Question

A normal man and a woman have a child with phenylketonuria, an autosomal recessive trait. The woman becomes pregnant again and in investigation she is carrying monozygotic twins. What is the probability that both of the couple’s twin will develop phenylketonuria
A. 1/16
B. 1/4
C. 1/2
D. 3/4

Answer 1

Hint: Humans have 23 pairs of chromosomes that include 22 pairs of autosomes and one pair of allosomes. Homologous chromosomes can be defined as a set of chromosomes that pair up with each other during fertilization of male and female gametes. It consists of one maternal and one paternal chromosome.

Complete answer:
In Phenylketonuria, the homozygous recessive autosomal alleles present on chromosome 12 cause absence of phenylalanine hydroxylase in liver. Enzyme deficiency results in the phenylalanine accumulation which is converted into phenyl pyruvic acid.
There is severe mental retardation, hypopigmentation of skin and hair, eczema, mousy odor of skin, hair and urine. The phenyl pyruvic acid shows poor absorption of the kidney. It is also an example of pleiotropic genes. This single gene mutation manifests itself through phenotypic expression as mental retardation and reduction in hair and skin pigmentation.
The parents must have the genotype Aa. The fraternal twins are formed by two different eggs fertilized to different zygotes and the genotype of the zygote to get the disease should be aa. For two zygotes, it would be ¼ X ¼ i.e. 1/16.

Note:
-So phenylketonuria is an inherited disorder caused due to error in the autosomal chromosome.
-Twins are of two types- maternal and fraternal. Maternal twins are formed from the same ovum and -thus they have similar genotypes while fraternal twins differ genotypically.