A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be the probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.

Introduction to colour blindness:

Colour blindness is the inability of an individual to differentiate between red and green colours.


Genotype of colour blindness:

Genotype of colour blindness:

• Colour blindness is a recessive X-linked trait.
• Females are affected by colour blindness only in homozygous condition (that is both X chromosomes carry the recessive allele, represented by the genotype $X^C{X}^C)$ and act as carrier in heterozygous condition $\left(X^{C}X\right)$.
• Males have only one X chromosome and show colour blindness due to the presence of a single recessive allele $\left(X^{C}Y\right)$.

Cross between a carrier woman and normal man for colour blindness :

A normal visioned woman may have XX or $X^{C}X$. In the given question, since her father is colour blind, she must have inherited his only XC chromosome and became a carrier of colour blindness $\left(X{X}^C\right)$.

When she marries a normal man (XY) without colour blindness, there is 50% probability that male progenies are colour blind as they inherit the colour blindness allele from their mother. Similarly, 50% of female progenies are carriers. Therefore, every daughter has a normal vision.

Carrier woman and normal man will produce 50% colour blind sons and 50% carrier daughters.



Pedigree chart:





Final answer:

Probability of colorblind sons = $50\%$
Probability of colorblind daughters= $0\%$