In a human family, the father and the mother are carriers for an autosomal

recessive disease. One of their sons is phenotypically normal. What is the

probability that this son is also a carrier?

1/2

1/3

1/4

2/3

The correct answer is 2/3. When both parents are carriers (heterozygous, Aa) for an autosomal recessive disease, their child has a 25% chance of being affected (aa), 50% chance of being a carrier (Aa), and 25% chance of being homozygous normal (AA). Since the son is phenotypically normal (not aa), the probability he is a carrier adjusts to 2/3 using conditional probability.

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Punnett Square Basics

Parents Aa × Aa produce: AA (1/4), Aa (1/2), aa (1/4). The normal phenotypes (AA or Aa) total 3/4 probability. Among these normals, carriers (Aa) make up 2/3 (since 1/2 divided by 3/4 = 2/3), while non-carriers (AA) are 1/3.

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Option Breakdowns

• 1/2: This is the unconditional probability of any child being a carrier, ignoring the normal phenotype info. It overlooks conditioning on the son not being affected.

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• 1/3: This matches the chance of AA among normals (1/4 divided by 3/4 = 1/3), but the question asks for carrier status (Aa), not non-carrier.

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• 1/4: This is the chance of AA overall, or affected (aa), but irrelevant for a known normal son.

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• 2/3: Correct, as it conditions on normal phenotype: P(carrier | normal) = P(Aa) / P(normal) = (1/2) / (3/4) = 2/3.

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