An X‑linked recessive disorder (e.g., haemophilia) is present in a carrier female (X^H X^h) who is crossed with a normal male (X^H Y). Determine (i) the genotype and phenotype proportions among male offspring, (ii) the probability that a randomly selected son will be affected, and (iii) explain why daughters are not affected.
Answer
Male offspring: 50% normal (X^H Y), 50% affected (X^h Y); probability a son is affected = 50%; daughters are either X^H X^H or X^H X^h and are phenotypically normal because the disorder is recessive and requires two copies in females.
Step-by-step solution
- Write the possible gametes of the carrier female: X^H and X^h; write the possible gametes of the normal male: X^H and Y.
- Form the Punnett square for the cross (female gamete × male gamete).
- For male offspring, the sperm contributes Y and the egg contributes either X^H or X^h, giving genotypes X^H Y (normal) and X^h Y (affected) in equal proportion (1:1).
- Thus 50% of male offspring are normal and 50% are affected.
- Since a randomly chosen son comes from this 1:1 distribution, the probability that he is affected is 1/2 = 50%.
- For daughters, the sperm contributes X^H and the egg contributes X^H or X^h, giving genotypes X^H X^H (homozygous normal) and X^H X^h (carrier). Both have at least one dominant normal allele, so they show the normal phenotype.
- Therefore daughters are not affected because the disorder is recessive and they lack two copies of the mutant allele.
Common mistake: Assuming daughters have a 50% chance of being affected instead of recognizing that they need two copies of the recessive allele.
Concept tested: Sex‑linked inheritance – X‑linked recessive
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