QuestionDecember 13, 2025

18.In humans, normal color vision (N) is dominant over color blindness (n) A man and woman with normal color vision produced two colorblind I sons and two daughters with normal vision. Choose from the diagram the correct parental genotypes. A. X^NY and X^NX^N B. X^NY and X^NX^n C. X^nY and X^NX^N D. X^nY and X^nX^n

18.In humans, normal color vision (N) is dominant over color blindness (n) A man and woman with normal color vision produced two colorblind I sons and two daughters with normal vision. Choose from the diagram the correct parental genotypes. A. X^NY and X^NX^N B. X^NY and X^NX^n C. X^nY and X^NX^N D. X^nY and X^nX^n
18.In humans, normal color vision (N) is dominant over color blindness (n) A man and woman with normal color vision produced two colorblind I sons and two daughters with normal vision. Choose from the diagram the correct parental genotypes. A. X^NY and X^NX^N B. X^NY and X^NX^n C. X^nY and X^NX^N D. X^nY and X^nX^n

Solution
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Answer

B. X^{N}Y and X^{N}X^{n} Explanation 1. Identify inheritance pattern Color blindness is X-linked recessive. Sons inherit their X from the mother and Y from the father. 2. Analyze sons' genotypes Colorblind sons must be X^nY, so mother must have at least one X^n allele. 3. Analyze daughters' genotypes Daughters are normal, so they must have at least one X^N. Since father is normal, he is X^NY; daughters get his X^N. 4. Determine mother's genotype Mother must be X^NX^n to produce both normal daughters (X^NX^N or X^NX^n) and colorblind sons (X^nY).

Explanation

1. Identify inheritance pattern<br /> Color blindness is X-linked recessive. Sons inherit their X from the mother and Y from the father.<br />2. Analyze sons' genotypes<br /> Colorblind sons must be $X^nY$, so mother must have at least one $X^n$ allele.<br />3. Analyze daughters' genotypes<br /> Daughters are normal, so they must have at least one $X^N$. Since father is normal, he is $X^NY$; daughters get his $X^N$.<br />4. Determine mother's genotype<br /> Mother must be $X^NX^n$ to produce both normal daughters ($X^NX^N$ or $X^NX^n$) and colorblind sons ($X^nY$).
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