Question

A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of the couple was a boy. This boy
A. Must have normal colour vision.
B. Will be partially colour blind since he is heterozygous for the colour blind mutant allele.
C. Must be colour blind.
D. May be colour blind or may be of normal vision.

Answer 1

Hint: Color blindness is simply defined as the inability to see or identify colours such as blue, green, or red. There are a few uncommon circumstances where a person is unable to detect or recognise any colours. A person with this disease also has trouble distinguishing between colours and tints. A colour vision problem or deficiency is another name for this syndrome.
Types of colour blindness- monochromacy and dichromacy.

Complete answer:
Option A: If the normal-visioned woman's father was colorblind, she could be a carrier, $X^C X^c$. The man will have the genotype Xc Y if he is colorblind.
The offspring: $X^C X^C$ (carrier daughter), $X^CY$ (normal son), $X^c X^c$ (colourblind daughter) and $X^c Y$ (colourblind son) (colourblind son). As a result, the son could be colorblind or have normal eyesight.
So, option A is incorrect.
Option B: as colour blindness is an X linked recessive trait, the women would be carriers of the disease.
So, the genotype of women is \[22AA + Xx\]. To be a coloublind dominant trait, a man's genotype should be \[22AA + xY\].
So, after crossing the genotype of the couple it is proven that the boy can be either colourblind or may be of normal vision.
So, option B is not correct.
Option C: If the lady has normal vision and her father is colorblind, she may be a carrier, $X^C X^c$. If a man is colorblind, he will have the genotype $X^c Y$.
$X^C X^C$ and $X^C Y$, $X^c X^c$ and $X^c Y$ are the children.
As a result, the son's eyesight may be normal or colorblind.
So, option C is not correct.
Option D:
Gametes \[22A + x\] \[22A + y\]
\[22A + X\] \[22AA + Xx\]- carrier female \[22AA + XY\] - normal male
\[22A + x\] \[22AA + xx\] - diseased female \[22AA + xY\]- diseased male
Because colour blindness is an X-Linked recessive illness, the woman should be a carrier of the disease because her father was colour blind.
As a result, the woman's genotype should be \[22AA + Xx\] (where 'X' stands for normal sex chromosome and 'x' stands for colour blindness).
To be colorblind, a man's genotype should have \[22AA + xY\].
As a result, the table shows the cross between \[22AA + Xx{\text{ }}X{\text{ }}22AA + xY\].
Male children have a \[50\% \] chance of being colorblind, and the ratio of diseased to normal male children is \[1:1\].
So, the youngster may be colorblind, or may have normal vision.

So, option D is the correct answer.

Additional information:
Color blindness is a condition that is passed down from both parents. It's a recessive sex-linked condition. X chromosomes contain genes for red-green colorblindness. Males acquire colour blindness only from their mothers because they only have one X chromosome. Females inherit two copies of the X chromosome from both parents.

Note:
1. Causes of colour blindness-
Damage to the brain or eye, as well as nerve cells.
Disorders of the genome.
Drugs have side effects.
Tobacco and alcohol use.
2. Treatment-
There is currently no treatment for this condition. To some extent, photographic frames or filters, as well as eyeglasses with contact lenses, can be utilised to improve the dimension between specific colours. Color blindness symptoms can be alleviated by eating a well-balanced diet.