What proportion of their sons would be color-blind and of normal height?

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A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man’s father was six feet tall, and both the woman’s parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.

2 Answers

  • since the mother is color-blind, both of her x chromosomes have the color blind gene. Since the father has to pass the Y-chromosome on to his son, the son’s only X chromosome comes from the mother. This means that 100% of their sons will inherit an X chromosome featuring the color blind gene from their mother, making them all color blind.

    As far as height is concerned, the father’s father was tall which means he was double recessive for the dwarfism gene, so the father must be a heterozygote, having gotten the dominant dwarfism gene from his mother. The woman is normal height which means she must also be homozygous recessive. Crossing a heterozygote with a homozygous recessive we get 2 offspring which are heterozygotes and 2 which are homozygous recessive, meaning 50% of their children (the heterozygotes) will be dwarfs, and 50% (the homozygous recessive) will be normal height.

    Combining the 100% color-blind with the 50% normal, we get 1.00*.50 or 50% of the children are both color-blind and normal height.

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