Answer:
The genotypic and phenotypic percentages of the offspring from a cross between a woman with color blindness (XCXc) and a man with normal color vision (XY) would be:
Genotypic percentages:
50% XCXC (homozygous dominant, normal color vision)
50% XCY (heterozygous, normal color vision)
Phenotypic percentages:
100% normal color vision
Step-by-step explanation:
Color blindness is caused by a recessive allele located on the X chromosome, which means that it is a sex-linked trait that is more commonly found in men than in women. Women are more likely to be carriers of the allele, as they have two X chromosomes, but they do not typically experience the symptoms of color blindness. Men have only one X chromosome, which means that they are more likely to be affected by the recessive allele if they inherit it from their mother.
In this cross, the woman with color blindness (XCXc) has one dominant allele for normal color vision (XC) and one recessive allele for color blindness (xc). The man with normal color vision (XY) has one dominant allele for normal color vision (X) and one recessive allele for sex determination (Y).
When these individuals have offspring, each child will inherit one X chromosome from each parent. This means that each child will have either XCXC, XCXc, or XCY genotype, depending on which X chromosome they inherit from each parent.
The genotypic percentages of the offspring would be:
50% XCXC (homozygous dominant, normal color vision)
50% XCY (heterozygous, normal color vision)
Since the dominant allele for normal color vision (XC) masks the recessive allele for color blindness (xc), all of the offspring would have normal color vision, regardless of their genotype.
The phenotypic percentages of the offspring would be:
100% normal color vision.