Final answer:
The mother's genotype is XcX, the father's genotype is XY, and the colorblind son's genotype is XcY. The mother is heterozygous, and the unaffected daughter doesn't express color blindness due to the dominance of the normal vision allele. A colorblind woman's parents must include a carrier mother and an affected father.
Step-by-step explanation:
The genotypes of the parents and the son regarding color blindness, an X-linked recessive trait, can be understood considering that the son has the condition while the parents do not. The son's genotype must be XcY, where Xc represents the X chromosome carrying the color blindness allele, and Y is the normal Y chromosome. Since color blindness is recessive and the mother appears normal but must have passed on the colorblind allele to her son, her genotype is XcX. The father, who cannot pass an X chromosome to his son (only the Y chromosome), must have a normal genotype XY. Because the son is affected and the daughter is unaffected, the pedigree for Color Blindness illustrates that the mother is a carrier (heterozygous) and the father has normal vision.
A test cross resulting in a 1:1 ratio of affected to unaffected offspring indicates that the parent is heterozygous, which applies to the mother in this scenario. Additionally, boys cannot be carriers of color blindness as they only have one X chromosome; hence, if they have the allele, they are affected.
Sex-linked inheritance explains why a son cannot inherit colorblindness from his father, because males can only pass their Y chromosome to their sons, which does not carry the gene for color vision. According to this understanding, a woman cannot have colorblindness if her father does not and her mother is not a carrier. If a woman is color blind but her sister is not, it indicates that their father must have the colorblind allele (XcY) and their mother must be a carrier (XcX) for the color blind woman to express this phenotype.