Final answer:
For a blue-eyed, color vision carrier mother and a heterozygous brown-eyed, colorblind father, daughters have a 50% chance of being brown-eyed carriers for colorblindness, and a 50% chance of being blue-eyed carriers. Sons have a 50% chance of being brown-eyed and colorblind and a 50% chance of being blue-eyed and colorblind.
Step-by-step explanation:
In humans, the gene for color vision (C) is dominant over the gene for red-green colorblindness (c), and this trait is sex-linked on the X chromosome. The gene for brown eyes (B) is dominant over the gene for blue eyes (b). For a blue-eyed woman (bb) who is a carrier for color vision (XCXc) and a heterozygous brown-eyed man (Bb) who is colorblind (XcY), we can determine the probable genotypes and phenotypes of their children.
Parental cross: XCXcbb (mother) x XcYBb (father)
The father can pass on either his X chromosome (resulting in a daughter) or his Y chromosome (resulting in a son), and his B or b allele for eye color. Daughters may inherit either XC or Xc from their mother, along with bb for eye color. Sons will inherit their X chromosome from their mother and will have blue eyes since the mother only has b alleles.
Genotype Probabilities:
- Daughters: 50% XCXcBb (brown-eyed, carrier for colorblindness), 50% XCXcbb (blue-eyed, carrier for colorblindness)
- Sons: 50% XcYBb (brown-eyed, colorblind), 50% XcYbb (blue-eyed, colorblind)
Phenotype Probabilities:
- Daughters: 50% carrier for colorblindness with brown eyes, 50% carrier for colorblindness with blue eyes
- Sons: 50% colorblind with brown eyes, 50% colorblind with blue eyes
As males only have one X chromosome, if that X chromosome carries the allele for colorblindness, they will express the trait. For eye color, the brown allele is dominant to the blue allele, therefore, children inheriting one B allele will have brown eyes, while only those inheriting two b alleles (from both parents) will have blue eyes.