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A couple has a daughter with Turner syndrome, a condition in which only a single copy...

A couple has a daughter with Turner syndrome, a condition in which only a single copy of the X chromosome is present. This results from nondisjunction, the failure of the X chromosome to segregate properly during meiosis.

During which meiotic division, and in which parent, could nondisjunction have occurred to produce a child with this condition?

Meiosis I in the mother
Meiosis II in the mother
Meiosis I in the father
Meiosis II in the father
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Answer #1

Daughter with Turner syndrome have only a single copy of X chromosome, 45 X. Generally daugher have a combination of XX, One X chromosome from mother and other from father. This condition arises when a daughter inherits nothing from father. This situation arises due to Nondisjunction (improper segregation of sex chromosomes) in father. So if father sex chromosomes undergone nondisjunction it will produce child with Turner syndrome. This
will take place in Meiosis I in father at the time of spermatogenesis. It should be noted that in this situation egg gets fertilized with sperm missing sex chromosome inside it. (due to nondisjunction)

Nondisjunction of sex chromosomes during spermatogenesis - 1st meiotic division XY nondisjunction XY XY XY fertilization +X X

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