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A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy -
A
Must have normal colour vision
B
May be colour blind or may be of normal vision
C
Will be partially colour blind since he is heterozygous for the colour blind mutant allele
D
Must be colour blind
Detailed Solution
Colour blindness is an X-linked recessive trait. Let $X^c$ carry the allele for colour blindness.
The woman has normal vision, but her father was colour blind, so she received $X^c$ from him. She is a carrier: $X^cX$.
The colour blind man is $X^cY$.
A son receives the Y chromosome from his father and one X chromosome from his mother.
The mother gives either $X^c$ or $X$ with equal probability, so each son has a 50 percent chance of being $X^cY$ (colour blind) and a 50 percent chance of being $XY$ (normal).
The birth order of the child has no effect on this probability.
A male has only one X chromosome, so he cannot be heterozygous for this trait.
So the boy may be colour blind or may have normal vision.
The woman has normal vision, but her father was colour blind, so she received $X^c$ from him. She is a carrier: $X^cX$.
The colour blind man is $X^cY$.
A son receives the Y chromosome from his father and one X chromosome from his mother.
The mother gives either $X^c$ or $X$ with equal probability, so each son has a 50 percent chance of being $X^cY$ (colour blind) and a 50 percent chance of being $XY$ (normal).
The birth order of the child has no effect on this probability.
A male has only one X chromosome, so he cannot be heterozygous for this trait.
So the boy may be colour blind or may have normal vision.