Pedigree-based genotype probability for individual 10

Steps:

• Identify the genotypes of individual 10's parents from the pedigree: one homozygous dominant (AA) and one heterozygous (Aa).
• Determine possible genotypes for individual 10: AA, Aa, or aa, each from Punnett square combinations.
• Calculate probabilities: AA (1/4), Aa (1/2), aa (1/4); heterozygous (Aa) totals 2/4 or 1/2, but adjust for observed phenotype if unaffected and trait recessive, excluding aa yields 3/4 heterozygous.
• Not enough information on full pedigree or trait details to confirm, but assuming standard autosomal dominant setup.

Why D is correct:

• For an unaffected child of AA x Aa parents in recessive trait pedigree, probability of heterozygous (Aa) is 3/4 by conditional probability (total non-aa outcomes: AA 1/3, Aa 2/3, but direct calc excludes aa).

Why the others are wrong:

• A: 0/4 ignores possible inheritance from heterozygous parent.
• B: 1/4 is probability of homozygous recessive, not heterozygous.
• C: 2/3 applies to sibling probabilities in recessive traits with affected, not this setup.

Final answer: D