For the following crosses, determine as accurately as po...

Question

For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-clotting disorder, are X-linked recessive traits. In each case, assume the parents have normal karyotypes (see Table 10.2).

A color-blind man and a woman who is wild type have a daughter with Turner syndrome (XO) who has normal color vision and blood clotting.

Accepted Answer

Hello, everyone and welcome to today's video. So which of the following is the most likely genotype of the color blind father in a family where a woman who is wild type homozygous and a man with color blindness have a daughter with Turner syndrome who has normal color vision and blood clotting. So as answer choice A, we have wall type X Y as answer choice B, we have mutant X Y as answer choice C, we have wall type X, mutant B or mutant X. And then as answer choice D we have both mutant X chromosomes. Now, in order to solve this problem, we first need to understand that the woman is going to be wild type homozygous for color blindness. So her chromosomes are going to look like this. We're told that the man has color blindness and this is affecting the X chromosome. And therefore this is going to be his genotype because it's going to be different from the women's and it's going to be affected by color blindness on its only X chromosome, which is going to be here. So this recessive or this mutant genotype is going to give rise to the color blindness. And this is represented by answer choice, the Mutant X and Y. I really hope this video helped you and I hope to see you on the next one.

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