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MTD52

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I don't know what the deal is but I don't see how they're getting to this answer...

Question:
A women carries a sex-linked lethal gene that causes spontaneous abortions. She has six children. How many of her children would you expect to be boys?

Answer:
Two


I'm at a loss here. Let (a) = the gene for abortion. If you do the cross of a carrier female, XaX, with a normal male, XY, you get:

XaX, XX, XaY, XY. So 50% are aborted. I don't know where to go from here.
 
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The gene is lethal, yet we have a mother that has it. Hence, the mother must be a carrier (and not a homozygous recessive). So, the mother's genotype is Aa. Since we have no info about the father, we can assume that he has the wildtype genotype AY. Now, draw a punnet square to calculate all the possible outcomes:
 
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dentrilla

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If 50% are males, and 50% of those males are not viable isn't it more like 1.5 males being rounded up --> 2? There is no reason that there should be more AY then aY
 

MTD52

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Joonkim - It's a sex linked recessive lethal gene. The fact that it is sex linked means that it is carried on the X chromosome. It is impossible for a male to have it and be living since he only has one X chromosome. Make sense?

Anyway, I don't like either of your approaches. Nze I think if you do it that way you're making an assumption that only 1 male has it. There's a 50% chance that a male will get it, so out of 3 boys, 2 could have it also.

The way I did it was this:

We already know that the mother is XaX, and the father must be XY (as described above). Each time they have children, it will come out as XaX, XX, XaY, XY. That means the XaY male will die, leaving 3 kids, 1 of them being a boy.

When she has 3 more kids, you have the same scenario. 1 boy + 1 boy = 2 boys total.
 
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Joonkim - It's a sex linked recessive lethal gene. The fact that it is sex linked means that it is carried on the X chromosome. It is impossible for a male to have it and be living since he only has one X chromosome. Make sense?

Anyway, I don't like either of your approaches. Nze I think if you do it that way you're making an assumption that only 1 male has it. There's a 50% chance that a male will get it, so out of 3 boys, 2 could have it also.

The way I did it was this:

We already know that the mother is XaX, and the father must be XY (as described above). Each time they have children, it will come out as XaX, XX, XaY, XY. That means the XaY male will die, leaving 3 kids, 1 of them being a boy.

When she has 3 more kids, you have the same scenario. 1 boy + 1 boy = 2 boys total.

Yea! Your approach seems more reasonable. I love punnet squares so I always use them to solve these problems, and most of the times it gives me the right answer. I'm not sure if the punnet square is the right way to go with this problem, but it does give the correct answer. Anyways, I like your explanation better.
 
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the questio really is how many of her children will be boys? the lethal gene thing complicates the question. the probability of having a son is just the same with or without the lethal gene.
 
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The gene is lethal, yet we have a mother that has it. Hence, the mother must be a carrier (and not a homozygous recessive). So, the mother's genotype is Aa. Since we have no info about the father, we can assume that he has the wildtype genotype AY. Now, draw a punnet square to calculate all the possible outcomes:
how come the aY son is not viable. the gene is supposedly affects childbirth. although he will manifest the condition because it will no way affect him because the "problem" is abortion. the son will not conceive in the first place :oops:
 

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how come the aY son is not viable. the gene is supposedly affects childbirth. although he will manifest the condition because it will no way affect him because the "problem" is abortion. the son will not conceive in the first place :oops:

The question doesn't ask how likely it would be that she has a male zygote. It asks of her children that exist currently, how many are boys?. The children that had the gene did not live.
 
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Then what happens if there are odd number of children but same situation?

The question would have to give you some multiple of 3 to give the relevant ratios of children. 1/3 boys, 2/3 girls. 2 boys out of 6, 3 boys out of 9, etc. Anything in between would be kinda...meh.
 
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The question would have to give you some multiple of 3 to give the relevant ratios of children. 1/3 boys, 2/3 girls. 2 boys out of 6, 3 boys out of 9, etc. Anything in between would be kinda...meh.


I mean let's say if she has 7 children. How many of her children would you expect to be boys?
 
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I mean let's say if she has 7 children. How many of her children would you expect to be boys?

That was my point. By probability ratios, you'd expect 2.33 boys. Obviously you can't have a third of a child, so the question isn't going to give you an oddball number like that. The best answer for 7 children would be 2 or 3 boys so unless that's an answer choice, don't expect to see it as a real question.
 
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