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Chapter 5 Β· Molecular Basis of Inheritance

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#1 SUB 2M

Question

Give types of nucleic acid and their functions.

Answer

Types:
DNA- deoxyribonucleic acid
RNA- Ribonucleic acid
Function:
  • DNA acts as the genetic material in most of the organisms.
  • RNA also acts as a genetic material in some viruses.
  • Mostly functions as a messenger.
  • RNA has additional roles as well.
  • It functions as adapter, structural, and in some cases as a catalytic molecule.
#2 SUB 2M

Question

The numbers of nucleotide in the length of DNA is different. Give example.

Answer

(i) ss bacteriophage Ο† Γ— 174 β†’ 5386 nucleotide
(ii) ds bacteriophage lambda β†’ 48502 base pairs (bp)
(iii) Escherichia coli (E.coli) β†’ 4.6 Γ— 106 base pairs (bp)
(iv) Human haploid cell (2n) β†’ 6.6 Γ— 109 base pairs (bp)
#3 SUB 2M

Question

What is the monomer of nucleic acid? Give its components?
OR
Explain: Nucleotide

Answer

Nucleotide is a monomer of nucleic acid.
There are three main components: Nitrogen base, pentose sugar, phosphate group

(i) Nitrogen base: There are two types of nitrogenous bases – Purines – Adenine and Guanine, and Pyrimidines – Cytosine, Uracil and Thymine. Cytosine is common for both DNA and RNA and Thymine is present in DNA. Uracil is present in RNA at the place of Thymine.

(ii) Pentose sugar: Deoxyribose sugar (C5H10O4) is present in DNA.

RNA contains ribose sugar (C5H10O5)
(iii) Phosphate group- H3PO4
#4 SUB 2M πŸ–Ό 1

Question

Draw a diagram of structure of polynucleotide strand and explain.

Answer

A nitrogenous base is linked to the OH of 1' C pentose sugar through a N-glycosidic linkage to form a nucleoside.
Two nucleotides are linked through 3'-5' phosphodiester linkage to form dinucleotide. More nucleotides can be joined in such a manner to form polynucleotide chain.
Similarly, at the other end of the polymer the sugar has a free OH of 3'C group which is referred to as 3'-end of the polynucleotide chain. The backbone of a polynucleotide chain is formed due to sugar and phosphates. The nitrogenous bases linked to sugar moiety project from the backbone.
#5 SUB πŸ–Ό 1

Question

Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer

Nitrogen base Nucleoside
β€’ Adenine β€’ Cytidine
β€’ Thymine β€’ Guanosine
β€’ Uracil
β€’ Cytosine
#6 SUB πŸ–Ό 1

Question

If a double stranded DNA has 20 percentage of cytosine, calculate the percentage of adenine in the DNA.

Answer

The ratio of purine and pyrimidines are always equal in a DNA molecule.
Which means the percentage of guanine and cytosine are same.
In DNA, 20% is cytosine, so 20% is guanine.
So, G+C = 40%
The remaining percentage is 60% which is for Adenine + Thymine in DNA.
A+T = 60%
So A = 30% and T = 30%
So the percentage of adenine is 30.
#7 SUB 2M πŸ–Ό 1

Question

Describe history of discovery of DNA.

Answer

DNA as an acidic substance present in nucleus was first identified by Friedrich Miescher in 1869.
He named it as β€˜Nuclein’. However, due to technical limitation in isolating such a long polymer intact, the elucidation of structure of DNA remained elusive for a very long period of time.
It was only in 1953 that James Watson and Francis Crick, based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple but famous Double Helix model for the structure of DNA.
One of the hallmarks of their proposition was base pairing between the two strands of polynucleotide chains.
However, this proposition was also based on the observation of Erwin Chargaff that for a double stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equals one.
#8 SUB 2M πŸ–Ό 1

Question

Describe central dogma of life.
OR
Give a diagrammatic presentation of central dogma.
OR
Write a note on central dogma of life.

Answer

The proposition of a double helix structure for DNA and its simplicity in explaining the genetic implication became revolutionary.
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA β†’ RNA β†’ Protein.
Genetic material expressed in the form of protein. There are two different stages:

(i) Transcription

(ii) Translation

In some viruses the flow of information is in reverse direction.
H. H. Temin and Baltimore stated that some virus has RNA as a genetic material and they form complementary DNA by replication, this is called reverse transcription.
#9 SUB 2M

Question

Calculate the length of double helix DNA present in the typical mammalian cell.

Answer

In a typical mammalian cell ds DNA has
Total base pairs = 6.6 Γ— 109 bp
Taken the distance between two consecutive base pairs as 0.34 nm = 0.34 Γ— 109 meter
Length of DNA = Total number of bp Γ— distance between two consecutive bp

= (6.6 Γ— 109 bp) Γ— (0.34 Γ— 10–9 m)

= 6.6 Γ— 0.34 Γ— 109 Γ— 10–9

= 2.244 meter

Length of DNA = 2.2 m
#10 SUB 4M πŸ–Ό 1

Question

How does nucleosome form in the eukaryotic cell?
OR
Describe: Nucleosome

Answer

There is a set of positively charged, basic proteins called histones.
A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.
Histones are rich in the basic amino acid residues lysine and arginine.
Both the amino acid residues carry positive charges in their side chains.
Histones are organised to form a unit of eight molecules called histone octamer. 2x (H2A, H2B, H3 and H4)
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
A typical nucleosome contains 200 bp of DNA helix.
#11 SUB 2M πŸ–Ό 1

Question

What are the characteristics of nuclear chromatin in a typical mammalian cell?

Answer

Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, thread-like stained (coloured) bodies seen in nucleus.
The nucleosomes in chromatin are seen as β€˜beads-on-string’ structure when viewed under electron microscope (EM).
The beads-on-string structure in chromatin is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes.
In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as euchromatin.
The chromatin that is more densely packed and stains dark are called as Heterochromatin.
Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is transcriptionally inactive.
The packaging of chromatin at higher level requires additional set of proteins that collectively are referred to as Non-histone Chromosomal (NHC) proteins.
#12 SUB 4M πŸ–Ό 1

Question

Explain the packaging of DNA.

Answer

Taken the distance between two consecutive base pairs as 0.34 nm (0.34 Γ— 109m), if the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with distance between two consecutive bp, that is, 6.6 Γ— 10–9 bp Γ—0.34 Γ— 10–9m/bp), it comes out to be approximately 2.2 metres.
A length that is far greater than the dimension of a typical nucleus (approximately 10–6 m).
In prokaryotes, such as, E. coli, though they do not have a defined nucleus, the DNA is not scattered throughout the cell.
DNA (being negatively charged) is held with some proteins (that have positive charges) in a region termed as β€˜nucleoid’.
The DNA in nucleoid is organised in large loops held by proteins.
In eukaryotes, this organisation is much more complex.
There is a set of positively charged, basic proteins called histones.
A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.
Histones are rich in the basic amino acid residues lysine and arginine.
Both the amino acid residues carry positive charges in their side chains.
Histones are organised to form a unit of eight molecules called histone octamer.
The negatively charged DNA is wrapped around the positively charged histone octamer to form
a structure called nucleosome.
A typical nucleosome contains 200 bp of DNA helix. Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, thread-like stained (coloured) bodies seen in nucleus.
The nucleosomes in chromatin are seen as β€˜beads-on-string’ structure when viewed under electron microscope (EM).
The beads-on-string structure in chromatin is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes.
The packaging of chromatin at higher level requires additional set of proteins that collectively are referred to as Non-histone Chromosomal (NHC) proteins.
In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as euchromatin.
The chromatin that is more densely packed and stains dark are called as Heterochromatin.
Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive.
#13 SUB 3M πŸ–Ό 1

Question

What are the salient features of the Double-helix structure of DNA?
OR
Describe any three salient features of the Double helix structure of DNA.

Answer

The salient features of the Double-helix structure of DNA are as follows:
It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and the bases project inside.
The two chains have anti-parallel polarity. It means, if one chain has the polarity 5’→ 3', the other has 3’→ 5'.
The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp).
Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.
Similarly, Guanine is bonded with Cytosine with three H-bonds. As a result, always a purine comes opposite to a pyrimidine.
This generates approximately uniform distance between the two strands of the helix.
The two chains are coiled in a right-handed fashion.
The pitch of the helix is 3.4 nm (a nanometre is one billionth of ametre, that is 10–9 m) and there are roughly 10 bp in each turn.
Consequently, the distance between a bp in a helix is approximately 0.34 nm.
The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confers stability of the helical structure.
#14 SUB πŸ–Ό 2

Question

Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Answer

While proposing the double helical structure of DNA. Watson and Crick has proposed a scheme for replication of DNA.
The scheme suggested that the two strands would separate and act as a template for the synthesis of new complementary strands.
After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand.
This scheme was termed as semi conservative DNA replication.
#15 SUB

Question

Give Biochemical Characterisation of Transforming Principle.

Answer

Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein.
They worked to determine the biochemical nature of β€˜transforming principle’ in Griffith's experiment.
They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells.
They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA.
Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material, but not all biologists were convinced.
#16 SUB 2M

Question

What are the criteria fulfilled by a nucleic acid molecule to act as a genetic material?
OR
What are the criteria of the genetic material?
OR
Describe criteria of a molecule which act as a genetic material.

Answer

A molecule that can act as a genetic material must fulfill the following criteria:
(i) It should be able to generate its replica (Replication).
(ii) It should be stable chemically and structurally.
(iii) It should provide the scope for slow changes (mutation) that are required for evolution.
(iv) It should be able to express itself in the form of 'Mendelian Characters’.
#17 SUB πŸ–Ό 2

Question

Explain Griffith’s transformation experiment.

Answer

In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumonia (bacterium responsible for pneumonia), witnessed a miraculous transformation in the bacteria.
During the course of his experiment, a living organism (bacteria) had changed in physical form.
When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R).
This is because the S strain bacteria have mucous (polysaccharide) coat, while R strain does not.
Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. Griffith was able to kill bacteria by heating them.
He observed that heat-killed S strain bacteria injected into mice did not kill them.
When he injected a mixture of heat-killed S and live R bacteria, the mice died.
Moreover, he recovered living S bacteria from the dead mice.
He concluded that the R strain bacteria had some how been transformed by the heat-killed S strain bacteria.
Transforming principle :
Some β€˜transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent. This must be due to the transfer of the genetic material. However, the biochemical nature of genetic material was not defined from his experiments.
#18 SUB 2M

Question

The DNA is better than RNA as a genetic material. - Explain.
OR
Discuss: DNA is chemically less reactive and structurally more stable when compared to RNA.

Answer

The genetic material should be stable enough not to change with different stages of life cycle, age or with change in physiology of the organism.
Stability as one of the properties of genetic material was very evident in Griffith’s β€˜transforming principle’ itself that heat, which killed the bacteria, at least did not destroy some of the properties of genetic material.
This now can easily be explained in light of the DNA that the two strands being complementary if separated by heating come together, when appropriate conditions are provided. Further,
2'-OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. RNA is also now known to be catalytic, hence reactive.
Therefore, DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material.
#19 SUB 4M πŸ–Ό 1

Question

How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
OR
Give experimental proof for DNA is a genetic material.
OR
Explain Hershey and Chase experiment in detail with diagram.
The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952).
They worked with viruses that infect bacteria called bacteriophages.
The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell.
The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more virus particles.
Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulfur.
Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur. Radioactive phages were allowed to attach to E.coli bacteria.
Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.
The virus particles were separated from the bacteria by spinning them in a centrifuge.
Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
Bacteria that were infected with viruses that had
radioactive proteins were not radioactive.
This indicates that proteins did not enter the bacteria from the viruses.
DNA is therefore the genetic material that is passed from virus to bacteria.

Answer

#20 SUB 2M

Question

How can we say that RNA is the first genetic material?

Answer

RNA was the first genetic material.
There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.), evolved around RNA.
RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein ezymes).
But, RNA being a catalyst was reactive and hence unstable.
Therefore, DNA has evolved from RNA with chemical modifications that make it more stable.
DNA being double stranded and having comple mentary strand further resists changes by evolving a process of repair.
#21 SUB 4M πŸ–Ό 1

Question

Explain the experiment performed by Meselson and Stahl to proof that DNA replication is semi conservative.
OR
Explain Matthew Meselson and Franklin Stahl experimental proof.

Answer

It is now proven that DNA replicates semiconservatively.
It was shown first in Escherichia coli and subsequently in higher organisms, such as plants and human cells. Matthew Meselson and Franklin Stahl performed the following experiment in 1958:
(i) They grew E. coli in a medium containing 15NH4Cl (15N is the heavy isotope of nitrogen) as the only nitrogen source for many generations.
The result was that 15N was incorporated into newly synthesised DNA (as well
as other nitrogen containing compounds).
This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (Please note that 15N is not a radioactive isotope, and it can be separated from 14N only based on densities).
(ii) Then they transferred the cells into a medium with normal 14NH4Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices.
The various samples were separated independently on CsCl gradients to measure the densities of DNA.
(iii) Thus, the DNA that was extracted from the culture one generation after the transfer from 15N to 14N medium [that is after 20 minutes; E. coli divides in 20 minutes] had a hybrid or intermediate density. DNA extracted from the culture after another generation [that is after 40 minutes, II generation] was composed of equal amounts (1:1) of this hybrid DNA and of β€˜light’ DNA.
If the culture of E. coli is taken after
80 minutes then the extracted DNA contains 1:7 of hybrid DNA and light DNA.
#22 SUB 4M πŸ–Ό 1

Question

Explain enzymes and mechanism of DNA replication.

Answer

In living cells, the process of replication requires a set of catalysts (enzymes). Furthermore, energetically replication is a very expensive process.
For long DNA molecules, since the two strands of DNA can not be separated in its entire length due to very high energy requirement.
The replication occur within a small opening of the DNA helix, referred to as replication fork.
The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5' β†’ 3'.
Consequently, on one strand (the template with polarity 3’ β†’ 5'), the replication is continuous, this is known as leading strand.
While on the other (the template with polarity 5’→ 3'), it is discontinuous. The discontinuously synthesised fragments are known as Okazaki fragments.
Later this are joined by the enzyme DNA ligase.
The DNA polymerases on their own cannot initiate the process of replication. Also the replication does not initiate randomly at any place in DNA. There is a definite region, such regions are termed as origin of replication.

Enzymes:

Helicase and gyrase : Helicase unzips the DNA by breaking the hydrogen bonds, separating the two DNA strands so they can be copied. DNA gyrase removes twisting and tension ahead of the replication fork.
RNA Polymerase : A short strand of RNA, complementary to the template DNA at its starting position, is called primer. After the RNA primer is formed, DNA polymerase-III is activated.
DNA polymerase III : It catalyses the polymerization of deoxyribonucleotides on the basis of DNA template.
Ligase : During replication of DNA, the Okazaki fragments of nucleotides are joined by DNA ligase.
#23 SUB 3M πŸ–Ό 1

Question

Give schematic structure of transcription unit and explain template and coding strand.

Answer

There is a convention in defining the two strands of the DNA in the structural gene of a transcription unit. Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the Polymerisation in only one direction, that is, 5'β†’3', the strand that has the polarity 3'β†’5' acts as a template, and is also referred to as template strand.
Coding strand: The other strand which has the polarity (5'β†’3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. Strangely, this strand (which does not code for anything) is referred to as coding strand.
All the reference point while defining a transcription unit is made with coding strand.
#24 SUB 2M

Question

Why do both the strand of DNA not participate in a transcription?

Answer

First, if both strands act as a template, they would code for RNA molecule with different sequences (Remember complementarity does not mean identical), and in turn, if they code for proteins, the sequence of amino acids in the proteins would be different.
Hence, one segment of the DNA would be coding for two different proteins, and this would complicate the genetic information transfer machinery.
Second, the two RNA molecules if produced simultaneously would be complementary to each other, hence would form a double stranded RNA.
This would prevent RNA from being translated into protein and the exercise of transcription would become a futile one.
#25 SUB πŸ–Ό 1 β–¦ 1

Question

Give difference between: Template strand and Coding strand.

Answer

Template Strand

Coding Strand

1.

Template strand of
DNA acts as a template for the synthesis of mRNA during replication

1.

A strand of DNA has the same sequence as mRNA. but contains thymine instead of Uracil.

2.

It's polarity is 3' β†’ 5'

2.

It's polarity is 5' β†’ 3'

3.

It has codons for protein synthesis

3.

It does not have codon for protein synthesis.
#26 SUB 2M πŸ–Ό 1

Question

Explain the scientific term Cistron, monocistronic, polycistronic, exons and introns given below:
OR
Describe : Exons.

Answer

Cistron, monocistronic, polycistronic, exons and introns.
Cistron: Cistron is a segment of DNA coding for a polypeptide, the structural gene in transcription unit.
Monocistronic: In eukaryotes, during transcription of mRNA only one gene participate so it is called monocistronic.
Polycistronic: Polycistronic, mostly in bacteria or prokaryotes, in which more than one gene participates during transcription of mRNA.
Example: Lac operon contains polycistronic regions.
Exons: In eukaryotes, the monocistronic structural genes have interrupted coding sequences – the genes in eukaryotes are split.
  • The coding sequences or expressed sequences are defined as exons.
  • Exons are said to be those sequence that appear in mature or processed RNA.
Introns: The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA. The split-gene arrangement further complicates the definition of a gene in terms of a DNA segment.
#27 SUB πŸ–Ό 1

Question

If the sequence of one strand of DNA is written as follows :
5'–ATGCATGCATGCATGCATGCATGCATGC–3'
Write down the sequence of complementary strand in 5' β†’ 3' direction.

Answer

5' – ATGCATGCATGCATGCATGCATGCATGC – 3'
The Sequence on complementary 3' β†’ 5' strand is
3'–TACGTACGTACGTACGTACGTACGTACG–5'Β 
Then, The sequence of complementary strand is
5'–GCATGCATGCATGCATGCATGCATGCAT–3'.
#28 SUB πŸ–Ό 1

Question

If the sequence of one coding strand in a transcription unit is written as follows :
5' – ATGCATGCATGCATGCATGCATGCATGC – 3'
Write down the sequence of mRNA.

Answer

If the coding strand has following sequence
5'–ATGCATGCATGCATGCATGCATGCATGC–3'
So, template strand sequence in 3' β†’ 5' will be
3'–TACGTACGTACGTACGTACGTACGTACG–5'
There is Uracil present instead of Thymine in m-RNA then, the sequence on m-RNA is
5'–AUGCAUGCAUGCAUGCAUGCAUGCAUGC–3'
(Note : The sequence of m-RNA is exactly similar to coding strand of DNA, but only the difference is, in m-RNA Uracil is present instead of thymine
#29 SUB πŸ–Ό 1

Question

Describe transcription.

Answer

The process of copying genetic information from one strand of the DNA into RNA is termed as transcription. Here also, the principle of complementarity governs the process of transcription, except the adenosine complements now forms base pair with uracil instead of thymine.
However, unlike in the process of replication, which once set in, the total DNA of an organism gets duplicated, in transcription only a segment of DNA and only one of the strands is copied into RNA.
There are mainly three parts of transcription unit:
Promoter, structural gene, terminator
#30 SUB πŸ–Ό 1

Question

Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

Answer

(1) DNA dependent DNA Polymerase :
Catalyse the polymerisation of deoxyribonucleic acid on the basis of template.
(2) DNA dependent RNA Polymerase :
Catalyse the transcription of all types of RNA in bacteria.
(3) DNA dependent RNA Polymerase I :
It transcribes 28S, 18S and 5.8S rRNA.
(4) DNA dependent RNA Polymerase II :
It transcribes hnRNA (heterogenous nuclear RNA) which is precursor of mRNA.
(5) DNA dependent RNA Polymerase III :
It is responsible for transcription of trna 5S rrna and snRNA (Small Nuclear RNAs)
#31 SUB πŸ–Ό 2

Question

Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons

Answer

(a) Promoter :
It acts as an initiation site.
RNA polymerase binds with this site and initiate transcription.
The promoter is said to be located towards 5'-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand).
It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that also defines the template and coding strands.
(b) tRNA:
Adaptive molecule, the tRNA, then called sRNA (soluble RNA).
tRNA has anti codon loop that has bases complementary to the codon.
It also has an amino acid acceptor end at 3’end to which it binds to amino acids. tRNAs are specific for each amino acid.
(c) Exon : Exons are said to be the sequence that Appear in mature or processed mRNA.
#32 SUB 4M πŸ–Ό 1

Question

Explain transcription unit with schematic diagram.

Answer

A transcription unit in DNA is defined primarily by the three regions in the DNA:
  • A Promoter
  • The Structural gene
  • A Terminator
(i) A Promoter:
It acts as an initiation site.
RNA polymerase binds with this site and initiate transcription.
The promoter is said to be located towards 5'-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand).
It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that also defines the template and coding strands.
(ii) The structural gene:
There is a convention in defining the two strands of the DNA in the structural gene of a transcription unit. Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction, that is, 5'β†’3', the strand that has the polarity 3'β†’5' acts as a template, and is also referred to as template strand.
The other strand which has the polarity (5' β†’ 3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. Strangely, this strand (which does not code for anything) is referred to as coding strand.
All the reference point while defining a transcription unit is made with coding strand.
3'-ATGCATGCATGCATGCATGCATGC-5' Template Strand
5'-TACGTACGTACGTACGTACGTACG-3' Coding Strand
(iii) Terminator:
The terminator is located towards 3'-end downstream of the coding strand and it usually defines the end of the process of transcription.
There are additional regulatory sequences that may be present further upstream or downstream to the promoter.
#33 SUB 4M πŸ–Ό 1

Question

Explain transcription in prokaryotes (bacteria) with diagram.

Answer

In bacteria, there are three major types of RNAs: mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA).
All three RNAs are needed to synthesise a protein in a cell.
The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation.
There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria.
RNA polymerase binds to promoter and initiates transcription (Initiation).
It uses nucleoside triphosphates as substrate and polymerises in a template dependent fashion following the rule of complementarity.
It somehow also facilitates opening of the helix and continues elongation.
Only a short stretch of RNA remains bound to the enzyme.
Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription.
An intriguing question is that how is the RNA polymerases able to catalyse all the three steps, which are initiation, elongation and termination.
The RNA polymerase is only capable of catalysing the process of elongation.
It associates transiently with initiation-factor (Οƒ) and termination-factor (ρ) to initiate and terminate the transcription, respectively.
Association with these factors alter the specificity of the RNA polymerase to either initiate or terminate
In bacteria, since the mRNA does not require any processing to become active, and also since transcription and translation take place in the same compartment (there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed.
Consequently, the transcription and translation can be coupled in bacteria.
#34 SUB 3M

Question

Explain transcription in eukaryotes with diagram.
OR
Explain splicing and tailing in eukaryotes. (Diagram is not required).
OR
Describe types of RNA and explain two additional complexity in the transcription for eukaryotes.

Answer

In eukaryotes, there are two additional complexities – There are at least three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles).
There is a clear cut division of labour.
The RNA polymerase I : It transcribes rRNAs (28S, 18S, and 5.8S).
RNA polymerase III : It is responsible for transcription of tRNA, 5Sr RNA, and snRNAs (small nuclear RNAs).
The RNA polymerase II : It transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).
The second complexity is that the primary transcripts contain both the exons and the introns and are non-functional.
Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order.
hnRNA undergoes additional processing called as capping and tailing.
In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA.
In tailing, adenylate residues (200-300) are added at 3'-end in a template independent manner. It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation.
The significance of such complexities is now beginning to be understood.
The split-gene arrangements represent probably an ancient feature of the genome.
The presence of introns is reminiscent of antiquity, and the process of splicing represents the dominance of RNA-world.
In recent times, the understanding of RNA and RNA-dependent processes in the living system have assumed more importance.
#35 SUB 2M β–¦ 1

Question

Give differences between DNA and RNA.

Answer

DNA

RNA

1.

It is made up of two polynucleotide strand.

1.

It contains only one polynucleotide strand

2.

It contains Adenine, Guanine, Cytosine and Thymine nitrogen base.

2.

RNA contains nitrogen base such as adenine, guanine, cytosine and uracil.

3.

In most of the organisms DNA is a genetic material.

3.

In certain viruses RNA is a genetic material.

4.

DNA is dependent on RNA for protein synthesis.

3.

RNA can directly have codon for protein synthesis.
#36 SUB πŸ–Ό 1 β–¦ 1

Question

Differentiate : m-RNA and t-RNA

Answer

m-RNA

t-RNA

1.

It has linear structure.

1.

It has clover leaf structure.

2.

m-RNA acts as a template for the translation because it contains codon.

2.

t-RNA carries specific amino acid on the basis of m-RNA codon for protein synthesis.

3.

m-RNA degenerate as soon as it's function is over.

3.

They are not degenerated after their work is finished.

4.

Synthesised by RNA polymerase II.

4.

It is synthesised by RNA polymerase III.
#37 SUB

Question

What are the salient features of genetic code?

Answer

The salient features of genetic code are as follows:
(i) The codon is triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons.
(ii) Some amino acids are coded by more than one codon, hence the code is degenerate.
(iii) The codon in mRNA is read in a contiguous fashion. There are no punctuations.
(iv) The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe). Some exceptions to this rule have been found in mitochondrial codons, and in some protozoans.
(v) AUG has dual functions. It codes for Methionine (met), and it also act as initiator codon.
(vi) UAA, UAG, UGA are stop terminator codons.
#38 SUB 2M

Question

Explain frame shift mutation and deletion.

Answer

Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
However, such mutations are referred to as frameshift insertion or deletion mutations.
Insertion or deletion of three or its multiple bases adds or removes one or multiple codon hence one or multiple amino acids are adds or deleted and reading frame remains unaltered from that point onwards.
#39 SUB 3M πŸ–Ό 1

Question

Explain t-RNA as an adaptive molecule.

Answer

The tRNA, then called sRNA (soluble RNA).
From the very beginning of the proposition of code, it was clear to Francis Crick that there has to be a mechanism to read the code and also to link it to the amino acids, because amino acids have no structural specialities to read the code uniquely.
However, its role as an adapter molecule was assigned much later.
tRNA has an anticodon loop that has bases complementary to the code, and it also has an amino acid acceptor end to which it binds to amino acids.
Example: AUG is a code for methionine. So its complementary code- anticodon is UAC present on the anticodon loop of tRNA.
tRNAs are specific for each amino acid.
For initiation, there is another specific tRNA that is referred to as initiator tRNA.
There are no tRNAs for stop codons.
The secondary structure of tRNA has been depicted that looks like a clover-leaf.
In actual structure, the tRNA is a compact molecule which looks like inverted L.
#40 SUB πŸ–Ό 1

Question

List two essential roles of ribosome during translation.

Answer

Essential roles of ribosome:
The cellular factory responsible for synthesising proteins is the ribosome.
In its inactive state, it exists as two subunits; a large subunit and a small subunit.
When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins.
There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond.
The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond.
#41 SUB 4M πŸ–Ό 2

Question

Explain the process refers to the polymerisation of amino acids to form a polypeptide.
OR
Explain the process for protein synthesis in detail.
OR
Explain the process of translation.

Answer

Translation refers to the process of polymerisation of amino acids to form a polypeptide.
The order and sequence of amino acids are defined by the sequence of bases in the mRNA.
The amino acids are joined by a bond which is known as a peptide bond.
Formation of a peptide bond requires energy. Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA – a process commonly called as charging of tRNA or amino acylation of tRNA to be more specific.
If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically.
The presence of a catalyst would enhance the rate of peptide bond formation.
The cellular factory responsible for synthesising proteins is the ribosome.
The ribosome consists of structural RNAs and about 80 different proteins.
In its inactive state, it exists as two subunits; a large subunit and a small subunit.
When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins.
There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond.
The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme - ribozyme) for the formation of peptide bond.
A translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for a polypeptide.
An mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR).
The UTRs are present at both 5'-end (before start codon) and at 3'-end (after stop codon).
They are required for efficient translation process.
For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA.
The ribosome proceeds to the elongation phase of protein synthesis.
During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one, translated into Polypeptide sequences dictated by DNA and represented by mRNA.
At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.
#42 SUB 4M πŸ–Ό 1

Question

Explain Lac operon.
OR
Explain Lac operon in the presence and absence of inducer.
OR
Explain lactose metabolism model given by Jacob and Monod in bacteria. (diagram is not required).
The elucidation of the lac operon was also a result of a close association between a geneticist, Francois Jacob and a biochemist, Jacque Monod.
They were the first to elucidate a transcriptionally regulated system.
In lac operon (here lac refers to lactose),
a polycistronic structural gene is regulated by a common promoter and regulator genes.
Such arrangement is very common in bacteria and is referred to as operon.

Answer

The lac operon consists of one regulatory gene (the i gene – here the term i does not refer to inducer, rather it is derived from the word inhibitor) and three structural genes (z, y, and a).
i- Gene: The i gene codes for the repressor of the lac operon.
The z gene codes for beta galactosidase (Ξ²-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose.
The y gene codes for permease, which increases permeability of the cell to Ξ²-galactosides.
The a gene encodes a transacetylase. Hence, all the three gene products in lac operon are required for metabolism of lactose.
In most other operons as well, the genes present in the operon are needed together to function in the same or related metabolic pathway.
Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon.
Hence, it is termed as inducer.
In the absence of a preferred carbon source such as glucose, if lactose is provided in the growth medium of the bacteria, the lactose is transported into the cells through the action of permease (Remember, a very low level of expression of lac operon has to be present in the cell all the time, otherwise lactose cannot enter the cells).
The lactose then induces the operon in the following manner.
The repressor of the operon is synthesised (all-the-time – constitutively) from the i gene.
The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.
In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.
This allows RNA polymerase access to the promoter and transcription proceeds.
Essentially, regulation of lac operon can also be visualised as regulation of enzyme synthesis by its substrate.
#43 SUB πŸ–Ό 1

Question

In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?

Answer

The lactose induces the operon in the following manner.
The repressor of the operon is synthesised (all-the-time – constitutively) from the i gene.
The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.
In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.
This allows RNA polymerase access to the promoter and transcription proceeds.
All three structural gene z, y, a is expressed and produce beta - galactosidase, permease and transacetylase.
All three enzyme metabolised lactose and convert it into glucose and galactose.
Essentially, regulation of lac operon can also be visualised as regulation of enzyme synthesis by its substrate.
In the absence of lactose, the repressor protein combines with operator and does not allow RNA polymerase to transcribe.
Thus the expression of lac operon is inhibited.
#44 SUB 2M

Question

Describe goals of HGP.

Answer

Goals of HGP
Some of the important goals of HGP were as follows:
(i) Identify all the approximately 20,000-25,000 genes in human DNA.
(ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA;
(iii) Store this information in databases;
(iv) Improve tools for data analysis;
(v) Transfer related technologies to other sectors, such as industries;
(vi) Address the ethical, legal, and social issues (ELSI) that may arise from the project.
#45 SUB πŸ–Ό 1

Question

Describe: Bioinformatics.

Answer

Bioinformatics is an important computer and statistical technique of molecular biology.
Bioinformatics enables the generation of biological data and the storage of a wide range of biological information.
It has developed a number of tools to make information easily and effectively accessible and usable.
By inventing new algorithms and statistical methods, bioinformatics can obtain information about protein structure and function.
With its help, practical problems arising in the analysis and management of biological data can be solved.
#46 SUB 3M

Question

What are the Salient Features of Human Genome?
OR
Describe any six characteristics of Human Genome Project.(MARCH 2025, June 2025)
OR
Describe any six features of HGP.

Answer

Salient Features of Human Genome
Some of the salient observations drawn from human genome project are as follows:
(i) The human genome contains 3164.7 million bp.
(ii) The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated at 30,000 – much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 percent) nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50 percent of the discovered genes.
(v) Less than 2 percent of the genome codes for proteins.
(vi) Repeated sequences make up very large portion of the human genome.
(vii) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
(viii) Chromosome 1 has most genes (2968), and the Y has the fewest (231).
(ix) Scientists have identified about 1.4 million locations where single-base DNA differences (SNPs – single nucleotide polymorphism, pronounced as β€˜snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.
#47 SUB πŸ–Ό 1

Question

Why is the Human Genome project called a mega project ?

Answer

Human Genome Project (HGP) was called a mega project. You can imagine the magnitude and the requirements for the project if we simply define the aims of the project as follows:
Human genome is said to have approximately
3 Γ— 109 bp, and if the cost of sequencing required is US $ 3per bp (the estimated cost in the beginning), the total estimated cost of the project would be approximately 9 billion US dollars.
Further, if the obtained sequences were to be stored in typed form in books, and if each page of the book contained 1000 letters and each book contained 1000 pages, then 3300 such books would be required to store the information of DNA sequence from a single human cell.
The enormous amount of data expected to be generated also necessitated the use of high speed computational devices for data storage and retrieval, and analysis.
HGP was closely associated with the rapid development of a new area in biology called Bioinformatics.
#48 SUB 4M

Question

Describe the methodology of human genome project.
OR
What are the different methodolog is used for HGP?

Answer

Methodologies : The methods involved two major approaches. One approach focused on identifying all the genes that are expressed as RNA (referred to as Expressed Sequence Tags (ESTs)).
The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).
For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors.
The cloning resulted into amplification of each piece of DNA fragment so that it subsequently could be sequenced with ease.
The commonly used hosts were bacteria and yeast, and the vectors were called as BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes).
The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger.
These sequences were then arranged based on some overlapping region spresent in them.
This required generation of overlapping fragments for sequencing.
Alignment of these sequences was manually not possible.
Therefore, specialised computer based programs were developed.
#49 SUB πŸ–Ό 1 β–¦ 1

Question

Give different between: Repetitive DNA and Satellite DNA.

Answer

Repetitive DNA

Satellite DNA

1.

A small stretch of DNA is repeated many times, these are called repetitive DNA.

1.

Satellite DNA contains a large number of repetitive DNA sequences.

2.

In CsCL density gradient analysis, light bands of repetitive DNA are seen.

2.

In CsCT density gradient analysis, dark bands of repetitive DNA are seen.
#50 SUB

Question

What are the steps of DNA fingerprinting?

Answer

(i) Isolation of DNA from the sample.
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.
(v) Hybridisation using labelled VNTR probe, and
(vi) Detection of hybridised DNA fragments by autoradiography.
#51 SUB 2M

Question

Describe: satellite DNA

Answer

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.
These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.
The bulk DNA forms a major peak and the other small peaks are referred to as satellite DNA. Depending on base composition (A : T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as micro-satellites, mini-satellites etc.
These sequences normally do not code for any proteins, but they form a large portion of human genome.
These sequence show high degree of polymorphism and form the basis of DNA fingerprinting.
#52 SUB πŸ–Ό 1

Question

What is DNA fingerprinting? What are its applications?

Answer

DNA fingerprinting can establish the identity of different individuals at the DNA level, and show the differences between them.
This method is based on the polymorphism and diversity in DNA sequence.
Applications:
To determine paternity and family relationships
To establish the identity of criminals in the field of forensic science
To identify and protect commercial varieties of crops and domestic animals
For determination of population and genetic diversity
#53 SUB πŸ–Ό 1

Question

Briefly describe the polymorphism seen in the DNA sequence.

Answer

The length of DNA segments as well as sequences containing repetitive sequences exhibit high levels of polymorphism.
Polymorphism found in DNA sequences is useful in DNA fingerprinting as well as genetic mapping of the human genome.
Polymorphism means variation on a genetic basis, caused by a mutation.
Sequence variation is traditionally called DNA polymorphism.
In simple words, if a hereditary disorder occurs more frequently in a population, it is called a DNA polymorphism.
The probability of this variation is higher in non-coding DNA.
A polymorphism of the same status is found in DNA obtained from every tissue (such as blood, hair follicle, skin, bone, saliva, sperm, etc.) of an individual.
Polymorphism is inherited from parent to offspring.
Polymorphisms are of various types, ranging from single nucleotide to large changes.
Polymorphism plays a very important role in evolution or speciation.
S src: Multiple Choice Questions (MCQs) match 85% type: 230 Q – Export ZIP
#54 MCQ 1M

Question

In a DNA strand the nucleotides are linked together by :

Options

  1. (A) glycosidic bonds
  2. (B) phosphodiester bonds
  3. (C) peptide bonds
  4. (D) hydrogen bonds

Answer

(B) phosphodiester bonds

#55 MCQ 1M

Question

A nucleoside differs from a nucleotide. It lacks the :

Options

  1. (A) base
  2. (B) sugar
  3. (C) phosphate group
  4. (D) hydroxyl group

Answer

(C) phosphate group

#56 MCQ 1M

Question

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

Options

  1. (A) trioses
  2. (B) hexoses
  3. (C) pentoses
  4. (D) polysaccharides

Answer

(C) pentoses

#57 MCQ 1M

Question

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

Options

  1. (A) the antiparallel nature
  2. (B) the semiconservative nature
  3. (C) uniform width throughout DNA
  4. (D) uniform length in all DNA

Answer

(C) uniform width throughout DNA

#58 MCQ 1M

Question

Which of the following is true with respect to AUG?

Options

  1. (A) It codes for methionine only
  2. (B) It is an initiation codon
  3. (C) It codes for methionine in both prokaryotes and eukaryotes
  4. (D) All of the above

Answer

(D) All of the above

#59 MCQ 1M

Question

The first genetic material could be :

Options

  1. (A) protein
  2. (B) carbohydrates
  3. (C) DNA
  4. (D) RNA

Answer

(D) RNA

#60 MCQ 1M

Question

Discontinuous synthesis of DNA occurs in one strand, because:

Options

  1. (A) DNA molecule being synthesized is very long.
  2. (B) DNA dependent DNA polymerase catalyses polymerisation only in one direction
    (5’ β†’ 3’).
  3. (C) It is a more efficient process.
  4. (D) DNA ligase joins the short stretches of DNA.

Answer

(B) DNA dependent DNA polymerase catalyses polymerisation only in one direction
(5’ β†’ 3’).

#61 MCQ 1M

Question

The net electric charge on DNA and histones is:

Options

  1. (A) both positive
  2. (B) both negative
  3. (C) negative and positive, respectively
  4. (D) zero

Answer

(C) negative and positive, respectively

#62 MCQ 1M

Question

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

Options

  1. (A) 3' (downstream) end and 5' (upstream) end, respectively of the transcription unit
  2. (B) 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit
  3. (C) the 5' (upstream) end
  4. (D) the 3' (downstream) end

Answer

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

#63 MCQ 1M

Question

The human chromosome with the highest and least number of genes in them are respectively:

Options

  1. (A) Chromosome 21 and Y
  2. (B) Chromosome 1 and X
  3. (C) Chromosome 1 and Y
  4. (D) Chromosome X and Y

Answer

(C) Chromosome 1 and Y

#64 MCQ 1M

Question

In some viruses, DNA is synthesized by using RNA as template. Such a DNA is called:

Options

  1. (A) A-DNA
  2. (B) B-DNA
  3. (C) cDNA
  4. (D) rDNA

Answer

(C) cDNA

#65 MCQ 1M

Question

With regards to mature mRNA in eukaryotes:

Options

  1. (A) exons and introns do not appear in the mature RNA
  2. (B) exons appear but introns do not appear in the mature RNA
  3. (C) introns appear but exons do not appear in the mature RNA
  4. (D) both exons and introns appear in the mature RNA

Answer

(B) exons appear but introns do not appear in the mature RNA

#66 MCQ 1M

Question

Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?

Options

  1. (A) Rosalind Franklin
  2. (B) Maurice Wilkins
  3. (C) Erwin Chargaff
  4. (D) Meselson and Stahl

Answer

(D) Meselson and Stahl

#67 MCQ 1M

Question

DNA is a polymer of nucleotides which are linked to each other by 3’-5’ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?

Options

  1. (A) Replace purine with pyrimidines
  2. (B) Remove\ Replace 3' OH group in deoxy ribose
  3. (C) Remove\ Replace 2' OH group with some other group in deoxy ribose
  4. (D) Both β€˜B’ and β€˜C’

Answer

(B) Remove\ Replace 3' OH group in deoxy ribose

#68 MCQ 1M

Question

Which of the following steps in transcription is catalysed by RNA polymerase?

Options

  1. (A) Initiation
  2. (B) Elongation
  3. (C) Termination
  4. (D) All of the above

Answer

(B) Elongation

#69 MCQ 1M

Question

Control of gene expression in prokaryotes take place at the level of:

Options

  1. (A) DNA-replication
  2. (B) Transcription
  3. (C) Translation
  4. (D) None of the above

Answer

(B) Transcription

#70 MCQ 1M

Question

Which of the following statements is correct about the role of regulatory proteins in transcription in prokaryotes?

Options

  1. (A) They only increase expression.
  2. (B) They only decrease expression.
  3. (C) They interact with RNA polymerase but do not affect the expression.
  4. (D) They can act both as activators and as repressors.

Answer

(D) They can act both as activators and as repressors.

#71 MCQ 1M

Question

Which was the last human chromosome to be completely sequenced ?

Options

  1. (A) Chromosome 1
  2. (B) Chromosome 11
  3. (C) Chromosome 21
  4. (D) Chromosome X

Answer

(A) Chromosome 1

#72 MCQ 1M

Question

The amino acid attaches to the tRNA at its:

Options

  1. (A) 5' - end
  2. (B) 3' - end
  3. (C) Anti codon site
  4. (D) DHU loop

Answer

(B) 3' - end

#73 MCQ 1M

Question

Which of the following is the function of RNA?

Options

  1. (A) It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
  2. (B) It carries amino acids to ribosomes.
  3. (C) It is a constituent component of ribosomes.
  4. (D) All of the above.

Answer

(D) All of the above.

#74 MCQ 1M

Question

While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that :

Options

  1. (A) It is a double stranded circular DNA.
  2. (B) It is single stranded DNA.
  3. (C) It is a double stranded linear DNA.
  4. (D) No conclusion can be drawn.

Answer

(B) It is single stranded DNA.

#75 MCQ 1M

Question

If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of N15/N15 : N15/N14 : N14/N14 containing DNA in the fourth generation would be :

Options

  1. (A) 1 : 1 : 0
  2. (B) 1 : 4 : 0
  3. (C) 0 : 1 : 3
  4. (D) 0 : 1 : 7

Answer

(B) 1 : 4 : 0

#76 MCQ 1M

Question

If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is :
5' - A T G A A T G - 3',
the sequence of bases in its RNA transcript would be;

Options

  1. (A) 5' - A U G A A U G - 3'
  2. (B) 5' - U A C U U A C - 3'
  3. (C) 5' - C A U U C A U - 3'
  4. (D) 5' - G U A A G U A - 3'

Answer

(A) 5' - A U G A A U G - 3'

#77 MCQ 1M

Question

The RNA polymerase holoenzyme transcribes:

Options

  1. (A) the promoter, structural gene and the terminator region
  2. (B) the promoter and the terminator region
  3. (C) the structural gene and the terminator region
  4. (D) the structural gene only

Answer

(D) the structural gene only

#78 MCQ 1M

Question

If the base sequence of a codon in mRNA is 5'-AUG-3', the sequence of tRNA pairing with it must be:

Options

  1. (A) 5' - UAC - 3'
  2. (B) 5' - CAU - 3'
  3. (C) 5' - AUG - 3'
  4. (D) 5' - GUA - 3'

Answer

(B) 5' - CAU - 3'

#79 MCQ 1M

Question

To initiate translation, the mRNA first binds to :

Options

  1. (A) The smaller ribosomal sub-unit
  2. (B) The larger ribosomal sub-unit
  3. (C) The whole ribosome
  4. (D) No such specificity exists.

Answer

(A) The smaller ribosomal sub-unit

#80 MCQ 1M

Question

In E.coli, the lac operon gets switched on when:

Options

  1. (A) lactose is present and it binds to the repressor.
  2. (B) repressor binds to operator.
  3. (C) RNA polymerase binds to the operator.
  4. (D) lactose is present and it binds to RNA polymerase.

Answer

(A) lactose is present and it binds to the repressor.

#81 MCQ 1M

Question

Chargaff's rules is regarding ________.

Options

  1. (A) Composition of bases in RNA
  2. (B) Number of chromosomes
  3. (C) Composition of bases in DNA
  4. (D) complexity of genetic material

Answer

(C) Composition of bases in DNA

#82 MCQ 1M

Question

If the total number of (T) is 77 then the total number of (C) will be : ________.

Options

  1. (A) 77
  2. (B) 100
  3. (C) 154
  4. (D) Can't say

Answer

(D) Can't say

#83 MCQ 1M

Question

Nucleotides are building blocks of nucleic acids. Each nucleotide is a composite molecule formed by

Options

  1. (A) (Base βˆ’ sugar βˆ’ phosphate) N
  2. (B) Base-sugar βˆ’OH
  3. (C) Base-sugar-phosphate
  4. (D) Sugar-phosphate

Answer

(C) Base-sugar-phosphate

#84 MCQ 1M

Question

Who among the following proposed rules regarding the composition of bases in DNA?

Options

  1. (A) Watson Crick
  2. (B) Erwin Chargaff
  3. (C) Dr. Hargovind Khorana
  4. (D) Griffith

Answer

(B) Erwin Chargaff

#85 MCQ 1M

Question

The first rule of Chargaff :

Options

  1. (A) A + G = T + C
  2. (B) A+ T = G + C
  3. (C) A + C = T + C
  4. (D) A= G

Answer

(A) A + G = T + C

#86 MCQ 1M πŸ–Ό 1

Question

It varies with the organism.

Options

  1. (A) A + G = T + C
  2. (B) A= G
  3. (C) A = T and G = C
  4. (D)

Answer

(D)

#87 MCQ 1M

Question

If the total amount of Adenine (A) and guanine (G) is 100 % then the total amount of thymine (T) and cytosine (C) will be ________.

Options

  1. (A) 50%
  2. (B) 75%
  3. (C) 25%
  4. (D) 100%

Answer

(D) 100%

#88 MCQ 1M

Question

In DNA if 10% guanine is present, how much is thymine present?

Options

  1. (A) 10%
  2. (B) 40%
  3. (C) 80%
  4. (D) 20%

Answer

(B) 40%

#89 MCQ 1M

Question

The unidirectional flow of genetic information from DNA β†’ m – RNA β†’ Protein replication is known as ________.

Options

  1. (A) Transcription
  2. (B) Replication
  3. (C) Central dogma
  4. (D) Translation

Answer

(C) Central dogma

#90 MCQ 1M

Question

Reverse transcription means ________.

Options

  1. (A) Formation of m- RNA from DNA
  2. (B) Formation of DNA from RNA
  3. (C) Formation of DNA from DNA
  4. (D) Formation of t-RNA from m-RNA

Answer

(B) Formation of DNA from RNA

#91 MCQ 1M

Question

DNA polymerase links nucleotide by forming which type of bond?

Options

  1. (A) Phosphodiester bond
  2. (B) Hydrogen bond
  3. (C) Hydrogen bond
  4. (D) Ester bond

Answer

(A) Phosphodiester bond

#92 MCQ 1M

Question

Which of the following is a nucleoside?

Options

  1. (A) Adenosine, Adenylic acid, Cytosine
  2. (B) Adenosine, Guanosine, Cytidine
  3. (C) Cytidylic acid, Adenosine, Adenylic acid
  4. (D) Guanylic acid, Cytosine, Adenosine

Answer

(B) Adenosine, Guanosine, Cytidine

#93 MCQ 1M

Question

Thirty percent of the bases in a sample of DNA extracted from eukaryotic cells is adenine. What percentage of cytosine is present in this DNA?

Options

  1. (A) 10%
  2. (B) 20%
  3. (C) 30%
  4. (D) 40%

Answer

(B) 20%

#94 MCQ 1M

Question

If a length of DNA has 45,000 base pairs,
how many complete turns will the DNA molecule take?

Options

  1. (A) 4,500
  2. (B) 45,000
  3. (C) 45
  4. (D) 450

Answer

(A) 4,500

#95 MCQ 1M

Question

If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?

Options

  1. (A) T : 20; G : 30; C : 20
  2. (B) T : 20; G : 20; C : 30
  3. (C) T : 30; G : 20; C : 20
  4. (D) T : 20; G : 25; C : 25

Answer

(C) T : 30; G : 20; C : 20

#96 MCQ 1M

Question

One turn of the helix in a B form DNA is approximately

Options

  1. (A) 2 nm
  2. (B) 20 nm
  3. (C) 0.34 nm
  4. (D) 3.4 nm

Answer

(D) 3.4 nm

#97 MCQ 1M

Question

Who among the following did the experiment of bacterial transformation?

Options

  1. (A) Avery, McCarty, Macleod
  2. (B) Griffith
  3. (C) Hershey, Chase
  4. (D) Watson, Crick

Answer

(B) Griffith

#98 MCQ 1M

Question

Griffith's experiment is famous for ________.

Options

  1. (A) viral transformation
  2. (B) pneumococcus transduction
  3. (C) bacterial transformation
  4. (D) bacterial recombination

Answer

(C) bacterial transformation

#99 MCQ 1M

Question

The organism used in Griffith's experiment was ________.

Options

  1. (A) Spirulina
  2. (B) Bacillus
  3. (C) Pneumococcus
  4. (D) Salmonella

Answer

(C) Pneumococcus

#100 MCQ 1M

Question

Griffith had performed series of experiments on

Options

  1. (A) Pneumococcus
  2. (B) Bacillus
  3. (C) Vibrio bacteria
  4. (D) Spirilium bacteria

Answer

(A) Pneumococcus

#101 MCQ 1M

Question

Which of the following is incorrect for Hershey - Chase experiment?

Options

  1. (A) Infection
  2. (B) Blending
  3. (C) Centrifugation
  4. (D) PCR

Answer

(D) PCR

#102 MCQ 1M

Question

Which of the following is not correct for molecule that act as a genetic material?

Options

  1. (A) It should be able to generate its replica
  2. (B) It should be structurally stable
  3. (C) It should be able to express itself in the form of Mendelian characters
  4. (D) It should be Heat resistant

Answer

(D) It should be Heat resistant

#103 MCQ 1M

Question

The unequivocal proof of DNA as the genetic material came from the studies on a

Options

  1. (A) bacterium
  2. (B) fungus
  3. (C) viroid
  4. (D) bacterial virus

Answer

(D) bacterial virus

#104 MCQ 1M

Question

Taylor conducted the experiments to prove semiconservative mode of chromosome replication on

Options

  1. (A) Vinca rosea
  2. (B) Vicia faba
  3. (C) Drosophila melanogaster
  4. (D) E. coli.

Answer

(B) Vicia faba

#105 MCQ 1M

Question

The term 'Nuclein' for the genetic material was used by

Options

  1. (A) Mendel
  2. (B) Franklin
  3. (C) Meischer
  4. (D) Chargaff

Answer

(C) Meischer

#106 MCQ 1M

Question

DNA strand is directly involved in the synthesis of all of the following except

Options

  1. (A) tRNA molecule
  2. (B) mRNA molecule
  3. (C) Another DNA strand
  4. (D) Protein synthesis

Answer

(D) Protein synthesis

#107 MCQ 1M

Question

The association of histone H1 with a nucleosome indicates that

Options

  1. (A) DNA replication is occurring
  2. (B) the DNA is condensed into a chromatin fibre
  3. (C) the DNA double helix is exposed
  4. (D) transcription is occurring.

Answer

(B) the DNA is condensed into a chromatin fibre

#108 MCQ 1M

Question

The expression of the genetic material which occurs generally through ________.

Options

  1. (A) production of carbohydrates
  2. (B) production of protein
  3. (C) accumulation of genetic material
  4. (D) production of lipids

Answer

(B) production of protein

#109 MCQ 1M

Question

Which scientist experimentally proved that DNA is the sole genetic material in bacteriophage?

Options

  1. (A) Beadle and Tautum
  2. (B) Meselson and Stahl
  3. (C) Hershey and Chase
  4. (D) Jacob and Monod

Answer

(C) Hershey and Chase

#110 MCQ 1M

Question

DNA is the genetic material because

Options

  1. (A) DNA is the chemically and structurally stable material.
  2. (B) It is having characteristic of self replication.
  3. (C) It is Expressed in the form of Mendelian characteristics.
  4. (D) All are the true.

Answer

(D) All are the true.

#111 MCQ 1M

Question

Which enzyme is responsible for reverse transcription ?

Options

  1. (A) Reverse polymerase
  2. (B) Reverse transcriptase
  3. (C) Reverse ligase
  4. (D) Reverse DNase

Answer

(B) Reverse transcriptase

#112 MCQ 1M

Question

An experiment was carried out in which a hybrid DNA was allowed to replicate for one generation in medium containing 15NH4Cl and for second generation in medium containing 14NH4Cl
Which of the given conclusions can be drawn based on above stated experiment?
A. Percentage of hybrid DNA obtained in both the generations are equal
B. 50% of heavy DNA are obtained in first generation.
C. 25% of light DNA are obtained in second generation
D. 25 % of heavy DNA are obtained in second generation

Options

  1. (A) A & B
  2. (B) B & C
  3. (C) C & D
  4. (D) A & D

Answer

(B) B & C

#113 MCQ 1M

Question

The total number of nitrogenous bases in human genome is estimated to be about

Options

  1. (A) 3.5 million
  2. (B) 35 thousand
  3. (C) 35 million
  4. (D) 3.1 billion

Answer

(C) 35 million

#114 MCQ 1M

Question

Reverse transcription is observed in _______.

Options

  1. (A) TMV
  2. (B) Tumor viruses
  3. (C) HIV
  4. (D) All of the above

Answer

(D) All of the above

#115 MCQ 1M

Question

Identify the incorrect statement about RNA.

Options

  1. (A) RNA was the first genetic material to evolve in the living systems.
  2. (B) Apart from being a genetic material, it also acts as catalyst.
  3. (C) DNA evolved from RNA with chemical modifications.
  4. (D) RNA being a catalyst is non-reactive and stable.

Answer

(D) RNA being a catalyst is non-reactive and stable.

#116 MCQ 1M

Question

Which of the following enzymes is non-proteinaceous ?

Options

  1. (A) Deoxyribonuclease
  2. (B) Ligase
  3. (C) Ribozyme
  4. (D) Lysozyme

Answer

(C) Ribozyme

#117 MCQ 1M

Question

Which of the following is found more widely in a cell?

Options

  1. (A) RNA
  2. (B) DNA
  3. (C) Sphaerosomes
  4. (D) Chloroplasts

Answer

(A) RNA

#118 MCQ 1M

Question

RNA contains repeating units of

Options

  1. (A) deoxyribonucleotides
  2. (B) ribonucleotides
  3. (C) deoxyribonucleosides
  4. (D) ribonucleosides

Answer

(B) ribonucleotides

#119 MCQ 1M

Question

RNA is present as a genetic material in ________.

Options

  1. (A) TMV virus
  2. (B) Cyanobacteria
  3. (C) Cladophora
  4. (D) Sieve tube cells

Answer

(A) TMV virus

#120 MCQ 1M

Question

Which of the following is not a protein?

Options

  1. (A) Abzyme
  2. (B) Ribozyme
  3. (C) DNA ligase
  4. (D) DNA gyrase

Answer

(B) Ribozyme

#121 MCQ 1M

Question

Consider the following statements,
(i) RNA was the first genetic material
(ii) RNA is more reactive
(iii) RNA is unstable as compared to DNA
Which statement is correct?

Options

  1. (A) (i)
  2. (B) (ii)
  3. (C) (i), (ii) and (iii)
  4. (D) None of the above

Answer

(C) (i), (ii) and (iii)

#122 MCQ 1M

Question

Choose the correct statement about the direction of coding strand.

Options

  1. (A) 3' β†’ 5'
  2. (B) 2' β†’ 5'
  3. (C) 5' β†’ 3'
  4. (D) 5' β†’ 5' and 3' β†’ 3'

Answer

(C) 5' β†’ 3'

#123 MCQ 1M

Question

It stimulates DNA replication :

Options

  1. (A) DNA polymerase
  2. (B) DNA Ligase
  3. (C) Transcriptase
  4. (D) Both (A) and (B)

Answer

(D) Both (A) and (B)

#124 MCQ 1M

Question

The direction and the type of DNA replication is ________.

Options

  1. (A) Bidirectional and conservative
  2. (B) Unidirectional and conservative
  3. (C) Bidirectional and semiconservative
  4. (D) Unidirectional and Semiconservative

Answer

(C) Bidirectional and semiconservative

#125 MCQ 1M

Question

Choose the correct option for formation of okazaki fragments.

Options

  1. (A) Replication
  2. (B) Transcription
  3. (C) Translation
  4. (D) Transduction

Answer

(A) Replication

#126 MCQ 1M

Question

What is a primer?

Options

  1. (A) long strand of DNA
  2. (B) short strand of DNA
  3. (C) long chain of RNA
  4. (D) short RNA chain

Answer

(D) short RNA chain

#127 MCQ 1M

Question

When DNA replication will start?

Options

  1. (A) When leading strand produce okazaki fragments.
  2. (B) When hydrogen bonds between two polynucleotides chains are sequentially broken with the help of proper enzymes.
  3. (C) When phosphodiester bonds between sequentially two nucleotides are broken.
  4. (D) When bonds between pentose sugar and Nitrogen base are broken.

Answer

(B) When hydrogen bonds between two polynucleotides chains are sequentially broken with the help of proper enzymes.

#128 MCQ 1M

Question

Enzyme that breaks the bond between DNA helix is known as ________.

Options

  1. (A) Helicase
  2. (B) Gyrase
  3. (C) Ligase
  4. (D) DNA polymerase

Answer

(A) Helicase

#129 MCQ 1M

Question

Ligase joins nucleotides with ________.

Options

  1. (A) glycosidic bond
  2. (B) ester bond
  3. (C) peptide bond
  4. (D) phosphodiester bond

Answer

(D) phosphodiester bond

#130 MCQ 1M

Question

Okazaki fragment were formed

Options

  1. (A) On the continuos strand at a time of DNA replication.
  2. (B) On the discontinuous strand at a time of DNA replication.
  3. (C) At a time of transcription.
  4. (D) From the introns at a time of transcription.

Answer

(B) On the discontinuous strand at a time of DNA replication.

#131 MCQ 1M

Question

Antiparallel strands of a DNA molecule means that

Options

  1. (A) one strand turns clockwise
  2. (B) one strand turns anticlockwise
  3. (C) the phosphate groups of twoDNA strands, at their ends, share the same position
  4. (D) the phosphate groups at the start of two DNA strands are in opposite position (pole).

Answer

(D) the phosphate groups at the start of two DNA strands are in opposite position (pole).

#132 MCQ 1M

Question

At the end of DNA replication, DNA molecule is having

Options

  1. (A) One strand of parental DNA and second is newly formed.
  2. (B) Both strands are newly synthesized.
  3. (C) Two strands are of parental DNA.
  4. (D) Both strands are different from parental DNA.

Answer

(A) One strand of parental DNA and second is newly formed.

#133 MCQ 1M

Question

In the DNA molecule,

Options

  1. (A) the proportion of adenine in relation to thymine varies with the organism
  2. (B) there are two strands which run antiparallel one in 5' β†’ 3' direction and other in 3' β†’ 5'
  3. (C) the total amount of purine nucleotides and pyrimidine nucleotides is not always equal
  4. (D) there are two strands which run parallel in the 5' β†’ 3' direction.

Answer

(B) there are two strands which run antiparallel one in 5' β†’ 3' direction and other in 3' β†’ 5'

#134 MCQ 1M

Question

During replication of a bacterial chromosome DNA synthesis starts from a replication origin site and

Options

  1. (A) Moves in one direction of the site
  2. (B) Moves in bi-directional way
  3. (C) RNA primers are involved
  4. (D) Move in unidirectional way

Answer

(B) Moves in bi-directional way

#135 MCQ 1M β–¦ 1

Question

Select the correct option.

Direction of

RNA Synthesis

Direction of reading of the template DNA

(a)

5' – 3'

3' – 5'

(b)

3' – 5'

5' – 3'

(c)

5' – 3'

5' – 3'

(d)

3' – 5'

3' – 5'

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(A) A and R both are correct, and R is correct explanation of A.

#136 MCQ 1M

Question

Which enzyme is responsible for transcription?

Options

  1. (A) DNA polymerase
  2. (B) RNA polymerase
  3. (C) Transcriptase
  4. (D) All of the above

Answer

(C) Transcriptase

#137 MCQ 1M

Question

Which RNA is synthesized by transcription?

Options

  1. (A) m-RNA
  2. (B) r-RNA
  3. (C) t-RNA
  4. (D) All of the above

Answer

(A) m-RNA

#138 MCQ 1M

Question

Enzyme responsible for the formation of phosphodiester bond in the process of transcription is ______.

Options

  1. (A) DNA polymerase-III
  2. (B) DNA Ligase
  3. (C) DNA helicase
  4. (D) None of the above

Answer

(D) None of the above

#139 MCQ 1M

Question

At the end of transcription m-RNA is translocated to ________.

Options

  1. (A) Nucleus
  2. (B) Cytoplasm
  3. (C) Mitochondria
  4. (D) One cell to another cell

Answer

(B) Cytoplasm

#140 MCQ 1M

Question

m-RNA is translocated to cytoplasm where it associates with _________ .

Options

  1. (A) Chloroplast
  2. (B) Mitochondria
  3. (C) Ribosomes
  4. (D) Golgibody

Answer

(C) Ribosomes

#141 MCQ 1M

Question

Nucleoid is present in

Options

  1. (A) Plant cell
  2. (B) Prokaryotic cell
  3. (C) Animal cell
  4. (D) Eukaryotic cell

Answer

(B) Prokaryotic cell

#142 MCQ 1M

Question

Which one of the following makes use of RNA as a template to synthesize DNA?

Options

  1. (A) Reverse transcriptase
  2. (B) DNA dependant RNA polymerase
  3. (C) DNA polymerase
  4. (D) RNA polymerase

Answer

(A) Reverse transcriptase

#143 MCQ 1M

Question

Choose the correct statement.

Options

  1. (A) Transcription and translation occur in the same compartment in prokaryotes
  2. (B) Monocistonic RNA - more than one structural genes under single promoter
  3. (C) Introns and exons both code for protein synthesis
  4. (D) In prokaryotes, splicing and tailing occurs before translation.

Answer

(A) Transcription and translation occur in the same compartment in prokaryotes

#144 MCQ 1M

Question

Which of the following RNAs picks up specific amino acid (from amino acid pool) in the cytoplasm to ribosome during protein synthesis?

Options

  1. (A) tRNA
  2. (B) mRNA
  3. (C) rRNA
  4. (D) All of the above

Answer

(A) tRNA

#145 MCQ 1M β–¦ 1

Question

Match the following RNA polymerase with their transcribed products
Select the correct option from the following

Column : I

Column : II

a

RNA polymerase I

i

tRNA

b

RNA polymerase II

ii

rRNA

c

RNA polymerase III

iii

hnRNA

Options

  1. (A) a – i, b – iii, c – ii
  2. (B) a – i, b – ii, c – iii
  3. (C) a – ii, b – iii, c – i
  4. (D) a – i, b – iii, c – ii

Answer

(C) a – ii, b – iii, c – i

#146 MCQ 1M

Question

In capping ________ is added to the 5' end of hnRNA.

Options

  1. (A) Methyl guanosine triphosphate
  2. (B) Methyl adenosine triphosphate
  3. (C) Methyl cytidine triphosphate
  4. (D) Methyl uridine triphosphate

Answer

(A) Methyl guanosine triphosphate

#147 MCQ 1M

Question

Which type of RNA is most abundant in cell?

Options

  1. (A) mRNA
  2. (B) tRNA
  3. (C) rRNA
  4. (D) catalytic RNA

Answer

(C) rRNA

#148 MCQ 1M

Question

The phase of gene expression is ________.

Options

  1. (A) Transcription
  2. (B) Translation
  3. (C) Transduction
  4. (D) Both (A) and (B)

Answer

(D) Both (A) and (B)

#149 MCQ 1M

Question

Removal of introns and joining the exons in a defined order in a transcription unit is called

Options

  1. (A) tailing
  2. (B) transformation
  3. (C) capping
  4. (D) splicing

Answer

(D) splicing

#150 MCQ 1M

Question

DNA-dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the

Options

  1. (A) template strand
  2. (B) coding strand
  3. (C) alpha strand
  4. (D) antistrand

Answer

(A) template strand

#151 MCQ 1M

Question

The secret of the blue print of genetic information lies in the arrangement of ________.

Options

  1. (A) definite linear sequence of nitrogen bases of a RNA
  2. (B) definite linear sequence of nitrogen bases of a DNA
  3. (C) on the quality of nitrogen bases
  4. (D) number of purine and pyrimidine bases

Answer

(B) definite linear sequence of nitrogen bases of a DNA

#152 MCQ 1M

Question

Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a "triplet?"

Options

  1. (A) Griffith
  2. (B) Watson, crick
  3. (C) Nirenberg, Matthaei and Khorana
  4. (D) Tschermak

Answer

(C) Nirenberg, Matthaei and Khorana

#153 MCQ 1M

Question

What is genetic code?

Options

  1. (A) The information present in the m-RNA is transcribed and carried by the DNA to the cytoplasm for protein synthesis.
  2. (B) The sequence of nitrogen base in t-RNA for synthesis of protein.
  3. (C) The sequence of nitrogen base in r-RNA molecule which contains information for Ribosomes.
  4. (D) The sequence of nitrogen bases in m-RNA molecule which contains the information for the synthesis of protein molecule.

Answer

(D) The sequence of nitrogen bases in m-RNA molecule which contains the information for the synthesis of protein molecule.

#154 MCQ 1M

Question

Any one codon specifies the position of one kind of amino-acid only ________.

Options

  1. (A) specific
  2. (B) universal
  3. (C) degenerate
  4. (D) non-sense

Answer

(A) specific

#155 MCQ 1M

Question

How many codons, codes for amino acids?

Options

  1. (A) 4
  2. (B) 61
  3. (C) 64
  4. (D) 3

Answer

(B) 61

#156 MCQ 1M

Question

Which codons cause chain termination of protein synthesis ?

Options

  1. (A) AUG, UAG, UGA
  2. (B) UAA, UGA, UAG
  3. (C) UAA, GAU, GUC
  4. (D) AUG, UAA, UCA

Answer

(B) UAA, UGA, UAG

#157 MCQ 1M

Question

How many non-sense codons are there?

Options

  1. (A) 20
  2. (B) 61
  3. (C) 3
  4. (D) 1

Answer

(C) 3

#158 MCQ 1M

Question

What is degenerate codons?

Options

  1. (A) One code which determine the same amino acid in all organisms.
  2. (B) Any one codon specifies the position of one kind of amino acid only.
  3. (C) A single amino acid may be specified by many codons.
  4. (D) Codons do not code for any amino-acid.

Answer

(C) A single amino acid may be specified by many codons.

#159 MCQ 1M

Question

Which of the following codon codes for methionine?

Options

  1. (A) AUG
  2. (B) UAG
  3. (C) CCC
  4. (D) AGU

Answer

(A) AUG

#160 MCQ 1M

Question

Which one of the following codons code for serine?

Options

  1. (A) UCU, UCC, UCA, UCG
  2. (B) UUU, UUC, UUA, UUG
  3. (C) CUU, CUC, CUA, CUG
  4. (D) CCU, CCC, CCA, CCG

Answer

(A) UCU, UCC, UCA, UCG

#161 MCQ 1M

Question

Which one of the following codons code for lysine ?

Options

  1. (A) AAA, AAG
  2. (B) AGU, AGC
  3. (C) GGU, GGC
  4. (D) UUA, UUG

Answer

(A) AAA, AAG

#162 MCQ 1M

Question

How many codons code for serine amino acid?

Options

  1. (A) 4
  2. (B) 2
  3. (C) 8
  4. (D) 6

Answer

(D) 6

#163 MCQ 1M

Question

The function of t-RNA is

Options

  1. (A) To provide information for synthesis of protein.
  2. (B) To transport Amino acid molecules of various kinds located in cytoplasm.
  3. (C) To transport Amino acid from cytoplasm to nucleus.
  4. (D) To act as a template strand in the process of m-RNA synthesis.

Answer

(B) To transport Amino acid molecules of various kinds located in cytoplasm.

#164 MCQ 1M

Question

How many effective nucleotides are present in t-RNA?

Options

  1. (A) 25 to 30
  2. (B) 80 to 90
  3. (C) 75 to 95
  4. (D) More than 95

Answer

(C) 75 to 95

#165 MCQ 1M

Question

How many types of t-RNA are present in cytoplasm?

Options

  1. (A) 61
  2. (B) 20
  3. (C) 64
  4. (D) 16

Answer

(A) 61

#166 MCQ 1M

Question

A specific nucleotide sequence attached to 3' end of t-RNA is ________.

Options

  1. (A) AAC
  2. (B) CCA
  3. (C) AUG
  4. (D) UUA

Answer

(B) CCA

#167 MCQ 1M

Question

The site for amino acid attachment on t-RNA is ________.

Options

  1. (A) Anticodon region
  2. (B) On TψC 100P
  3. (C) On D LOOP
  4. (D) At 3–- OH end

Answer

(D) At 3–- OH end

#168 MCQ 1M

Question

During protein synthesis in an organism, at one point the process comes to a halt. Select the group of the three codons from the following, from which anyone of the three could bring about this halt.

Options

  1. (A) UUU, UCC, UAU
  2. (B) UUC, UUA, UAC
  3. (C) UAG, UGA, UAA
  4. (D) UUG, UCA, UCG

Answer

(C) UAG, UGA, UAA

#169 MCQ 1M

Question

Select the correct option regarding genetic code of Glycine.

Options

  1. (A) GUU, GUC, GUA
  2. (B) GAU, GAC, GAA
  3. (C) GGU, GGA, GGC
  4. (D) GGU, GGA,GCU

Answer

(C) GGU, GGA, GGC

#170 MCQ 1M

Question

Which form of RNA has a structure resembling clover leaf?

Options

  1. (A) mRNA
  2. (B) tRNA
  3. (C) rRNA
  4. (D) hn RNA

Answer

(B) tRNA

#171 MCQ 1M

Question

Which one of the following group of codons it is not called as degenerate codons?

Options

  1. (A) UAA, UAG and UGA
  2. (B) GUA, GUG and GCA
  3. (C) UUC, UUG and CCU,
  4. (D) UUA, UUG and CUU

Answer

(A) UAA, UAG and UGA

#172 MCQ 1M

Question

Out of 64 codons, 61 codons code for 20 types of amino acid. It is called

Options

  1. (A) Wobbling of codon
  2. (B) Overlapping of gene
  3. (C) Universility of codons
  4. (D) Degeneracy of genetic code

Answer

(D) Degeneracy of genetic code

#173 MCQ 1M

Question

Which of the following is the simplest amino acid?

Options

  1. (A) Tyrosine
  2. (B) Proline
  3. (C) Glycine
  4. (D) Glutamic acid

Answer

(C) Glycine

#174 MCQ 1M

Question

Anticodon is

Options

  1. (A) Paired triplet of bases on messenger RNA
  2. (B) Unpaired triplet of bases on rRNA
  3. (C) Paired triplet of bases on rRNA
  4. (D) An unpaired triplet of bases in an exposed position of tRNA

Answer

(D) An unpaired triplet of bases in an exposed position of tRNA

#175 MCQ 1M

Question

Identify the basic amino acid from the following.

Options

  1. (A) Valine
  2. (B) Tyrosine
  3. (C) Glutamic Acid
  4. (D) Lysine

Answer

(D) Lysine

#176 MCQ 1M

Question

There are 64 types of codons in genetic code dictionary because

Options

  1. (A) There are 64 types of tRNA’s found in cell
  2. (B) There are 44 meaningless and 20 codons for amino acids
  3. (C) There are 64 amino acids for coding
  4. (D) Genetic code is triplet

Answer

(D) Genetic code is triplet

#177 MCQ 1M

Question

Which one of the following pairs is correctly matched with regard to the codon and the amino acid coded by it?

Options

  1. (A) UUA-Valine
  2. (B) AAA-Lysine
  3. (C) AUG-Cysteine
  4. (D) CCC-Alanine

Answer

(B) AAA-Lysine

#178 MCQ 1M

Question

Which one of the following is not correct option for t βˆ’ RNA?

Options

  1. (A) It has and anticodon
  2. (B) It has an amino acid acceptor end.
  3. (C) t βˆ’ RNAs are not specific for each amino acid.
  4. (D) There are no t βˆ’ RNAs for stop codons.

Answer

(C) t βˆ’ RNAs are not specific for each amino acid.

#179 MCQ 1M

Question

________ is common for both DNA and RNA but ________ is present only in DNA

Options

  1. (A) Uracil, thymine
  2. (B) Cytosine, Uracil
  3. (C) Guanine, thymine
  4. (D) cytosine, uracil

Answer

(C) Guanine, thymine

#180 MCQ 1M

Question

Which one of the following pairs is correctly matched with regards to the codon and the amino acid coded by it?

Options

  1. (A) UUA βˆ’ valine
  2. (B) AUG – lysine
  3. (C) AUG βˆ’ cysteine
  4. (D) CCC βˆ’ alanine

Answer

(A) UUA βˆ’ valine

#181 MCQ 1M

Question

Which is stop codon?

Options

  1. (A) CUA
  2. (B) AUG
  3. (C) UGG
  4. (D) UAG

Answer

(D) UAG

#182 MCQ 1M

Question

Genetic code consists of

Options

  1. (A) 4 codons, each with two nucleotides
  2. (B) 16 codons, each with four nucleotides
  3. (C) 64 codons, each with two nucleotides
  4. (D) 64 codons, each with three nucleotides

Answer

(D) 64 codons, each with three nucleotides

#183 MCQ 1M

Question

Which RNA contains information for synthesis of protein?

Options

  1. (A) m-RNA
  2. (B) r-RNA
  3. (C) t-RNA
  4. (D) All of the above

Answer

(A) m-RNA

#184 MCQ 1M

Question

A unit of three successive nucleotides indicates the position of particular ________in the constitution of protein.

Options

  1. (A) Monomer of carbohydrates
  2. (B) Fatty acids
  3. (C) Nucleotides
  4. (D) Amino acids

Answer

(D) Amino acids

#185 MCQ 1M

Question

The Sequence of nitrogen bases in a particular region of DNA molecule was found to be CAG, CCC, GAT. What would be the sequence of nitrogen bases in the m-RNA?

Options

  1. (A) CAG, CCC, CAU
  2. (B) GUC, GGG, CUA
  3. (C) GUC, CCC, GAT
  4. (D) GAC, TAG, CTA

Answer

(B) GUC, GGG, CUA

#186 MCQ 1M

Question

Main components for translation process are ___.

Options

  1. (A) ribosome, t-RNA, amino acids
  2. (B) ribosome m-RNA, DNA
  3. (C) DNA m-RNA, t-RNA
  4. (D) ribosome, cytoplasm, t-RNA, r-RNA

Answer

(A) ribosome, t-RNA, amino acids

#187 MCQ 1M

Question

Which one of the following amino acids always initiate the synthesis of translation?

Options

  1. (A) Serine
  2. (B) Valine
  3. (C) Methionine
  4. (D) Tryptophan

Answer

(C) Methionine

#188 MCQ 1M

Question

Which enzyme is essential for transportation of amino acids ?

Options

  1. (A) m-RNA synthetase
  2. (B) DNA polymerase-III
  3. (C) Transcriptase
  4. (D) t-RNA synthetase

Answer

(D) t-RNA synthetase

#189 MCQ 1M

Question

The code of nitrogen bases in a particular region of m-RNA is AUG, then the anticodon will be on t-RNA is ________.

Options

  1. (A) AUC
  2. (B) TAC
  3. (C) UAC
  4. (D) TAG

Answer

(C) UAC

#190 MCQ 1M

Question

Out of the following which has anticodon?

Options

  1. (A) m-RNA
  2. (B) DNA
  3. (C) r-RNA
  4. (D) t-RNA

Answer

(D) t-RNA

#191 MCQ 1M

Question

Which is helpful in the process of elongation?

Options

  1. (A) GTP as energy source, t-RNA synthetase.
  2. (B) GTP as energy source, elongation factors.
  3. (C) ATP as energy source, elongation factors.
  4. (D) ATP as energy source, t-RNA synthetase.

Answer

(B) GTP as energy source, elongation factors.

#192 MCQ 1M

Question

The ribosome moves along m-RNA in 3' direction by distance of ________.

Options

  1. (A) Two codons
  2. (B) Three codons
  3. (C) Four codons
  4. (D) One codon

Answer

(D) One codon

#193 MCQ 1M

Question

What is the function of non-sense codon?

Options

  1. (A) It translocate m-RNA to cytoplasm from DNA.
  2. (B) Release the synthesized polypeptide chain from ribosomes.
  3. (C) It carrying specific amino acid molecule on m-RNA.
  4. (D) It release polypeptide chain from methionine.

Answer

(B) Release the synthesized polypeptide chain from ribosomes.

#194 MCQ 1M

Question

Who can directly code for synthesis of proteins?

Options

  1. (A) DNA
  2. (B) RNA
  3. (C) Gene
  4. (D) Nucleotide

Answer

(B) RNA

#195 MCQ 1M

Question

The process of translation of mRNA to proteins begins as soon as :

Options

  1. (A) The larger subunit of ribosome encounters mRNA
  2. (B) Both the subunits join together to bind with mRNA
  3. (C) The tRNA is activated and the larger subunit of ribosome encounters mRNA
  4. (D) The small subunit of ribosome encounters mRNA

Answer

(D) The small subunit of ribosome encounters mRNA

#196 MCQ 1M

Question

Polysome is formed by

Options

  1. (A) a ribosome with several subunits
  2. (B) ribosomes attached to each other in a linear arrangement
  3. (C) several ribosomes attached to a single mRNA
  4. (D) many ribosomes attached to a strand of endoplasmic reticulum.

Answer

(C) several ribosomes attached to a single mRNA

#197 MCQ 1M

Question

Expressed Sequence Tages (ESTs) refers to

Options

  1. (A) Genes expressed as RNA
  2. (B) Polypeptide expression
  3. (C) DNA polymorphism
  4. (D) None of the above.

Answer

(A) Genes expressed as RNA

#198 MCQ 1M

Question

The first phase of translation is

Options

  1. (A) Recognition of an anti-codon
  2. (B) Binding of mRNA to ribosome
  3. (C) Recognition of DNA molecule
  4. (D) Aminoacylation of tRNA

Answer

(D) Aminoacylation of tRNA

#199 MCQ 1M

Question

In mRNA, UTRs are present at

Options

  1. (A) 5' end before start codon only
  2. (B) Both ends before start codon and after stop codon
  3. (C) 3' end after stop codon only
  4. (D) 5'end after start codon and 3' end before stop codon

Answer

(B) Both ends before start codon and after stop codon

#200 MCQ 1M

Question

Choose the correct statement

Options

  1. (A) Transcription and translation occur in same compartment for prokaryotes
  2. (B) Monocistonic RNA - more than one structural genes under single promoter
  3. (C) Introns and exons both code for protein synthesis
  4. (D) In prokaryotes, splicing and tailing occurs before translation.

Answer

(A) Transcription and translation occur in same compartment for prokaryotes

#201 MCQ 1M

Question

Which of the following step of translation does not consume a high energy phosphate bond?

Options

  1. (A) Translocation
  2. (B) Amino acid activation
  3. (C) Peptidyl transferase reaction
  4. (D) Aminoacyl tRNA binding to A-site

Answer

(A) Translocation

#202 MCQ 1M

Question

The process in which amino acids are activated in the presence of ATP and linked to their cognate tRNA is commonly called

Options

  1. (A) Aminoacylation of tRNA
  2. (B) Aminoacylation of amino acid
  3. (C) Discharging of tRNA
  4. (D) Charging of mRNA

Answer

(A) Aminoacylation of tRNA

#203 MCQ 1M

Question

In mRNA, untranslated regions (UTRs) are required for

Options

  1. (A) Efficient translation process
  2. (B) Synthesising rRNA
  3. (C) Sequencing the codes for polypeptide formation
  4. (D) Binding of ribosome with tRNA

Answer

(A) Efficient translation process

#204 MCQ 1M

Question

In an inducible operon, the genes are

Options

  1. (A) usually not expressed unless a signal turns them β€œon”.
  2. (B) usually expressed unless a signal turns them β€œoff”.
  3. (C) never expressed
  4. (D) always expressed.

Answer

(A) usually not expressed unless a signal turns them β€œon”.

#205 MCQ 1M

Question

Who discovered a lac-operon?

Options

  1. (A) Jacob and Monad
  2. (B) Watson and Crick
  3. (C) Gamov and Khorana
  4. (D) Avery and McCarty

Answer

(A) Jacob and Monad

#206 MCQ 1M

Question

Name the gene which can produce a product which will block operator.

Options

  1. (A) Promoter gene
  2. (B) Structural gene
  3. (C) Repressor gene
  4. (D) Control gene

Answer

(C) Repressor gene

#207 MCQ 1M

Question

The Repressor protein is a product of ________.

Options

  1. (A) promoter gene
  2. (B) repressor gene
  3. (C) operator gene
  4. (D) structural gene

Answer

(B) repressor gene

#208 MCQ 1M

Question

The sugar which was used by Jacob and Monad for explanation of regulation of gene expression is...

Options

  1. (A) glucose
  2. (B) lactose
  3. (C) sucrose
  4. (D) Both (A) and (B)

Answer

(D) Both (A) and (B)

#209 MCQ 1M

Question

Which bacteria has been used regarding regulation of gene expression ?

Options

  1. (A) Streptococcus
  2. (B) Bacillus
  3. (C) E-coli
  4. (D) Salmonella

Answer

(C) E-coli

#210 MCQ 1M

Question

Which enzyme is necessary to break down the lactose into Glucose and galactose ?

Options

  1. (A) Ξ²-glucosidase
  2. (B) Permease
  3. (C) Transacetylase
  4. (D) All of the above

Answer

(D) All of the above

#211 MCQ 1M

Question

The segment of DNA which exercises a control of an enzyme which is responsible for digestion of lactose is ________.

Options

  1. (A) Operator gene
  2. (B) Promoter gene
  3. (C) Structural gene
  4. (D) Regulator gene

Answer

(A) Operator gene

#212 MCQ 1M

Question

Which is responsible for switching on the lac-operon in bacteria?

Options

  1. (A) lactose
  2. (B) numbers of bacteria
  3. (C) structural gene
  4. (D) RNA polymerase

Answer

(A) lactose

#213 MCQ 1M

Question

Operon contains :

Options

  1. (A) Structural gene + promoter gene + operator gene
  2. (B) Regulator gene + operator gene
  3. (C) Regulator gene + promoter gene + operator gene
  4. (D) Regulator gene + promoter gene+ operator gene+ structural gene

Answer

(D) Regulator gene + promoter gene+ operator gene+ structural gene

#214 MCQ 1M

Question

It controls the rate of m-RNA synthesis.

Options

  1. (A) Operator gene
  2. (B) Regulator gene
  3. (C) Promoter gene
  4. (D) Structural gene

Answer

(C) Promoter gene

#215 MCQ 1M

Question

It exercises a control over transcription.

Options

  1. (A) Operator gene
  2. (B) Promoter gene
  3. (C) Termination gene
  4. (D) Regulator gene

Answer

(A) Operator gene

#216 MCQ 1M

Question

In operon concept, regulatory gene functions as

Options

  1. (A) Repressor
  2. (B) Regulator
  3. (C) Inhibitor
  4. (D) All of these

Answer

(A) Repressor

#217 MCQ 1M

Question

In lac - operon if mutation occurs in the y gene of the 'Structural gene' then ________.

Options

  1. (A) Permease will not be synthesized.
  2. (B) b-Galactosidase will not be synthesized.
  3. (C) Transacetylase will not be synthesized
  4. (D) Lactose digestion will be rapid.

Answer

(A) Permease will not be synthesized.

#218 MCQ 1M

Question

Genes that are involved in turning on or off the transcription of a set of structural genes are called

Options

  1. (A) Polymorphic genes
  2. (B) Operator genes
  3. (C) Redundant genes
  4. (D) Regulatory genes

Answer

(B) Operator genes

#219 MCQ 1M

Question

At what level gene expression does not occur?

Options

  1. (A) Formation of primary transcript
  2. (B) Processing level
  3. (C) Translation level
  4. (D) Replication level

Answer

(D) Replication level

#220 MCQ 1M

Question

Which one of the following is not a part of a transcription unit in DNA ?

Options

  1. (A) The inducer
  2. (B) A terminator
  3. (C) A promoter
  4. (D) The structural gene

Answer

(A) The inducer

#221 MCQ 1M β–¦ 1

Question

Match the following genes of the Lac operon with their respective products.
Select the correct option.

Column : I

Column : II

a

i gene

i

bβˆ’galactosidase

b

z gene

ii

Permease

c

a gene

iii

Repressor

d

y gene

iv

Transacetylase

Options

  1. (A) a – i, b – iii, c – ii, d – iv
  2. (B) a – iii, b – i, c – ii, d – iv
  3. (C) a – iii, b – i, c – iv, d – ii
  4. (D) a – iii, b – iv, c – i, d – ii

Answer

(C) a – iii, b – i, c – iv, d – ii

#222 MCQ 1M

Question

Y gene in lac operon responsible for _____.

Options

  1. (A) Codes for b - galactosidase
  2. (B) Primarily responsible for hydrolysis of the disaccharide
  3. (C) Encodes for transacetylase.
  4. (D) Codes for permease.

Answer

(D) Codes for permease.

#223 MCQ 1M

Question

In Operon concept, the regulator gene regulates chemical reactions in the cell by

Options

  1. (A) Inactivating enzymes in the reaction
  2. (B) Inhibiting transcription of mRNA
  3. (C) Inhibiting migration of mRNA into cytoplasm
  4. (D) Inhibiting the substrate in the reaction

Answer

(A) Inactivating enzymes in the reaction

#224 MCQ 1M

Question

Regulation of lac operon by repressor is referred to as

Options

  1. (A) Positive regulation
  2. (B) Induced regulation
  3. (C) Negative regulation
  4. (D) Autoregulation

Answer

(C) Negative regulation

#225 MCQ 1M

Question

Select the two correct statements out of the four (i - iv) statements given below about lac operon
(i) Glucose or galactose may bind with the repressor and inactivate it
(ii) In the absence of lactose, the repressor binds with the operator region.
(iii) The z gene codes for permease.
(iv) This was elucidated by Francois Jacob and Jacques Monod.
The correct statements are

Options

  1. (A) (ii) and (iii)
  2. (B) (i) and (iii)
  3. (C) (ii) and (iv)
  4. (D) (i) and (ii)

Answer

(C) (ii) and (iv)

#226 MCQ 1M

Question

Genes that are involved in turning on or off the transcription of a set of structural genes are called

Options

  1. (A) Polymorphic genes
  2. (B) Operator genes
  3. (C) Redundant genes
  4. (D) Regulatory genes

Answer

(B) Operator genes

#227 MCQ 1M

Question

Wild type E.coli cells are growing in normal medium with glucose. There are transferred to a medium containing only lactose as the sugar. Which one of the following changes take place ?

Options

  1. (A) The lac-operon is prepressed
  2. (B) All Operons are induced
  3. (C) E.coil cells stop dividing
  4. (D) The lac-Operon is induced

Answer

(D) The lac-Operon is induced

#228 MCQ 1M

Question

In lac operon, lactose works as inducer as

Options

  1. (A) It is responsible for the binding of repressor to the operator region
  2. (B) In presence of lactose there is transcription of structural gene
  3. (C) It initiates the transcription of repressor mRNA
  4. (D) It prevents RNA polymerase from transcribing the operon

Answer

(B) In presence of lactose there is transcription of structural gene

#229 MCQ 1M

Question

It is not a HGP goal.

Options

  1. (A) To prepare a genetic map.
  2. (B) The deciphering of nucleotide sequence and store is referred to data base.
  3. (C) To understand ELSI.
  4. (D) To understand genetic mutation

Answer

(D) To understand genetic mutation

#230 MCQ 1M

Question

When was HGP completed?

Options

  1. (A) April - 2001
  2. (B) July - 2003
  3. (C) April - 2000
  4. (D) April - 2003

Answer

(D) April - 2003

#231 MCQ 1M

Question

Human Genome size is ________.

Options

  1. (A) > 3 billion
  2. (B) > 5 billion
  3. (C) > 10 billion
  4. (D) > 1 billion

Answer

(A) > 3 billion

#232 MCQ 1M

Question

The deciphering of nucleotide sequence and its store is referred to as ________.

Options

  1. (A) genome
  2. (B) data size
  3. (C) data base
  4. (D) genetic map

Answer

(C) data base

#233 MCQ 1M

Question

How much genome codes for proteins?

Options

  1. (A) 10%
  2. (B) 2%
  3. (C) 99%
  4. (D) 50%

Answer

(B) 2%

#234 MCQ 1M

Question

In humans most genes are present on which chromosome.

Options

  1. (A) first chromosome
  2. (B) fifth chromosome
  3. (C) X chromosome
  4. (D) Y chromosome

Answer

(A) first chromosome

#235 MCQ 1M

Question

In humans, fewest genes are present on chromosome ________ .

Options

  1. (A) X
  2. (B) Y
  3. (C) First
  4. (D) Both X and Y

Answer

(B) Y

#236 MCQ 1M

Question

How many gene are present on Y chromosome?

Options

  1. (A) 231
  2. (B) 2968
  3. (C) 3000
  4. (D) 900

Answer

(A) 231

#237 MCQ 1M

Question

Chromosome 1 has _______ genes Y has the ____.

Options

  1. (A) 2768, 236
  2. (B) 2968, 231
  3. (C) 2163, 336
  4. (D) 1753, 721

Answer

(B) 2968, 231

#238 MCQ 1M

Question

The total number of nitrogenous bases in human genome is estimated to be about

Options

  1. (A) 3.5 million
  2. (B) 35 thousand
  3. (C) 35 million
  4. (D) 3.1 billion

Answer

(D) 3.1 billion

#239 MCQ 1M

Question

Human Genome Project (HGP) is closely associated with the rapid development of a new area in biology called as

Options

  1. (A) biotechnology
  2. (B) bioinformatics
  3. (C) biogeography
  4. (D) bioscience

Answer

(B) bioinformatics

#240 MCQ 1M β–¦ 1

Question

Identify the following statements as true (T) or false (F) w.r.t. goals of Human Genome Project and select the correct option.
A : Address the ethical, legal and social issues that may arise from the project.
B : Identify approximately 3 billion genes in human DNA.

A

B

(A)

F

F

(B)

T

T

(C)

F

T

(D)

T

F

Answer

(D)

T

F

#241 MCQ 1M

Question

Select the incorrect match w.r.t Human Genome Project.

Options

  1. (A) SNP – Single nucleotide polymerase
  2. (B) BAC – Bacteria artificial chromosome
  3. (C) YAC – Yeast artificial chromosome
  4. (D) ELSI – Ethical, legal and social issues

Answer

(A) SNP – Single nucleotide polymerase

#242 MCQ 1M

Question

Read the following statements and select the correct option.
A : Less than 2 percent of human genome codes for proteins.
B : Chromosome Y in humans has the fewest genes.

Options

  1. (A) Only statement A is correct
  2. (B) Only statement B is correct
  3. (C) Both the statements are correct
  4. (D) Both the statements are incorrect

Answer

(C) Both the statements are correct

#243 MCQ 1M β–¦ 1

Question

Identify the following statements as true (T) or false (F) and select the correct option.
A : Frederick Sanger is credited for developing the method for determination of amino acid sequences in proteins.
B : DNA of Drosophila has not yet been sequenced till now.

A

B

(A)

T

T

(B)

F

F

(C)

F

T

(D)

T

F

Answer

(D)

T

F

#244 MCQ 1M

Question

Which of the following is correct about DNA segment?

Options

  1. (A) Code for proteins
  2. (B) Intervening sequences
  3. (C) VNTR
  4. (D) All of the above

Answer

(D) All of the above

#245 MCQ 1M

Question

In DNA fingerprinting, separation of DNA fragments occur by ________.

Options

  1. (A) Gas - chromatography
  2. (B) TLC
  3. (C) Electrophoresis
  4. (D) PCR

Answer

(C) Electrophoresis

#246 MCQ 1M

Question

It is known as DNA scissors.

Options

  1. (A) RE (Restriction Enzyme)
  2. (B) Ligase
  3. (C) DNA polymerase
  4. (D) RNA polymerase

Answer

(A) RE (Restriction Enzyme)

#247 MCQ 1M

Question

Who discovered DNA fingerprinting ?

Options

  1. (A) Alec Jeffery
  2. (B) Jacob Monad
  3. (C) Herbert Boyer
  4. (D) Stanley Cohen

Answer

(A) Alec Jeffery

#248 MCQ 1M

Question

What is the use of radioactive DNA probe ?

Options

  1. (A) The DNA probe binds to specific DNA sequence on the nylon membrane.
  2. (B) Separation of DNA fragments.
  3. (C) It makes x-ray film.
  4. (D) Extraction of DNA fragments.

Answer

(A) The DNA probe binds to specific DNA sequence on the nylon membrane.

#249 MCQ 1M

Question

Satellite DNA is a useful tool for

Options

  1. (A) Organ transplantation
  2. (B) Gender detection
  3. (C) Forensic science
  4. (D) Mutation detection

Answer

(C) Forensic science

#250 MCQ 1M

Question

The grouping of human chromosomes is based on

Options

  1. (A) Secondary constrictions alone
  2. (B) Dot-like satellites aone
  3. (C) Banding patterns alone
  4. (D) All of the above

Answer

(C) Banding patterns alone

#251 MCQ 1M

Question

What is it that forms the basis of DNA fingerprinting?

Options

  1. (A) The relative proportions of purines and pyrimidnes in DNA
  2. (B) The relative difference in the DNA occurrence in blood, skin and saliva.
  3. (C) The relative amount of DNA in the ridges and grooves of the fingerprints.
  4. (D) Satellite DNA occurring as highly repeated short DNA segments.

Answer

(D) Satellite DNA occurring as highly repeated short DNA segments.

#252 MCQ 1M

Question

When does radioactive DNA pattern become visible?

Options

  1. (A) While bombarding with DNA probes.
  2. (B) While radioactive DNA pattern is transferred to X-ray film by direct exposure.
  3. (C) While washing DNA fragments.
  4. (D) While DNA fragments are separated by electrophoresis.

Answer

(B) While radioactive DNA pattern is transferred to X-ray film by direct exposure.

#253 MCQ 1M

Question

DNA polymorphism forms the basis of :

Options

  1. (A) DNA finger printing
  2. (B) Both genetic mapping and DNA finger printing
  3. (C) Translation
  4. (D) Genetic mapping

Answer

(B) Both genetic mapping and DNA finger printing

#254 MCQ 1M

Question

Satellite DNA is important because it

Options

  1. (A) does not code for proteins and is same inall members of the population
  2. (B) codes for enzymes needed for DNA-replication
  3. (C) codes for proteins needed in cell cycle
  4. (D) shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable from parents to children.

Answer

(D) shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable from parents to children.

#255 MCQ 1M β–¦ 1

Question

The technique of DNA fingerprinting involves use of __A__ as __B__ that shows very high degree of polymorphism.
The correct option for A and B is

(A)

A - satellite DNA,

B - probe

(B)

A - repetitive DNA,

B - coding DNA

(C)

A - probe,

B - polymerase

(D)

A - short tandem repeats,

B - repetitive DNA

Answer

(A)

A - satellite DNA,

B - probe

#256 MCQ 1M

Question

All of the following are the basis for classifying satellite DNA into different categories, except

Options

  1. (A) Length of segment
  2. (B) Presence of repetitive units
  3. (C) Base composition
  4. (D) Number of repetitive units

Answer

(B) Presence of repetitive units

#257 MCQ 1M

Question

Polymorphism in DNA arises due to

Options

  1. (A) Translation
  2. (B) DNA sequencing
  3. (C) Transcription
  4. (D) Mutation

Answer

(D) Mutation

#258 MCQ 1M

Question

Assertion : Replication and transcription occur in the nucleus but translation takes place in the cytoplasm.
Reason : mRNA is transferred from the nucleus into cytoplasm where ribosomes and amino acids are available for protein synthesis.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(A) A and R both are correct, and R is correct explanation of A.

#259 MCQ 1M

Question

Assertion : m-RNA has terminator codons at the end.
Reason : Terminator codons are present on m-RNA otherwise protein synthesis will not be completed.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(A) A and R both are correct, and R is correct explanation of A.

#260 MCQ 1M

Question

Assertion : AUG is the first codon in the transcription process.
Reason : AUG is an initiation codon.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(A) A and R both are correct, and R is correct explanation of A.

#261 MCQ 1M

Question

Assertion : In eukaryotes, hnRNA is formed at the end of transcription.
Reason : In eukaryotes genes are interrupted.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(A) A and R both are correct, and R is correct explanation of A.

#262 MCQ 1M

Question

Assertion : New strand of DNA in replication is synthesized with the help of RNA polymerase.
Reason : RNA polymerase forms short strand of RNA which is complementary to DNA-template which joins DNA at initiation site - this is called primer.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(D) A is wrong, but R is correct

#263 MCQ 1M

Question

Assertion : Avery and McCarty proved that DNA replication is semiconservative.
Reason : In replication of DNA, DNA ligase joins the DNA fragments with phosphodiester bond.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(D) A is wrong, but R is correct

#264 MCQ 1M

Question

Assertion : Genetic codon refers to a sequence of nitrogen bases on m-RNA.
Reason : A genetic codon encodes information for the protein synthesis.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(B) A and R both are correct, but R is not correct explanation of A.

#265 MCQ 1M

Question

Assertion : DNA replication takes place in the S-phase.
Reason : DNA replication is semiconservative type.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(B) A and R both are correct, but R is not correct explanation of A.

#266 MCQ 1M

Question

Assertion : DNA polymerase breaks hydrogen bond between DNA strands in the process of transcription.
Reason : In bacterial cell, one type of DNA polymerase is seen.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(D) A is wrong, but R is correct

#267 MCQ 1M

Question

Assertion : Catalytic functions were assigned to RNA molecule during evolution.
Reason : The rate of mutation is quite fast in RNA.

Options

  1. (A) A and R both are correct, and R is correct explanation of A.
  2. (B) A and R both are correct, but R is not correct explanation of A.
  3. (C) A is correct, but R is wrong
  4. (D) A is wrong, but R is correct

Answer

(B) A and R both are correct, but R is not correct explanation of A.

#268 MCQ 1M πŸ–Ό 1

Question

Identify x and y in a given diagram.
x
y
Y - H1 Histone
Y - Histone octamer

Options

  1. (A) X - H1 Histone, Y = DNA
  2. (B) X - DNA, Y - H1 Histone
  3. (C) X - Histone octamer,
  4. (D) X - DNA,

Answer

(B) X - DNA, Y - H1 Histone

#269 MCQ 1M πŸ–Ό 1

Question

Which stage is represented by following diagram of transcription?
3'
5'
Οƒ
5'
3'

Options

  1. (A) Elongation
  2. (B) Initiation
  3. (C) Termination
  4. (D) Both (B) and (C)

Answer

(A) Elongation

#270 MCQ 1M πŸ–Ό 1

Question

What is X and Y in a given diagram.

Options

  1. (A) Replication, Protein
  2. (B) Cell division, Protein
  3. (C) Replication, Amino acid
  4. (D) t-RNA, r-RNA

Answer

(A) Replication, Protein

#271 MCQ 1M πŸ–Ό 1

Question

The DNA replication diagram is given, choose the correct option regarding this.

Options

  1. (A) Strand (i) Shows direction of DNA replication
  2. (B) Strand (ii) Shows direction of DNA replication
  3. (C) Strand (i) Shows direction of DNA replication for leading strand.
  4. (D) Strand (ii) Shows direction of DNA replication for leading strand.

Answer

(C) Strand (i) Shows direction of DNA replication for leading strand.

#272 MCQ 1M πŸ–Ό 1

Question

Which process is represented by given diagram?

Options

  1. (A) Replication
  2. (B) Transcription
  3. (C) Translation
  4. (D) DNA fingerprinting

Answer

(C) Translation

#273 MCQ 1M πŸ–Ό 1

Question

Which process is shown by given diagram ?

Options

  1. (A) Replication in prokaryotes
  2. (B) Transcription in prokaryotes
  3. (C) Replication in eukaryotes
  4. (D) Transcription in eukaryotes

Answer

(D) Transcription in eukaryotes

#274 MCQ 1M πŸ–Ό 1

Question

Identify 'x' in a given diagram.

Options

  1. (A) Terminator
  2. (B) Template strand
  3. (C) RNA polymerase
  4. (D) Promoter

Answer

(D) Promoter

#275 MCQ 1M πŸ–Ό 1

Question

What is in a given diagram ?

Options

  1. (A) Inducer
  2. (B) Repressor
  3. (C) Regulatory gene
  4. (D) m-RNA

Answer

(A) Inducer

#276 MCQ 1M πŸ–Ό 1

Question

What is E, F, G and H in a given diagram ?

E

F

G

H

(A)

Ξ² -

galacto sidase

Permease

Trans-acetylase

Lactose

(B)

Permease

Ξ² -

galacto sidase

Trans-acetylase

Lactose

(C)

Lactose

Ξ² -

galacto sidase

Permease

Transacetylase

(D)

Lactose

Trans-acetylase

Permease

Ξ² -

galacto sidase

Answer

(C)

Lactose

Ξ² -

galacto sidase

Permease

Transacetylase

#277 MCQ 1M πŸ–Ό 1

Question

Identify the given model.
5'
GC
GC
A
TA
T
TA
G
GC
AT
CG
AT
C
T
A
GC
AT
TA
CGΒ 
GC
AT
AT
GC
T
CG
TA
A
CG
AT
GC
AT
G
C
TA
TA
A
T
GC
GC
3'
conservative DNA
replication
for semiconservative
DNA replication
for semi-conservative
DNA transcription
for conservative
DNA transcription

Options

  1. (A) Watson-Crick model for
  2. (B) Watson-Crick model
  3. (C) Watson-Crick model
  4. (D) Watson-Crick model

Answer

(B) Watson-Crick model

#278 MCQ 1M πŸ–Ό 4

Question

Which of the following diagram is correct regarding DNA replication ?

Options

  1. (A)
  2. (B)
  3. (C)
  4. (D)

Answer

(D)

#279 MCQ 1M πŸ–Ό 3

Question

The given diagram shows ________
Y
Z

Options

  1. (A) Nucleoid
  2. (B) Nucleosome
  3. (C) Chromatin material
  4. (D) Histone octamer

Answer

(B) Nucleosome

#280 MCQ 1M πŸ–Ό 1

Question

Which experiment is shown by given diagram?

Options

  1. (A) Reproduction in bacteria
  2. (B) Reproduction of virus
  3. (C) Viral DNA / genetic material pass in to bacteria
  4. (D) Viral capsid passes in to bacteria

Answer

(C) Viral DNA / genetic material pass in to bacteria

#281 MCQ 1M πŸ–Ό 1

Question

What is 'A' in a given diagram ?
A

Options

  1. (A) Phosphate
  2. (B) Sugar
  3. (C) Phosphate-Sugar
  4. (D) Base pair

Answer

(C) Phosphate-Sugar

#282 MCQ 1M πŸ–Ό 1

Question

Nucleotide structure is given below. Which bonds are presented by P and Q ?

Options

  1. (A) P-N-Glycosidic bond, Q-phosphoester bond
  2. (B) P-Phosphoester bond, Q-N - Glycosidic bond
  3. (C) P-Phosphodiester bond, Q-N-Glycosidic bond
  4. (D) N-Glycosidic bond, Q-Phosphodiester bond

Answer

(A) P-N-Glycosidic bond, Q-phosphoester bond

#283 MCQ ⚠ needs answer review 1M πŸ–Ό 2

Question

The Hershey - Chase experiment is given. What processes are shown by P, Q and R ?

Answer

not detected

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#284 CS

Question

DNA as an acidic substance present in nucleus was first identified by Friedrich Miescher in 1869. He named it as β€˜Nuclein’. However, due to technical limitation in isolating such a long polymer intact, the elucidation of structure of DNA remained elusive for a very long period of time. It was only in 1953 that James Watson and Francis Crick, based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple but famous Double Helix model for the structure of DNA. One of the hallmarks of their proposition was base pairing between the two strands of polynucleotide chains. However, this proposition was also based on the observation of Erwin Chargaff that for a double stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equals one.
(i) What is nuclein?
(ii) Give name of nitrogen base present in DNA.
(iii) What is not present in a DNA from the following?
(i) Phosphate (ii) Pentose sugar
(iii) Uracil (iv) Hydrogen bond
(iv) Who gave the DNA double helix model?
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#285 CS πŸ–Ό 1

Question

Translation refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. The amino acids are joined by a bond which is known as a peptide bond. Ribosome takes part in the protein synthesis. The complete process takes place in the cytoplasm.
(i) Which ion is required for the joining of ribosomal subunit during translation?
(i) ca 2+ (ii) mg2+ (iii) zn2+ (iv) cl –
(ii) Some sequence on the mRNA remains untranslated, this is known as___.
(iii) Which enzyme catalyzes the reaction of amino acylation of tRNA?
(iv) Which bond is present between amino acids of peptide chain?
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#286 CS

Question

DNA replication is a complex and multi step process, which require enzymes, protein and metal ions. In eukaryotes the replication of DNA occurs in the S phase of cell cycle and it takes place in the nucleoplasm. In prokaryotes, replication occurs in the nucleus. It takes place before the cell division. Nucleoid is a made up of DNA molecule. It can be linear or circular. In virus, nucleic acid can be DNA or RNA.
(i) Which enzymes are required for DNA replication?
(ii) Where does DNA replication occur in bacteria and eukaryotes?
(iii) From where DNA replication start?
(iv) How many types of DNA polymerase enzymes are present?
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#287 CS

Question

There are at least three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles). There is a clear cut division of labour. The RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible for transcription of tRNA, 5SrRNA, and snRNAs (small nuclear RNAs). The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).
The second complexity is that the primary transcripts contain both the exons and the introns and are non-functional. Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order. hnRNA undergoes additional processing called as capping and tailing. In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA. In tailing addition of adenylate residue to the 3' end of messanger RNA (m-RNA) occurs.
(i) How many types of RNA polymerase is seen in eukaryotes?
(ii) What is splicing?
(iii) How do hn RNA and mRNA differe from each other in eukaryotes?
(iv) What is the function of RNA polymerase I, II and III?
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#288 CS πŸ–Ό 1

Question

DNA is a long polymer of deoxyribonucleotides. The length of DNA is usually defined as number of nucleotides (or a pair of nucleotide referred to as base pairs) present in it. This is also the characteristic of and organism. For example, a bacteriophage known as Γ— 174 has 5386 nucleotides, Bacteriophage lambda has 48,502 base pairs (bp), Escherichia coli has 4.6 Γ— 106 bp, and haploid content of human DNA is 3.3Γ—109 bp.
(i) What is the monomer unit of DNA ?
(a) Ribose Sugar
(b) Deoxyribonucleotides
(c) Nucleotide
(d) Ribonucleotide
(ii) How many nucleotides are there in bacteriophage lambda ?
(a) 5386 nucleotide (b) 48,502 bp
(c) 4.6 Γ— 106 bp (d) 109 bp
(iii) What is the length of haploid DNA in human ?
(a) 1.1 cm (b) 1.1 m
(c) 2.2 m (d) 1.36 mm
(iv) How many base pairs are there in the E-coli DNA ?
(a) 4.6 Γ— 109 bp (b) 4.6 Γ— 106 bp
(c) 5386 bp (d) 48,502 bp
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#289 CS πŸ–Ό 1

Question

A structural unit of nucleotide is called nucleotide in a biological system. A nucleotide has three components - a nitrogenous base, a pentose sugar (ribose in case of RNA, and deoxyribose for DNA), and a phosphate group. There are two types of nitrogenous bases – Purines (Adenine and Guanine), and Pyrimidines (Cytosine, Uracil and Thymine). Cytosine is common for both DNA and RNA and Thymine is present in DNA. Uracil is present RNA at the place of Thymine. A nitrogenous base is linked to the OH of 1'C pentose sugar through a N-glycosidic linkage to form a nucleoside.
(i) What is the key difference between nucleoside and nucleotide ?
(a) Nitrogen (b) Phosphate group
(c) Pentose sugar (d) Hydrogen bond
(ii) Which of the following group is of pyrimidine ?
(a) Adenine and Guanine
(b) Guanine and Cytosine
(c) Cytosine, Thymine and Uracil
(d) Adenine and Thymine
(iii) Which nitrogen base is present in RNA, instead of thymine ?
(a) Guanine (b) Uracil
(c) Adenine (d) Cytosine
(iv) Where does N-glycosidic bond formed in following ?
(a) Between two nitrogen base
(b) Between Sugar and phosphate
(c) Between nitrogen base and pentose sugar
(d) Between amino acid and sugar
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#290 CS

Question

Human genome is said to have approximately
3 Γ— 109 bp, and if the cost of sequencing required is 3 US $ bp (the estimated cost in the beginning), the total estimated cost of the project would be approximately 9 billion US dollars. Further, if the obtained sequences were to be stored in typed form in books, if the obtained sequences were to be stored in typed form in books. and if each page of the book contained 1000 letters and each book contained 1000 pages, then 3300 such books would be required to store the information of DNA sequence from a single human cell. The enormous amount of data expected to be generated also necessitated the use of high speed computational devices for data storage and retrieval, and analysis. HGP was closely associated with the rapid development of a new area in biology called Bioinformatics.
(i) How many base pairs are present in human genome ?
(a) 3Γ—106 bp (b) 6Γ—109 bp
(c) 3Γ—109 bp (d) 9Γ—106 bp
(ii) If each page of the book contained 1000 letters and each book contained 1000 pages, then, how many such books would be required to store the information of DNA sequence from a Single human cell ?
(a) 330 (b) 3300
(c) 3000 (d) 33000
(iii) Which area of biology was rapidly developed associated with Human Genome Project (HGP)?
(a) Biotechnology
(b) Biochemistry
(c) Bioinformatics
(d) Genetics
(iv) If the cost of sequencing required is US $ 3 per bp, then what will be the total estimated cost of the project ?
(a) 3 billion US dollars
(b) 6 billion US dollars
(c) 9 million US dollars
(d) 9 billion US dollars

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