In the pedigree below, the rare trait is autosomal recessive. If the frequency o

Question

In the pedigree below, the rare trait is autosomal recessive. If the frequency of heterozygotes for the trait in the general population is 1/15, then what is the probability of II-2 having an affected child with his phenotypically normal wife (she is not shown)?

4. In the pedigree below, the rare trait is autosomal recessive. If the frequency of heterozygotes for the trait in the general population is 1/15, then what is the probability of Il-2 having an affected child with his phenotypically normal wife (she is not shown)? (3 pts)

Explanation / Answer

In the given pedigree chart, II-1 is affected by a rare recessive autosomal disease. Hence his parents are heterozygous for the trait. The second progeny of the F2 generation, II-2 is normal and marries a girl who is also normal.

For II-2 and his wife to have an affected child, both of them need to be heterozygous.

P (II-2 being heterozygous) = 2/3 x 1/15 = 2/45

Now, chances the couple will have an affected child = 1/4

Therefore P( parents being heterozygous and child born is affected) = 2/45 x 1/4 = 1/90

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