Question

A diseased man marries a normal woman. They bear 3 daughters and 5 sons. All the daughters were diseased and the sons were normal. The gene of this disease is

• A. sex limited character
• B. sex linked recessive
• C. sex linked dominant
• D. autosomal dominant

Answer 1

The correct option is C sex linked dominant

Following is a pedigree chart constructed based on the details given in the question:
From the pedigree chart that has been drawn, we can conclude that it is not a Y-linked trait. Had it been one, the sons would have been affected and not the daughters. So, it is either an autosomal or sex linked trait.
If it was an autosomal trait, sons and daughters would have had almost an equal probability of getting affected. But here we can see that only daughters are affected and all the sons are normal. So, this rules out the possibility of this being an autosomal trait. This is definitely an X-linked trait.
Only the father is affected and the mother is unaffected. The daughters receive one X chromosome from each parent. Since the mother has normal alleles, she would pass on them to the daughters. The daughters would receive defective alleles from the father. Since the trait is exhibiting itself even in the presence of a single mutated allele, it is a dominant trait.
Hence, the above pedigree shows the transmission of an X-linked dominant trait.
It’s not a sex limited character because both males and females are seen to have been affected. Examples of sex limited traits could be the secondary sexual characteristics in males and females that are markedly different. The genes for the development of secondary sexual characteristics are present in both males and females, but their expression is based on the level of hormones.