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Step-by-Step Solution
Step 1: Recall Basic Genetic Principle of Color Blindness
Color blindness (especially red-green type) is commonly an X-linked recessive trait. Males (genetically $XY$) have only one X chromosome, so a single defective X (written as $X^c$) makes them color blind. Females (genetically $XX$) must have both X chromosomes carrying the defect ($X^cX^c$) to be color blind. If they have only one defective X chromosome ($X^cX$), they are carriers but do not express the trait.
Step 2: Determine the Father's Genotype
The man is normal-visioned but his father was color blind. This means:
The color-blind grandfather had genotype $X^cY$.
The normal-visioned father must have inherited the Y chromosome from that color-blind grandfather (otherwise, if he had inherited $X^c$, he would himself be color blind).
Therefore, the father has a normal X chromosome (denote it by $X$) and a normal Y chromosome, giving him genotype $XY$ (normal).
Step 3: Determine the Mother's Genotype
The woman’s father was also color blind ($X^cY$). Therefore, she must inherit the $X^c$ (color-blind) chromosome from her father. However, whether she is a carrier or color blind depends on her mother's contribution:
If her mother gave a normal X chromosome ($X$), the woman would be a carrier ($X^cX$) but not color blind.
If her mother also gave a defective chromosome ($X^c$), the woman would be color blind ($X^cX^c$). Since the question suggests she is marrying and typically individuals are assumed to be phenotypically normal unless stated, we often assume she is just a carrier ($X^cX$). But the problem statement does not explicitly clarify her phenotype.
Step 4: Predict Outcome for Their Daughter
Their first child is a daughter, which means:
The daughter receives one X chromosome from her father. Since the father has a normal X chromosome ($X$), he will pass a normal $X$ to any daughter.
The daughter receives one X chromosome from her mother. The mother could pass either a normal $X$ or a defective $X^c$ if she is a carrier.
Hence, the possible genotypes for the daughter are:
$XX$ (completely normal) if she receives the normal chromosome from the mother.
$X^cX$ (carrier) if she receives the defective chromosome from the mother.
In both cases, the daughter has at least one normal X from her father and therefore will not be color blind. A female needs both X chromosomes to be defective ($X^cX^c$) to express color blindness.
Step 5: Final Probability
Since there is no scenario here in which the daughter could inherit two defective X chromosomes (the father only has a normal X to give), the probability that the daughter will actually be color blind is:
$$ 0\% $$
Even if the mother is a carrier, the daughter would at most become a carrier (one defective X), but not color blind.
Conclusion
The correct answer is “Zero percent” chance that their daughter would be color blind.
