Question

A normal couple produces half the sons as haemophilic and half the daughters as carriers. Choose the option that correctly indicates the chromosome on which the gene for this trait is located.

• A. \( \text{X-chromosome of father} \)
• B. \( \text{Y-chromosome of father} \)
• C. \( \text{One X-chromosome of mother} \)
• D. \( \text{Both the X-chromosomes of the mother} \)

Hint:

For X-linked recessive traits, affected sons usually have carrier mothers. If a mother has an affected son, she must be at least a carrier.

Answer 1

The Correct Option is C

Haemophilia is a recessive X-linked trait. Let \(X^H\) represent the normal allele and \(X^h\) represent the allele for haemophilia. A normal couple has a haemophilic son, which means the son's genotype is \(X^hY\). He inherited the \(X^h\) allele from his mother. Since the mother is normal but has an affected son, her genotype must be heterozygous carrier, \(X^HX^h\). The father is normal, so his genotype is \(X^HY\). Let's consider the possible genotypes of their offspring: \begin{itemize} \item Daughters: \(X^HX^H\) (normal) or \(X^HX^h\) (carrier). There is a 50% chance for each. The question states half the daughters are carriers, which aligns with the mother being \(X^HX^h\). \item Sons: \(X^HY\) (normal) or \(X^hY\) (haemophilic). There is a 50% chance for each. The question states half the sons are haemophilic, which also aligns with the mother being \(X^HX^h\). \end{itemize} The gene for haemophilia is located on the X-chromosome. The mother, being a carrier (\(X^HX^h\)), passes on the \(X^h\) allele to half of her sons, making them haemophilic, and passes on the \(X^h\) allele to half of her daughters, making them carriers (\(X^HX^h\)). The father's X-chromosome (\(X^H\)) contributes to normal sons and normal or carrier daughters. Therefore, the mother carries the recessive allele on one of her X-chromosomes.