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Question: What proportion of their sons will have hyperphosphatemia? Hyperphosphatemia, a disease that causes abnormal bone growth and development, is inherited as an X-linked dominant trait. A woman who is heterozygous for the disease allele marries a normal man. What proportion of their sons will have hyperphosphatemia? What proportion of their daughters will have hyperphosphatemia? If hyperphosphatemia were an X-linked recessive trait, how would your answer differ?

Answer: Ok.... let's think this through. First we are given a disease caused by a dominant trait, where one parent (female) is heterozygous for the disease (and therefore has it), and her male companion is normal (doesn't have it.). We are told it's on the X chromosome. So, we concentrate on the female, and 1/2 of her children are going to "inherit" her X chromosome with the disease allele, and thus 1/2 of her daughters and sons will have the disease. Now if the allele for the disease were recessive, that would mean the parent (female) would not show the disease, and again, 1/2 of her children whether male or female would have that chromosome with the recessive allele. But now the situation is different. The mother is a carrier.... because she carriers a recessive disease causing allele, and therefore 1/2 of her daughters will also be carriers. As to her sons, again 1/2 of them will get the X chromosome with the disease allele, but because that son with the recessive trait does not have a corresponding spot on the Y chromosome for that gene, will show the disease. In sex-linked recessive traits, it only takes one allele in males (on the X chromosome) to show the disease.

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