Question

If a colour blind man marries a normal visioned woman, what is the percentage of offsprings showing colour blindness phenotypically?

• A. 50%
• B. 0%
• C. 25%
• D. 100%

Hint:

In X-linked recessive traits, daughters of an affected father and normal mother are usually carriers, while sons are normal if the mother is not a carrier.

Answer 1

The Correct Option is B

Step 1: Understanding the genetic nature of colour blindness.
Colour blindness is a sex-linked recessive trait, carried on the X-chromosome. A male has only one X-chromosome, so if it carries the defective gene, the trait is expressed. Females have two X-chromosomes, so the trait is expressed only if both X-chromosomes carry the defective gene.
Step 2: Writing the genotypes of parents.
Colour blind man: \( X^cY \)
Normal visioned woman (homozygous): \( X^NX^N \)
Step 3: Formation of offspring.
All daughters receive one normal X-chromosome from the mother and the colour blind X-chromosome from the father, so daughters are \( X^NX^c \) (carriers but phenotypically normal).
All sons receive the Y-chromosome from the father and a normal X-chromosome from the mother, so sons are \( X^NY \) (normal).
Step 4: Conclusion.
Since none of the offspring express colour blindness phenotypically, the percentage of colour blind offspring is zero.