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  #1  
Old 04-24-2010
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Genetics Lethal X Linked Recessive Disorder?

A normal woman is a carrier of an X linked recessive disorder that is lethal in utero (spontaneously aborted). Which of the following is the expected ratio of her viable female to male children?

A- 1:1
B- 1:2
C- 2:1
D- 2:3
E- 3:2
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  #2  
Old 04-24-2010
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i would think c. 2:1

cause 50 percent of sons will be aborted

and 100 percent of her daughters will survive.
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Old 04-24-2010
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Default B

For some reason when i worked this out i got 1:2

we don't know what the paternal genome is. He could be Xay or X (A)y (so 1/2 chance)

As for the daughters since they get one X from mom and one X from dad there is a 1/2 chance that one daughter will be homozygous recessive or Heterzygous, while the other will for sure be Heterzygous so she'll survive.

If we keep in mind that they're only asking about viable ones
so : females (1/2)(1/2)= 1/4
males (1/2)

so ratio is 1:2

if my thinking is wrong, I would appreciate it if someone corrected me.
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Old 04-24-2010
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i think the answer should be (c)-2:1
as housewannable said, i don't think its necessary to think abt paternal genome coz the disease is x-linked recessive and when it expresses its lethal,so there would be no male having that disease...so i think male would normal XY in every case.
i might be wrong so would like to hav ur reply.
thanx
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Old 04-24-2010
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Thumbs Up Correct answer is C

drgroverahul and convalescence are correct.

We don't need to care about the father because if he was to have the abnormal X then he'd die in utero and doesn't live to mate with our case

Now let's take the mother
-If she gives a normal X chromosome then there's 50% chance of making a normal XX female (viable) and 50% chance of making a normal XY male (viable).
-If she gives her abnormal X' chromosome then there's 50% chance of making a carrier X'X female (viable) and 50% chance of making a lethal X'Y male (non viable).
So out of four possibilities we have 2 viable females and one viable male and one non viable male and therefore the viable females to males ratio is 2:1
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  #6  
Old 04-25-2010
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Quote:
Originally Posted by mary View Post
drgroverahul and convalescence are correct.

We don't need to care about the father because if he was to have the abnormal X then he'd in utero and doesn't live to mate with our case.
Thank you that's what i forgot to factor in
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Old 04-26-2010
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Quote:
Originally Posted by mary View Post
drgroverahul and convalescence are correct.

We don't need to care about the father because if he was to have the abnormal X then he'd die in utero and doesn't live to mate with our case

Now let's take the mother
-If she gives a normal X chromosome then there's 50% chance of making a normal XX female (viable) and 50% chance of making a normal XY male (viable).
-If she gives her abnormal X' chromosome then there's 50% chance of making a carrier X'X female (viable) and 50% chance of making a lethal X'Y male (non viable).
So out of four possibilities we have 2 viable females and one viable male and one non viable male and therefore the viable females to males ratio is 2:1
Excellent explanation.Thanks
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