Question

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Explanation:

Step1: Identify Genotypes

The female is a carrier for red - green color blindness, so her genotype is $X^{B}X^{b}$ (where $B$ represents the normal allele and $b$ represents the color - blind allele). The normal male has the genotype $X^{B}Y$.

Step2: Fill Punnett Square

• For the top - left cell (mother's $X^{B}$ and father's $X^{B}$): The genotype is $X^{B}X^{B}$, and the phenotype is normal vision (female).
• For the top - middle cell (mother's $X^{b}$ and father's $X^{B}$): The genotype is $X^{B}X^{b}$, and the phenotype is carrier (female, normal vision but carries the allele).
• For the bottom - left cell (mother's $X^{B}$ and father's $Y$): The genotype is $X^{B}Y$, and the phenotype is normal vision (male).
• For the bottom - middle cell (mother's $X^{b}$ and father's $Y$): The genotype is $X^{b}Y$, and the phenotype is red - green color blind (male).

Step3: Phenotype of Sons

Sons inherit the $Y$ chromosome from the father and an $X$ chromosome from the mother. So, 50% of the sons will have the genotype $X^{B}Y$ (normal vision) and 50% will have $X^{b}Y$ (red - green color blind).

Answer:

• Punnett Square cells (from top - left to bottom - middle):
• $X^{B}X^{B}$ (normal female)
• $X^{B}X^{b}$ (carrier female)
• $X^{B}Y$ (normal male)
• $X^{b}Y$ (color - blind male)
• Phenotype of the sons: 50% normal vision, 50% red - green color blind.