Molecular Genetics

Class 12 | Pure Science — Zoology | Chapter 5

Class 12 Unit 5 Pure_Science

UNIT 2 CHAPTER 5 — MOLECULAR GENETICS

New life for Woolly Mammoth — DNA researchers can now re-create the genes of mammoth and study the proteins they encoded

Chapter Outline

5.1 Gene as the functional unit of inheritance | 5.2 In search of the genetic material | 5.3 DNA is the genetic material | 5.4 Chemistry of nucleic acids | 5.5 RNA world | 5.6 Properties of genetic material | 5.7 Packaging of DNA helix | 5.8 DNA Replication | 5.9 Transcription | 5.10 Genetic code | 5.11 tRNA the adapter molecule | 5.12 Translation | 5.13 Regulation of Gene expression | 5.14 Human Genome Project (HGP) | 5.15 DNA finger printing technique

Learning Objectives

• Identifies DNA as the genetic material.
• Understands the organization of prokaryotic and eukaryotic genome.
• Learns to differentiate the nucleotides of DNA and RNA.
• Understands gene expression — Replication, Transcription and Translation.
• Learns about codons and the salient features of genetic code.
• Understands the gene regulation through Lac operon model.
• Realizes the importance of Human Genome Project.
• Illustrates the applications of DNA finger printing technique.

Mendel's theory dispelled the mystery of why traits seemed to appear and disappear magically from one generation to the next. Mendel's work reveals the patterns of heredity and reflect the transmission of evolved information from parents to offspring. This information is located on the chromosomes. One of the most advanced realizations of human knowledge was that our unique characteristics are encoded within molecules of DNA. The discovery that DNA is the genetic material left several questions unanswered. How is the information in DNA used? Scientists now know that DNA directs the construction of proteins. Proteins determine the shapes of cells and the rate of chemical reactions, such as those that occur during metabolism and photosynthesis.

The hereditary nature of every living organism is defined by its genome, which consists of a long sequence of nucleic acids that provide the information needed to construct the organism. The genome contains the complete set of hereditary information for any organism. The genome may be divided into a number of different nucleic acid molecules. Each of the nucleic acid molecule may contain large number of genes. Each gene is a sequence within the nucleic acid that represents a single protein. In this chapter we will discuss the structure of DNA, its replication, the process of making RNA from DNA (transcription), the genetic code that determines the sequence of amino acid in protein synthesis (translation), regulation of gene expression and the essentials of human genome sequencing.

5.1 Gene as the Functional Unit of Inheritance

A gene is a basic physical and functional unit of heredity. The concept of the gene was first explained by Gregor Mendel in 1860's. He never used the term 'gene'. He called it 'factor'. In 1909, the Danish biologist Wilhelm Johannsen, coined the term 'gene', that was referred to discrete determiners of inherited characteristics. According to the classical concept of gene introduced by Sutton in 1902, genes have been defined as discrete particles that follow Mendelian rules of inheritance, occupy a definite locus in the chromosome and are responsible for the expression of specific phenotypic character.

They show the following properties:

• Number of genes in each organism is more than the number of chromosomes; hence several genes are located on the same chromosome.
• The genes are arranged in a single linear order like beads on a string.
• Each gene occupies a specific position called locus.
• Genes may exist in several alternate forms called alleles.
• Genes may undergo sudden change in positions and composition called mutations.
• Genes are capable of self-duplication producing their own copies.

One Gene – One Enzyme Hypothesis

The experiments of George Beadle and Edward Tatum in the early 1940's on Neurospora crassa (the red bread mould) led them to propose one gene–one enzyme hypothesis, which states that one gene controls the production of one enzyme.

One Gene – One Polypeptide Hypothesis

It was observed that an enzyme may be composed of more than one polypeptide chain and a gene can code for only one polypeptide chain. Thus one gene–one polypeptide hypothesis states that one gene controls the production of only one polypeptide chain of an enzyme molecule.

5.2 In Search of the Genetic Material

As early as 1848, Wilhelm Hofmeister, a German botanist, had observed that cell nuclei organize themselves into small, rod like bodies during mitosis called chromosomes. In 1869, Friedrich Miescher, a Swiss physician, isolated a substance from the cell nuclei and called it as nuclein. It was renamed as nucleic acid by Altman (1889), and is now known as DNA. By 1920, it became clear that chromosomes are made up of proteins and DNA. Many experiments were carried out to study the actual carriers of genetic information.

Griffith's experiment proved that DNA is the genetic material which has been dealt in class XI. Bacterial transformation experiments provided the first proof that DNA is the genetic material. However, he could not understand the cause of bacterial transformation, and the biochemical nature of genetic material was not defined from his experiments. Later, Oswald Avery, Colin Macleod and Maclyn McCarty in 1944 repeated Griffith's experiments in an in vitro system in order to identify the nature of the transforming substance responsible for converting a nonvirulent strain into virulent strain. They observed that the DNA, RNA and proteins isolated from the heat killed S-strain when added to R-strain changed their surface character from rough to smooth and also made them pathogenic (Figure 5.1).

But when the extract was treated with DNase (an enzyme which destroys DNA) the transforming ability was lost. RNase (an enzyme which destroys RNA) and proteases (an enzyme which destroys protein) did not affect the transformation. Digestion with DNase inhibited transformation suggesting that the DNA caused the transformation. These experiments suggested that DNA and not proteins is the genetic material.

Transformation

The phenomenon, by which DNA isolated from one type of cell (S strain), when introduced into another type (R strain), is able to retain some of the properties of the S strain is referred to as transformation.

5.3 DNA is the Genetic Material

Many biologists despite the earlier experiments of Griffith, Avery and others, still believed that protein, not DNA, was the hereditary material in a cell. As eukaryotic chromosomes consist of roughly equal amounts of protein and DNA, it was said that only a protein had sufficient chemical diversity and complexity to encode the information required for genetic material. In 1952, however, the results of the Hershey-Chase experiment finally provided convincing evidence that DNA is the genetic material.

5.3.1 Hershey and Chase Experiment on T2 Bacteriophage

Alfred Hershey and Martha Chase (1952) conducted experiments on bacteriophages that infect bacteria. Phage T2 is a virus that infects the bacterium Escherichia coli. When phages (virus) are added to bacteria, they adsorb to the outer surface, some material enters the bacterium, and then later each bacterium lyses to release a large number of progeny phage. Hershey and Chase wanted to observe whether it was DNA or protein that entered the bacteria.

All nucleic acids contain phosphorus, and proteins contain sulphur (in the amino acid cysteine and methionine). Hershey and Chase designed an experiment using radioactive isotopes of Sulphur (³⁵S) and phosphorus (³²P) to keep separate track of the viral protein and nucleic acids during the infection process. The phages were allowed to infect bacteria in culture medium which containing the radioactive isotopes ³⁵S or ³²P. The bacteriophage that grew in the presence of ³⁵S had labelled proteins and bacteriophages grown in the presence of ³²P had labelled DNA. The differential labeling thus enabled them to identify DNA and proteins of the phage.

Hershey and Chase mixed the labeled phages with unlabeled E. coli and allowed bacteriophages to attack and inject their genetic material. Soon after infection (before lysis of bacteria), the bacterial cells were gently agitated in a blender to loosen the adhering phage particles. It was observed that only ³²P was found associated with bacterial cells and ³⁵S was in the surrounding medium and not in the bacterial cells. When phage progeny was studied for radioactivity, it was found that it carried only ³²P and not ³⁵S (Figure 5.2).

These results clearly indicate that only DNA and not protein coat entered the bacterial cells. Hershey and Chase thus conclusively proved that it was DNA, not protein, which carries the hereditary information from virus to bacteria.

5.4 Chemistry of Nucleic Acids

Having identified the genetic material as the nucleic acid DNA (or RNA), we proceed to examine the chemical structure of these molecules. Generally nucleic acids are a long chain or polymer of repeating subunits called nucleotides. Each nucleotide subunit is composed of three parts: a nitrogenous base, a five-carbon sugar (pentose) and a phosphate group.

Pentose Sugar

There are two types of nucleic acids depending on the type of pentose sugar. Those containing deoxyribose sugar are called Deoxyribo Nucleic Acid (DNA) and those with ribose sugar are known as Ribonucleic Acid (RNA). DNA is found in the nucleus of eukaryotes and nucleoid of prokaryotes. The only difference between these two sugars is that there is one oxygen atom less in deoxyribose.

Nitrogenous Bases

The bases are nitrogen containing molecules having the chemical properties of a base (a substance that accepts H⁺ ion or proton in solution). DNA and RNA both have four bases (two purines and two pyrimidines) in their nucleotide chain. Two of the bases, Adenine (A) and Guanine (G) have double carbon nitrogen ring structures and are called purines. The bases, Thymine (T), Cytosine (C) and Uracil (U) have single ring structure and these are called pyrimidines. Thymine is unique for DNA, while Uracil is unique for RNA.

The Phosphate Functional Group

It is derived from phosphoric acid (H₃PO₄), has three active OH groups of which two are involved in strand formation. The phosphate functional group (PO₄) gives DNA and RNA the property of an acid (a substance that releases an H⁺ ion or proton in solution) at physiological pH, hence the name nucleic acid. The bonds that are formed from phosphates are esters. The oxygen atom of the phosphate group is negatively charged after the formation of the phosphodiester bonds. This negatively charged phosphate ensures the retention of nucleic acid within the cell or nuclear membrane.

Nucleoside and Nucleotide

The nitrogenous base is chemically linked to one molecule of sugar (at the 1' carbon of the sugar) forming a nucleoside. When a phosphate group is attached to the 5' carbon of the same sugar, the nucleoside becomes a nucleotide. The nucleotides are joined (polymerized) by condensation reaction to form a polynucleotide chain. The hydroxyl group on the 3' carbon of a sugar of one nucleotide forms an ester with the phosphate of another nucleotide. The chemical bonds that link the sugar components of adjacent nucleotides are called phosphodiester bond (5' to 3'), indicating the polarity of the strand.

The ends of the DNA or RNA are distinct. The two ends are designated by the symbols 5' and 3'. The symbol 5' refers to carbon in the sugar to which a phosphate (PO₄) functional group is attached. The symbol 3' refers to carbon in the sugar to which hydroxyl (OH) functional group is attached. In RNA, every nucleotide residue has an additional hydroxyl group at 2' position in the ribose. Understanding the 5'→3' direction of a nucleic acid is critical for understanding the aspects of replication and transcription.

Based on the X-ray diffraction analysis of Maurice Wilkins and Rosalind Franklin, the double helix model for DNA was proposed by James Watson and Francis Crick in 1953. The highlight was the base pairing between the two strands of the polynucleotide chain. This proposition was based on the observations of Erwin Chargaff that Adenine pairs with Thymine (A=T) with two hydrogen bonds and Guanine pairs with Cytosine (G≡C) with three hydrogen bonds. The ratios between Adenine with Thymine and Guanine with Cytosine are constant and equal. The base pairing confers a unique property to the polynucleotide chain. They are said to be complementary to each other, that is, if the sequence of bases in one strand (template) is known, then the sequence in the other strand can be predicted.

5.5 RNA World

A typical cell contains about ten times as much RNA as DNA. The high RNA content is mainly due to the variety of roles played by RNA in the cell. Fraenkel-Conrat and Singer (1957) first demonstrated that RNA is the genetic material in RNA containing viruses like TMV (Tobacco Mosaic Virus) and they separated RNA from the protein of TMV viruses. Three molecular biologists in the early 1980's (Leslie Orgel, Francis Crick and Carl Woese) independently proposed the 'RNA world' as the first stage in the evolution of life, a stage when RNA catalysed all molecules necessary for survival and replication. The term 'RNA world' first used by Walter Gilbert in 1986, hypothesizes RNA as the first genetic material on earth.

There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing etc.) evolved around RNA. RNA has the ability to act as both genetic material and catalyst. There are several biochemical reactions in living systems that are catalysed by RNA. This catalytic RNA is known as ribozyme. But, RNA being a catalyst was reactive and hence unstable. This led to evolution of a more stable form of DNA, with certain chemical modifications. Since DNA is a double stranded molecule having complementary strand, it has resisted changes by evolving a process of repair. Some RNA molecules function as gene regulators by binding to DNA and affect gene expression. Some viruses use RNA as the genetic material. Andrew Fire and Craig Mellow (recipients of Nobel Prize in 2006) were of the opinion that RNA is an active ingredient in the chemistry of life.

5.6 Properties of Genetic Material

The experiment by Hershey and Chase clearly indicates that it is DNA that acts as a genetic material. However, in some viruses like Tobacco mosaic virus (TMV), bacteriophage θB, RNA acts as the genetic material. A molecule that can act as a genetic material should have the following properties:

• Self Replication: It should be able to replicate. According to the rule of base pairing and complementarity, both nucleic acids (DNA and RNA) have the ability to direct duplications. Proteins fail to fulfill this criterion.
• Stability: It should be stable structurally and chemically. 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 property of genetic material was clearly evident in Griffith's transforming principle. Heat which killed the bacteria did not destroy some of the properties of genetic material. In DNA the two strands being complementary, if separated (denatured) by heating can come together (renaturation) when appropriate condition is provided. Further 2'–OH group present at every nucleotide in RNA is a reactive group that makes RNA labile and easily degradable. RNA is also known to be catalytic and reactive. Hence, DNA is chemically more stable and chemically less reactive when compared to RNA. Presence of thymine instead of uracil in DNA confers additional stability to DNA.
• Information storage: It should be able to express itself in the form of 'Mendelian characters'. RNA can directly code for protein synthesis and can easily express the characters. DNA, however depends on RNA for synthesis of proteins. Both DNA and RNA can act as a genetic material, but DNA being more stable stores the genetic information and RNA transfers the genetic information.
• Variation through mutation: It should be able to mutate. Both DNA and RNA are able to mutate. RNA being unstable, mutates at a faster rate. Thus viruses having RNA genome with shorter life span can mutate and evolve faster.

The above discussion indicates that both RNA and DNA can function as a genetic material. DNA is more stable, and is preferred for storage of genetic information.

5.7 Packaging of DNA Helix

The distance between two consecutive base pairs is 0.34 nanometre (0.34×10⁻⁹ metre) of the DNA double helix in a typical mammalian cell. When the total number of base pairs is multiplied with the distance between two consecutive base pairs (6.6 × 10⁹ × 0.34 ×10⁻⁹ m/base pair), the length of DNA double helix is approximately 2.2 metre. If the length of Escherichia coli DNA is 1.36 mm, the number of base pairs in E. coli is 4 × 10⁶ bp. The length of the DNA double helix is far greater than the dimension of a typical mammalian nucleus (approximately 10⁻⁶ m). How is such a long DNA polymer packaged in a cell?

Chromosomes are carriers of genes which are responsible for various characters from generation to generation. Du Praw (1965) proposed a single stranded model (unineme), as a long coiled molecule which is associated with histone proteins in eukaryotes. Plants and animals have more DNA than bacteria and must fold this DNA to fit into the cell nucleus.

In prokaryotes such as E. coli though they do not have 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 called the nucleoid. The DNA as a nucleoid is organized into large loops held by protein. DNA of prokaryotes is almost circular and lacks chromatin organization, hence termed genophore.

In eukaryotes, this organization is much more complex. Chromatin is formed by a series of repeating units called nucleosomes. Kornberg proposed a model for the nucleosome, in which 2 molecules of the four histone proteins H2A, H2B, H3 and H4 are organized to form a unit of eight molecules called histone octamere. The negatively charged DNA is wrapped around the positively charged histone octamere to form a structure called nucleosome. A typical nucleosome contains 200 bp of DNA helix. The histone octameres are in close contact and DNA is coiled on the outside of nucleosome. Neighbouring nucleosomes are connected by linker DNA (H1) that is exposed to enzymes. The DNA makes two complete turns around the histone octameres and the two turns are sealed off by an H1 molecule.

Chromatin lacking H1 has a beads-on-a-string appearance in which DNA enters and leaves the nucleosomes at random places. H1 of one nucleosome can interact with H1 of the neighbouring nucleosomes resulting in the further folding of the fibre. The chromatin fiber in interphase nuclei and mitotic chromosomes have a diameter that vary between 200–300 nm and represents inactive chromatin. 30 nm fibre arises from the folding of nucleosome chains into a solenoid structure having six nucleosomes per turn. This structure is stabilized by interaction between different H1 molecules. DNA is a solenoid and packed about 40 folds (Figure 5.3).

Additional set of proteins are required for packing of chromatin at higher level and are referred to as non-histone chromosomal proteins (NHC). In a typical nucleus, some regions of chromatin are loosely packed (lightly stained) and are referred to as euchromatin. The chromatin that is tightly packed (stained darkly) is called heterochromatin. Euchromatin is transcriptionally active and heterochromatin is transcriptionally inactive.

5.8 DNA Replication

Replication of DNA takes place during the S phase of cell cycle. During replication, each DNA molecule gives rise to two DNA strands, identical to each other as well as to the parent strand. Three hypotheses of DNA replication have been proposed. They are conservative replication, dispersive replication, and semiconservative replication.

• In conservative replication, the original double helix serves as a template. The original molecule is preserved intact and an entirely new double stranded molecule is synthesized.
• In dispersive replication, the original molecule is broken into fragments and each fragment serves as a template for the synthesis of complementary fragments. Finally two new molecules are formed which consist of both old and new fragments.
• Semi-conservative replication was proposed by Watson and Crick in 1953. This mechanism of replication is based on the DNA model. They suggested that the two polynucleotide strands of DNA molecule unwind and start separating at one end. During this process, covalent hydrogen bonds are broken. The separated single strand then acts as template for the synthesis of a new strand. Subsequently, each daughter double helix carries one polynucleotide strand from the parent molecule that acts as a template and the other strand is newly synthesized and complementary to the parent strand (Figure 5.4).

5.8.1 Experimental Proof of DNA Replication

The mode of DNA replication was determined in 1958 by Meselson and Stahl. They designed an experiment to distinguish between semi conservative, conservative and dispersive replications. In their experiment, they grew two cultures of Escherichia coli for many generations in separate media. The 'heavy' culture was grown in a medium in which the nitrogen source (NH₄Cl) contained the heavy isotope ¹⁵N and the 'light' culture was grown in a medium in which the nitrogen source contained light isotope ¹⁴N for many generations.

At the end of growth, they observed that the bacterial DNA in the heavy culture contained only ¹⁵N and in the light culture only ¹⁴N. The heavy DNA could be distinguished from light DNA (¹⁵N from ¹⁴N) with a technique called Cesium Chloride (CsCl) density gradient centrifugation. In this process, heavy and light DNA extracted from cells in the two cultures settled into two distinct and separate bands (hybrid DNA) (Figure 5.5).

The heavy culture (¹⁵N) was then transferred into a medium that had only NH₄Cl, and took samples at various definite time intervals (20 minutes duration). After the first replication, they extracted DNA and subjected it to density gradient centrifugation. The DNA settled into a band that was intermediate in position between the previously determined heavy and light bands. After the second replication (40 minutes duration), they again extracted DNA samples, and this time found the DNA settling into two bands, one at the light band position and one at intermediate position. These results confirm Watson and Crick's semi conservative replication hypothesis.

5.8.2 Enzymes and Mechanism of Replication

In prokaryotes, replication process requires three types of DNA polymerases (DNA polymerase 1, 2, and 3). DNA polymerase III is the main enzyme involved in DNA replication. DNA polymerase I (also known as Kornberg enzyme) and DNA polymerase II are involved in DNA repair mechanism. Eukaryotes have five types of DNA polymerases that catalyses the polymerization of nucleotides at the 3' hydroxyl of the new strand within a short period of time. Escherichia coli that has 4.6 × 10⁶ base pairs completes its replication process within 18 minutes; that means the average rate of polymerization has to be approximately 2000 base pairs per second.

Replication takes place faster at the same time accurately. Any error will lead to mutation. However replication errors are corrected by repair enzymes such as nucleases. Deoxynucleotide triphosphate acts as substrate and also provides energy for polymerization reaction.

Replication begins at the initiation site called the site of 'origin of replication' (ori). In prokaryotes, there is only one origin of replication, whereas in eukaryotes with giant DNA molecules, there can be several origins of replication (replicons). Since the two strands of DNA cannot be separated throughout at a time (due to large requirement of energy) the replication occurs within a small opening of the DNA helix called as replication fork. Unwinding of the DNA strand is carried out by DNA helicase.

Thus, in one strand (template strand with polarity 3' to 5') the replication is continuous and is known as the leading strand while in the other strand (coding strand with polarity 5' to 3') replication is discontinuous, known as the lagging strand (Figure 5.6). The discontinuously synthesized fragments of the lagging strand (called the Okazaki fragments) are joined by the enzyme DNA ligase.

As they move away in both directions, newly synthesized complementary nucleotides are paired with the existing nucleotides on the parent strand and covalently bonded together by DNA polymerase. Formation of new strand requires a primer (a short stretch of RNA) for initiation. The primer produces a 3' hydroxyl end on the sequence of ribonucleotides, to which deoxyribonucleotides are added. The RNA primer is ultimately removed leaving a gap in the newly synthesized DNA strand. It is removed from 5' end one by one by the exonuclease activity of DNA polymerase. Finally, when all the nucleotides are in position, gaps are sealed by the enzyme DNA ligase.

At the point of origin of replication, the helicases and topoisomerases (DNA gyrase) unwind and pull apart the strands, forming a Y-shaped structure called the replication fork. There are two replication forks at each origin. The two strands of a DNA helix have an antiparallel orientation. The enzyme DNA polymerase can only catalyse the addition of a nucleotide to the new strands in the 5' to 3' direction, as it can only add nucleotides to the 3' carbon position.

5.9 Transcription

Francis Crick proposed the Central dogma of protein synthesis in molecular biology which states that genetic information flows as follows:

Central Dogma

DNA → RNA → Protein (Replication → Transcription → Translation)

The process of copying genetic information from one strand of DNA into RNA is termed transcription. This process takes place in presence of DNA dependent RNA polymerase. In some retroviruses that contain RNA as the genetic material (example, HIV), the flow of information is reversed. RNA synthesizes DNA by reverse transcription, then transcribed into mRNA by transcription and then into proteins by translation.

For a cell to operate, its genes must be expressed. This means that the gene products, whether proteins or RNA molecules must be made. The RNA that carries genetic information encoding a protein from genes into the cell is known as messenger RNA (mRNA). For a gene to be transcribed, the DNA which is a double helix must be pulled apart temporarily, and RNA is synthesized by RNA polymerase. This enzyme binds to DNA at the start of a gene and opens the double helix. Finally, RNA molecule is synthesized. The nucleotide sequence in the RNA is complementary to the DNA template strand from which it is synthesized.

Both the strands of DNA are not copied during transcription for two reasons:

1. If both the strands act as a template, they would code for RNA with different sequences. This in turn would code for proteins with different amino acid sequences. This would result in one segment of DNA coding for two different proteins, hence complicate the genetic information transfer machinery.
2. If two RNA molecules were produced simultaneously, double stranded RNA complementary to each other would be formed. This would prevent RNA from being translated into proteins.

5.9.1 Transcription Unit and Gene

A transcriptional unit in DNA is defined by three regions: a promoter, the structural gene and a terminator. The promoter is located towards the 5' end of the coding strand. It is a DNA sequence that provides binding site for RNA polymerase. The presence of promoter in a transcription unit, defines the template and coding strands. The terminator region located towards the 3' end of the coding strand contains a DNA sequence that causes the RNA polymerase to stop transcribing.

In eukaryotes the promoter has AT rich regions called TATA box (Goldberg-Hogness box) and in prokaryotes this region is called Pribnow box. Besides promoter, eukaryotes also require an enhancer.

The two strands of the DNA in the structural gene of a transcription unit have opposite polarity. DNA dependent RNA polymerase catalyses the polymerization in only one direction, the strand that has the polarity 3' to 5' acts as a template, and is called the template strand. The other strand which has the polarity 5' to 3' has a sequence same as RNA (except thymine instead of uracil) and is displaced during transcription. This strand is called coding strand (Figure 5.7).

The structural gene may be monocistronic (eukaryotes) or polycistronic (prokaryotes). In eukaryotes, each gene transcribes a single mRNA and encodes information for only a single protein and is called monocistronic mRNA. In prokaryotes, clusters of related genes, known as operon, often found next to each other on the chromosome are transcribed together to give a single mRNA and hence are polycistronic.

Before starting transcription, RNA polymerase binds to the promoter, a recognition sequence in front of the gene. Bacterial (prokaryotic) RNA polymerase consists of the core enzyme and the sigma subunit. The core enzyme (2α, β, β' and ω) is responsible for RNA synthesis whereas a sigma subunit is responsible for recognition of the promoter. Promoter sequences vary in different organisms. RNA polymerase opens up the DNA to form the transcription bubble. The core enzyme moves ahead, manufacturing RNA leaving the sigma subunit behind at the promoter region. The end of a gene is marked by a terminator sequence that forms a hair pin structure in the RNA. The sub-class of terminators require a recognition protein, known as rho (ρ), to function.

5.9.2 Process of Transcription

In prokaryotes, there are three major types of RNAs: mRNA, tRNA, and rRNA. All three RNAs are needed to synthesize 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 a single DNA dependent RNA polymerase that catalyses transcription of all types of RNA. It binds to the promoter and initiates transcription (Initiation). The polymerases binding sites are called promoters. It uses nucleoside triphosphate as substrate and polymerizes in a template dependent fashion following the rule of complementarity.

After the initiation of transcription, the polymerase continues to elongate the RNA, adding one nucleotide after another to the growing RNA chain. Only a short stretch of RNA remains bound to the enzyme, when the polymerase reaches a terminator at the end of a gene, the nascent RNA falls off, so also the RNA polymerase. The RNA polymerase is only capable of catalyzing the process of elongation. The RNA polymerase associates transiently with initiation factor sigma (σ) and termination factor rho (ρ) to initiate and terminate the transcription, respectively (Figure 5.8).

In bacteria, since the mRNA does not require any processing to become active and also since transcription and translation take place simultaneously in the same compartment (since there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. This is because the genetic material is not separated from other cell organelles by a nuclear membrane; consequently, transcription and translation can be coupled in bacteria.

In Eukaryotes, there are at least three RNA polymerases in the nucleus (in addition to RNA polymerase found in the organelles). There is a clear 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, 5S rRNA and snRNA. The RNA polymerase II transcribes precursor of mRNA, the hnRNA (heterogenous nuclear RNA).

In eukaryotes, the monocistronic structural genes have interrupted coding sequences known as exons (expressed sequences) and non-coding sequences called introns (intervening sequences). The introns are removed by a process called splicing. hnRNA undergoes additional processing called capping and tailing. In capping an unusual nucleotide, methyl guanosine triphosphate is added at the 5' end, whereas adenylate residues (200–300) (Poly A) are added at the 3' end in tailing (Figure 5.9). Thereafter, this processed hnRNA, now called mRNA is transported out of the nucleus for translation.

The split gene feature of eukaryotic genes is almost entirely absent in prokaryotes. Originally each exon may have coded for a single polypeptide chain with a specific function. Since exon arrangement and intron removal are flexible, the exon coding for these polypeptide subunits act as domains combining in various ways to form new genes. Single genes can produce different functional proteins by arranging their exons in several different ways through alternate splicing patterns, a mechanism known to play an important role in generating both protein and functional diversity in animals.

Introns would have arisen before or after the evolution of eukaryotic gene. If introns arose late how did they enter eukaryotic gene? Introns are mobile DNA sequences that can splice themselves out of, as well as into, specific 'target sites' acting like mobile transposon-like elements (that mediate transfer of genes between organisms — Horizontal Gene Transfer HGT). HGT occurs between lineages of prokaryotic cells, or from prokaryotic to eukaryotic cells and between eukaryotic cells. HGT is now hypothesized to have played a major role in the evolution of life on earth.

5.10 Genetic Code

DNA is the genetic material that carries genetic information in a cell and from generation to generation. The translation of proteins follows the triplet rule; a sequence of three mRNA bases (a codon) designates one of the 20 different kinds of amino acids used in protein synthesis. Genetic code is the sequence relationship between nucleotide in genes (or mRNA) and the amino acids in the proteins they encode. There are 64 possible triplets, and 61 of them are used to represent amino acids. The remaining three triplet codons are termination signals for polypeptide chains. Since there are only 20 amino acids involved in protein synthesis, most of them are encoded by more than one triplet.

Two things make this multiple (degenerate) coding possible. First, there is more than one tRNA for most amino acids. Each tRNA has a different anticodon. Second, this pairing is highly specific for the first two portions of the codon, permitting Watson and Crick base pairs (A–U and G–C) to be formed. But at the third position there is a great deal of flexibility as to which base pairs are acceptable. Most part of the genetic code is universal, being the same in prokaryotes and eukaryotes. The order of base pairs along DNA molecule controls the kind and order of amino acids found in the proteins of an organism.

Marshall Nirenberg, Severo Ochoa (enzyme polynucleotide phosphorylase called Ochoa's enzyme), Hargobind Khorana, Francis Crick and many others have contributed significantly to decipher the genetic code. The order in which bases are arranged in mRNA decides the order in which amino acids are arranged in proteins. Finally a checker board for genetic code was prepared (Table 5.1).

Salient Features of Genetic Code

• The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).
• The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms.
• A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUUGUC represents only two codons.
• It is comma less, which means that the message would be read directly from one end to the other — no punctuation is needed between two codes.
• A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.
• Non-ambiguous code means that one codon will code for one amino acid.
• The code is always read in a fixed direction, that is from 5' to 3' direction called polarity.
• AUG has dual functions. It acts as an initiator codon and also codes for the amino acid methionine.
• UAA, UAG and UGA codons are designated as termination (stop) codons and also are known as "non sense" codons.

5.10.1 Mutation and Genetic Code

Comparative studies of mutations (sudden change in a gene) and corresponding alteration in amino acid sequence of specific protein have confirmed the validity of the genetic code. The relationship between genes and DNA are best understood by mutation studies. The simplest type of mutation at the molecular level is a change in nucleotide that substitutes one base for another. Such changes are known as base substitutions which may occur spontaneously or due to the action of mutagens.

A well studied example is sickle cell anaemia in humans which results from a point mutation of an allele of β haemoglobin gene (β Hb). A haemoglobin molecule consists of four polypeptide chains of two types, two α chains and two β chains. Each chain has a heme group on its surface. The heme groups are involved in the binding of oxygen. The human blood disease, sickle cell anaemia is due to abnormal haemoglobin. This abnormality in haemoglobin is due to a single base substitution at the sixth codon of the beta globin gene from GAG to GTG in β chain of haemoglobin. It results in a change of amino acid glutamic acid to valine at the 6th position of the β chain. This is the classical example of point mutation that results in the change of amino acid residue glutamic acid to valine (Figure 5.10). The mutant haemoglobin undergoes polymerisation under oxygen tension causing the change in the shape of the RBC from biconcave to a sickle shaped structure.

Wobble Hypothesis

Wobble Hypothesis is a hypothesis proposed by Crick (1966) which states that tRNA anticodon has the ability to wobble at its 5' end by pairing with even non-complementary base of mRNA codon. According to this hypothesis, in codon–anticodon pairing the third base may not be complementary. The third base of the codon is called wobble base and this position is called wobble position. The actual base pairing occurs at first two positions only. The importance of Wobbling hypothesis is that it reduces the number of tRNAs required for polypeptide synthesis and it overcomes the effect of code degeneracy. In the above example though the codon and the anticodon do not match perfectly, yet the required amino acid is brought perfectly. This enables the economy of tRNA. GUU, GUC, GUA and GUG code for the amino acid Valine.

Frame Shift Mutations

The effect of point mutation can be understood by the following example:

A B C D E F G H I J K L

If we insert a letter O between DEF and GHI the arrangement would be:

A B C D E F O G H I J K L

If we insert OQ at the same place the arrangement would be:

A B C D E F O Q G H I J K L

The above information shows that insertion or deletion of one or two bases, changes the reading frame from the point of insertions or deletions. Such mutations are referred to as frame shift insertion or deletion mutations. This forms the genetic basis of proof that codon is a triplet and is read in a continuous manner.

5.11 tRNA — The Adapter Molecule

The transfer RNA (tRNA) molecule of a cell acts as a vehicle that picks up the amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick. The two dimensional clover leaf model of tRNA was proposed by Robert Holley. The secondary structure of tRNA (Figure 5.11) looks like a clover leaf. In actual structure, the tRNA is a compact molecule which looks like an inverted L.

The clover leaf model of tRNA shows the presence of three arms namely DHU arm, middle arm and TΨC arm. These arms have loops such as amino acyl binding loop, anticodon loop and ribosomal binding loop at their ends. In addition it also shows a small lump called variable loop or extra arm. The amino acid is attached to one end (amino acid acceptor end) and the other end consists of three anticodon nucleotides. The anticodon pairs with a codon in mRNA ensuring that the correct amino acid is incorporated into the growing polypeptide chain. Four different regions of double-stranded RNA are formed during the folding process. Modified bases are especially common in tRNA. Wobbling between anticodon and codon allows some tRNA molecules to read more than one codon.

The process of addition of amino acid to tRNA is known as aminoacylation or charging and the resultant product is called aminoacyl tRNA (charged tRNA). Without aminoacylation tRNA is known as uncharged tRNA (Figure 5.12). If two such tRNAs are brought together peptide bond formation is favoured energetically. Numbers of amino acids are joined by peptide bonds to form a polypeptide chain. This aminoacylation is catalyzed by an enzyme aminoacyl tRNA synthetase. This is an endothermic reaction and is associated with ATP hydrolysis. 20 different aminoacyl tRNA synthetases are known. The power to recognize codon on the mRNA lies in the tRNA and not in the attached amino acid molecule.

5.12 Translation

Translation refers to the process of polymerization of amino acids to form polypeptide chain. The decoding process is carried out by ribosomes that bind mRNA and charged tRNA molecules. The mRNA is translated, starting at the 5' end. After binding to mRNA, the ribosomes move along it, adding new amino acids to the growing polypeptide chain each time it reads a codon. Each codon is read by an anticodon on the corresponding tRNA. Hence the order and sequence of amino acids are defined by the sequence of bases in the mRNA.

5.12.1 Mechanism of Translation

The cellular factory responsible for synthesizing protein is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In inactive state, it exists as two subunits; large subunit and small subunit. When the subunit encounters an mRNA, the process of translation of the mRNA to protein begins. The prokaryotic ribosome (70S) consists of two subunits, the larger subunit (50S) and smaller subunit (30S). The ribosomes of eukaryotes (80S) are larger, consisting of 60S and 40S subunits. 'S' denotes the sedimentation coefficient which is expressed as Svedberg unit (S).

One of the alternative ways of dividing up a sequence of bases in DNA or RNA into codons is called reading frame. Any sequence of DNA or RNA, beginning with a start codon and which can be translated into a protein is known as an Open Reading Frame (ORF). A translational unit in mRNA is the sequence of RNA that is flanked by the start codon and the stop codon and codes for polypeptides. mRNA also have some additional sequences that are not translated and are referred to as Untranslated Regions (UTR). UTRs are present at both 5' end (before start codon) and at 3' end (after stop codon).

The start codon (AUG) begins the coding sequence and is read by a special tRNA that carries methionine (met). The initiator tRNA charged with methionine binds to the AUG start codon. In prokaryotes, N-formyl methionine (fmet) is attached to the initiator tRNA whereas in eukaryotes unmodified methionine is used. The 5' end of the mRNA of prokaryotes has a special sequence which precedes the initial AUG start codon of mRNA. This ribosome binding site is called the Shine Dalgarno sequence or SD sequence. This sequence base pairs with a region of the 16S rRNA of the small ribosomal subunit facilitating initiation.

The subunits of the ribosomes (30S and 50S) are usually dissociated from each other when not involved in translation (Figure 5.13a). Initiation of translation in Escherichia coli begins with the formation of an initiation complex, consisting of the 30S subunits of the ribosome, a messenger RNA and the charged N-formyl methionine tRNA (fmet-tRNA^fmet), three proteinaceous initiation factors (IF1, IF2 and IF3), GTP (Guanine Tri Phosphate) and Mg²⁺.

The components that form the initiation complex interact in a series of steps. IF3 binds to the 30S and allows the 30S subunit to bind to mRNA. Another initiation protein (IF2) then enhances the binding of charged formyl methionine tRNA to the small subunit in response to the AUG triplet. This step 'sets' the reading frame so that all subsequent groups of three ribonucleotides are translated accurately. The assembly of ribosomal subunits, mRNA and tRNA represent the initiation complex. Once initiation complex has been assembled, IF3 is released and allows the initiation complex to combine with the 50S ribosomal subunit to form the complete ribosome (70S). In this process a molecule of GTP is hydrolyzed providing the required energy and the initiation factors (IF1, IF2 and GDP) are released (Figure 5.13b).

Elongation is the next phase of translation. Once both subunits of the ribosomes are assembled with the mRNA, binding sites for two charged tRNA molecules are formed. The sites in the ribosome are referred to as the aminoacyl site (A site), the peptidyl site (P site) and the exit site (E site). The charged initiator tRNA binds to the P site. The next step in prokaryotic translation is to position the second tRNA at the 'A' site of the ribosome to form hydrogen bonds between its anticodon and the second codon on the mRNA (step 1). This step requires the correct transfer RNA, another GTP and two proteins called elongation factors (EF-Ts and EF-Tu).

Once the charged tRNA molecule is positioned at the A site, the enzyme peptidyl transferase catalyses the formation of peptide bonds that link the two amino acids together (step 2). At the same time, the covalent bond between the amino acid and tRNA occupying the P site is hydrolyzed (broken). The product of this reaction is a dipeptide which is attached to the 3' end of tRNA still residing in the A site. For elongation to be repeated, the tRNA attached to the P site, which is now uncharged is released from the large subunit. The uncharged tRNA moves through the 'E' site on the ribosome. The entire mRNA–tRNA–aa1–aa2 complex shifts in the direction of the 'P' site by a distance of three nucleotides (step 3). This step requires several elongation factors (EFs) and the energy derived from hydrolysis of GTP. This results in the third triplet of mRNA to accept another charged tRNA into the A site (step 4). The sequence of elongation is repeated over and over (step 5 and step 6). An additional amino acid is added to the growing polypeptide, each time mRNA advances through the ribosome. Once a polypeptide chain is assembled, it emerges out from the base of the large subunit (Figure 5.13c).

Termination is the final phase of translation. Termination of protein synthesis occurs when one of the three stop codons appears in the 'A' site of the ribosome. The terminal codon signals the action of GTP dependent release factor, which cleaves the polypeptide chain from the terminal tRNA releasing it from the translational complex (step 1). The tRNA is then released from the ribosome, which then dissociates into its subunits (step 2) (Figure 5.13d).

5.13 Regulation of Gene Expression

We have previously established how DNA is organized into genes, how genes store genetic information, and how this information is expressed. We now consider the most fundamental issues in molecular genetics. How is genetic expression regulated? Evidence in support of the idea that genes can be turned on and off is very convincing. Regulation of gene expression has been extensively studied in prokaryotes, especially in Escherichia coli. Gene expression can be controlled or regulated at transcriptional or post transcriptional or translational level. Usually, small extracellular or intracellular metabolites trigger initiation or inhibition of gene expression. The clusters of gene with related functions are called operons. They usually transcribe single mRNA molecules. In Escherichia coli, nearly 260 genes are grouped into 75 different operons.

Antibiotics and Protein Synthesis

Many antibiotics do not allow pathogenic bacteria to flourish in animal host because they inhibit one or the other stage of bacterial protein synthesis.

• Tetracycline inhibits binding between aminoacyl tRNA and mRNA.
• Neomycin inhibits the interaction between tRNA and mRNA.
• Erythromycin inhibits the translocation of mRNA along the ribosome.
• Streptomycin inhibits the initiation of translation and causes misreading.
• Chloramphenicol inhibits peptidyl transferase and formation of peptide bonds.

Structure of the Operon

Each operon is a unit of gene expression and regulation and consists of one or more structural genes and an adjacent operator gene that controls transcriptional activity of the structural gene.

1. The structural gene codes for proteins, rRNA and tRNA required by the cell.
2. Promoters are the signal sequences in DNA that initiate RNA synthesis. RNA polymerase binds to the promoter prior to the initiation of transcription.
3. The operators are present between the promoters and structural genes. The repressor protein binds to the operator region of the operon.

The Lac (Lactose) Operon

The metabolism of lactose in E. coli requires three enzymes — permease, β-galactosidase (β-gal) and transacetylase. The enzyme permease is needed for entry of lactose into the cell, β-galactosidase brings about hydrolysis of lactose to glucose and galactose, while transacetylase transfers acetyl group from acetyl CoA to β-galactosidase.

The lac operon consists of one regulator gene ('i' gene refers to inhibitor) promoter sites (p), and operator site (o). Besides these, it has three structural genes namely lac z, y and lac a. The lac 'z' gene codes for β-galactosidase, lac 'y' gene codes for permease and 'a' gene codes for transacetylase.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in E. coli. In lac operon, a polycistronic structural gene is regulated by a common promoter and regulatory gene.

When the cell is using its normal energy source as glucose, the 'i' gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. It binds to the operator region of the operon and prevents transcription; as a result, β-galactosidase is not produced.

In the absence of glucose, if lactose is available as an energy source for the bacteria then lactose enters the cell as a result of permease enzyme. Lactose acts as an inducer and interacts with the repressor to inactivate it. The repressor protein binds to the operator of the operon and prevents RNA polymerase from transcribing the operon. In the presence of inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to bind to the promotor site and transcribe the operon to produce lac mRNA which enables formation of all the required enzymes needed for lactose metabolism (Figure 5.14). This regulation of lac operon by the repressor is an example of negative control of transcription initiation. Lac operon is also under the control of positive regulation as well.

5.14 Human Genome Project (HGP)

The international human genome project was launched in the year 1990. It was a mega project and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and is the first vertebrate genome to be completed. Human genome is said to have approximately 3 × 10⁹ base pairs. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

5.14.1 Goals and Methodologies of Human Genome Project

Main Goals of Human Genome Project

• Identify all the genes (approximately 30000) in human DNA.
• Determine the sequence of the three billion chemical base pairs that make up the human DNA.
• To store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, such as industries.
• Address the ethical, legal and social issues (ELSI) that may arise from the project.

The methodologies of the Human Genome Project involved two major approaches. One approach was focused on identifying all the genes that are expressed as RNA (ESTs — Expressed Sequence Tags). The other approach was sequence annotation. Here, sequencing the whole set of genomes was taken, that contains all the coding and non-coding sequences and later assigning different regions in the sequences with functions.

For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes and cloned in suitable hosts using specialized vectors. This cloning results in amplification of pieces of DNA fragments so that it could subsequently be sequenced with ease. Bacteria and yeast are two commonly used hosts and these vectors are called as BAC (Bacterial Artificial Chromosomes) and YAC (Yeast Artificial Chromosomes). The fragments are sequenced using automated DNA sequencers (developed by Frederick Sanger). The sequences are then arranged based on few overlapping regions, using specialized computer-based programs. These sequences were subsequently annotated and are assigned to each chromosome.

The genetic and physical maps on the genome are assigned using information on polymorphism of restriction endonuclease recognition sites and some repetitive DNA sequences, called microsatellites. The latest method of sequencing even longer fragments is by a method called Shotgun sequencing using super computers, which has replaced the traditional sequencing methods.

5.14.2 Salient Features of Human Genome Project

• The human genome contains 3 billion nucleotide bases.
• An average gene consists of 3000 bases, the largest known human gene being dystrophin with 2.4 million bases.
• The chromosomal organization of human genes shows diversity.
• Approximately 30,000 genes are present in human genome and almost 99.9% nucleotide bases are exactly the same in all people.
• Functions for over 50 percent of the discovered genes are unknown.
• Less than 2 percent of the genome codes for proteins.
• Repeated sequences make up very large portion of the human genome. Repetitive sequences have no direct coding functions but they shed light on chromosome structure, dynamics and evolution (genetic diversity).
• Chromosome 1 has 2968 genes whereas chromosome 'Y' has 231 genes.
• Scientists have identified about 1.4 million locations where single base DNA differences (SNPs — Single Nucleotide Polymorphism, pronounced as 'snips') occur in humans. Identification of 'SNiPs' is helpful in finding chromosomal locations for disease associated sequences and tracing human history.

5.14.3 Applications and Future Challenges

The mapping of human chromosomes is possible to examine a person's DNA and to identify genetic abnormalities. This is extremely useful in diagnosing diseases and to provide genetic counselling to those planning to have children. This kind of information would also create possibilities for new gene therapies. Besides providing clues to understand human biology, learning about non-human organisms, DNA sequences can lead to an understanding of their natural capabilities that can be applied towards solving challenges in healthcare, agriculture, energy production and environmental remediation. A new era of molecular medicine, characterized by looking into the most fundamental causes of disease than treating the symptoms will be an important advantage.

• Once genetic sequence becomes easier to determine, some people may attempt to use this information for profit or for political power.
• Insurance companies may refuse to insure people at 'genetic risk' and this would save the companies the expense of future medical bills incurred by 'less than perfect' people.
• Another fear is that attempts are being made to "breed out" certain genes of people from the human population in order to create a 'perfect race'.

Do You Know?

Pharmacogenomics is the study of how genes affect a person's response to drugs. This relatively new field combines pharmacology (the science of drugs) and genomics (the study of genes and their functions) to develop effective, safe medications and doses that will be tailored to a person's genetic makeup.

5.15 DNA Fingerprinting Technique

The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985 (Recipient of the Royal Society's Copley Medal in 2014). Each of us have the same chemical structure of DNA. But there are millions of differences in the DNA sequence of base pairs. This makes the uniqueness among us so that each of us except identical twins is different from each other genetically. The DNA of a person and finger prints are unique. There are 23 pairs of human chromosomes with 1.5 million pairs of genes.

It is a well-known fact that genes are segments of DNA which differ in the sequence of their nucleotides. Not all segments of DNA code for proteins, some DNA segments have a regulatory function, while others are intervening sequences (introns) and still others are repeated DNA sequences. In DNA fingerprinting, short repetitive nucleotide sequences are specific for a person. These nucleotide sequences are called as variable number tandem repeats (VNTR). The VNTRs of two persons generally show variations and are useful as genetic markers.

DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called 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 sub-categories such as micro-satellites, minisatellites, etc. These sequences do not code for any proteins, but they form a large portion of human genome. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting (Figure 5.15).

DNA isolated from blood, hair, skin cells, or other genetic evidences left at the scene of a crime can be compared through VNTR patterns, with the DNA of a criminal suspect to determine guilt or innocence. VNTR patterns are also useful in establishing the identity of a homicide victim, either from DNA found as evidence or from the body itself.

Steps in DNA Fingerprinting Technique (Figure 5.16)

1. Extraction of DNA: The process of DNA fingerprinting starts with obtaining a sample of DNA from blood, semen, vaginal fluids, hair roots, teeth, bones, etc.
2. Polymerase Chain Reaction (PCR): In many situations, there is only a small amount of DNA available for DNA fingerprinting. If needed many copies of the DNA can be produced by PCR (DNA amplification).
3. Fragmenting DNA: DNA is treated with restriction enzymes which cut the DNA into smaller fragments at specific sites.
4. Separation of DNA by electrophoresis: During electrophoresis in an agarose gel, the DNA fragments are separated into bands of different sizes. The bands of separated DNA are sieved out of the gel using a nylon membrane (treated with chemicals that allow for it to break the hydrogen bonds of DNA so there are single strands).
5. Denaturing DNA: The DNA on gels is denatured by using alkaline chemicals or by heating.
6. Blotting: The DNA band pattern in the gel is transferred to a thin nylon membrane placed over the 'size fractionated DNA strand' by Southern blotting.
7. Using probes to identify specific DNA: A radioactive probe (DNA labeled with a radioactive substance) is added to the DNA bands. The probe attaches by base pairing to those restriction fragments that are complementary to its sequence. The probes can also be prepared by using either 'fluorescent substance' or 'radioactive isotopes'.
8. Hybridization with probe: After the probe hybridizes and the excess probe washed off, a photographic film is placed on the membrane containing 'DNA hybrids'.
9. Exposure on film to make a genetic or DNA Fingerprint: The radioactive label exposes the film to form an image (image of bands) corresponding to specific DNA bands. The thick and thin dark bands form a pattern of bars which constitutes a genetic fingerprint.

Applications of DNA Fingerprinting

• Forensic analysis: It can be used in the identification of a person involved in criminal activities, for settling paternity or maternity disputes, and in determining relationships for immigration purposes.
• Pedigree analysis: Inheritance pattern of genes through generations and for detecting inherited diseases.
• Conservation of wild life: Protection of endangered species. By maintaining DNA records for identification of tissues of the dead endangered organisms.
• Anthropological studies: It is useful in determining the origin and migration of human populations and genetic diversities.

Summary

In the twentieth century, one of the landmark discoveries in biology was the identification of DNA, as genetic material of living organisms. Gene may be defined as a segment of DNA which is responsible for inheritance and expression of a particular character. In 1953, James Watson and Francis Crick proposed DNA structure based on X-ray crystallographic studies provided by Maurice Wilkins and Rosalind Franklin. Nucleotides are the structural units of nucleic acids. Each nucleotide has three components: 1) pentose sugar 2) nitrogenous base and 3) phosphate. DNA and RNA are polynucleotides. DNA has double stranded helical structure while RNA is a single stranded structure. DNA acts as genetic material of almost all the living organism except few viruses. The nongenetic RNAs are of three types; mRNA, rRNA and tRNA. They help in protein synthesis. DNA has capacity of replication, while the three types of RNA are transcribed on DNA.

Meselson and Stahl (1958) proved experimentally the semi-conservative nature of DNA replication using heavy isotope of nitrogen N¹⁵ in E. coli. In 1958 Crick proposed that DNA determines the sequence of amino acids in a polypeptide (protein) through mRNA, and proposed the central dogma of protein synthesis which involves transcription and translation. The process of copying genetic information from one strand of DNA into RNA is termed transcription. The DNA transcribed RNA molecules serve as a template for the synthesis of polypeptides by a process termed translation. Each amino acid in a polypeptide chain is represented by a sequence of three nucleotides in the RNA known as the genetic code. RNA transfers genetic message from nucleus to the cytoplasm. DNA is always present in the nucleus and synthesis is also confined to the nucleus.

Jacob and Monod proposed the classical model of Lac operon to explain gene expression and regulation in Escherichia coli. In lac operon a polycistronic structural gene is regulated by a common promoter and regulator. It is an example of negative control of transcription initiation. Human genome project, a mega project was aimed to sequence every gene in the human genome. Polymerase chain reaction is an in vitro method of synthesis of nucleic acids wherein, a specific DNA segment is amplified rapidly without concomitant replication of the rest of the DNA molecule. DNA fingerprinting is a technique to identify variations in individuals of a population at the DNA level. It has immense applications in the field of forensic analysis, pedigree analysis, anthropological studies, and conservation of wild life.

Evaluation

Exercise 1. MCQ

1. Hershey and Chase experiment with bacteriophage showed that

a) Protein gets into the bacterial cells b) DNA is the genetic material c) DNA contains radioactive Sulphur d) Viruses undergo transformation

2. DNA and RNA are similar with respect to

a) Thymine as a nitrogen base b) A single-stranded helix shape c) Nucleotide containing sugars, nitrogen bases and phosphates d) The same sequence of nucleotides for the amino acid phenylalanine

3. A mRNA molecule is produced by

a) Replication b) Transcription c) Duplication d) Translation

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

a) 3.5 million b) 35000 c) 35 million d) 3.1 billion

5. Escherichia coli cell grown on ¹⁵N medium are transferred to ¹⁴N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?

a) One high and one low density band. b) One intermediate density band. c) One high and one intermediate density band. d) One low and one intermediate density band.

6. What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?

a) Origin of replication occurs only at the 5' end of the molecules. b) DNA ligase works only in the 3' to 5' direction. c) DNA polymerase can join new nucleotides only to the 3' end of the growing strand. d) Helicases and single strand binding proteins that work at the 5' end.

7. Which of the following is the correct sequence of event with reference to the central dogma?

a) Transcription, Translation, Replication b) Transcription, Replication, Translation c) Duplication, Translation, Transcription d) Replication, Transcription, Translation

8. Which of the following statements about DNA replication is not correct?

a) Unwinding of DNA molecule occurs as hydrogen bonds break. b) Replication occurs as each base is paired with another exactly like it. c) Process is known as semi conservative replication because one old strand is conserved in the new molecule. d) Complementary base pairs are held together with hydrogen bonds.

9. Which of the following statements is not true about DNA replication in eukaryotes?

a) Replication begins at a single origin of replication. b) Replication is bidirectional from the origins. c) Replication occurs at about 1 million base pairs per minute. d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.

10. The first codon to be deciphered was _____ which codes for _____

a) AAA, proline b) GGG, alanine c) UUU, Phenylalanine d) TTT, arginine

11. Meselson and Stahl's experiment proved

a) Transduction b) Transformation c) DNA is the genetic material d) Semi-conservative nature of DNA replication

12. Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for _____ and the larger subunit has two binding sites for two _____.

 

13. An operon is a:

a) Protein that suppresses gene expression b) Protein that accelerates gene expression c) Cluster of structural genes with related function d) Gene that switched other genes on or off

14. When lactose is present in the culture medium:

a) Transcription of lac y, lac z, lac a genes occurs. b) Repressor is unable to bind to the operator. c) Repressor is able to bind to the operator. d) Both (a) and (b) are correct. Exercise 15. Short Answer

15. Give reasons: Genetic code is 'universal'.

Exercise 16. Short Answer

16. Name the parts marked 'A' and 'B' in the given transcription unit.

Exercise 17. Short Answer

17. Differentiate Leading strand and Lagging strand.

Exercise 18. Short Answer

18. Differentiate Template strand and Coding strand.

Exercise 19. Short Answer

19. Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.

Exercise 20. Short Answer

20. State any three goals of the human genome project.

Exercise 21. Short Answer

21. In Escherichia coli, three enzymes β-galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.

Exercise 22. Short Answer

22. Distinguish between structural gene, regulatory gene and operator gene.

Exercise 23. Short Answer

23. A low level of expression of lac operon occurs at all the time. Justify the statement.

Exercise 24. Long Answer

24. HGP is the windows for treatment of various genetic disorders. Justify the statement.

Exercise 25. Short Answer

25. Why the human genome project is called a mega project?

Exercise 26. Long Answer

26. From their examination of the structure of DNA, what did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?

Exercise 27. Short Answer

27. Why tRNA is called an adapter molecule?

Exercise 28. Short Answer

28. What are the three structural differences between RNA and DNA?

Exercise 29. Short Answer

29. Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.

Codon	Anticodon
AAU
CGA
UAU
GCA

Exercise 30. Long Answer

30. a) Identify the figure given below.

b) Redraw the structure as a replicating fork and label the parts.

c) Write the source of energy for this replication and name the enzyme involved in this process.

d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.

Exercise 31. Short Answer

31. If the coding sequence in a transcription unit is written as follows:

5' T G C A T G C A T G C A T G C A T G C A T G C A T G C 3'

Write down the sequence of mRNA.

Exercise 32. Short Answer

32. How is the two-stage process of protein synthesis advantageous?

Exercise 33. Long Answer

33. Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?

Exercise 34. Long Answer

34. Explain the formation of a nucleosome.

Exercise 35. Long Answer

35. It is established that RNA is the first genetic material. Justify giving reasons.

ICT Corner — Molecular Genetics

Procedure:

1. Step 1: Use the URL or scan the QR Code to launch the "Gene Expression Essentials" activity page.
2. Step 2: Click "Expression" pick the genetic material from the Biomolecule Toolbox, understand the changes for the three different genes.
3. Step 3: Click "mRNA" and slide through the slider in Positive Transcription factors and Negative Transcription factors such as Concentration, Affinity. Also Slide through "Affinity" in RNA polymerase.
4. Step 4: Click "Multiple Cells" and find the average protein level vs Time in the graph indicated above.

URL: https://phet.colorado.edu/sims/html/gene-expression-essentials/latest/gene-expression-essentials_en.html

*Pictures are indicative only. *Allow flash player.

Book-back Exercises — Questions & Answers

Answers generated strictly from the Samacheer Kalvi textbook content. Attempt each question first, then click "Show Answer" to check.

I. Choose the Correct Answer

1. Hershey and Chase experiment with bacteriophage showed that

a) Protein gets into the bacterial cells b) DNA is the genetic material c) DNA contains radioactive Sulphur d) Viruses undergo transformation

Answer: b) DNA is the genetic material

Explanation: Hershey and Chase used radioactive isotopes of Sulphur (35S) and phosphorus (32P). Only 32P was found in bacterial cells and phage progeny, conclusively proving that DNA, not protein, carries hereditary information.

2. DNA and RNA are similar with respect to

a) Thymine as a nitrogen base b) A single-stranded helix shape c) Nucleotide containing sugars, nitrogen bases and phosphates d) The same sequence of nucleotides for the amino acid phenylalanine

Answer: c) Nucleotide containing sugars, nitrogen bases and phosphates

Explanation: Each nucleotide subunit in both DNA and RNA is composed of three parts: a nitrogenous base, a five-carbon sugar (pentose) and a phosphate group.

3. A mRNA molecule is produced by

a) Replication b) Transcription c) Duplication d) Translation

Answer: b) Transcription

Explanation: The process of copying genetic information from one strand of DNA into RNA is termed transcription. The RNA polymerase 2 transcribes precursor of mRNA.

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

a) 3.5 million b) 35000 c) 35 million d) 3.1 billion

Answer: d) 3.1 billion

Explanation: The human genome contains 3 billion nucleotide bases (approximately 3 × 10⁹ base pairs) as stated in the salient features of the Human Genome Project.

5. Escherichia coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?

a) One high and one low density band b) One intermediate density band c) One high and one intermediate density band d) One low and one intermediate density band

Answer: d) One low and one intermediate density band

Explanation: After the second replication (two generations), Meselson and Stahl found the DNA settling into two bands — one at the light band position and one at intermediate position, confirming semi-conservative replication.

6. What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?

a) Origin of replication occurs only at the 5' end of the molecules b) DNA ligase works only in the 3' to 5' direction c) DNA polymerase can join new nucleotides only to the 3' end of the growing strand d) Helicases and single strand binding proteins that work at the 5' end

Answer: c) DNA polymerase can join new nucleotides only to the 3' end of the growing strand

Explanation: The enzyme DNA polymerase can only catalyse the addition of a nucleotide to the new strands in the 5' to 3' direction, as it can only add nucleotides to the 3' carbon position, which results in continuous synthesis on one strand (leading) and discontinuous on the other (lagging).

7. Which of the following is the correct sequence of event with reference to the central dogma?

a) Transcription, Translation, Replication b) Transcription, Replication, Translation c) Duplication, Translation, Transcription d) Replication, Transcription, Translation

Answer: d) Replication, Transcription, Translation

Explanation: Francis Crick proposed the Central dogma of protein synthesis which states that genetic information flows from DNA (replication) → RNA (transcription) → Protein (translation).

8. Which of the following statements about DNA replication is not correct?

a) Unwinding of DNA molecule occurs as hydrogen bonds break b) Replication occurs as each base is paired with another exactly like it c) Process is known as semi conservative replication because one old strand is conserved in the new molecule d) Complementary base pairs are held together with hydrogen bonds

Answer: b) Replication occurs as each base is paired with another exactly like it

Explanation: In DNA replication, each base pairs with its complementary base (A with T, G with C), not with the same base. Adenine pairs with Thymine and Guanine pairs with Cytosine based on Chargaff's rules.

9. Which of the following statements is not true about DNA replication in eukaryotes?

a) Replication begins at a single origin of replication b) Replication is bidirectional from the origins c) Replication occurs at about 1 million base pairs per minute d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time

Answer: a) Replication begins at a single origin of replication

Explanation: In eukaryotes with giant DNA molecules, there can be several origins of replication (replicons), unlike prokaryotes which have only one origin of replication.

10. The first codon to be deciphered was ___ which codes for ___

a) AAA, proline b) GGG, alanine c) UUU, Phenylalanine d) TTT, arginine

Answer: c) UUU, Phenylalanine

Explanation: The mRNA codon UUU codes for phenylalanine in all cells of all organisms, and this was the first codon to be deciphered. For example, the mRNA (UUU) codon codes for phenylalanine in all organisms.

11. Meselson and Stahl's experiment proved

a) Transduction b) Transformation c) DNA is the genetic material d) Semi-conservative nature of DNA replication

Answer: d) Semi-conservative nature of DNA replication

Explanation: Meselson and Stahl (1958) designed an experiment using 15N and 14N isotopes and CsCl density gradient centrifugation that confirmed Watson and Crick's semi-conservative replication hypothesis.

12. Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for ___ and the larger subunit has two binding sites for two ___.

Answer: mRNA; tRNA

Explanation: The smaller subunit binds to mRNA, while the larger subunit has sites (A site, P site, and E site) for tRNA molecules during translation.

13. An operon is a:

a) Protein that suppresses gene expression b) Protein that accelerates gene expression c) Cluster of structural genes with related function d) Gene that switched other genes on or off

Answer: c) Cluster of structural genes with related function

Explanation: The clusters of genes with related functions are called operons. They usually transcribe single mRNA molecules. In E. coli, nearly 260 genes are grouped into 75 different operons.

14. When lactose is present in the culture medium:

a) Transcription of lac y, lac z, lac a genes occurs b) Repressor is unable to bind to the operator c) Repressor is able to bind to the operator d) Both (a) and (b) are correct

Answer: d) Both (a) and (b) are correct

Explanation: In the presence of inducer lactose, the repressor is inactivated and becomes unable to bind to the operator. This allows RNA polymerase to transcribe the operon (lac z, y, a genes), producing the enzymes needed for lactose metabolism.

II. Answer Briefly

15. Give reasons: Genetic code is 'universal'.

Answer: The genetic code is universal because all known living systems — both prokaryotes and eukaryotes — use nucleic acids and the same three base codons (triplet codons) to direct the synthesis of proteins from amino acids. For example, the mRNA codon UUU codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes, but similarities are more common than differences.

16. Name the parts marked 'A' and 'B' in the given transcription unit.

Answer: A transcriptional unit in DNA is defined by three regions. 'A' refers to the Promoter — located towards the 5' end of the coding strand, it provides the binding site for RNA polymerase. 'B' refers to the Terminator — located towards the 3' end of the coding strand, it contains a DNA sequence that causes RNA polymerase to stop transcribing.

17. Differentiate Leading strand and Lagging strand.

Answer:

Leading Strand	Lagging Strand
Template strand with polarity 3' to 5'	Coding strand with polarity 5' to 3'
Replication is continuous	Replication is discontinuous
Synthesized in one continuous piece	Synthesized as Okazaki fragments joined by DNA ligase

18. Differentiate Template strand and Coding strand.

Answer:

Template Strand	Coding Strand
Has polarity 3' to 5'	Has polarity 5' to 3'
Acts as template for RNA synthesis	Is displaced during transcription
Sequence is complementary to mRNA	Has the same sequence as RNA (except thymine instead of uracil)

19. Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.

Answer: (1) Identification of SNPs is helpful in finding chromosomal locations for disease-associated sequences, enabling diagnosis of genetic diseases and providing genetic counselling. (2) SNPs are useful in tracing human history and understanding migration patterns of human populations, contributing to anthropological studies.

20. State any three goals of the Human Genome Project.

Answer: Three goals of the Human Genome Project are: (1) To identify all the genes (approximately 30,000) in human DNA. (2) To determine the sequence of the three billion chemical base pairs that make up the human DNA. (3) To store this information in databases and improve tools for data analysis.

21. In Escherichia coli, three enzymes β galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.

Answer: When the cell is using glucose as its normal energy source (in the absence of lactose), the 'i' gene transcribes a repressor mRNA and after its translation, a repressor protein is produced. This repressor protein binds to the operator region of the lac operon and prevents RNA polymerase from transcribing the structural genes (lac z, y, a). As a result, β-galactosidase, permease and transacetylase are not produced. Lactose acts as an inducer that inactivates the repressor; without lactose, the repressor remains active and blocks transcription.

22. Distinguish between structural gene, regulatory gene and operator gene.

Answer: Structural gene codes for proteins, rRNA and tRNA required by the cell. Regulatory gene (i gene) transcribes a repressor mRNA whose product (repressor protein) controls the expression of structural genes. Operator gene is present between the promoter and structural genes; the repressor protein binds to the operator region to prevent or allow transcription of the structural genes.

23. A low level of expression of lac operon occurs at all the time. Justify the statement.

Answer: A low level of expression of lac operon occurs at all times because lactose must enter the cell to act as an inducer, and this entry requires permease enzyme. For lactose to enter the cell initially, a small amount of permease must be present. This requires a basal (low) level of transcription of the lac operon even in the absence of lactose, so that the cell is always ready to respond when lactose becomes available.

24. HGP is the windows for treatment of various genetic disorders. Justify the statement.

Answer: The mapping of human chromosomes makes it possible to examine a person's DNA and identify genetic abnormalities. This is extremely useful in diagnosing diseases and providing genetic counselling to those planning to have children. This information would also create possibilities for new gene therapies. A new era of molecular medicine, characterized by looking into the most fundamental causes of disease rather than treating the symptoms, will be an important advantage of HGP.

25. Why is the Human Genome Project called a mega project?

Answer: The Human Genome Project is called a mega project because it was launched in 1990 and took 13 years to complete. The human genome is about 25 times larger than the genome of any organism sequenced to date and was the first vertebrate genome to be completed. It involved sequencing approximately 3 × 10⁹ base pairs, identifying approximately 30,000 genes, and required the development of new fields like bioinformatics for data analysis, making it an enormous scientific undertaking.

26. From their examination of the structure of DNA, what did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?

Answer: Watson and Crick proposed semi-conservative replication in 1953, suggesting that the two polynucleotide strands unwind and each separated strand acts as a template for a new complementary strand. Regarding coding capability, the specific sequence of base pairs along the DNA molecule controls the kind and order of amino acids in proteins. Regarding mutation, sudden changes (mutations) in nucleotide positions or composition alter the base sequence, which changes the amino acid sequence, as seen in sickle cell anaemia where a single base substitution (GAG to GTG) changes glutamic acid to valine.

27. Why is tRNA called an adapter molecule?

Answer: The tRNA molecule acts as a vehicle that picks up amino acids scattered through the cytoplasm and also reads specific codes of mRNA molecules. Hence it is called an adapter molecule. This term was postulated by Francis Crick. The amino acid is attached to one end (amino acid acceptor end) and the other end consists of three anticodon nucleotides. The anticodon pairs with a codon in mRNA ensuring that the correct amino acid is incorporated into the growing polypeptide chain.

28. What are the three structural differences between RNA and DNA?

Answer: (1) Sugar: DNA contains deoxyribose sugar whereas RNA contains ribose sugar; the only difference is that deoxyribose has one oxygen atom less. (2) Nitrogenous bases: DNA contains Thymine (T) while RNA contains Uracil (U) instead of Thymine; Thymine is unique to DNA and Uracil is unique to RNA. (3) Strand structure: DNA is a double-stranded helical structure while RNA is generally a single-stranded structure. Also, RNA has an additional hydroxyl group at 2' position in the ribose.

29. Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.

Answer: The anticodon is complementary and antiparallel to the codon (read 3' to 5' on tRNA for a 5' to 3' codon on mRNA):

• AAU → anticodon: UUA
• CGA → anticodon: GCU
• UAU → anticodon: AUA
• GCA → anticodon: CGU

(The anticodon is complementary to the codon, following the rule of base pairing: A pairs with U, G pairs with C.)

III. Answer in Detail

30. a) Identify the figure given below b) Redraw the structure as a replicating fork and label the parts c) Write the source of energy for this replication and name the enzyme involved in this process d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.

Answer:

a) The figure represents the DNA double helix (double stranded DNA structure).

b) The replicating fork shows: the parental DNA strands unwinding at the origin of replication (ori), helicase unwinding the strands, forming a Y-shaped replication fork. Parts to label: Origin of replication, Replication fork, Leading strand (continuous synthesis, 5' to 3'), Lagging strand (discontinuous — Okazaki fragments, 5' to 3'), RNA primer, DNA polymerase, DNA ligase.

c) The source of energy is Deoxyribonucleotide triphosphate (dNTP) — it acts as substrate and also provides energy for polymerization reaction. The main enzyme involved is DNA polymerase III. Other enzymes include: helicase (unwinding), topoisomerase/DNA gyrase (relieving tension), DNA ligase (joining Okazaki fragments), and primase (synthesizing RNA primer).

d) On the template strand with polarity 3' to 5', replication is continuous (leading strand); the new strand is synthesized in 5' to 3' direction. On the template strand with polarity 5' to 3', replication is discontinuous (lagging strand); short Okazaki fragments are synthesized in 5' to 3' direction and joined by DNA ligase.

31. If the coding sequence in a transcription unit is written as follows: 5' TGCATGCATGCATGCATGCATGCATGC 3' Write down the sequence of mRNA.

Answer

📝 Summary

1. DNA is the genetic material that encodes unique characteristics of living organisms; it directs the construction of proteins, which determine cell shapes and the rate of chemical reactions such as metabolism and photosynthesis.
2. The genome contains the complete set of hereditary information for any organism, divided into nucleic acid molecules containing large numbers of genes, where each gene represents a single protein.
3. The concept of the gene was first explained by Gregor Mendel in the 1860s as a "factor"; the term "gene" was coined in 1909 by Danish biologist Wilhelm Johannsen to refer to discrete determiners of inherited characteristics.
4. According to the classical concept introduced by Sutton in 1902, genes occupy a definite locus on chromosomes, are arranged in a single linear order, may exist in alternate forms called alleles, and are capable of self-duplication producing their own copies.
5. Molecular genetics encompasses key processes including DNA replication, transcription (making RNA from DNA), translation (protein synthesis guided by the genetic code), regulation of gene expression, and human genome sequencing.