Genetics-i Miscellaneous Easy Questions and Answers | Page - 3

Genetics-i Miscellaneous

PKU is one of the best known hereditary disorders in amino acid metabolism. The defect is attributed to a lesion in one of the following enzymatic activities

1. Phenylalanine ammonia lyase
1. Phenylalanine hydroxylase
1. Tryosine hydroxylase
1. Phenylalanine transaminase

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Phenylketonuria (PKU) is an autosomal recessive metabolic genetic disorder characterized by a mutation in the gene for the hepatic enzyme phenylalanine hydroxylase (PAH), rendering it nonfunctional.

Correct Option: B

Phenylketonuria (PKU) is an autosomal recessive metabolic genetic disorder characterized by a mutation in the gene for the hepatic enzyme phenylalanine hydroxylase (PAH), rendering it nonfunctional.

A disease is inherited by a child with a probability of 1 .
4

In a family with two children, the probability that exactly one sibling is affected by this disease is

1. 1
  4
1. 3
  8
1. 7
  16
1. 9
  16

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Given : P(E) = 1 , P(E) = 3
4 4

P(x = 1) = 2C₁ × 1 × 3 = 3
4 4 8

Correct Option: B

Given : P(E) = 1 , P(E) = 3
4 4

P(x = 1) = 2C₁ × 1 × 3 = 3
4 4 8

Diploid Drosophila has eight chromosomes. Which one of the following terms should NOT be used to describe Drosophila with sixteen numbers of chromosomes?

1. Polyploid
1. Aneuploid
1. Euploid
1. Tetraploid Common

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Aneuploidy is an error in cell division that results in the daughter cells having the wrong number of chromosomes. It is the second major category of chromosomes mutation in which chromosome number is abnormal. An aneuploid is an individual organism whose chromosome number differs from the wild type by part of a chromosome set. Generally, the aneuploid chromosome set differs from wild type by one or small number of chromosomes. Aneuploids have chromosome number either greater or smaller than of wild type. Aneuploid nomenclature is based on the number of copies of the specific chromosome in the aneuploid state. So, for drosophila with 16 chromosomes, it is not aneuploid in condition.

Correct Option: B

Aneuploidy is an error in cell division that results in the daughter cells having the wrong number of chromosomes. It is the second major category of chromosomes mutation in which chromosome number is abnormal. An aneuploid is an individual organism whose chromosome number differs from the wild type by part of a chromosome set. Generally, the aneuploid chromosome set differs from wild type by one or small number of chromosomes. Aneuploids have chromosome number either greater or smaller than of wild type. Aneuploid nomenclature is based on the number of copies of the specific chromosome in the aneuploid state. So, for drosophila with 16 chromosomes, it is not aneuploid in condition.

Direction: Red-green colour blindness is inherited as a recessive X-linked trait.

What will be the probability of having the colour-blind son to a woman with phenotypically normal parents and a colour-blind brother, and married to a normal man? (Assume that she has no previous children)

1. 100 %
1. 50 %
1. 25 %
1. 12.5 %

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The probability of having the colorblind son to woman with phenotypically normal parent and a colorblind brother, and married to a normal man than may be the woman is a carrier of blind trait or else not. Then we can say mother can be XRX a carrier or XX normal and have a cross with a normal man.

Therefore, there is 1 chance of having a colorblind son out of 4 possible progenies.
Hence, probability of one child to be colorblind = ¼ = 0.25 = 25% .
Again 2 sons could be born, out of which 1 could be colorblind;
Probability for colorblind son = ½ = 50%

Correct Option: B

The probability of having the colorblind son to woman with phenotypically normal parent and a colorblind brother, and married to a normal man than may be the woman is a carrier of blind trait or else not. Then we can say mother can be XRX a carrier or XX normal and have a cross with a normal man.

Therefore, there is 1 chance of having a colorblind son out of 4 possible progenies.
Hence, probability of one child to be colorblind = ¼ = 0.25 = 25% .
Again 2 sons could be born, out of which 1 could be colorblind;
Probability for colorblind son = ½ = 50%

What will be the probability of having the colour-blind daughter to a phenotypically normal woman, who already had one colour-blind son, and is married to a colour-blind man?

1. 75 %
1. 50 %
1. 25 %
1. 15 %

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Let say woman can be a carrier or normal and married to a colorblind man.
So, woman can be either X^RX or XX and man is X^RY
Therefore, X^RX * X^RY and XX * X^RY
X^R X X X
X^R X^RX^R X^RX X^R X^RX X^RX
Y X^RY XY Y XY XY

Therefore, out of 4 children, there is only one chance of having a colorblind daughter.
The probability for 1 out 4 children = ¼ = 0.25 = 25% ;
Now there are 2 probable daughters.
Probability for daughter to be colorblind = ½ = 50%

Correct Option: B

Let say woman can be a carrier or normal and married to a colorblind man.
So, woman can be either X^RX or XX and man is X^RY
Therefore, X^RX * X^RY and XX * X^RY
X^R X X X
X^R X^RX^R X^RX X^R X^RX X^RX
Y X^RY XY Y XY XY

Therefore, out of 4 children, there is only one chance of having a colorblind daughter.
The probability for 1 out 4 children = ¼ = 0.25 = 25% ;
Now there are 2 probable daughters.
Probability for daughter to be colorblind = ½ = 50%