Question

A colorblind man marries the daughter of another colorblind man whose wife had a normal genotype for color vision. In their progeny-
A) All the children would colorblind.
B) All their sons are colorblind.
C) None of the daughters would be colorblind.
D) Half of their sons and half of their daughters would be colorblind.

Answer 1

Hint: Colorblindness is a genetic disease in which a person has a reduced ability to distinguish between certain types of colors. Like between green and red. It is a very common and self-diagnosable condition. There are three types of color blindness -red - green color blindness, blue - yellow color blindness, total color blindness.

Complete answer:
Color blindness is a genetic condition caused by a difference in how one or more of the light sensitive cells found in the retina of the eye responds to certain colors. These cells are called rods and cones. People with color blindness are able to see things as other people but they are unable to see fully red, green or blue light. It affects the day today activities, like choosing clothes, seeing traffic lights and selecting ripe fruits. Color blindness is a recessive sex - linked disorder. The genotype of the woman will be \[{X^N}{X^n}\] as she is the daughter of color-blind man.
Option A) All the children would colorblind. This is not possible as the woman is the carrier. The woman married the man who is colorblind so the genotype of man will be\[{X^n}Y\].
So, option A is incorrect
Option B) All their sons are colorblind. This option is also possible as we know that in this case the woman is the carrier so the dominant allele is of the man who is colorblind. So, the ratio of sons getting colorblind is\[1\].
Option B is in correct
Option C) None of the daughters would be colorblind. As we discussed, the woman here is the carrier so ultimately no daughters can be fully disease free. They can be carriers if not diagnosed with the disease as one of the Parents or the \[X\] gene is carrier.
So, the option is incorrect.
Option D) Half of their sons and half of their daughters would be colorblind-
The offspring combination will be as follows -\[{X^N}{X^n}\] - one normal (carrier) daughter because there is one dominant and one recessive allele, so the girl is a carrier. \[{X^N}Y\]- one normal son because there is a dominant allele. \[{X^n}{X^n}\]- one color blind daughter because there are two recessive alleles so the girl is colorblind. \[{X^n}Y\]- one color blind son as there is a recessive allele which is \[X\]- linked. Hence as from the above result it is concluded that half of the sons and half of daughters have the probability to be colorblind.

So, the correct answer is option D.

Note:
Color blindness affects more male than females as we know that male have the dominant gene called \[Y\] gene.
Females either can be carrier or colorblind but chances are less.
Colorblindness is a sex linked disorder.