Parental genotypes determine recessive trait probability Steps:

• Sickle cell anemia requires genotype aa; the affected child confirms both parents carry the a allele and are likely Aa (assuming parents unaffected).
• Unaffected child is AA or Aa, but this does not alter parental genotypes.
• For Aa × Aa cross, each child independently inherits a from mother (probability 1/2) and a from father (1/2).
• Thus, probability third child is aa = (1/2) × (1/2) = 1/4 = 25%.

Why C is correct:

• Punnett square for Aa × Aa shows aa offspring in 1/4 of cases, per Mendel's law of independent assortment.

Why the others are wrong:

• A: 0% ignores the established 1/4 chance from carrier parents.
• B: 12.5% misapplies probabilities, perhaps confusing with joint events, but each child is independent at 25%.
• D: 50% applies to Aa × aa cross, but data (unaffected parent implied) fits Aa × Aa better.

Final answer: C