Molecular Basis of Inheritance Test

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Molecular Basis of Inheritance Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

In a DNA strand the nucleotides are linked together by:

A
Glycosidic bonds

B
Phosphodiester bonds

C
Peptide bonds

D
Hydrogen bonds

Solution

In a DNA strand the nucleotides are linked together by 3'-5' phosphodiester linkage(bonds) to form a dinucleotide. More nucleotides can be joined in such a manner to form a polynucleotide chain.
Question 2
1 / -0

A nucleoside differs from a nucleotide. It lacks the:

A
Base

B
Sugar

C
Phosphate group

D
Hydroxyl group

Solution

A nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a nucleoside, i.e., Nucleoside=Nitrogen base Pentose sugar. When a phosphate group is attached to the 5'-OH of a nucleoside through phosphodiester linkage, a nucleotide is formed, i.e., Nucleotide = Nitrogen base+ Pentose sugar +Phosphate \((PO_4)\). So, a nucleoside differs from a nucleotide as it lacks the phosphate group.
Question 3
1 / -0

Both deoxyribose and ribose belong to a class of sugars called:

A
Trioses

B
Hexoses

C
Pentoses

D
Polysaccharides

Solution

Ribose is the pentose sugar in RNA, while deoxyribose is the pentose sugar in DNA.
Question 4
1 / -0

The fact that a purine base always pairs through hydrogen bonds with a pyrimidine base in the DNA double helix leads to:

A
The antiparallel nature

B
The semiconservative nature

C
Uniform width throughout DNA

D
Uniform length in all DNA

Solution

The diameter of the strand is always constant due to a pairing of purine (adenine and guanine) and pyrimidine (cytosine and thymine). This specific bonding gives uniform width to the DNA.
Question 5
1 / -0

The net electric charge on DNA and histones is :

A
Both positive

B
Both negative

C
Negative and positive, respectively

D
Zero

Solution

The negatively charged DNA is wrapped around positively charged histone; making a structure called nucleosome. This explains the efficient packing of DNA in such a small space inside the nucleus.
Question 6
1 / -0

The promoter site and the terminator site for transcription are located at :

A
3' (downstream) end and 5' (upstream) end, respectively of the transcription unit

B
5' (upstream) end and 3' (downstream) end, respectively of the transcription unit

C
The 5' (upstream) end

D
The 3' (downstream) end

Solution

5' (upstream) end and 3' (downstream) end, respectively of the transcription unit.
Question 7
1 / -0

Which of the following statements is the most appropriate for sickle cell anaemia?

A
It cannot be treated with iron supplements

B
It is a molecular disease

C
It confers resistance to acquiring malaria

D
All of the above

Solution

Sickle cell anaemia is a genetic disorder and hence cannot be treated with iron supplements. Altered shape of RBCs confer resistance to malaria in people suffering from sickle cell anaemia.
Question 8
1 / -0

One of the following is true with respect to AUG

A
It codes for methionine only

B
It is an initiation codon

C
It codes for methionine in both prokaryotes and eukaryotes

D
All of the above

Solution

Polypeptide synthesis is signaled by two initiation codons commonly AUG or methionine codon and rarely GUG or valine codon. Since there are 64 triplet codons and only 20 amino acids, the incorporation of some amino acids must be influenced by more than one codon. Only tryptophan (UGG) and methionine (AUG) are specified by single codons. AUG codes for methioninein both prokaryotes and eukaryotes.
Question 9
1 / -0

The first genetic material could be:

A
Protein

B
Carbohydrates

C
DNA

D
RNA

Solution

There are many conclusive proofs which show that RNA was the first genetic material. But RNA being a catalyst was reactive and unstable. So, this paved the way for DNA as genetic material in the living world.
Question 10
1 / -0

With regard to mature mRNA in eukaryotes:

A
Exons and introns do not appear in the mature RNA

B
Exons appear but introns do not appear in the mature RNA

C
Introns appear but exons do not appear in the mature RNA

D
Both exons and introns appear in the mature RNA

Solution

In eukaryotes, the monocistronic structural genes have interrupted coding sequences i.e., the genes in eukaryotes are split. The coding sequences or expressed sequences are defined as exons. These sequences (exons) appear in mature or processed RNA, The exons are interrupted by introns or intervening sequences which do not appear in mature or processed RNA.

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Re-Attempt Weekly Quiz Competition

Molecular Basis of Inheritance Test - 20

Result

Molecular Basis of Inheritance Test - 20

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SHARING IS CARING

Question 1

Glycosidic bonds

Phosphodiester bonds

Peptide bonds

Hydrogen bonds

Solution

Question 2

Base

Sugar

Phosphate group

Hydroxyl group

Solution

Question 3

Trioses

Hexoses

Pentoses

Polysaccharides

Solution

Question 4

The antiparallel nature

The semiconservative nature

Uniform width throughout DNA

Uniform length in all DNA

Solution

Question 5

Both positive

Both negative

Negative and positive, respectively

Zero

Solution

Question 6

3' (downstream) end and 5' (upstream) end, respectively of the transcription unit

5' (upstream) end and 3' (downstream) end, respectively of the transcription unit

The 5' (upstream) end

The 3' (downstream) end

Solution

Question 7

It cannot be treated with iron supplements

It is a molecular disease

It confers resistance to acquiring malaria

All of the above

Solution

Question 8

It codes for methionine only

It is an initiation codon

It codes for methionine in both prokaryotes and eukaryotes

All of the above

Solution

Question 9

Protein

Carbohydrates

DNA

RNA

Solution

Question 10

Exons and introns do not appear in the mature RNA

Exons appear but introns do not appear in the mature RNA

Introns appear but exons do not appear in the mature RNA

Both exons and introns appear in the mature RNA

Solution

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