Wererew

2+ Year Member
Jun 1, 2017
103
27
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Pre-Dental
Hey guys! I need some more explanation to get to the correct answer, 1/8.

This is what I got so far, I thought the answer was 1/2 *1/2 = 1/4 because theres a chance of hemophilia,son = 1/2 and cb,son = 1/2 from each.

The question asks (from Bootcamp) :

A color blind man without hemophilia (both X-linked traits) marries a woman who is a carrier for both traits. What is the probability they will have a son with both color blindness and hemophilia? Hemophilia and color blindness are unlinked genes.

I didn't understand their solutions that "the man must donate the Y chromosome (1/2 probability), the woman must donate Xh (1/2 probability), and she must also donate Xc (1/2 probability). We just multiply these chances together to find (1/2)(1/2)(1/2) = 1/8."
 

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May 17, 2008
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I'll try to run it down in a short summary, but I think the 'Further Details' explanation on this one goes into a lot more detail.

The man is color blind and without hemophilia. The mother is a carrier for both traits. The question wants probability of:

1. Having a son
2. The son having color blindness
3. The son having hemophilia

We have to account for all three of these outcomes. The probability of having a son is (1/2) - that would be the chance of the father passing on his Y chromosome rather than his X chromosome (otherwise they'd have a daughter).

Next is the more confusing part: we're told the mother is a carrier for two X-linked traits: hemophilia and colorblindness. Does she have color blindness on one of her X chromosomes and hemophilia on the other X chromosome? Or does she have hemophilia and colorblindness on the same X-chromosome? There's a 1/2 probability of either outcome. If they are to have a child that has both color blindness AND hemophilia, we have to assume that it's the latter - she carries both colorblindness and hemophilia on one X-chromosome - there's a 1/2 chance of that being the case. There's no other way for the son to inherit both conditions, since they're X-linked.

Finally, if we know she has both conditions on one X-chromosome, that means her other X-chromosome is unaffected. If their child is to have both conditions, it means he has to inherit the affected chromosome rather than a normal one - there's a 1/2 chance of that.

So three major probabilities we consider:
(chance of having a son) * (chance that mother carries both conditions on one X-chromosome) * (chance that the son inherits his mother's affected X-chromosome)

= 1/2 * 1/2 * 1/2
=1/8
 
Last edited:
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Wererew

2+ Year Member
Jun 1, 2017
103
27
Status
Pre-Dental
I'll try to run it down in a short summary, but I think the 'Further Details' explanation on this one goes into a lot more detail.

The man is color blind and without hemophilia. The mother is a carrier for both traits. The question wants probability of:

1. Having a son
2. The son having color blindness
3. The son having hemophilia

We have to account for all three of these outcomes. The probability of having a son is (1/2) - that would be the chance of the father passing on his Y chromosome rather than his X chromosome (otherwise they'd have a daughter).

Next is the more confusing part: we're told the mother is a carrier for two X-linked traits: hemophilia and colorblindness. Does she have color blindness on one of her X chromosomes and hemophilia on the other X chromosome? Or does she have hemophilia and colorblindness on the same X-chromosome? There's a 1/2 probability of either outcome. If they are to have a child that has both color blindness AND hemophilia, we have to assume that it's the latter - she carries both colorblindness and hemophilia on one X-chromosome - there's a 1/2 chance of that being the case. There's no other way for the son to inherit both conditions, since they're X-linked.

Finally, if we know she has both conditions on one X-chromosome, that means her other X-chromosome is unaffected. If their child is to have both conditions, it means he has to inherit the affected chromosome rather than a normal one - there's a 1/2 chance of that.

So three major probabilities we consider:
(chance of having a son) * (chance that mother carries both conditions on one X-chromosome * (chance that the son inherits his mother's affected X-chromosome)

= 1/2 * 1/2 * 1/2
=1/8
Thank you Feralis. I was doing a "random biology test" and further details weren't included with those so I missed reading that
 
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