Question

A carrier daughter ($X^N X^c$) of a colour blind father marries a normal male ($X^N Y$). What is the probability of a colour blind son?

• A. 0%
• B. 25%
• C. 50%
• D. 75%

Hint:

X-linked recessive disorders mostly affect males because males possess only one X chromosome. Examples include: Colour blindness, Haemophilia, and Duchenne muscular dystrophy.

Answer 1

The Correct Option is B

Concept: Colour blindness is an example of an X-linked recessive genetic disorder. The gene responsible for colour vision is located on the X chromosome.
• Males have XY chromosomes. Since they possess only one X chromosome, a single recessive allele can express the disorder.
• Females have XX chromosomes. They must inherit two recessive alleles to express colour blindness. A carrier female has one normal allele and one recessive allele but does not show the disorder. Given:
• Carrier female: $X^N X^c$
• Normal male: $X^N Y$

Step 1: Determine the possible gametes. Female gametes: \[ X^N, \; X^c \] Male gametes: \[ X^N, \; Y \]

Step 2: Construct the Punnett square for the cross. \[ \begin{array}{c|cc} & X^N & Y \hline X^N & X^N X^N & X^N Y X^c & X^N X^c & X^c Y \end{array} \]

Step 3: Interpret the offspring.
• $X^N X^N$ → Normal daughter
• $X^N X^c$ → Carrier daughter
• $X^N Y$ → Normal son
• $X^c Y$ → Colour blind son Out of the four possible offspring combinations, only one results in a colour blind son. \[ \text{Probability} = \frac{1}{4} = 25\% \] Thus, the probability of a colour blind son is 25%.