A colour blind male $(X^cY)$ marries a carrier female $(X^CX^c)$. Possible genotype of daughters will be______?
a. $X^cX^c$ only
b. $X^CX^C$ only
c. $X^CX^C$ and $X^cX^c$
d. $X^CX^c$ and $X^cX^c$

Hint: Colour blind is defined as the inability to see colour. It is a type of $X$ linked or sex-linked recessive disease where when both alleles have carrier recessive $X$ and a person gets affected by the disease if the person has the homozygous recessive alleles together in the same cell. It is typically referred to as congenital colour vision disorders.

Complete answer:
According to Mendel’s law of inheritance with together of the law of segregation, the law of independent assortment the combination of the alleles will take place $X^cY$ (man) mate with $X^CX^c$ (female) then the outcome of girl genotype will be Four alleles will be present:
$X^c, Y, X^C, X^c$
Possible outcomes will be :
$X^cX^C, X^cX^c, YX^C, YX^c$
From these combination daughter genotype can be :
$X^cX^C, X^cX^c$

Hence there are 50-50 chances that the child daughter can be colour blind as for being colour blind genotype present should be $X^cX^c$.
$X^CX^c$ daughter genotype will be the carrier of the disease, she can transfer the disease but will not get herself affected by the disease. Probability for a daughter child is 2:1.

Hence, the correct answer is option (D).

Note: Colour blindness is a disease where a person able to see colour is decreased, he/she won't be able to differentiate between the colour. A person with 100 percent colour-blindness is known as achromatopsia. The diagnostic method used for testing the disease is known as the Ishihara colour test. It is a common problem that affects around 1 in 12 men and 1 in 200 women.