In the case of red-green color blindness being a recessive sex-linked trait, the gene responsible for this condition is located on the X chromosome. Here’s how we can determine the genotypes:
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Male Parent (Colorblind): Since colorblindness is recessive and he is colorblind, his genotype must be \( X^cY \), where \( X^c \) is the X chromosome carrying the colorblind allele and \( Y \) is the Y chromosome.
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Female Parent (Normal Vision): Since the mother is normal and color blindness is recessive, her genotype must be either \( XX \) (homozygous normal) or \( X^cX \) (heterozygous carrier). However, since they have a colorblind daughter, she must carry the colorblind allele, so her genotype is \( X^cX \).
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Daughter (Colorblind): The daughter must inherit one X chromosome from each parent. She can inherit \( X^c \) from her father (because he can only pass on his \( X^c \) chromosome) and must inherit \( X^c \) from her mother. Thus, her genotype is \( X^cX^c \).
Summary: The genotype for their daughter is \( X^cX^c \) (homozygous colorblind).