ( >> )
WWW (all 3 children are affected)

So, out of 8 possible outcomes, there are three among them satisfying the condition in question, i.e. the possibility that EXACTLY one of the three children is affected is 3/8.

What the h*** am I doing wrong again

 

[doctorF]

u r ryt ath.

probabilities that one of the three children is affected are:






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 which is approximately 1/3


ans is 1/3 D

 

[Seetal]

i dont like this question!!

so lemmie get this straight,, everytime i come a across a question like this i need to think about how many children they WANT (whether it is 2 ,3 or 4), and then each of them have a 1/2 probability (for autosomal dominant), and then times it with the probability of the chances that they will have one affected child and 2 none (in this case obviously 3 times) so 3/8 approximates 1/3.. am i making sense?

 

[doctorF]

actually this is a stupid question in one of the banks i was going thruu... so i put it here... but yes genetics questions are kind of confusing for most

 

[Seetal]

yeah

they did mention somewhere that beh.science is kinda difficult for most imgs. i bought this high-yield behavioral science book and i intend to do it many times!! im gonna get this and get it good.

 

[ath.pantelis]

Although HY Behavioral Science is a very concise, well-organized book, it is not enough for the USMLE, at least not for Step 1. I don't say that I studied sth more than Kaplan, FA & HY, as well as the high yield topics found in UWorld, but the questions that appeared in the actual exam were kinda different, out of the ordinary (as usual)!!!

 

[Seetal]

thank u ath.pantelis

well in that case ill master the material that i have and pretend like im a GP when i answer beh.sci questions. ...that way ill have to apply logic based on what ive read of course.

 

[ath.pantelis]

That's right! And it's true for EVERY USMLE question... When every piece knowledge seems to be non-applicable or even deleted, then the last resort is your logic and your powerful clinical sensorium!