Question

A colour blind man marries a woman with normal sight who has no history of color blindness in her family. What is the probability of their grandson through daughter being color blind?

• A. 0.25
• B. 0.5
• C. 1
• D. 0.33

Answer 1

Answer: (B)

0.5

• The genotype for normal vision may be symbolized by (XX), and colour blindness by (XX). X indicates the sex-linked recessive gene for colour blindness. If a colour blind man 0(Y) marries a normal woman (XX), in the F${}_1$ generation all-male progeny (sons) will be normal (XY).
• The female progeny (daughters) though will show normal phenotype, but genetically they will be heterozygous (XX). Since these daughters bear the recessive gene of colour blindness, they are the carriers of the trait.
• If such a carrier woman with normal vision (heterozygous for colour blindness) marries a normal man (XY), the following progeny may be expected in the F${}_2$ generation: among the daughters, 50% is normal and 50% are carriers for the diseases; among sons, 50% are colour blind and 50% are with normal vision.
• All sex-linked genes, therefore, pass from male to female and then come back to a male of F${}_2$ generation (grand-children generation). That is there is a chance of 1 in 2 (50% or 0.5) of her grandson being colour blind.