S1

//M2//QN3//SUB//DL0

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

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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

//M2//QN4//SUB//DL0//EQ

Draw a diagram of structure of polynucleotide strand and explain.

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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.

//M0//QN5//SUB//DL0//EQ

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

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Nitrogen base Nucleoside

• Adenine • Cytidine

• Thymine • Guanosine

• Uracil

• Cytosine

//M0//QN6//SUB//DL0//EQ

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

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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.

//M2//QN7//SUB//DL0//EQ

Describe history of discovery of DNA.

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DNA as an acidic substance present in nucleus was first identified by Friedrich Miescher in 1869.

He named it as ‘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.

//M2//QN8//SUB//DL0//EQ

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

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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 → RNA → 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.

//M2//QN9//SUB//DL0

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

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In a typical mammalian cell ds DNA has

Total base pairs = 6.6 × 109 bp

Taken the distance between two consecutive base pairs as 0.34 nm = 0.34 × 109 meter

Length of DNA = Total number of bp × distance between two consecutive bp

= (6.6 × 109 bp) × (0.34 × 10–9 m)

= 6.6 × 0.34 × 109 × 10–9

= 2.244 meter

Length of DNA = 2.2 m

//M4//QN10//SUB//DL0//EQ

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

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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.

//M2//QN11//SUB//DL0//EQ

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

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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 ‘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.

//M4//QN12//SUB//DL0//EQ

Explain the packaging of DNA.

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Taken the distance between two consecutive base pairs as 0.34 nm (0.34 × 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 × 10–9 bp ×0.34 × 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 ‘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 ‘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.

//M3//QN13//SUB//DL0//EQ

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.

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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.

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

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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.

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Give Biochemical Characterisation of Transforming Principle.

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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 ‘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.

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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.

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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’.

//M0//QN17//SUB//DL0//EQ

Explain Griffith’s transformation experiment.

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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 ‘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.

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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.

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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 ‘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.

//M4//QN19//SUB//DL0//EQ

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.

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

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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.

//M4//QN21//SUB//DL0//EQ

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.

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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 ‘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.

//M4//QN22//SUB//DL0//EQ

Explain enzymes and mechanism of DNA replication.

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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' → 3'.

Consequently, on one strand (the template with polarity 3’ → 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.

//M3//QN23//SUB//DL0//EQ

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

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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'→3', the strand that has the polarity 3'→5' acts as a template, and is also referred to as template strand.

Coding strand: The other strand which has the polarity (5'→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.

//M2//QN24//SUB//DL0

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

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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.

//M0//QN25//SUB//DL0//EQ

Give difference between: Template strand and Coding strand.

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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' → 5'	2.	It's polarity is 5' → 3'
3.	It has codons for protein synthesis	3.	It does not have codon for protein synthesis.

//M2//QN26//SUB//DL0//EQ

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

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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.

//M0//QN27//SUB//DL0//EQ

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

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5' – ATGCATGCATGCATGCATGCATGCATGC – 3'

The Sequence on complementary 3' → 5' strand is
3'–TACGTACGTACGTACGTACGTACGTACG–5'

Then, The sequence of complementary strand is
5'–GCATGCATGCATGCATGCATGCATGCAT–3'.

//M0//QN28//SUB//DL0//EQ

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

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If the coding strand has following sequence

5'–ATGCATGCATGCATGCATGCATGCATGC–3'

So, template strand sequence in 3' → 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

//M0//QN29//SUB//DL0//EQ

Describe transcription.

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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

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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.

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(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)

//M0//QN31//SUB//DL0//EQ

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

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(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.

//M4//QN32//SUB//DL0//EQ

Explain transcription unit with schematic diagram.

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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'→3', the strand that has the polarity 3'→5' acts as a template, and is also referred to as template strand.

The other strand which has the polarity (5' → 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.

//M4//QN33//SUB//DL0//EQ

Explain transcription in prokaryotes (bacteria) with diagram.

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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 (σ) 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.

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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.

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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.

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Give differences between DNA and RNA.

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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.

//M0//QN36//SUB//DL0//EQ

Differentiate : m-RNA and t-RNA

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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.

//M0//QN37//SUB//DL0

What are the salient features of genetic code?

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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.

//M2//QN38//SUB//DL0

Explain frame shift mutation and deletion.

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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.

//M3//QN39//SUB//DL0//EQ

Explain t-RNA as an adaptive molecule.

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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.

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List two essential roles of ribosome during translation.

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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.

//M4//QN41//SUB//DL0//EQ

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.

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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.

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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.

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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 (β-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 β-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.

//M0//QN43//SUB//DL0//EQ

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?

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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.

//M2//QN44//SUB//DL0

Describe goals of HGP.

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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.

//M0//QN45//SUB//DL0//EQ

Describe: Bioinformatics.

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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.

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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.

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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 ‘snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

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Why is the Human Genome project called a mega project ?

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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 × 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.

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

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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.

//M0//QN49//SUB//DL0//EQ

Give different between: Repetitive DNA and Satellite DNA.

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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.

//M0//QN50//SUB//DL0

What are the steps of DNA fingerprinting?

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(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.

//M2//QN51//SUB//DL0

Describe: satellite DNA

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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.

//M0//QN52//SUB//DL0//EQ

What is DNA fingerprinting? What are its applications?

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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

//M0//QN53//SUB//DL0//EQ

Briefly describe the polymorphism seen in the DNA sequence.

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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.