A woman with normal colorvision has a daughter with a colorblind man. The woman's father is colorblind and the man's mother is colorblind.

What is the probability that their  daughter is colorblind? ___________

What is the probability that their daughter will have colorvision but be a carrier? ___________

What is the probability that their daughter will have colorvision and not be a carrier for colorblindness?

		50% ; 50% ; 0%
		75% ; 50% ; 25%
		25% ; 0% ; 50%
		0% ; 50% ; 50%

Answer:

Colorblindness is X linked recessive disorder. So for female to be colour blind both x chromosome should be defective and for male to be colour blind the x chromosome should be defective.

According to the question, woman's father is colour blind so the genotype of him is X^c Y. Therefore genotype of woman is X X^c ( because in question it is mentioned that woman has normal colour vision so one chromosome must be dominant).

Man's mother is colour blind so her genotype is X^c X^c ??????. Therefore genotype of men is X^c Y

Now the offspring may be XX^c ??????, XY, X^c X^c , X^c Y.

So the daughter born may have genotype X X^c ?????? or X^c X^c.

The probability that the daughter born will be colour blind is 50%.

The probability that the daughter born will have normal vision but a carrier is 50% .

The probability that their daughter will have colour vision and not be a carrier for colour blindness is 0%.

So the correct answer is option A.