Q.53 If a colour-blind woman marries a normal man, the chance that their boy child will be
colour-blind is ________%.

100%

Color blindness follows X-linked recessive inheritance, where a color-blind woman (genotype X^c X^c) passes the recessive allele to all offspring, while a normal man (X^N Y) contributes either X^N to daughters or Y to sons. All boy children thus inherit X^c Y and express color blindness.

Punnett Square Analysis

The cross is X^c X^c (mother) × X^N Y (father):

• Daughters (X^c X^N): All carriers with normal vision.

• Sons (X^c Y): All color-blind.
  This confirms 100% probability for boy children.

Option Explanations

Typical CSIR NET options might include 0%, 25%, 50%, 100%:

• 0%: Incorrect; assumes no inheritance from mother, ignoring X-linked pattern.

• 25%: Incorrect; applies to carrier mother × normal father (sons: 50% affected).

• 50%: Incorrect; fits heterozygous carrier mother, not homozygous affected.

• 100%: Correct; all sons receive mother’s X^c and father’s Y.

Colour blindness inheritance is a key topic in genetics for CSIR NET Life Sciences, especially when a colour-blind woman marries a normal man. The chance that their boy child will be colour-blind is 100%, due to X-linked recessive patterns. This detailed guide breaks down the Punnett square, genotypes, and why sons are always affected.

Genetics of Colour Blindness

Colour blindness is X-linked recessive: females need two recessive alleles (X^c X^c) to express it, males need one (X^c Y). A colour-blind woman is X^c X^c; normal man is X^N Y. Sons inherit her X^c and his Y, making them colour-blind. Daughters get X^c X^N (carriers). Affects 8% males vs. 0.5% females globally.

Punnett Square Breakdown

Visualize the cross:

• 100% sons colour-blind (X^c Y).

• 100% daughters carriers (normal vision).

CSIR NET Exam Insights

Questions test X-linked probability: carrier mother gives 50% son risk; affected mother gives 100%. No autosomal confusion—focus sex chromosomes. Practice criss-cross inheritance.