+1 vote
in Class 12 by kratos

Inheritance of traits in humans is studied by analyzing the presence or absence of a trait in a family. This is then depicted as a chart showing different generations and indicating individuals who show the trait. The following is a pedigree for a rare genetic disorder representing parents (top row) and their three ****.

The pedigree suggests that the genetic disorder could be either autosomal recessive or X-linked recessive.

The following statements were made about the genotypes (with respect to the gene responsible for the disorder) of the 5 individuals in this pedigree:

a. The mother is heterozygous.

b. The daughters could be either homozygous or heterozygous.

c. The daughters are only homozygous.

(I) Which of the above statement(*) is/ are correct if the genetic disorder is

i. Autosomal recessive ?

ii. X-linked recessive ?

(II) Assuming that both the parents carry the allele responsible for the disorder, what is the probability that the first daughter would carry the same allele?

(a) 1/4

(b) 2/4

(c) 1/3

(d) 2/3

(III) Assume that the trait is X-linked recessive. The affected son marries a woman who does not carry the allele for the disorder. When they have a child, what is the probability that child will carry the allele, if the child is a

i. Son

ii. Daughter

1 Answer

+4 votes
by kratos
 
Best answer

(I) (i) Statements a and b are correct as the mother and father both need to be heterozygous for a recessive trait to appear in next generation. Daughters, however can be either AA/Aa homozygous / heterozygous.

(ii) Statements a & b are correct as mother has to be carrier X6X if the male child is affected (Criss –Cross inheritance). Both daughters however can receive either XC or X from mother, Therefore, can be homozygous / heterozygous.

(II) If both parents carry allele for disorder.

Probability that first daughter would carry same allele is 2/4 (b) option.

(III) If the affected son marries a woman, who does not carry the allele of the disorder, then child can be

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