Question

Red–green color blindness is a human X-linked recessive disorder. A young man with a 47, XXY karyotype (Klinefelter syndrome) is color blind. His 46,XY brother also is color blind. Both parents have normal color vision. Where did the nondisjunction that gave rise to the young man with Klinefelter syndrome take place? (Assume that no crossing over took place in prophase I of meiosis.)

Answer 1

Answer:

At meiosis II in the mother

Explanation:

Both the parents have normal vision but both the sons are colorblind. Since colorblindness is X linked recessive disorder, the sons have obtained the allele for colorblindness from mother. This makes the mother carrier for colorblindness. The genotype of the mother is X^cX. The young man with Klinefelter syndrome is colorblind which means that he is homozygous recessive for the allele of colorblindness. His genotype is X^cX^cY. Since this man has obtained two copies of the allele of colorblindness, the mother must have experienced nondisjunction at meiosis II during gamete formation.

Anaphase-II of meiosis II separates and segregates the sister chromatids (now called daughter chromosomes) to opposite poles. Failure of segregation of two copies of X^c chromosome of mother to opposite poles and their distribution to the same egg cell resulted in the formation of an egg with two copies of X^c chromosome. Fertilization of this egg with a sperm carrying "Y" chromosome as sex chromosome resulted in a zygote with X^cX^cY that developed into the man with Klinefelter syndrome and colorblindness.