A married man and woman, both of whom are deaf, carry some recessive mutant alle

Question

A married man and woman, both of whom are deaf, carry some recessive mutant alleles in three different, unlinked and autosomal ''hearing'' genes: dl is recessive to Dl, d2 is recessive to D2, and d3 is recessive to D3. Homozygosity for a mutant allele at any one of these three genes causes deafness. In addition, homozygosity for any two of the three genes together in the same genome will cause prenatal lethality (and spontaneous abortion) with a penetrance of 25%. Furthermore, homozygosity for the mutant alleles of all three genes will cause prenatal lethality with a penetrance of 75%. The genotypes of the mother and father are indicated below. Mother: D1d1; D2d2; d3d3 Father: d1d1; D2d2; D3d3 Part A What is the likelihood that a live-born child will be deaf? Enter your answer as a reduced fraction. For example: 1/4 not 2/8 or 0.25.

Explanation / Answer

Answer:

2 heterozygotes in the mother x 2 heterozygotes in the father = 2^2 x 2^2 = 16.

7/16 will be homozygous recessive for 1 gene, 15/64 will be homozygous recessive for 2 genes and live,

and 1/64 will be homozygous recessive for 3 genes and live.

7/16 + 15/64 + 1/64 = 44/64 (11/16)

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