Question

A carrier daughter \((X^NX^c)\) of a colour blind father marries a normal male \((X^NY)\). What is the probability of a colour blind son?

• A. \(0\%\)
• B. \(25\%\)
• C. \(50\%\)
• D. \(75\%\)

Hint:

Sex-linked recessive disorders such as colour blindness and haemophilia are more common in males because males have only one X chromosome. A single defective allele on the X chromosome will express the disorder.

Answer 1

The Correct Option is B

Concept: Colour blindness is a sex-linked recessive disorder controlled by a gene located on the X-chromosome.
• \(X^N\) = Normal vision allele
• \(X^c\) = Colour blindness allele Important points:
• Males have one X and one Y chromosome (\(XY\)).
• Females have two X chromosomes (\(XX\)).
• A recessive trait like colour blindness appears in males if their single X chromosome carries the defective gene.
• A female must have two defective alleles to be colour blind, otherwise she becomes a carrier.

Step 1: Identify the parental genotypes. The given cross is: \[ \text{Carrier female} \ (X^NX^c) \ \times \ \text{Normal male} \ (X^NY) \] Gametes produced:
• Female gametes: \(X^N\), \(X^c\)
• Male gametes: \(X^N\), \(Y\)

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

Step 3: Interpret the offspring genotypes.
• \(X^NX^N\) → Normal daughter
• \(X^NX^c\) → Carrier daughter
• \(X^NY\) → Normal son
• \(X^cY\) → Colour blind son Thus, out of the four possible offspring combinations, only one represents a colour blind son.

Step 4: Calculate the probability. \[ \text{Probability of colour blind son} = \frac{1}{4} = 25\% \] Therefore, the probability of a colour blind son is \(25\%\).