Preview:

• As the name suggests, this chapter revolves around the molecular basis of inheritance which is DNA and RNA. Thus, we shall study various processes that govern the synthesis and functioning of these elements. We will find that these strands of DNA and RNA are made of nucleotide chains; thus, forming a polynucleotide chain.
• We will begin our study with the structure of one such polynucleotide chain. We shall state an important rule that governs the pairing of these nucleotides, called, Chargaff s Rule.
• Then, we will study the Watson and Crick Model of DNA as a double-helix structure. We shall also see how this double helix structure is held together by nucleosome model of DNA. Further, we will discuss an important experiment by Frederick Griffith that ultimately led to the confirmation that DNA is the genetic material. We shall describe another experiment in this regard by Hershey and Chase.
• We will see that DNA replication is a semi-conservative process. That is, out of two strands present in the molecule, one of the strands is newly synthesized. We shall describe an experiment by Matthew Meselson and Franklin Stahl offering proof for the same.
• We will then define Transcription as the process of copying genetic information from one strand of the DNA into RNA is termed as transcription. We shall study in detail, the functions and structures of the three types of RNA: mRNA, tRNA and rRNA.
• We will define another important process that occurs after transcription called, Translation. Translation refers to the process of polymerisation of amino acids to form a polypeptide chain.
• We will then look at Lac Operon concept which explains Lactose metabolism in E. coli.
• Then, Human Genome Project shall be dealt with in some detail a project undertaken by scientists to map the genome in man to make a detailed data analysis of the entire DNA (number of nucleotides, genes causing genetic disorders, functional genes and so on).
• Lastly, we shall look at the most interesting concept: DNA Fingerprinting. DNA Fingerprinting involves determining a nucleotide sequence of certain regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. We shall study the procedure for it and its applications.

6.0 Introduction:

• Chromosome is a component present in gametes. It is these chromosomes that transmit heredity characters. The chromosomes are chemically composed of nucleic acids and proteins. The nucleic acids which carry genetic information are DNA.
• However, the nucleic acids are of two types: Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA). While, DNA acts as the genetic material in most of the organisms, RNA acts as a genetic material in some viruses.
• In this chapter, we are going to discuss the structure of DNA, its replication, the process of making RNA from DNA (transcription), the genetic code that determines the sequences of amino acids in proteins, the process of protein synthesis (translation) and elementary basis of their regulation.

Questions for the section:

1. Name the two types of nucleic acids present.

6.1 The DNA

• DNA was first identified by Friedrich Meischer. He named it as Nuclein . However, it was Altmann who identified the acidic properties of this material and called it nucleic acid .
• DNA is a long polymer of deoxyribonucleotides. The length of DNA is usually defined as number of nucleotides present in it. The length of DNA is characteristic of an organism.

Note box: A pair of nucleotides is called as a base pair (bp).

Example box:   A bacteriophage known as Φ 174 has 5386 nucleotides, Bacteriophage lambda has 48502 base pairs (bp), Escherichia coli has 4.6 106 bp, and haploid content of human DNA is 3.3 109 bp.

6.1.1 Structure of Polynucleotide Chain

• We know that, DNA and RNA are made of Polynucleotide chain.
  Each nucleotide has three components:

a) A nitrogenous base: Two kinds of nitrogenous bases are present: Purines (double ring structures) and Pyramidines (single ring structures). Adenine and Guanine are purines. Cytosine, Uracil and Thymine are pyramidines.

b) A pentose sugar: The pentose sugar is ribose in case of RNA, and deoxyribose in case of DNA),

c) A phosphate group.

• DNA consists of Adenine, Guanine, Thymine and Cytosine. Whereas, RNA consists of Adenine, Guanine, Uracil and Cytosine.
• A nitrogenous base is linked to the pentose sugar through an N-glycosidic linkage to form a nucleoside. Nucleosides are made of Sugar and Organic Nitrogenous base. Examples for nucleosides: Adenosine or deoxyadenosine, Guanosine or deoxyguanosine, Cytidine or deoxycytidine and Thymidine or deoxythymidine.
• A nucleotide is made of a Nucleoside and a $PO_4$ group.
  In DNA, a phosphate group is linked to 5′-OH of a nucleoside through phosphoester linkage, a corresponding nucleotide or deoxynucleotide is formed, depending upon the type of sugar present.
• Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a dinucleotide. Many such nucleotides can be joined in such a manner to form a polynucleotide chain.
• Referring figure 6.1, a polymer thus formed has a free phosphate group at 5′-end of ribose sugar, which is also called as 5 -end of polynucleotide chain.
• Similarly, at the other end of the polymer, the ribose has a free 3′-OH group which is referred to as 3′-end of the polynucleotide chain.

In RNA, every nucleotide residue has an additional OH group present at 2′-position in the ribose. Also, in RNA the uracil is found at the place of thymine (5-methyl uracil, another chemical name for thymine).

Chargaff s Rule:

Definition box:   Chargaff s Rule states that, the number of purines is equal to number of pyramidenes. Adenine is complementary to Thymine and; Guanine is complementary to Cytosine.

Explanation:

A pairs with T: the purine adenine (A) always pairs with the pyrimidine thymine (T).
C pairs with G: the pyrimidine cytosine (C) always pairs with the purine guanine (G).

$$A + G = T + C $$

Watson and Crick Model of DNA (or) Double-helix structure of DNA:

Salient features:

The DNA molecule consists of two helically twisted strands connected together by base pairs as shown in figure 6.2.

The two chains have anti-parallel polarity. It means that, if one chain has the polarity 5′ $ \vec{}$ 3′, the other has 3′ $\vec$ 5′.
Referring figure 6.3, the pentose sugar is linked to the nitrogenous base on the $1^{st}$ carbon atom and to $PO_4$ group on $5^{th}$ carbon atom of sugar. The linkage between the sugar and $PO_4$ molecule is called phosphodiester-bond. The $PO_4$ group is attached to next sugar at third carbon atom.

• The bases in two strands are paired through hydrogen bond (H-bonds); thus, forming base pairs (bp). Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa. Similarly, Guanine bonds 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 (as shown in Figure 6.2), called pitch.
• The two chains are coiled in a right-handed fashion.
• The pitch of the helix is 0.34 nm or 3.4 $A^0$.
• There are roughly 10 bp in each turn. Consequently, the distance between a bp in a helix is approximately equal to pitch = 0.34 nm.
• The length of a complete turn of helix is called, a gyre and measures 3.4 nm or 34 $A^0$.

Note box: Francis Crick proposed the Central dogma in molecular biology, which states that: the genetic information flows from DNA RNA Protein .

6.1.2 Packaging of DNA Helix

Length of DNA: Taken the distance between two consecutive base pairs as 0.34 nm (0.34 $10^{ 9}$ m), the length of DNA double helix in a typical mammalian cell can be calculated (simply by multiplying the total number of bp with distance between two consecutive bp, that is, 6.6 $10^9$ bp 0.34 $10^{-9}$m/bp), it comes out to be approximately 2.2 metres.

Packaging in Prokaryotes: 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.

Packaging in Eukaryotes:

• In eukaryotes, 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 lysines and arginines. Both the amino acid residues carry positive charges in their side chains.
  Histones are organised to form a unit of eight molecules called as histone octamer.
  The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (Figure 6.5 shown above). 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 (figure 6.6 shown above).
• The beads-on-string structure in chromatin is packaged to form chromatin fibres 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.

Note box:   Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive.

Note box:

Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive.

Questions for section 6.1:

1. By whom was DNA first identified?
2. How is length of DNA defined?
3. What is a pair of nucleotides called?
4. Explain the structure of a polynucleotide chain with diagram.
5. State and explain Chargaff s rule.
6. List the salient features of Watson and Crick Model of DNA.
7. State the Central Dogma theory with a flow diagram.
8. Explain packaging of DNA in eukaryotes.

6.2 The Search for Genetic Material

It took a long time to discover that DNA is the genetic material. Previous discoveries by Gregor Mendel, Walter Sutton, Thomas Hunt Morgan and numerous other scientists had narrowed the search to identifying the chromosomes.

Transforming Principle

• In 1928, Frederick Griffith, in a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia), witnessed a miraculous transformation in the bacteria. During the 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 a 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 when 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.

Conclusions:

• He concluded that, the R strain bacteria had somehow been transformed by the heat-killed S-strain-bacteria.
• Due to some transforming principle , that is, the heat-killed S strain had transferred some material to enable 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.

Biochemical Characterisation of Transforming Principle

Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44) worked to determine the biochemical nature of transforming principle in Griffith’s experiment.
They purified and separated biochemicals (proteins, DNA, RNA) 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. As expected, Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation.
Thus, they concluded that DNA is the hereditary material.

6.2.1 The Genetic Material is DNA

The accurate 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.

Experimental observations:

• The bacteriophage attaches to the bacteria and injects its genetic material into the bacterial cell.
• The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more viral genetic material.
• Hershey and Chase worked to discover whether it was a 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 sulphur.
• 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 that were 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 was therefore the genetic material that was passed from virus to bacteria (Figure 6.7).

6.2.2 Properties of Genetic Material (DNA versus RNA)

Note box:   RNA is the genetic material in some viruses like: Tobacco Mosaic viruses and QB bacteriophage.

A molecule that can act as a genetic material must fulfil the following criteria:

(1) It should be able to generate its replica (called replication).

(2) It should chemically and structurally be stable.

(3) It should provide the scope for slow changes (mutation) that are required for evolution.

(4) It should be able to express itself in the form of ‘Mendelian Characters .

Why DNA is a better genetic material?

Stability as one of the properties of genetic material was very evident from Griffith s transforming principle that heat; though killed the bacteria, did not destroy some of the properties of genetic material. This is because, the two strands of DNA 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. 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.
In fact, the presence of thymine at the place of uracil also confers additional stability to DNA.

Comparison between DNA and RNA:

Both DNA and RNA mutate. In fact, RNA being more unstable mutates at a faster rate. Consequently, viruses having RNA genome and having shorter lifespan mutate and evolve faster.
RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins.

Note box:   The above discussion indicate that both RNA and DNA can function as genetic material, but DNA being more stable is preferred for storage of genetic information. For the transmission of genetic information however, RNA is better.

Questions for section 6.2:

1. Explain Griffith s experiment of transforming principle on Streptococcus pneumonia. List the conclusions drawn from it.
2. Explain the biochemical characterization of transforming principle.
3. With figure, explain the Hershey-Chase experiment.
4. List the criteria for a substance to qualify as a genetic material.
5. Give reasons as to why DNA is a better genetic material than RNA.

6.3 RNA World

RNA was the first genetic material. Essential life processes such as metabolism, translation and splicing evolved around RNA. RNA used to act as a genetic material as well as a catalyst. 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 complementary strand further resists changes by evolving a process of repair.

6.4 Replication

Replication of DNA takes place in S phase of interphase. It is identified as an autocatalytic property of DNA. It was confirmed by Messelson and Stahl.
It is a semi-conservative process. That is, out of two strands present in the molecule, one of the strands is newly synthesized; while, the other is parental strand.
Replication takes place with the help of enzymes.

6.4.1 The Experimental Proof

Matthew Meselson and Franklin Stahl performed the following experiment to show that DNA replication is a semi-conservative process:

(i) They grew E. coli in a medium containing $^{15}NH_4Cl$ ($^{15}N$ is the heavy isotope of nitrogen). The result was that, $^{15}N$ was incorporated into newly synthesised DNA. This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

(ii) Then they transferred the cells into a medium with normal $^{14}NH_4Cl$ 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 (Figure 6.8).

(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 since 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 of this hybrid DNA and of light DNA.

6.4.2 The Machinery and the Enzymes

Mechanism of Replication:

• DNA replication begins at certain fixed unique points called origin of replication.
• The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template (or strand) to catalyse the polymerisation of deoxynucleotides.
• Replication begins with a nick or incision made by endonuclease. This opening is called, the replication fork.
• The two strands unwind with the help of helicase. Thus, breaking the hydrogen bonds between complementary nitrogenous bases. This process is called melting.
• The unwinding of the strands imposes tension at the distal end of the DNA molecule. The tension is eased by super-helix-relaxing-proteins which bind to the DNA strand. This protein is called gynase or topoisomerase or swivelling protein.
• Initiation of DNA synthesis always requires a small segment of RNA primer. The length of primer varies between 10-60 nucleotides. Synthesis of RNA is catalysed by primase or RNA polymerase.
• The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is, 5′ $\vec{}$ 3′. Consequently, in the template strand (having the polarity 3′ $\vec{}$ 5′), the replication is continuous, while on the coding strand (having polarity 5′ $\vec{}$ 3′), it is discontinuous. The discontinuously synthesised fragments and are called Okazaki fragments. These fragments are later joined by the enzyme DNA ligase.

6.5 Transcription

Definition box: Transcription: The process of copying genetic information from one strand of the DNA into RNA is termed as transcription.

Q and A:

Why are both strands not copied during transcription?

As we know, the DNA strand involved in transcription is called template DNA and the other strand which does not take part in transcription is called coding strand. If both strands participate in transcription and each acts as a template, they would code for RNA molecule with different sequences. Hence, if these RNA strands code for proteins, they would be of different types. This means, one DNA strand would code for different proteins and this would result in complication in terms of the genetic information transfer mechanism. Besides, if two RNA molecules are produced simultaneously, they would complement each other and would form a double stranded RNA. This would prevent RNA from being translated into protein.

6.5.1 Transcription Unit

A transcription unit in DNA is defined by three regions in the DNA:

(i) A Promoter
(ii) The Structural gene
(iii) A Terminator

• Since, the two strands have opposite polarity (5′ $\vec{}$ 3′ and 3′ $\vec{}$ 5′) and because the DNA-dependent RNA polymerase catalyse the polymerisation in only one direction, that is, 3′ $\vec{}$ 5′ , this strand acts as the template strand. The other strand which has the polarity: 5′ $\vec{}$ 3′ is referred to as coding strand.
• The promoter, the terminator flank and the structural gene forms a transcription unit.
• The promoter is said to be located towards 5′-end (upstream) of the structural gene (with respect to coding strand). The promoter is a DNA sequence that provides binding site for RNA polymerase. Also, it is the presence of a promoter in a transcription unit which defines the template and coding strands. Thus, by switching the position of promoter with terminator, the definition of coding and template strands could be reversed.
• The terminator is located towards 3′-end (downstream) of the coding strand and it usually defines the end of the process of transcription (Figure 6.10).

6.5.2 Transcription Unit and the Gene

Definition box:   Gene: A gene is defined as the functional unit of inheritance.

Note box:   1. A section of a DNA or RNA molecule that codes for a specific polypeptide in protein synthesis is called a Cistron. 2. The structural gene or cistron in a transcription unit of eukaryotes is said to be monocistronic and in bacteria or prokaryotes it is, polycistronic. 3. In eukaryotes, the monocistronic structural genes have interrupted coding sequences or splits called Exons. The exons are interrupted by introns. Introns are nothing but intervening sequences in exons. 4. Exons appear in mature or processed RNA and such RNA does not have introns.

6.5.3 Types of RNA and the process of Transcription

Types of RNA: In bacteria, there are three major types of RNAs. All three RNAs are needed to synthesise a protein in a cell. They are:

1. mRNA (messenger RNA): Messenger RNA is a single stranded RNA. The mRNA carries genetic information from DNA to Ribosome present in cytosol, where it is used as a template for protein synthesis. Messenger RNA is made of a sequence of codons. Such a single stranded mRNA codes for more than one protein and is thus considered polycistronic m-RNA. As discussed in the previous section, eukaryotic mRNA remain monocistronic (each mRNA codes only for one protein).

Structure of mRNA:

• The structure of mRNA was first discovered by Volkin and later coined by Jacob and Monad. The mRNA is synthesized in the nucleus which later migrates into the cytoplasm. A eukaryotic mRNA has the following features:
  • Average size of eukaryotic mRNAs is 1500 to 2000 nucleotides.
  • Cap region: It is the initial segment of the polynucleotide strand. It has Methyl guanosine nucleotide. This region assists in attachment to ribosome.
  • Initiator codon: RNA molecule has an initiator codon, AUG. It is the first sensible codon that codes for amino acid, methionine. It acts as the start signal for protein synthesis.
  • Leader sequence: Upstream of the initiator codon contains non-translatable sequences called leader which consists of 100 to 150 nucleotides.
  • Coding region: This region carries sensible information (triplet codes) of amino acid sequence. Accordingly, the type of protein is synthesized.
  • Chain terminator: The three triplet codons: UGA, UAA and UAG do not recognize or code for any amino acids. They terminate the process of protein synthesis. They are also called, Punctuation codon/ stop codon/ nonsense codon/ terminator codon.
  • Non coding region 2: It is composed of 100-150 nucleotides.

2. tRNA (transfer RNA): The tRNA binds to amino acids and reads the genetic code. (refer section 6.6.2 for detailed structure and function)

3. rRNA (ribosomal RNA):

• It is a class of un-branched RNA molecule found associated with proteins and is organized into spherical bodies called ribosomes.
• It is also called as insoluble RNA and constitutes 80% of the total cellular RNA. It is one of the non-genetic RNA molecules found as a relatively stable molecule and occasionally may exhibit pseudo helical structure.
• It coordinates the function of protein synthesis.

Types of rRNA:

• The eukaryotic cells have four kinds of rRNA molecules; namely, 28-rRNA, 18S rRNA, 5.8S rRNA and 5S rRNA.
• The eukaryotic cell has 80S type ribosome with two subunits: 60S and 40S types. The 60S is made of 28S type, 5.8S type and 5S subunits while 40S unit is made of 18S rRNA subunits.
• In prokaryotic cells, 70S type of ribosome with two subunits: 50S and 30S is present. The 50S is made of 23S and 5S subunits while, 30S is made of 16S rRNA subunits.

Transcription of a gene takes place in three stages: initiation, elongation, and termination as described:

1. Initiation: RNA polymerase binds to a sequence of DNA called the promoter, found near the beginning of a gene. Each gene has its own promoter. Once bound, RNA polymerase separates the DNA strands, providing the single-stranded template needed for transcription.

2. Elongation: One strand of DNA, the template strand, acts as a template for RNA polymerase. As it reads this template one base at a time, the polymerase builds an RNA molecule out of complementary nucleotides, making a chain that grows from 5′ to 3′. The RNA transcript carries the same information as the non-template (coding) strand of DNA, but it contains the base uracil (U) instead of thymine (T).

3. Termination: Sequences called terminators signal that the RNA transcript is complete. Once they are transcribed, they cause the transcript to be released from the RNA polymerase.

Note box:   The RNA polymerase is only capable of catalysing the process of elongation. It associates transiently with initiation-factor (s) and termination-factor (r) to initiate and terminate the transcription, respectively. These factors specify the RNA polymerase to either initiate or terminate (Figure 6.11).

In eukaryotes, there are two complexities during transcription

(i) 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).

(ii) 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 an 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 (Figure 6.12).

Questions for section 6.4 and 6.5:

1. Describe Matthew Meselson and Franklin Stahl s experiment to show that DNA replication is a semi-conservative process.
2. Explain with a diagram, the mechanism of replication.
3. Define transcription. What are the three units of transcription?
4. Define the terms: a) Gene b) Cistron c) Exons d) Introns 4. Write a note on function and structure of mRNA.
5. Write a note on function and structure of tRNA.
6. What is rRNA? List the types.
7. Describe the stages in transcription.

6.6 Genetic Code

During replication and transcription, a nucleic acid is copied to form another nucleic acid. The process of translation requires transfer of genetic information from a polymer of nucleotides to a polymer of amino acids.

Checker-board for genetic code was prepared which is as given in figure 6.1:

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) One codon codes for only one amino acid, hence, it is unambiguous and specific.

(iii) Some amino acids are coded by more than one codon (that is, more than one triplet), hence the code is degenerate.

(iv) The codon is read in mRNA in a contiguous (adjacent) fashion. There are no punctuations.

(v) 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.

(vi) AUG has dual functions. It codes for Methionine (met), and it also act as initiator codon.

6.6.1 Mutations and Genetic Code

The effect of point mutations will be explained here.

Example box:   An example of point mutation is a change of single base pair in the gene that codes for beta globin chain. When there is a change of amino acid: glutamate to valine, it results into a diseased condition called as sickle cell anemia.

6.6.2 tRNA the Adapter Molecule

• Francis Crick suggested 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.
• He postulated the presence of an adapter molecule that would on one hand read the code and on other hand would bind to specific amino acids.
• The role of tRNA as an adapter molecule was assigned much later.

Structure of tRNA:

The structure of tRNA is referred to as the clover-leaf model. It exhibits 5 arms of which, one is short (called as variable arm or mini loop) with few nucleotides. Out of the other four arms, 3 possess loops while one does not. The 5 arms are as described:

a) DHU arm: It is the second arm from 5 end. This arm has a loop that contains 8-12 unpaired bases with a special nitrogen base called Di-hydroxyl uridine. This arm of tRNA is linked to an amino acid. The enzyme, Amino acyl synthetase is responsible for covalently linking amino acids to the tRNA. This enzyme is also responsible for activation of amino acids prior to their transport.

b) Anticodon arm: This arm is located opposite to the amino acid acceptor arm. It has a loop with 7 unpaired bases. Three bases form the Anticodon arm or the NODOC arm. The anticodon arm remains complementary to a specific codon found on mRNA which assists in synthesis of a protein.

c) Variable arm or Mini loop: A small extra loop having variable number of nucleotides is present between anticodon arm and pseudo-uridine arm.

d) Pseudo-uridine arm: This arm is located near 3 end of tRNA molecule. It has a loop with seven unpaired bases with a special base, namely, pseudo uridine. This arm functions as the ribosome attachment region during protein synthesis.

In figure 6.13, 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)

Questions from section 6.6:

1. List the salient features of genetic code.
2. Explain the structure of tRNA.

6.7 Translation

Concept box:   Proteins are polymerized compounds built of amino acids. These polymers are synthesized over specific ribosomes in the cytoplasm. The number and sequence of amino acids for a given protein is finite and remains independent of the other.

Definition box:   Translation refers to the process of polymerisation of amino acids to form a polypeptide chain (as shown in Figure 6.14 below).

Process of Translation:

Step 1 – Binding of mRNA with ribosome:

The transcribed m-RNA gets released into the cytoplasm and subsequently adheres to 30s (s stands for smaller sub-unit) of ribosome. This bondage provides a stable medium for m-RNA molecule to go for decoding. This attachment is by the 5 end of the m-RNA molecule.

Step 2 – Activation of Amino acid:

The process of activation is induced by amino acyl synthetase (identified by DHU arm of t-RNA). Indeed, it is the activated amino acid which develops bondage with t-RNA at CCA (3 end). The phenomenon of activation involves the participation of one ATP molecule as depicted below:

Activated Amino acid + ATP + Enzyme $\vec{}$ AA-AMP enzyme + PP
(Activated AA-enzyme complex)
As indicated in the equation, the complex built is unstable.

Step 3 – Binding of activated AA with tRNA:

AA AMP enzyme + tRNA $\vec{}$ AA-tRNA + AMP + Enzyme
(AA- tRNA complex)

As presented in the equation, the activated amino acid develops a bondage with tRNA molecule at the CCA (3 end) getting organized as AA-tRNA complex.

Step 4 – Chain Initiation

Process of polymerization of amino acid commences with the arrival of first tRNA molecule (with anticodon as UAC) carrying activated methionine. Incidentally, the triplet, AUG the first sensible codon would have got placed at p site of the ribosome (decoding area of ribosome). The t-RNA with amino acid now interacts with AUG by its anticodon (UAC). This interaction facilitates the formation of initiation complex. Further, this event is influenced by the participation of certain intrinsic factors (proteins) such a s $IF_1, IF_2$, and $IF_3$. It is at this instance, the larger sub-unit of ribosome (50s) gets bound to form a dimer (full-fledged ribosome).

Step 5 – Chain Elongation

This stage involves the decoding of all sensible codons of the mRNA ribbon. It is during this process that a definite polypeptide is synthesized at p site of the ribosome. It is to be emphasized that this process is accomplished due to the jerk of ribosome on the mRNA favouring the respective codons to be brought at p site, every time the ribosome takes a jerk towards 3 direction of m-RNA. The process of chain elongation thus involves the arrival of fresh t-RNA amino acid complex at A site and its subsequent shift to P site of the ribosome. The amino acids are left at P site when the tRNA molecules make their exit at E-site of the ribosome.
Polymerization of amino acids at P site gets extended till the last sensible codon gets decoded. The process of this polymerization continues with formation of peptide bonds after every 2 amino acids.

Step 6 – Chain termination

With the decoding of last sensible codon, the terminator codon (UAA or UAG or UGA) offers stop signal to process of polymerization (polypeptide linkage). It is to be emphasized that this codon does not possess its complementary anticodon. Thus, it gets distinguished as chain terminator only (that is, nonsense codon).

Concept box:   Polyribosome: It is the chain of ribosomes on a single mRNA strand. In other words, the mRNA molecules feature more than one ribosome which makes the system to qualify as polysome.

Gene and Gene concept:

The term gene was coined by Johannsen to a chromosomal unit involved in heredity and character expression. Beadle and Tatum described the gene as an elementary heredity unit known for its ability of undertaking transcription and subsequent translation to generate a protein (enzyme).

Characteristics of a gene:

1. Gene is a component of genetic material which contains coded information for the specific inheritable characteristic.
2. It is able to express its effect by regulating structure or metabolism.
3. It is capable of replication so that its replica or copy can be transmitted to a daughter cell or an offspring.
4. It can undergo mutation to form an allele with a different expression. It is basic to biological variations that are essential for adaptation and evolution.
5. New genes develop due to re-shuffling of components of existing genes.
6. It occupies a specific locus over the chromosome. The change in its position can change its expression.
7. There is a system of differential expression of genes. It governs development and differentiation.

According to Seymour Benzer, the gene can be defined as an elementary unit of chromosome promoting heredity is composed of DNA. However, the gene can exist in any of the following forms:

1) Replication: Property of the given gene is to go for replication and to produce daughter helices (autocatalytic).

2) Cistron: It is the functional state of the gene involving transcription and translation. It is by this property that the gene regulated an activity through enzyme production.

3) Recon: It indicates the ability of the given gene to undergo recombination.

4) Muton: The smallest element of genetic material capable of undergoing mutation usually identified as a single pair of nucleotides.

Questions from section 6.7:

1. Define translation. Explain the steps in translation.
2. By who was the term, gene coined? 3. List the characteristics of a gene.
3. List the three forms in which genes exist in.

6.8 Regulation of Gene Expression

Regulation of gene expression occurs at various levels. Gene expression results in the formation of a polypeptide as we now know. In eukaryotes, the regulation could be exerted at:

(i) transcriptional level (formation of primary transcript),

(ii) processing level (regulation of splicing),

(iii) transport of mRNA from nucleus to the cytoplasm,

(iv) translational level

The genes in a cell are expressed to perform a particular function or a set of functions as explained below.

Example box:   1. For example, if an enzyme called beta-galactosidase is synthesised by E. coli, it is used to catalyse the hydrolysis of a disaccharide, lactose into galactose and glucose (the bacteria use them as a source of energy). If the bacteria do not have lactose around them to be utilised for energy source, they would no longer require the synthesis of the enzyme beta-galactosidase. Therefore, in simple terms, it is the metabolic, physiological or environmental conditions that regulate the expression of a gene. 2. The development and differentiation of embryo into adult organisms are also a result of the coordinated regulation of expression of several sets of genes

The operon model:

The gene constitution of living cells is known for its full complement. However, at a given occasion, all genes may not strike. In other words, some of them would not go for transcription. This specific gene action could be attributed to the phenomenon of gene regulation. The selective operation of DNA molecule is analysed as Operon Model. In short, operon model suggests that there is coordinated activity with respect to functional genes.

6.8.1 The Lac operon

Operon concept explaining the role of structural genes and controlled by regulatory genes as a coordinated gene activity was proposed by Jacob and Monad.

Definition box:   The term operon is coined to an integrated assembly of structural genes controlled by a set of regulatory genes with operator as main switch.

In cell system, there exist a number of operons promoting a number of different enzymes required for the metabolism of different substances. We shall study the lac operon of E. coli.

• Lac Operon helps enzymes to coordinate and brings about, lactose metabolism.
• Metabolism of lactose in E. coli needs 3 independent enzymes namely, $\Beta$ -Galactosidase, permease and trans-acetylase. After all, these 3 enzymes are proteins and have been the products of gene activity of 3 structural genes presently marked as: Z, Y and A.
• The activity ZYA and their transcription to produce polycistronic mRNA depends on the state of operator-gene found in proximity to ZYA. However, the activity of operator is controlled by regulator gene.
• The regulator gene synthesizes Repressor protein which in turn binds operator and makes it inactive. Thus, the structural genes fail to go for transcription. This state of operon is the reflection of absence of lactose in the cell.
• The operator gene gets into active state (switched on state) with the influx of lactose molecule. In other words, the lactose molecules qualify as inducers. The operon is in a switched on state when the repressor protein gets bound by lactose molecule. In such a state, the repressor protein fails to attack the operator. Hence, the operator gene keeps ZYA active and makes them go for transcription and subsequent translation. As a result, all three enzymes required for lactose metabolism is produced.
• Transcription of three structural genes is influenced by mRNA polymerase ofcourse identified by promoter gene (P). In fact, the mRNA polymerase slides over the operator and in turn makes ZYA to go for transcription.
• In short, the Lac Operon concept illustrates the coordinated and organized activity of the given set of structural genes influenced by regulatory genes such as O,R and P.

Note box: Negative regulation: Regulation of lac operon by repressor is referred to as negative regulation.

Negative regulation: Regulation of lac operon by repressor is referred to as negative regulation.

Questions from section 6.8:

1. List the 4 levels at which regulation of a gene expression can occur.
2. Explain with an example how a gene is responsible for a particular function and how gene regulation also depends on environmental factors.
3. Explain the concept of Lac operon with a diagram.
4. What is negative regulation?

6.9 Human Genome Project (HGP)

Each and every individual has a unique identity due to his or her own genetic make-up. No two individuals will have the same nucleotide sequence except monozygotic twins . It therefore became the ambition of scientists to map the genome in man. Advanced techniques in genetic engineering have made it possible to map the human genome.

Fact box:   If the obtained sequences of DNA 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 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.

Goals of HGP

Some of the important goals of HGP were as follows:

(1) Identify all the approximately 20,000-25,000 genes in human DNA;

(2) Determine the sequences of the 3 billion chemical base pairs that make up human DNA and to store this information in databases.

(3) Improve tools for data analysis.

(4) Transfer related technologies to other sectors, such as industries.

(5) Address the ethical, legal, and social issues (ELSI) that may arise from the project.

(6) Recognize the genes causing genetic disorders from the functional genes.

Fact box:   Many non-human model organisms, such as bacteria, yeast, Caenorhabditis elegans (a free living non-pathogenic nematode), Drosophila (the fruit fly) and plants (rice and Arabidopsis) have been sequenced. Besides providing clues to understanding human biology, it helped understanding DNA sequence. This could in turn help solve problems in health care, agriculture, energy production and in environment remediation.

Methodologies: Two major approaches at analysis of HGP were done:

a) One approach focused on identifying all the genes that are expressed as RNA (referred to as Expressed Sequence Tags (ESTs).

b) Sequencing the whole set of genome and later assigning different regions in the sequence with functions is called Sequence Annotation.

Procedure:

• For such sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes.
• Next, they are 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). These hosts were responsible for cloning.
• The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger these sequences were arranged based on some overlapping regions present in them.
• These sequences were subsequently annotated and were assigned to each chromosome from 1 to 22 and, X and Y.

6.9.1 Salient Features of Human Genome

Some of the salient observations drawn from human genome project are as follows:

(1) The human genome contains 3.1647 billion nucleotide bases.

(2) The average gene consists of 3000 base pairs, but sizes vary greatly, with the largest known human gene being dystrophin with 2.4 million bps (located on x chromosome).

(3) The total number of genes estimated is 30,000. Almost all, that is, 99.9 per cent nucleotide bases are exactly the same in all people.

(4) The functions are unknown for over 50 per cent of the discovered genes.

(5) Less than 2 per cent of the genome codes for proteins.

(6) Repeated sequences make up very large portion of the human genome.

(7) 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 are believed to be involved in chromosome structure, dynamics and evolution.

(8) Chromosome 1 has most genes (2968), and the Y has the fewest (231).

(9) Scientists have identified about 1.4 million locations where singlebase DNA differences (SNPs single nucleotide polymorphism, pronounced as snips ) occur in humans.

6.9.2 Applications and Future Challenges

1. Efforts are in progress to determine genes causing cancer.
2. To design drugs and genetically modified diets.
3. Possibility of the study of various genes and proteins working together in an inter connected network.

Questions from section 6.9:

1. What is HGP? List the goals of HGP.
2. Explain the two approaches at analysis of HGP.
3. Explain the procedure for sequence annotation.
4. List the salient features of human genome.
5. List the applications of HGP.

6.10 DNA Fingerprinting

Definition box: Principle of DNA Fingerprinting: DNA Fingerprinting involves determining a nucleotide sequence of certain regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.   These repetitive DNA sequences are unique to an individual.

Note box: It was discovered by Alec Jeffrey. In India, Dr. VK Kashyap and Dr. Laljit Singh started DNA fingerprinting.

Description:

• The repetitive DNA fragments 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. These sequences normally do not code for any proteins, but they form a large portion of human genome.
• The repetitive sequence are basically short nucleotide repeats specific to each individual and varies in number from person to person but is inherited. These are: Variable number of Tandem repeats (VNTR). Since they are inherited, half the VNTR sequences resemble the mother and other half resembles the father.

Note box: If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.

Concept box: Polymerase chain reaction (PCR) is a technique used in molecular biology to amplify a single copy or a few copies of a segment of DNA to produce thousands to millions of copies of a particular DNA sequence.

Steps in DNA fingerprinting:

Figure 6.16 Schematic representation of DNA fingerprinting: Few representative chromosomes have been shown to contain different copy number of VNTR. For the sake of understanding different colour schemes have been used to trace the origin of each band in the gel. The two alleles (paternal and maternal) of a chromosome also contain different copy numbers of VNTR. It is clear that the banding pattern of DNA from crime scene matches with individual B, and not with A.

1. The DNA is extracted from nuclei of white blood cells or hair follicle or spermatozoa.
2. The DNA fragments are cut into fragments using Restriction endonucleases (REN). These fragments also contain VNTR.
3. The fragments are separated according to size by gel electrophoresis (In this method, DNA is separated by subjecting it to field of electricity).
4. Fragments of a particular size having VNTR are multiplied through PCR technique. Then, they are treated with alkaline chemicals to split them into single stranded DNA.
5. Transferring (or) blotting: The DNA fragments are separated and transferred to synthetic membranes, such as nitrocellulose or nylon.
6. Radioactive DNA probes living repeated base sequences complementary to possible VNTRs are poured over nylon membrane. This method of hybridization of DNA probes with single stranded DNA on nylon membrane is called Southern blotting.
7. Then, the nylon membrane is washed to remove the extra probes left after hybridization.
8. An X-ray is exposed to nylon membrane to look where the radioactive probes have bound to the DNA. These regions appear as dark bands when the X-ray film is developed. This is called auto radiography.
9. The dark bands represent the DNA fingerprints or DNA profile.

Applications:

1. Individuality: Helps in distinguishing one individual from another (except in monozygotic twins).
2. Paternity and maternity issues can be resolved.
3. Helps identify genes that cause genetic disorders.
4. Useful in forensic sciences.
5. Human Lineage can be traced.
6. Can be used to identify racial groups their origin, historical migration and invasion.

Questions from section 6.10:

1. State the principle behind DNA fingerprinting technique.
2. What is VNTR?
3. Explain the steps in DNA finger-printing. Also, state its applications.