Nondisjunction question--Turner's syndrome and colorblindness

[fluff]

Hi,

This is a question from EK Bio lecture 2 (one of the 30min passage questions in the back of the book).

Turner's syndrome occurs due to nondisjunction at the sex chromosome resulting in an individual with one X and no Y chromosome. Color-blindness is a sex-linked recessive trait. A color-blind man marries a healthy woman. They have two children both with Turner's syndrome. One of the children is color-blind. Which one of the following is true?

A. Nondisjunction occurred in the father for both children.
B. Nondisjunction occurred in the mother for both children.
C. Nondisjunction occurred in the mother for the color-blind child and in the father for the child with normal vision.
D. Nondisjunction occurred in the father for the color-blind child and in the mother for the child with normal vision.

C is the correct answer.


I understand that C is the right answer if the genotype of the mother were big-A big-A. However, a "healthy woman" does not necessary mean big-A big-A, correct? Could be big-A little-a, right?

If the mother were big-A little-a, then non-disjunction could have occurred in either the mother or the father for the child with color-blindness and the father for the child with normal vision...ya?

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[Akarat]

Hi,

This is a question from EK Bio lecture 2 (one of the 30min passage questions in the back of the book).

Turner's syndrome occurs due to nondisjunction at the sex chromosome resulting in an individual with one X and no Y chromosome. Color-blindness is a sex-linked recessive trait. A color-blind man marries a healthy woman. They have two children both with Turner's syndrome. One of the children is color-blind. Which one of the following is true?

A. Nondisjunction occurred in the father for both children.
B. Nondisjunction occurred in the mother for both children.
C. Nondisjunction occurred in the mother for the color-blind child and in the father for the child with normal vision.
D. Nondisjunction occurred in the father for the color-blind child and in the mother for the child with normal vision.

C is the correct answer.


I understand that C is the right answer if the genotype of the mother were big-A big-A. However, a "healthy woman" does not necessary mean big-A big-A, correct? Could be big-A little-a, right?

If the mother were big-A little-a, then non-disjunction could have occurred in either the mother or the father for the child with color-blindness and the father for the child with normal vision...ya?

If nondisjunction occurred in the mother, then yes the child could still be color-blind because the father passed on the trait for color-blindness. However, it would then not be considered Turner's Syndrome. For this case, you would have a color-blind girl with the color-blind trait coming from the father's X chromosome.

In regards to the "healthy woman" issue, I guess they mean healthy to be absolutely nothing wrong. A carrier doesn't feel any negative effects, but I guess because there is still technically something "wrong" with one of her chromosomes, she is considered to be not healthy.

 

[Econ2MD]

Yes, the mother could be a carrier for colorblindness (i.e. Aa) Nondisjunction in the mother would mean either 0 or xx from mom, x from dad (to end up with x0).
Only possible way to get xo (Turner's) is 0 from mom and x from dad = colorblind child

If nondisjunction in dad, dad gives an xy or an 0. mom gives an x (A or a). So dad gives an o and mom gives an x to get Turner's.
In this case, you could get a normal or colorblind child.

I think the correct answer isn't shown. You're right, unless we're both wrong 🙂

 

[BennieBlanco]

Hi,

This is a question from EK Bio lecture 2 (one of the 30min passage questions in the back of the book).

Turner's syndrome occurs due to nondisjunction at the sex chromosome resulting in an individual with one X and no Y chromosome. Color-blindness is a sex-linked recessive trait. A color-blind man marries a healthy woman. They have two children both with Turner's syndrome. One of the children is color-blind. Which one of the following is true?

A. Nondisjunction occurred in the father for both children.
B. Nondisjunction occurred in the mother for both children.
C. Nondisjunction occurred in the mother for the color-blind child and in the father for the child with normal vision.
D. Nondisjunction occurred in the father for the color-blind child and in the mother for the child with normal vision.

C is the correct answer.


I understand that C is the right answer if the genotype of the mother were big-A big-A. However, a "healthy woman" does not necessary mean big-A big-A, correct? Could be big-A little-a, right?

If the mother were big-A little-a, then non-disjunction could have occurred in either the mother or the father for the child with color-blindness and the father for the child with normal vision...ya?

actually guys this isn't correct.

If the mother were healthy, as you stated she could be Aa or AA. Lets say she was Aa. Now non-disjunction occurs and she has offspring with turner's syndrome (i.e. only 1 X chromosome) then there is no possible way non-disjunction occurred in the mother.

If it did, she either would give her offspring 2 X chromosomes or none (referring specifically to the non-disjunction eggs). If they have 2 then they do NOT have turner's.

Therefore this is not possible:

then non-disjunction could have occurred in either the mother or the father for the child with color-blindness

I'm not so sure how well offspring survives with 3 X's, I forgot.

 

[Econ2MD]

actually guys this isn't correct.

If the mother were healthy, as you stated she could be Aa or AA. Lets say she was Aa. Now non-disjunction occurs and she has offspring with turner's syndrome (i.e. only 1 X chromosome) then there is no possible way non-disjunction occurred in the mother.

If it did, she either would give her offspring 2 X chromosomes or none (referring specifically to the non-disjunction eggs). If they have 2 then they do NOT have turner's.

Therefore this is not possible:

C is the only answer that is close

I'm not so sure how well offspring survives with 3 X's, I forgot.

Bennie,
You're right that if nondisjunction occurs in the mother that the x comes from the father and the mother can only contribute an 0. Father is colorblind, so Turner's child is also.

BUT...If nondisjunction occurs in the father, mother contributes the x (which in a mother heterozygous for colorblindness can result in either a normal or colorblind Turner's child)

i.e. A colorblind Turner's child can result from nondisjunction in either the father or the mother. A normal Turner's child can only result from nondisjunction in the father.

The mother can provide a 0 or xx and both are nondisjunction. Dad can contribute and x or y. mom 0, dad x = Turner's child

And I don't think 3 x's is survivable, but XXY is a Kleinfelter male.

 

[engineeredout]

I'm not so sure how well offspring survives with 3 X's, I forgot.

And I don't think 3 x's is survivable, but XXY is a Kleinfelter male.

XXX is triple X syndrome and is not only survivable but also pretty asymptomatic. IIRC there are XXXX and even XXXXX (Though with more Xs problems begin to develop).

 

[Econ2MD]

XXX is triple X syndrome and is not only survivable but also pretty asymptomatic. IIRC there are XXXX and even XXXXX (Though with more Xs problems begin to develop).

Thanks for the correction.