Question

A couple without an ancestral history of Tay-Sachs disease (a rare autosomal recessive) have two normal children and an infant affected with Tay-Sachs. The brother of the husband wants to marry the sister of the wife; in such a marriage, what would be the probability of their first child having Tay-Sachs? Assume that all the adults are phenotypically normal.

Answer 1

Answer:

The probability is 25%.

Explanation:

• Let the dominant allele for the disease be 'T' and recessive allele be 't'.

• The couple with normal phenotype produces an infant with the Tay Sach's disease.
• As the disease is autosomal recessive in nature, it will get expressed in a homozygous recessive condition, that is, when the genotype is : tt.
• As the normal parents are producing a diseased offspring, the both parents must be heterozygous dominant (Tt) for the disease allele.
• The brother of the husband and the sister of the wife, both having a normal phenotype, can have the following genotypes, TT or Tt.
• If either of the individual (brother or sister) has TT genotype and the other has other has either TT or Tt genotype, there is zero probability that they will produce an offspring with the disease.

                     T          T                        and                       T            T

             T     TT        TT                                             T     TT         TT

                     (N)        (N)                                                    (N)        (N)

             T      TT        TT                                             t      Tt          Tt        

                     (N)        (N)                                                    (N)         (N)

• If both the individuals have a Tt genotype, the probability to produce  diseased offspring is one-fourth or 25%.

                                               T             t

                                   T        TT           Tt

                                             (N)           (N)

                                    t        Tt             tt

                                             (N)           (D)

• (N) stands for normal and (D) stands for diseased phenotype.