Question

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? Hint: It may be helpful to sketch a diagram depicting the outcome of nondisjunction in meiosis I and II of both parents to solve this problem.

Answer 1

Nondisjunction can occur during either meiosis I or II. In the case of Turner syndrome, where only a single copy of the X chromosome is present, the nondisjunction likely occurred in the mother during meiosis I.

Step-by-step explanation:

Nondisjunction can occur during either meiosis I or II. If homologous chromosomes fail to separate during meiosis I, the result is two gametes that lack that particular chromosome and two gametes with two copies of the chromosome. If sister chromatids fail to separate during meiosis II, the result is one gamete that lacks that chromosome, two normal gametes with one copy of the chromosome, and one gamete with two copies of the chromosome.

In the case of a couple having a daughter with Turner syndrome, where only a single copy of the X chromosome is present, the nondisjunction likely occurred in the mother during meiosis I. This resulted in one normal gamete with one copy of the X chromosome, one gamete lacking the X chromosome, one gamete with two copies of the X chromosome, and one normal gamete with one copy of the X chromosome. Fertilization of the gamete lacking the X chromosome with the gamete with two copies of the X chromosome resulted in a child with Turner syndrome, where only a single copy of the X chromosome is present.