Question
Asked Sep 16, 2019
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A woman who is homozygous for normal color vision mates with a colorblind male. they produce a female child how has only one X chromosome and who is also color blind. In which parent did the nondisjuction take place? Can you specify which meiotic division?

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Expert Answer

Step 1

Color blindness is an X-linked recessive genetic disorder. In this disease, the person has less ability to see respective colors and the difference between the colors. Apart from genetic variations, color blindness can also occur due to any physical or chemical damage to the optic nerve of the eye. Males are more prone to color-blindness than female, while females mostly act as a carrier of this disease.

Step 2

Given-

The female is homozygous on-colorblind. Hence, her genotype is XX.

The male has color blindness. Hence, his genotype is XcY.

When the male and the female mate, then they produce a girl with only one X chromosome, and moreover, she is colorblind in nature. Therefore, her genotype is Xc.

The above mating is an example of non-disjunction. The failure of separation of homozygous chromosomes or sister chromatids during the cell division is termed as nondisjunction. There are three types of nondisjunction, that occur in meiosis I, meiosis II, or mitosis. Further, aneuploidy, presence of abnormal chromosome number, occurs due to...

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