Two brothers have X‑linked red–green colorblind vision, and their parents have normal color vision. The first brother's karyotyp
e is 47,XXY (Klinefelter syndrome), and the second brother's karyotype is 46,XY. In which parent and in what cell division phase did the first brother's chromosomal nondisjunction occur? Assume no recombination. (A) in the father in meiosis I
(B) in either parent in somatic cell mitosis
(C) in the father in meiosis II
(D) in the mother in meiosis II
(E) in the mother in meiosis I
The correct answer is option C, that is, in the mother in meiosis II.
Explanation:
A nondisjunction in meiosis I produce all the aneuploid gametes, while nondisjunction at meiosis II produces 50 percent aneuploid gametes and the 50 percent normal gametes comprising haploid number of chromosomes.
As it is mentioned that one of the brothers possess normal number of chromosomes (46), the mother encounters nondisjunction at meiosis II resulting in the formation of one n+1 gamete exhibiting two copies of X chromosome. Fertilization of this n+1 gamete with sperm results in the creation of zygote (47, XXY).
Nondisjunction is the failure of homologous or sister chromatids to separate properly during cell division .There are 3 forms of non junctions.1)failure of a pair of homologous chromosome to separate in meisis 1
