X-linked question

[arlo]

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I had this question in Kaplan FL and I have a question on the terminology.

If a woman who is a carrier for hemophilia A marries a man who does not have the disease, what is the probability their first daughter will be a carrier?

I know hemophilia is X-linked and I know mom is XH Xh and dad is XH Y which gives XH XH, XhXH, XH Y, and Xh Y. I got this wrong because I did 1/2 (chance of child being female) x 1/2 (chance of female being Xh XH), but the answer is just 50%. I'm wondering the terminology for when I would have to factor in the chance of the first child being a certain sex for these types of questions.

 

[aldol16]

Yeah, this is an excellent question. Kaplan tends to trick you on the wording whereas the MCAT usually doesn't do that. So the wording here is the probability that the first daughter will be a carrier. In other words, it's the probability that a child will be a carrier, given that she is female. It's an application of Bayesian statistics, formally speaking. What you calculated was the probability of a child being female AND a carrier, which doesn't take into account known information.

This is surprisingly difficult for many students to realize. Initially, the chances of a child being female and a carrier is 25%. But once you're given more information - that is, that the first child is a daughter - then that becomes certain and it becomes 1*0.5 = 0.5. If you're interested in this sort of thing, you can look up the Monty Hall problem.

 

[To be MD]

I had this question in Kaplan FL and I have a question on the terminology.

If a woman who is a carrier for hemophilia A marries a man who does not have the disease, what is the probability their first daughter will be a carrier?

I know hemophilia is X-linked and I know mom is XH Xh and dad is XH Y which gives XH XH, XhXH, XH Y, and Xh Y. I got this wrong because I did 1/2 (chance of child being female) x 1/2 (chance of female being Xh XH), but the answer is just 50%. I'm wondering the terminology for when I would have to factor in the chance of the first child being a certain sex for these types of questions.

^ above poster answered your Q, but I'll try breaking it down a little more. Find the probabilities based off the information you are given.

Problem 1: What is the probability that the first child will be a female carrier?
- Here you have to calculate the probability of (1) being female x of (2) a carrier.

Problem 2: What is the probability that the first daughter will be a carrier?
- Here you already know the daughter is female, so it is just the probability of a XX offspring being a carrier.

Problem 3: What is the probability that the first son will be a carrier?
- Now you know you have a son, so it's the probability of an XY offspring being a carrier.

 

[arlo]

Thanks for breaking that down. That was really helpful.

 

[NextStepTutor_2]

I had this question in Kaplan FL and I have a question on the terminology.

If a woman who is a carrier for hemophilia A marries a man who does not have the disease, what is the probability their first daughter will be a carrier?

I know hemophilia is X-linked and I know mom is XH Xh and dad is XH Y which gives XH XH, XhXH, XH Y, and Xh Y. I got this wrong because I did 1/2 (chance of child being female) x 1/2 (chance of female being Xh XH), but the answer is just 50%. I'm wondering the terminology for when I would have to factor in the chance of the first child being a certain sex for these types of questions.

Hi @arlo ! This is actually a great question (though tricky) as it forces you to avoid just mindlessly plugging and chugging on the MCAT. The "trick" is meant to train you to pay attention and think about the information you have been given before you go jumping into an answer. The AAMC does and has made questions like this, I know this because I used to fall for them when I studied for the MCAT. Then I saw a few like them when I took both the old and the new MCAT.

The way the Q is worded does not ask what are the odds of their child being a girl and a carrier, the Q simply asks what are the odds their first daughter (regardless of what, if any births came before) will be a carrier.

Of the potential daughters, only 1 of them could be a carrier. There is no real trick here, its just making sure you can recognize and make the logical deductions necessary for the science on the MCAT.

Here is another example of similar thinking:

Tay-Sachs is an autosomal recessive disease. If a male carrier has a son with a female carrier, what is the probability that their son is a carrier?

What would you choose as your answer? 50%, 25% 0%?
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Take a look at the Punnet square you might make.

We have AA, Aa, Aa and aa as potential outcomes for the genotype of the son. We KNOW the son cannot be aa as the pedigree shows a phenotypically normal male. thus, of the 3 remaining choices, only 2/3 are carriers. the correct answer is 2/3 or 66%, not 50%. If you had not been given the pedigree then your 50% answer would be correct. One tweak to the information, and it has not changed the question, just the correct answer.

For all recessive diseases, you would only need to "check" people that appear phenotypically normal for carrier status, as if they are homozygous then they will express the disease and their genotype is quite obvious. Here we see that the son is normal phenotypically, so he cannot be affected. When reading MCAt questions, make sure you pause and make sure you know what it wants, as these are not typical science Qs as you would see in you undergraduate courses.


Hope his helps, good luck!

 

[arlo]

Hi @arlo ! This is actually a great question (though tricky) as it forces you to avoid just mindlessly plugging and chugging on the MCAT. The "trick" is meant to train you to pay attention and hink about the information you have been given before you go jumping into an answer. The AAMC does and has made questions like this, I know this because I used to fall for them when I studied for the MCAT. Then I saw a few like them when I took both the old and the new MCAT.

The way the Q is worded does not ask what are the odds of their child being a girl and a carrier, the Q simply asks what are the odds their first daughter (regardless of what, if any births came before) will be a carrier.

View attachment 202103

Of the potential daughters, only 1 of them could be a carrier. There is no real trick here, its just making sure you can recognize and make the logical deductions necessary for the science on the MCAT.

Here is another example of similar thinking:

Tay-Sachs is an autosomal recessive disease. If a male carrier has a son with a female carrier, what is the probability that their son is a carrier?

View attachment 202104​

What would you choose as your answer? 50%, 25% 0%?
.
.
.
.
.
.
.
.
.
.
.
.
Take a look at the Punnet square you might make.

We have AA, Aa, Aa and aa as potential outcomes for the genotype of the son. We KNOW the son cannot be aa as the pedigree shows a phenotypically normal male. thus, of the 3 remaining choices, only 2/3 are carriers. the correct answer is 2/3 or 66%, not 50%. If oyu had not been given the pedigree then your 50% answer would be correct. One tweak to the information, and it has not changed the question, just the correct answer.

For all recessive diseases, you would only need to "check" people that appear phenotypically normal for carrier status, as if they are homozygous then they will express the disease and their genotype is quite obvious. Here we see that the son is normal phenotypically, so he cannot be affected. When reading MCAt questions, make sure you pause and make sure you know what it wants, as these are not typical science Qs as you would see in you undergraduate courses.


Hope his helps, good luck!

That is a good question. I can see myself falling for the 50% just by picking Aa from the possibilities, instead of seeing that the only possibilities for the phenotypically normal son are AA and Aa.