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UW genetics q

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lakshya_0_7
Forum Elite

Topics: 19
Posts: 240

04/08/08 - 02:32 AM

If the mother and father are heterozygous carriers for Cystic fibrosis, and they have an affected son, what is the probability that the other daughter will be a carrier?

1. 2/3

2. 1/4

3.1/2

i have confusion with this question. This is not the exact question on UW. Its one part of the question...where they showed a pedigree with 3 generations of CF and asked to calculate the probability in the 3rd generation...i cant upload the complete q here sorry! if anyone remembers doing this question and can post the q here it would be very helpful.....

Tiff
Forum Guru

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04/08/08 - 12:27 PM

The probability of the daughter being a carrier is 1/2.

The probability of the daughter having the disease is the same for the son. 1/4

lakshya_0_7
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04/08/08 - 01:46 PM

well tiff....i thought the same....but UW says its 2/3....

Tiff
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04/08/08 - 05:09 PM

I'll try to find the question and see what I can make of it.

Tiff
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04/08/08 - 05:50 PM

I think I found the question you're referring to. A couple wants to have a baby and dad has the disease and they want to know if the child will inherit it. The pedigree suggests that the mother may or may not be a carrier. She's unshaded on the pedigree, but has a siblng with the disease. That means her parents were both heterozygotes. Because mom was unshaded we know she didn't have the disease. Therefore there was no possibility of her having homozygous recessive genes. Hence, she was either a heterozygous carrier or a homozygous dominant(no disease/non carrier). So the possibility of her being a carrier would be 2/3 (instead of 2/4, because we're excluding the homozygous recessive possibility). Now, the mother has a 2/3 chance of being a carrier and the dad has the homozygous recessive disease. Out of 12 posssible outcomes, there's only 4 where their kid would be a homozygous recessive. Therefore the possiblity of the couple having a child with the disease would be 1/3.

akm
Forum Newbie
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Topics: 5
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04/08/08 - 07:42 PM

This is an easy way of looking at it:

with two heterozygote parents, the offspring will be :
1/4- affected, 1/4- normal, 2/4- carriers

since one child (the brother) is affected,...one can count that out.....

out of the remaining 3 kids 1 will be normal and 2 will be carriers

thus, the carrier chance is 2/3.....and being not affected is 1/3

Tiff
Forum Guru

Topics: 50
Posts: 505

04/08/08 - 08:20 PM

akm wrote:
This is an easy way of looking at it:

with two heterozygote parents, the offspring will be :
1/4- affected, 1/4- normal, 2/4- carriers

since one child (the brother) is affected,...one can count that out.....

out of the remaining 3 kids 1 will be normal and 2 will be carriers

thus, the carrier chance is 2/3.....and being not affected is 1/3

No. No offense, but you can't do probability like that. Just because one brother is affected doesn't mean you can just count him out. Probability is applied to each offspring. In this case, seeing the pedigree chart is essential because if I'm right with referring to the particular question I mentioned above (which is what I think the original poster was referring to), it shows that the person already doesn't have the disease. If we didn't know anything about this person, we'd have to assume it would be 1/2. But since the chart suggests that she's non-diseased, that would make it 2/3.

lakshya_0_7
Forum Elite

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04/09/08 - 06:07 AM

excellent Tiff ..i got it now....thanks smiling face

sigh
Forum Senior

Topics: 15
Posts: 152

04/09/08 - 06:12 AM

Tiff you are right! I remember this question. The daughter does not have the disease and we can eliminate the last quarter of affected homozygote which will leave us with 2/3 as carrier and 1/3 as not affected homozygote.

akm
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04/09/08 - 06:41 AM

#10

Yeah i agree......disregard my last post...

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