A colour‑blind male (genotype XcY) marries a woman who is a carrier for colour‑blindness (genotype XCXc). What proportion of their daughters will be colour‑blind?

Step-by-step solution

1. List the possible gametes from the father: Xc (for a daughter) or Y (for a son).
2. List the possible gametes from the mother: XC or Xc.
3. Form the possible daughter genotypes by combining the father's Xc with each maternal X chromosome:
4. 1) Xc (father) + XC (mother) = XCXc (heterozygous, normal vision).
5. 2) Xc (father) + Xc (mother) = XcXc (colour‑blind).
6. Since the mother contributes XC or Xc with equal probability, half the daughters are XCXc and half would be XcXc.
7. However, a male contributes only Xc, and the mother’s Xc is already present in the carrier; the daughter receives one normal allele (XC) from the mother half the time, making her a carrier, not affected.
8. Because the mother is a carrier, the probability of a daughter being XcXc is 0 (the father provides only Xc, not a normal allele).
9. Thus, none of the daughters will be colour‑blind.

Concept tested: X‑linked recessive inheritance – daughters

Published 25 August 2026 · How this solution was prepared and checked

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