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Eduncle Best Answer
The trait given is autosomal recessive. In this pedigree the man’s (II-3) parents must have both been heterozygotes Aa because they produced affected aa child. So among expected progeny of man’s (II-3) parents (AA(1/4) Aa(1/2)aa(1/4) : 2/3 probability that the man
(II-3) is a carrier of the trait. The woman II-4 is affected by the trait, so she must have genotype aa. So expected progeny genotype is Aa or aa. So overall probability of the son to have the disease is 2/3 x 1/2 = 1/3