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A color-blind woman with Turner syndrome (XO) has a father who is color blind. Given that the

gene for the color-blind condition is recessive and X-linked, provide a likely explanation for the
origin of the color-blind and cytogenetic conditions in the woman.
A) nondisjunction in the mother in meiosis I and in the father in meiosis II
B) nondisjunction in the father at meiosis I
C) nondisjunction in the other in meiosis I or II
D) nondisjunction in the father in meiosis I or II

User M Hadadi
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Final answer:

The color-blind woman with Turner syndrome must have inherited the color blind allele from her color blind father and the Turner syndrome resulted from nondisjunction in her mother during meiosis.

Step-by-step explanation:

In the case of the color-blind woman with Turner syndrome (XO), her condition can be explained by two separate events. Her color blindness is explained by X-linked recessive inheritance.

Since colorblindness is recessive, she would have to inherit two alleles for the condition to express it. However, because Turner syndrome is characterized by having only one X chromosome (XO instead of XX), the presence of Turner syndrome likely resulted from nondisjunction during her mother's meiosis, where one of her gametes did not receive any of the X chromosomes.

Her father being color blind indicates that his X chromosome carries the allele for colorblindness. Since her father can only pass his X chromosome to his daughters, she must have inherited the colorblind allele from him. Hence, the likely explanation for her conditions is option B) nondisjunction in the father at meiosis I for the colorblind gene inheritance and option C) nondisjunction in the mother in meiosis I or II for Turner syndrome. The father's nondisjunction would not result in Turner syndrome, as he only carries one X chromosome and any failure to separate would result in an XY or an OY gamete, which are not viable or do not result in Turner syndrome.

User Miyasudokoro
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