An X‑linked recessive disorder (e.g., haemophilia) is present in a carrier female (X⁽ʰ⁾X⁽⁺⁾) crossed with a normal male (X⁽⁺⁾Y). State the genotype and phenotype ratios among male offspring, and give the probability that a randomly selected son will be affected.
Answer
Among male offspring, 50 % are normal (X⁽⁺⁾Y) and 50 % are affected (X⁽ʰ⁾Y); probability a son is affected = 1/2.
The correct answer is Among male offspring, 50 % are normal (X⁽⁺⁾Y) and 50 % are affected (X⁽ʰ⁾Y); probability a son is affected = 1/2. — Both genotypes occur with equal frequency (1:1), giving a 50 % chance of an affected son.
Step-by-step solution
- Write the possible gametes from the carrier female: X⁽ʰ⁾ and X⁽⁺⁾.
- Write the male gametes: X⁽⁺⁾ and Y.
- Form the male offspring genotypes by combining female X gametes with male Y: X⁽ʰ⁾Y (affected) and X⁽⁺⁾Y (normal).
- Both genotypes occur with equal frequency (1:1), giving a 50 % chance of an affected son.
Common mistake: Assuming females contribute a Y chromosome to sons
Concept tested: X‑linked recessive inheritance
Published · How this solution was prepared and checked
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