Is hermaphroditism a form of asexual reproduction
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Answer:
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Explanation:
The genotype of a color-blind man is X^cY and the genotype of the heterozygous carrier female is X^cX. A cross between X^cY and X^cX would produce a progeny in following ratio=
25% colorblind daughter: 25% colorblind son: 25% carrier daughters with normal vision: 25% normal son.
Therefore, the couple is likely to have 50% normal son and 50% affected son. Likewise, the couple is likely to have 50% normal daughters and 50% colorblind daughters.
