Class 12 Biology - Chapter 5 : Molecular Basis of Inheritance

 

 

Chapter 5

Molecular Basis of Inheritance

Syllabus: Search for genetic material and DNA as genetic material; Structure of DNA and RNA; DNA packaging; DNA replication; Central Dogma; transcription, genetic code, translation; gene expression and regulation - lac operon; Genome, Human and rice genome projects; DNA fingerprinting.

Introduction

1.     At the time of Mendel, the nature of those ‘factors’ regulating the pattern of inheritance was not clear.

2.     Over the next hundred years, the nature of the putative (accepted) genetic material was investigated culminating (completion) in the realisation that DNA – deoxyribonucleic acid – is the genetic material, at least for the majority of organisms.

3.     Nucleic acids are polymers of nucleotides.

4.     Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the two types of nucleic acids found in living systems.

5.     DNA acts as the genetic material in most of the organisms. It is the largest bio molecule. It is also called as polynucleotide.

6.     RNA though it also acts as a genetic material in some viruses, mostly functions as a messenger.

 

5.1 THE DNA

1.     DNA is a long polymer of deoxyribonucleotides.

2.     The length of DNA is usually defined as number of nucleotides (or a pair of nucleotide referred to as base pairs) present in it.

 

3.     This also is the characteristic of an organism.

4.     For example, a bacteriophage known as φ ×174 has 5386 nucleotides, Bacteriophage lambda has 48502 base pairs (bp), Escherichia coli has ds DNA 4.6 × 10⁶ bp, and haploid content of human dsDNA is 3.3 × 10⁹ bp.

 

5.1.1 Structure of Polynucleotide Chain

1.     A nucleotide has three components – a nitrogenous base, a pentose sugar (ribose in case of RNA, and deoxyribose for DNA), and a phosphate group.

2.     Structure of DNA.

3.     There are two types of nitrogenous bases –

i).     Purines (Adenine and Guanine)

Ø It has 9 members and double ring structure.

 

ii).   Pyrimidines (Cytosine, Uracil and Thymine).

Ø It has 6 members and single ring structure. (Nam bade darshan chhote and y letter present in their names except Uracil.)

iii). Uracil is present in RNA at the place of Thymine. RNA is first discovered nucleic acid.

iv).            RNA is very reactive due to the presence of Uracil which is very reactive.

v).   Structure of RNA.

 

4.     A nucleoside is a combination of Sugar and nitrogenous Base.

5.     A nucleotide is a combination of Sugar, nitrogenous base and Phosphoric acid.       

6.     Nucleotide in case of DNA are as follows:

 

 

Deoxyribose sugar	Nitrogenous Base		Phosphate group
Nucleoside
Nucleotide
Sugar	A	deoxyadenosine	Deoxyadenosine mono phosphate dAMP
Sugar	G	Deoxyguanosine	Deoxyguanosine mono phosphate dGMP
Sugar	C	Deoxycytidine	Deoxycytidine mono phosphate dCMP
Sugar	T	deoxythymidine	deoxythymidine mono phosphate dTMP

 

7.     Nucleotide in case of RNA are as follows:

Deoxyribose sugar	Nitrogenous Base		Phosphate group
Nucleoside
Nucleotide
S	A	adenosine	adenosine mono phosphate  AMP
S	G	guanosine	guanosine mono phosphate  GMP
S	C	cytidine	cytidine mono phosphate      CMP
S	U	uridine	uridine mono phosphate        UMP

 8.     Bond between Deoxyribose sugar and nitrogenous base.

 

·        Glycosidic linkage is due to the involvement of sugar in the formation of bond.

·        This structure is nucleoside.

·        In the case of Pyrimdine it is 1’(sugar) -----1 (pyridmidine) glycosidic linkage.

9.     A nitrogenous base is linked to the OH of 1' Carbon of pentose sugar through a N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.

10.When a phosphate group is linked to OH of 5' C of a nucleoside through phosphodiester linkage, a corresponding nucleotide (or deoxynucleotide depending upon the type of sugar present) is formed.

i).     Two nucleotides are linked through 3'-5' phosphodiester linkage to form a dinucleotide.

ii).   More nucleotides can be joined in such a manner to form a polynucleotide chain.

 

11.A polymer thus formed has at one end a free phosphate moiety (part) at 5' -end of sugar, which is referred to as 5’-end of polynucleotide chain.

12.Similarly, at the other end of the polymer the sugar has a free OH of 3'C group which is referred to as 3' -end of the polynucleotide chain.

13.The backbone of a polynucleotide chain is formed due to sugar and phosphates.

14.The nitrogenous bases linked to sugar moiety project from the backbone (Figure 5.1).

 

15.Double stranded DNA structure is given below:

 

16.In RNA, every nucleotide residue has an additional –OH group present at 2' -position in the ribose.

 

17.Also, in RNA the uracil is found at the place of thymine (5-methyl uracil, another chemical name for thymine).

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

19.He named it as ‘Nuclein’. However, due to technical limitation in isolating such a long polymer intact, the elucidation of structure of DNA remained elusive for a very long period of time.

20.It was only in 1953 that James Watson and Francis Crick, based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin, proposed a very simple but famous Double Helix model for the structure of DNA.

21.One of the hallmarks of their proposition was base pairing between the two strands of polynucleotide chains.

22.However, this proposition was also based on the observation of Erwin Chargaff that for a double stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equals one.

23.The base pairing confers a very unique property to the polynucleotide chains.

24.They are said to be complementary to each other, and therefore if the sequence of bases in one strand is known then the sequence in other strand can be predicted.

25.The salient features of the Double-helix structure of DNA are as follows:

(i)                It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and the bases project inside.

(ii)              The two chains have anti-parallel polarity. It means, if one chain has the polarity 5'------3', the other has 3'-------5'.

(iii)            The bases in two strands are paired through hydrogen bond (H-bonds) forming base pairs (bp).

Ø    Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.

Ø    Similarly, Guanine is bonded with Cytosine with three H-bonds.

Ø    As a result, always a purine comes opposite to a pyrimidine.

Ø    This generates approximately uniform distance between the two strands of the helix (Figure 5.2).

 

(iv)            The two chains are coiled in a right-handed fashion.

Ø    The pitch (distance between two turns) of the helix is 3.4 nm (a nanometre is one billionth of a metre, that is 10-⁹ m) and there are roughly 10 bp in each turn.

Ø    Consequently, the distance between a bp in a helix is approximately 0.34 nm.

 

(v)              The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confers stability of the helical structure (Figure 5.3).

 

 

 

The central dogma

1.     The central dogma of molecular biology is a fundamental concept that describes the flow of genetic information within a biological system.

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

3.     It outlines the processes by which genetic information is transferred from DNA to RNA and then to protein, guiding the functioning of living organisms.

4.     The central dogma can be summa rized in three main steps:

 

Replication: DNA replication is the process by which a DNA molecule makes an identical copy of itself. This occurs during cell division and ensures that each daughter cell receives a complete set of genetic information.

Transcription: Transcription is the process by which a specific segment of DNA is used as a template to synthesize a complementary RNA molecule. The enzyme RNA polymerase reads the DNA template and assembles a corresponding RNA strand. This RNA molecule is known as messenger RNA (mRNA) and carries the genetic information from the DNA to the next step.

Translation: Translation is the process by which the information encoded in the mRNA molecule is used to synthesize a specific protein. This occurs at ribosomes, where transfer RNA (tRNA) molecules bring the appropriate amino acids in sequence, according to the instructions provided by the mRNA. Amino acids are linked together to form a polypeptide chain, which then folds into a functional protein.

In some viruses the flow of information is in reverse direction, that is, from RNA to DNA. This process called as reverse transcription or teminism based on the scientist Temin and Baltimore who discovered this process. Ex:- Raw Sarcoma Virus and HIV (Human Immuno deficiency virus).

 

 

5.1.2 Packaging of DNA Helix

1.     Taken the distance between two consecutive base pairs as 0.34 nm (0.34×10^–9 m), if the length of DNA double helix in a typical mammalian cell is calculated (simply by multiplying the total number of bp with distance between two consecutive bp, that is, 6.6 × 10⁹ bp × 0.34 × 10^-9 m/bp), it comes out to be approximately 2.2 metres.

2.     A length that is far greater than the dimension of a typical nucleus (approximately 10^–6 m).

3.     How is such a long polymer packaged in a cell?

4.     If the length of E. coli DNA is 1.36 mm, can you calculate the number of base pairs in E.coli?

Key Information: 1.     Length of E.coli DNA: 1.36 mm. 2.     Length of one base pair: The distance between successive base pairs is about 0.34 nanometers (nm). Conversion Factors: 1.     1 mm = 1×106 nm. 2.     Length of DNA per base pair: 0.34 nm. Calculation Steps: 1.     Convert the DNA length from mm to nm: 1.36        mm×1,000,000 nm=1,360,000 nm 2.     Determine the number of base pairs by dividing the total length in nanometers by the length of each base pair: Thus, the number of base pairs in the E.coliE. coliE.coli DNA is approximately 4,000,000 base pairs.

 

5.     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 due to the presence of phosphate group) is held with some proteins (that have positive charges) in a region termed as ‘nucleoid’.

6.     In prokaryotes only non histone proteins are used but in eukaryotes both histone and non histone proteins are used.

 

 

 

Proteins are made up of amino acids, each of which has a side chain (R group) that can be acidic, basic, neutral, or hydrophobic. 1.     The name "histone" is believed to be derived from the Greek word “which means “to stand” or “to make stand.” 2.     This Greek root reflects the histones’ role as structural proteins that "stand" or serve as the backbone around which DNA is wound to form chromatin.   The Role of Histones Histones are fundamental to many cellular processes: i).     DNA Packaging: Histones form nucleosomes by allowing DNA to wrap around them, thereby compacting the DNA into chromatin and fitting it within the nucleus. ii).   Chromatin Structure: Histones help maintain the overall structure of chromatin, influencing both its compaction and accessibility.

 

7.     The DNA in nucleoid is organised in large loops held by proteins.

8.     In eukaryotes, this organisation is much more complex.

9.     There is a set of positively charged, basic proteins (BAHL, B-BASIC, A-ARGININE, H- HYSTIDINE, L-LYSINE) called histones. These are H1, H2A, H2B, H3, H4. H2A and H2B are rich in lysine and H3 and H4 are rich in argininie.

(i)                A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.

(ii)              Histones are rich in the basic amino acid residues lysine and arginine.

(iii)            Both the amino acid residues carry positive charges in their side chains.

(iv)            Histones H2A, H2B, H3, H4 duplicated and are organised to form a unit of eight molecules called histone octamer.

 

10.The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (Figure 5.4 a).

A nucleosome is the basic structural unit of chromatin, which is the complex of DNA and proteins that makes up the chromosomes in eukaryotic cells.

 

11.A typical nucleosome contains 200 bp of DNA helix.

(i)    Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, thread like stained (coloured) bodies seen in nucleus.

(ii)  The nucleosomes in chromatin are seen as ‘beads-on-string’ structure when viewed under electron microscope (EM) (Figure 5.4 b).

 

(iii)                        The beads-on-string structure in chromatin is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes.

(iv)                        The packaging ratio is typically calculated by comparing the length of the fully extended DNA molecule (i.e., if it were stretched out linearly) to the actual length of DNA present in the nucleus.

(v)  The actual length of DNA is much shorter due to the various levels of compaction, including the formation of nucleosomes, higher-order chromatin structures, and chromosome organization.

 

(vi)                        The packaging of chromatin at higher level (more condensing stage) 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.

(vii)                      The chromatin that is more densely packed and stains dark are called as Heterochromatin.

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

Heterochromatin: Maintains chromosome structure, contributes to genome stability, prevents unwanted gene expression, and can include permanent or reversible silencing of genes. Euchromatin: Facilitates gene expression by providing an accessible environment for transcription factors and RNA polymerases, supporting various cellular processes including gene transcription, replication, and DNA repair.   Transcription is a fundamental biological process that involves the synthesis of RNA molecules based on a DNA template. It is the first step in gene expression, through which the genetic information stored in DNA is used to produce functional RNA molecules, such as messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA).

 

 

5.2 THE SEARCH FOR GENETIC MATERIAL

1.     Even though the discovery of nuclein by Meischer and the proposition for principles of inheritance by Mendel were almost at the same time, but that the DNA acts as a genetic material took long to be discovered and proven.

2.     By 1926, the quest to determine the mechanism for genetic inheritance had reached the molecular level.

3.     Previous discoveries by Gregor Mendel, Walter Sutton, Thomas Hunt Morgan and numerous other scientists had narrowed the search to the chromosomes located in the nucleus of most cells.

4.     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.

5.     During the course of his experiment, a living organism (bacteria) had changed in physical form.

6.     When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R).

7.     This is because the S strain bacteria (strain means subtype or variant of bacteria) have a mucous (polysaccharide) coat, while R strain does not.

8.     Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia.

Both the strains or bacteria have antigens attached to their cell wall but S strain has mucous layer so  not attacked by antibody of the mice hence mice died and R strain do not have mucous layer over antigens hence antibodies of mice will kill the R strain bacteria and mice will survive. A microorganism (such as a bacterium or virus) is considered virulent if it has the ability to infect a host organism and cause significant harm, illness, or damage to the host's cells or tissues.

 

9.     Griffith was able to kill bacteria by heating them. He observed that heat-killed S strain bacteria injected into mice did not kill them. When he injected a mixture of heat-killed S and live R bacteria, the mice died.

 

10.Moreover, he recovered living S bacteria from the dead mice. (It may be possible here that R strain virulent absorb some chemical from heat killed S strain virulent and developed mucous layer over them and transformed to S strain bacteria so able to kill mice).

11.He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria.

12.Some ‘transforming principle’, transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent.

13.This must be due to the transfer of the genetic material.

14.However, the biochemical nature of genetic material was not defined from his experiments.

15.Biochemical Characterisation of Transforming Principle Prior to the work of Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), the genetic material was thought to be a protein.

16.They worked to determine the biochemical nature of ‘transforming principle’ in Griffith's experiment.

17.They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells.

18.They discovered that DNA alone from S bacteria caused R bacteria to become transformed.

19.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.

20.Digestion with DNase (DNA digesting enzyme) did inhibit transformation, suggesting that the DNA caused the transformation.

21.They concluded that DNA is the hereditary material, but not all biologists were convinced.

 

5.2.1 The Genetic Material is DNA

1.     The unequivocal (clear) proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952).

2.     They worked with viruses that infect bacteria called bacteriophages.

3.     The bacteriophage attaches to the bacteria and its genetic material then enters the bacterial cell.

4.     The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more virus particles.

5.     Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.

6.     They grew some viruses on a medium that contained radioactive phosphorus (P 32) and some others on medium that contained radioactive sulfur (S 35).

7.     Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.

8.     Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur.

9.     Radioactive phages were allowed to attach to E. coli bacteria.

10.Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender.

11.The virus particles were separated from the bacteria by spinning them in a centrifuge.

12.Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

13.Bacteria that were infected with viruses that had radioactive proteins were not radioactive.

14.This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria (Figure 5.5).

 

5.2.2 Properties of Genetic Material (DNA versus RNA)

1.     From the foregoing discussion, it is clear that the debate between proteins versus DNA as the genetic material was unequivocally resolved from Hershey-Chase experiment.

2.     It became an established fact that it is DNA that acts as genetic material.

3.     However, it subsequently became clear that in some viruses, RNA is the genetic material (for example, Tobacco Mosaic viruses, QB bacteriophage, etc.).

4.     A molecule that can act as a genetic material must fulfill the following criteria:

(i)                It should be able to generate its replica (Replication).

(ii)              It should be stable chemically and structurally.

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

(iv)            It should be able to express itself in the form of 'Mendelian Characters’.

 

Why DNA is a better genetic material	Why RNA is not a better genetic material
(i)    Both the nucleic acids (DNA and RNA) have the ability to direct their duplications. Also the bases are between the two strands so highly stable.	(i)    Both the nucleic acids (DNA and RNA) have the ability to direct their duplications. Bases are open so reactive.
(ii)  Stability as one of the properties of genetic material was very evident in Griffith’s ‘transforming principle’ itself that heat, which killed the bacteria, at least did not destroy some of the properties of genetic material. In fact, the presence of thymine at the place of uracil also confers additional stability to DNA.	(ii) Further, 2'-OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable. RNA is also now known to be catalytic, hence reactive. Also Uracil is less stable and can be converted to Thymine by methylation.
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.

 

5.     Both DNA and RNA are able to mutate.

6.     In fact, RNA being unstable, mutate at a faster rate.

7.     Consequently, viruses having RNA genome and having shorter life span mutate and evolve faster.

8.     RNA can directly code for the synthesis of proteins, hence can easily express the characters.

9.     DNA, however, is dependent on RNA for synthesis of proteins.

10.The protein synthesising machinery has evolved around RNA.

11.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.

12.For the transmission of genetic information, RNA is better.

 

5.3 RNA WORLD

1.     RNA was the first genetic material.

2.     There is now enough evidence to suggest that essential life processes (such as metabolism, translation, splicing, etc.), evolved around RNA.

3.     RNA used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzymes).

4.     But, RNA being a catalyst was reactive and hence unstable.

5.     Therefore, DNA has evolved from RNA with chemical modifications that make it more stable.

6.     DNA being double stranded and having complementary strand further resists changes by evolving a process of repair.

 

5.4 REPLICATION

1.     While proposing the double helical structure for DNA, Watson and Crick had immediately proposed a scheme for replication of DNA.

2.     To quote their original statement that is as follows: ‘‘It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material’’ (Watson and Crick, 1953).

3.     The scheme suggested that the two strands would separate and act as a template for the synthesis of new complementary strands. This process of forming its own copy also known as autocatalytic/replication.

 

4.     After the completion of replication, each DNA molecule would have one parental and one newly synthesized strand. Therefore called semi conservative DNA.

5.     DNA replication takes place in the interphase of mitosis.

What happens during Interphase? Interphase is the longest phase of the cell cycle, where the cell:          i.            Grows        ii.            Performs normal functions       iii.            Prepares for cell division It has three sub-phases: 1.     G1 Phase (Gap 1) – Cell grows and performs normal activities. 2.     S Phase (Synthesis) – DNA is replicated (copied). 3.     G2 Phase (Gap 2) – Cell prepares for mitosis by making proteins and checking for DNA errors. Ø  Interphase occurs before mitosis. Ø  It's the preparation phase for cell division.

 

6.     This scheme was termed as semi conservative DNA replication (Figure 5.6).

 

6.4.1     The Experimental Proof

 

1.     It is now proven that DNA replicates semi conservatively.

2.     It was shown first in Escherichia coli and subsequently in higher organisms, such as plants and human cells.

3.     Matthew Meselson and Franklin Stahl performed the following experiment in 1958:

(i)                They grew E. coli in a medium containing ¹⁵NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as the only nitrogen source for many generations.

(ii)              The result was that ¹⁵N was incorporated into newly synthesized DNA (as well as other nitrogen containing compounds).

(iii)            This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient (means take the solution of CsCl solution to centrifuge DNA)(Please note that ¹⁵N is not a radioactive isotope, and it can be separated from ¹⁴N only based on densities).

density gradient refers to the gradual variation in the density of the CsCl solution as you move from the top to the bottom of the centrifuge tube.

 

(iv)            Then they transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices.

(v)              The various samples were separated independently on CsCl gradients to measure the densities of DNA (Figure 5.7).

 

(vi)            Thus, the DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium [that is after 20 minutes; E. coli divides in 20 minutes] had a hybrid or intermediate density.

(vii)          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.

 

4.     Very similar experiments involving use of radioactive thymidine (he used  ³₁H) to detect distribution of newly synthesised DNA in the chromosomes was performed on Vicia faba (faba beans) by Taylor and colleagues in 1958.

 

5.     The experiments proved that the DNA in chromosomes also replicate semi conservatively.

 

5.4.2     The Machinery and the Enzymes (Process of Replication)

 

1.     In living cells, such as E. coli, the process of replication requires a set of catalysts (enzymes).

2.     The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerisation of deoxynucleotides.

3.     These enzymes are highly efficient enzymes as they have to catalyse polymerisation of a large number of nucleotides in a very short time. E. coli that has only 4.6 ×10⁶ bp (compare it with human whose diploid content is 6.6 × 10⁹ bp), completes the process of replication within 18 minutes; that means the average rate of polymerisation has to be approximately 2000 bp per second.

4.     Not only do these polymerases have to be fast, but they also have to catalyse the reaction with high degree of accuracy.

5.     The DNA polymerases on their own cannot initiate the process of replication.

6.     Also the replication does not initiate randomly at any place in DNA.

7.     There is a definite region in E. coli DNA where the replication originates. Such regions are termed as origin of replication.

8.     It is because of the requirement of the origin of replication that a piece of DNA if needed to be propagated during recombinant DNA procedures, requires a vector.

In simple terms, a vector in the context of recombinant DNA technology is a carrier or delivery system used to introduce a specific piece of DNA into a host cell so that it can be propagated or replicated. Think of a vector as a “molecular taxi” that transports the DNA you want to study or use into a cell where it can be reproduced and expressed.

 

9.     The vectors provide the origin of replication.

10.In eukaryotes, the replication of DNA takes place at S-phase of the cell-cycle.

11.The replication of DNA and cell division cycle should be highly coordinated. A failure in cell division after DNA replication results into polyploidy(a chromosomal anomaly).

 

12.Any mistake during replication would result into mutations.

13.Furthermore, energetically replication is a very expensive process.

14.We know that a nucleotide consists of a deoxyribose sugar, nitrogenous base and a phosphate group but this process needs high energy so here Deoxyribonucleoside triphosphates serve dual purposes.

15.In addition to acting as substrates (a substrate is the molecule or substance that interacts with an enzyme's active site to undergo a chemical reaction, or in a more general sense, the starting material in a chemical reaction catalyzed by any type of catalyst.), they provide energy for polymerisation reaction (the two terminal phosphates in a deoxynucleoside triphosphates are high-energy phosphates, same as in case of ATP).

 

16.In addition to DNA-dependent DNA polymerases, many additional enzymes are required to complete the process of replication with high degree of accuracy.

Additional enzymes are as follows:-

i).     DNA Helicase: This is the first enzyme used in this process. It will act on the point of origin. Helicases are enzymes responsible for unwinding the double-stranded DNA molecule by breaking the hydrogen bonds between the base pairs. This unwinding is essential for the replication fork to progress.

ii).   Single-Strand DNA Binding Proteins (SSBs): These proteins stabilize the single-stranded DNA regions that are exposed after the helicase unwinds the DNA. SSBs prevent the single strands from re-forming base pairs and tangling.

iii). Topoisomerases: DNA unwinding during replication creates tension ahead of the replication fork. Topoisomerases relieve this tension by introducing reversible breaks in the DNA strands, allowing them to rotate and relieve the stress.

iv). Primase: Primase is an enzyme that synthesizes short RNA primers complementary to the DNA template strand. These primers provide a starting point for DNA synthesis by DNA polymerase. DNA polymerase is basically the important enzyme which starts synthesis of DNA. Its function is to bring the complementary nucleotide depends on the DNA so it is a DNA dependent polymerase.

v).   Exonucleases: The primers and the incorrect nucleotides in the fragments removed by the exonuclease.

vi). Ligase: In DNA replication, the enzyme DNA ligase plays a crucial role in joining together fragments of DNA. This process is essential for the synthesis of the new DNA strand to be continuous and complete.

 

 

17.For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occur within a small opening of the DNA helix, referred to as replication fork.

18.DNA polymerase enzyme is responsible for the synthesis of DNA but this enzyme cannot initiate DNA replication on its own. It requires primer (a small RNA strand) to initiate the process with the help of enzyme primase.

19.The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5'----3'. This creates some additional complications at the replicating fork.

 

 

20.Consequently, on one strand (the template with polarity 3'----5' ), the replication is continuous, while on the other, it is discontinuous. Here we will consider the starting position of template is from replicating fork to the other end.

21.The primers and the incorrect nucleotides in the fragments removed by the exonuclease.

22.The discontinuously synthesised fragments are later joined by the enzyme DNA ligase (Figure 5.8).

 

 

 

5.5          TRANSCRIPTION

1.     The process of copying genetic information from one strand of the DNA into RNA is termed as transcription.

2.     Here also, the principle of complementarity governs the process of transcription, except the adenosine complements now forms base pair with uracil instead of thymine.

3.     However, unlike in the process of replication, which once set in, the total DNA of an organism gets duplicated, in transcription only a segment of DNA and only one of the strands is copied into RNA.

4.     This necessitates defining the boundaries that would demarcate the region and the strand of DNA that would be transcribed.

5.     Why both the strands are not copied during transcription has the simple answer.

i).     First, if both strands act as a template, they would code for RNA molecule with different sequences (Remember complementarity does not mean identical), and in turn, if they code for proteins, the sequence of amino acids in the proteins would be different. Hence, one segment of the DNA would be coding for two different proteins, and this would complicate the genetic information transfer machinery.

ii).   Second, the two RNA molecules if produced simultaneously would be complementary to each other, hence would form a double stranded RNA. This would prevent RNA from being translated into protein and the exercise of transcription would become a futile one.

 

5.5.1     Transcription Unit

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

(i)                A Promoter

(ii)              The Structural gene

(iii)            A Terminator

2.     Requirement for the formation of RNA or for the process of transcription are as follows:-

i).     DNA template (3’ ---------- 5’)

ii).   RNA polymerase

iii). In case of prokaryotes, Initiation-factor (σ, Sigma factor) and termination-factor (ρ, Rho factor) are required, no such factors are required in eukaryotes.

 

iv). In case of eukaryotes Methyl guanosine triphosphate is required for caping at the 5'-end and adenylate (adenosine monophosphate) for tailing at the 3’ end.

 

3.     There is a convention in defining the two strands of the DNA in the structural gene of a transcription unit.

4.     Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction, that is, 5'→3', the strand that has the polarity 3'→5' acts as a template, and is also referred to as template strand.

5.     The other strand which has the polarity (5'→3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. The coding strand is "displaced" during transcription, meaning it is not involved in the base-pairing process of RNA synthesis

6.     Strangely, this strand (which does not code for anything) is referred to as coding strand.

7.     All the reference point while defining a transcription unit is made with coding strand because the sequence of the coding strand directly correlates with the RNA.

8.     To explain the point, a hypothetical sequence from a transcription unit is represented below:

3' -ATGCATGCATGCATGCATGCATGC-5' Template Strand

5' -TACGTACGTACGTACGTACGTACG-3' Coding Strand

Can you now write the sequence of RNA transcribed from the above DNA?

 

9.     The promoter and terminator flank (border) the structural gene in a transcription unit.

10.The promoter is said to be located towards 5' -end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand).

11.It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that also defines the template and coding strands.

12.By switching its position with terminator, the definition of coding and template strands could be reversed.

13.The terminator is located towards 3' -end (downstream) of the coding strand and it usually defines the end of the process of transcription. There are additional regulatory sequences (these regulate genes) that may be present further upstream or downstream to the promoter.

Enhancing or suppressing transcription: Regulatory sequences can bind proteins called transcription factors, which either enhance (activators) or suppress (repressors) the activity of RNA polymerase at the promoter, thus influencing how much mRNA is made from a gene.

 

5.5.2 Transcription Unit and the Gene

1.     A gene is defined as the functional unit of inheritance. Though there is no ambiguity (doubt) that the genes are located on the DNA, it is difficult to literally define a gene in terms of DNA sequence. It is a specific sequence of nucleotide bases (adenine, thymine, cytosine, and guanine) that encodes the instructions for synthesizing a functional molecule, usually a protein or a non-coding RNA (rRNA and tRNA are non coding RNA). mRNA is a coding RNA which carries the instructions from DNA to make a protein.

2.     The DNA sequence coding for tRNA or rRNA molecule also define a gene.

3.     However by defining a cistron as a segment of DNA coding for a polypeptide, the structural gene in a transcription unit could be said as monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).

A cistron is a more specific part of a gene. It is the segment of DNA that codes for one polypeptide (protein chain). It’s basically the coding part of a gene (also called the structural gene). Since each cistron codes for one polypeptide, we say: Eukaryotic structural genes are mostly monocistronic because their mRNA carries the code for just one polypeptide. Prokaryotic structural genes are often polycistronic because their mRNA carries the code for several polypeptides

 

4.     Monocistronic genes are genes that are transcribed into a single messenger RNA (mRNA) molecule, which codes for a single protein or polypeptide.

5.     Polycistronic Gene Expression in Prokaryotes: Multiple genes are transcribed into a single mRNA, and each gene is translated into a different protein.

 

6.     In eukaryotes, the monocistronic structural genes have interrupted coding sequences – the genes in eukaryotes are split.

i)       The coding sequences or expressed sequences are defined as exons. Exons are said to be those sequence that appear in mature or processed RNA.

ii)     The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA.

iii)   The split-gene arrangement further complicates the definition of a gene in terms of a DNA segment.

 

7.     Inheritance of a character is also affected by promoter and regulatory sequences of a structural gene. Hence, sometime the regulatory sequences are loosely defined as regulatory genes, even though these sequences do not code for any RNA or protein (normal structural protein).

8.     Regulatory genes produce regulatory molecules (proteins or non-coding RNAs, these proteins only used in regulating other genes).These regulatory molecules then affect the expression of other genes by turning them on or off, or by modulating their expression levels.

i)       The promoter is a crucial regulatory DNA sequence located at the beginning of a gene that controls the initiation of transcription, the process by which RNA is synthesized from a DNA template.

 

 

5.6          TRANSCRIPTION

 

1.     The process of copying genetic information from one strand of the DNA into RNA is termed as transcription. Or in simple terms, formation of RNA from DNA is transcription.

 

2.     In eukaryotes, it is done in nucleus and in prokaryotes, it is done in cytoplasm.

3.     Requirement for the formation of RNA are as follows:-

v).   DNA template (3’ ---------- 5’)

vi). RNA polymerase

vii).                       Ribonucleoside triphosphate- ATP(Adenosine triphosphate) or APPP, GTP, UTP and CTP

viii).                    Some factors like Mg²⁺, Mn²⁺, initiation-factor (σ, Sigma factor) and termination-factor (ρ, Rho factor) in case of prokaryotes, no such factors are required in eukaryotes.

 

Process of transcription

In Prokaryotes

i).     In these only one RNA polymerase is required for all the functions.

ii).  RNA polymerase needs to attach with the promoter as its recognition sequence is present in promoter. Attachment of RNA polymerase needs sigma factor so after attaching, sigma factor will be removed from the RNA polymerase.

 

 

 

 

 

 

Steps in Transcription

 

i).     RNA polymerase will attach with the promoter with the help of sigma factor as promoter has its recognition sequence.

ii).   As the process initiates sigma factor separates from the polymerase. Now RNA polymerase will initiate the process of transcription. This phase is initiation phase. Here RNA polymerase also unwinds DNA strand.

iii). Elongation phase: It uses nucleoside triphosphates as substrate and polymerises in a template depended fashion following the rule of complementarity. It somehow also facilitates opening of the helix and continues elongation.

iv). Once the polymerases reaches the terminator region, Rho factor attaches to the polymerase due to which the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription.

 

An intriguing question is that how is the RNA polymerases able to catalyse all the three steps, which are initiation, elongation and termination. The RNA polymerase is only capable of catalysing the process of elongation. It associates transiently (quickly) with initiation-factor (σ) and termination-factor (ρ) to initiate and terminate the transcription, respectively. Association with these factors alter the specificity of the RNA polymerase to either initiate or terminate.

 

Difference between prokaryotic transcription and eukaryotic transcription

prokaryotic transcription	eukaryotic transcription
i).     Occurs in Cytoplasm ii).   Requires RNA polymerase for the formation of mRNA, rRNA and tRNA. iii). Sigma factor and Rho factor are required.	i).     Occurs in Nucleus. ii).   Requires three RNA polymerase for different RNA which are as follows: a)     RNA Polymerase I: for rRNA (28S, 18S, 2.8S). b)     RNA Polymerase II: for mRNA c)     RNA Polymerase III: for tRNA, 5srRNA (small regulatory), snRNA (small nuclear) iii). Sigma factor and Rho factor are not required.
Note: S stands for svedberg unit on the name of scientist. It is the rate of sedimentation of RNA in a centrifugal tube.

 

 In Eukaryotes

Steps of transcription in eukaryotes (Process of formation of mRNA from hnRNA)

 

i).     Both rRNA and tRNA forms directly with the help of their respective polymerase I and III. But in case of mRNA precursor of mRNA that is hn RNA (Heterogeneous nuclear RNA) is formed in the direction 5’ to 3’. mRNA is formed with RNA poly. II.

It is called heterogeneous because it consists of a mix of RNA molecules of different lengths and sequences

ii).   Before sending to cytoplasm post transcription modification is required to protect RNA (from the RNAase present in cytoplasm) which includes  Caping at 5’ , tailing at 3’ and splicing in mRNA.

 

A.    Caping

i).   In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA.

 

 

B.     Tailing

i).     In tailing, adenylate (nucleotide form of adenosine with a single phosphate group, that is adenosine monophosphate ) residues (200-300) are added at 3'-end in a template independent manner because DNA is not required here. It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation.

ii).   Here we can see exon and introns. Exons are the functional part which will make protein and Introns are non functional part which should be removed from the hnRNA.

 

C.     Splicing

i).     Splicing is done in the form of loops.

ii).   Introns cut off and exons joined with the help of enzyme RNA ligase.

 

 

 

5.7          GENETIC CODE

 

1.     During replication and transcription a nucleic acid was copied to form another nucleic acid.

2.     Hence, these processes are easy to conceptualise on the basis of complementarity.

3.     The process of translation requires transfer of genetic information from a polymer of nucleotides to synthesise a polymer of amino acids.

4.     Neither does any complementarity exist between nucleotides and amino acids, nor could any be drawn theoretically.

5.     This led to the proposition of a genetic code that could direct the sequence of amino acids during synthesis of proteins. It is the relationship between sequence of amino acids (in polypeptide) and sequence of nucleotide (in DNA/RNA).

The Central Dogma of Molecular Biology: DNA → RNA → Protein 1.     Transcription (in the nucleus): o    DNA's sequence of nucleotides (A, T, C, G) is copied into mRNA (A, U, C, G). 2.     Translation (in the cytoplasm/ribosome): o    The mRNA is read in triplets (groups of 3 nucleotides), called codons. o    Each codon codes for a specific amino acid. o    For example: §  AUG = Methionine (start codon) §  UUU = Phenylalanine §  UAA = Stop codon 3.     tRNA helps by bringing the correct amino acid to the ribosome, based on the codon. ✅ So, how is the relationship established? ·         The sequence of nucleotides in DNA/RNA determines the order of codons in mRNA. ·         Each codon specifies a particular amino acid. ·         Thus, the nucleotide sequence controls the amino acid sequence in a protein.

 

6.     It was George Gamow, a physicist, who argued that since there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases.

7.     He suggested that in order to code for all the 20 amino acids, the code should be made up of three nucleotides.

8.     This was a very bold proposition, because a permutation combination of 4³ (4 × 4 × 4) would generate 64 codons; generating many more codons than required.

We choose 3 bases, and each position can be A, U, G, or C. So: 4 choices × 4 choices × 4 choices = 64 codons These 64 combinations are like: Codon Example Combination 1 AAA 2 AAU 3 AAG 4 AAC And so on for the rest of bases

 

9.     He concluded that the genetic codes are present in triplets. For example 3 nucleotide will code for 1 amino acid. The group of three nucleotides is known as codon.

10.Providing proof that the codon was a triplet, was a more daunting task.

11.Marshall Nirenberg’s cell-free system for protein synthesis finally helped the code to be deciphered (help in understanding). In his experiment he also found that the genetic code is present in triplet. He used uridine tri phosphate (UTP) as homopolymer. In this polymer he added severo ochoa enzyme and found that codes are UUUUUUUUU (Because here only Uracil is present ) which codes for phenylalanine (Phe) which was the first code cracked.

12.The chemical method developed by Har Gobind Khorana was instrumental in synthesising RNA molecules with defined combinations of bases (homopolymers and copolymers (a kind of heteropolymers)). He used UTP and CTP with the enzyme, RNA was formed which had coding CUCUCUCUCUCU (Because here only Cytosine and Uracil is present ), here CUC codes for leucine and UCU codes for serine. Similarily they discovered all the codes and got nobel prize.

 

13.The salient features of genetic code are as follows:

(i)                The codon is triplet.

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

(iii)            UAA, UAG, UGA are stop terminator codons. These are also known as non sense codon as these do not code for any amino acids.

(iv)            61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons. There are total 64 codons.

(v)              The codon is read in mRNA in a contiguous (continuous)  fashion Like                 AUGUUUUUCUUCUUU.

(vi)            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.

Other examples:- UGA, which is a stop codon in the standard genetic code, codes for the amino acid tryptophan (Trp) in mitochondria.

AUA, which codes for isoleucine (Ile) in the standard genetic code, codes for methionine (Met) in mitochondria.

(vii)          Polarity of the sequence of amino acids is 5’ to 3’.

(viii)        Generally 1 codon codes form 1 amino acid. This property is called non ambiguous (specific). Exceptions:- GUG codes for valine but in prokaryotes codes for methionine and acts as initiator codon.

(ix)            Some amino acids are coded by more than one codon, hence the code is degenerate. For example Synonymous Codons: So degenerate means that some amino acids can be coded for by more than one codon. For example, both UUU and UUC code for phenylalanine.

(x)               If following is the sequence of nucleotides in mRNA, predict the sequence of amino acid coded by it (take help of the checkerboard): -

AUG UUU UUC UUC UUU UUU UUC.

Met- Phe- Phe-  Phe- Phe-  Phe- Phe

 

5.6.1 Mutations and Genetic Code

1.     Effects of large deletions and rearrangements in a segment of DNA are easy to comprehend (understand).

 

2.     It may result in loss or gain of a gene and so a function.

3.     The effect of point mutations will be explained here.

4.     A classical example of point mutation is a change of single base pair in the gene for beta globin chain that results in the change of amino acid residue glutamate to valine.

5.     GAG is a triplet codon for glutamic acid. Here A is replaced by U due to which codon will become GUG which codes for valine.

6.     It results into a diseased condition called as sickle cell anemia. In this disease, shape of RBC is changed to sickle shape due to which its oxygen carrying capacity will reduce.

7.     Effect of point mutations that inserts or deletes a base in structural gene can be better understood by following simple example.

 

RAM HAS RED CAP
Adding	INSERTION	Deleting	DELETION
	RAM HAS RED CAP		RAM HAS RED CAP
B	RAM HAS BRE DCA P	R	RAM HAS EDC AP
I	RAM HAS BIR EDC AP	E	RAM HAS DCA P
G	RAM HAS BIG RED CAP	D	RAM HAS CAP

 

8.     Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion. Such mutations are referred to as frameshift insertion or deletion mutations.

9.     Insertion or deletion of three or its multiple bases insert or delete in one or multiple codon hence one or multiple amino acids, and reading frame remains unaltered from that point onwards.

5.6.2 tRNA– the Adapter Molecule

1.     Two types of tRNA structure are given which are as follows:-

i).     2-D structure- Clover leaf structure

Here due to the presence of some complementary codons they get close to each other and forms a loop like structure as given below.

ii).   3-D structure – It is inverted L shape structure.

We will study 2-D structure.

 

2.     From the very beginning of the proposition of code, it was clear to Francis Crick that there has to be a mechanism to read the code and also to link it to the amino acids, because amino acids have no structural specialities to read the code uniquely.

3.     He postulated the presence of an adapter molecule (tRNA) that would on one hand read the code and on other hand would bind to specific amino acids.

4.     The tRNA, then called sRNA (soluble RNA), was known before the genetic code was postulated. However, its role as an adapter molecule (a molecule that helps bridge the gap between two different types of biological information) was assigned much later.

5.     tRNA has an anticodon loop that has bases complementary to the code, and it also has an amino acid acceptor end to which it binds to amino acids.

 

6.     In the given fig. tRNA has anticodon UCA for the AGU, here AGU codes for Ser, so by reading codon tRNA with the respective anticodon brings respective amino acid which is already attached with the respective tRNA.

Ø Aminoacyl-tRNA synthetase (an enzyme) binds the correct amino acid to its corresponding tRNA before translation starts. This forms what we call aminoacyl-tRNA or charged tRNA. Ø This charged tRNA then comes to the ribosome during translation. Ø The anticodon of the tRNA matches with the codon on the mRNA. Ø Once the codon-anticodon match is confirmed, the ribosome adds the already attached amino acid to the growing polypeptide chain.

 

7.     tRNAs are specific for each amino acid (Figure 5.12).

8.     For initiation, there is another specific tRNA that is referred to as initiator tRNA.

9.     There are no tRNAs for stop codons.

10.In figure 5.12, the secondary structure of tRNA has been depicted that looks like a clover-leaf. It is a 2-D structure.

11.In actual structure, the tRNA is a compact molecule which looks like inverted L (3-D structure).

 

5.8          TRANSLATION

1.     Translation refers to the process of polymerisation of amino acids to form a polypeptide (Figure 5.13).

2.     It can also defined as the process of synthesis of protein from mRNA with the help of ribosome.

3.     Here mRNA is act as a template. rRNA and tRNA also assists in the translation.

4.     There is a translation unit which consist of mRNA (5’ to 3’). Various regions on mRNA together constitutes translation unit.

5.     Various regions of translation unit are as follows:-

i).     mRNA (5’ to 3’)

ii).   Start codon--- AUG (methionine)

iii). Region coding for polypeptide.

iv). Stop codon—UAA, UAG, UGA. These also don’t code of any protein.

v).   UTR – Untranslated regions are those regions which don’t code for protein.

 

6.     Structure of ribosome.

 

Prokaryotes	Eukaryotes
In mRNA caping and tailing is absent.	In mRNA caping is done at 5’ and tailing at 3’.
There is 70S ribosome, larger is 50S and smaller is 30S.	There is 80S ribosome, larger 60S and smaller is 40S
Here S is Svedberg’s unit which is the rate of sedimentation of ribosome during centrifugation.

 

7.     Larger subunit has three sites.

i).     A site – for the entry of new tRNA.

ii).   P site – has tRNA containing the growing peptide chain.

iii). E site – through which a tRNA molecule after delivery of amino acid exists the ribosome.

 

8.     As we studied earlier that DNA forms mRNA in the nucleus, this mRNA moves outside in the cytoplasm, so we can say that the main site of translation is cytoplasm.

9.     There are amino acids all over in the cytoplasm and mRNA. mRNA has codes. The order and sequence of amino acids are defined by the sequence of bases in the mRNA.

10. "tRNA, which acts as an adaptor, binds to a specific amino acid in the cytoplasm and then reads the codon on the mRNA during translation."

 

11.There are three steps in the process of translation:-

i).     Initiation

ii).   Elongation

iii). Termination

 

A.    Initiation

i).     There are two steps in the initiation:-

a)     Activation of amino acid

b)     Transfer of Amino acid to tRNA.

 

a)     Activation of amino acid

i).     In this step one amino acid is attached to Enzyme aminoacyl tRNA synthetase with the help of ATP.

ii).   Two phosphates from are removed from the ATP and thus we get AA-AMP-E complex. Now we can say that the amino acid is activated by combining with AMP and enzyme.

 

b)     Transfer of Amino acid to tRNA.

i).     Here Activated amino acid formed in the above step joined with tRNA.

ii).   After this AMP and Enzyme separates from the complex.

iii). This is known as charging of tRNA. So we can say that tRNA is charged due to the attachment of amino acid.

 

B.     ELONGATION

i).     In the first step smaller subunit of ribosome binds to mRNA.

Smaller Sub Unit of ribosomes
Prokaryotes	Eukaryotes
30S	40S
rRNA is 16S rRNA	rRNA is 18S rRNA

 

1.      Eukaryotic Small Ribosomal Subunit (40S): In eukaryotic cells, such as those found in animals, plants, and fungi, the small ribosomal subunit is referred to as the "40S" subunit. It contains both ribosomal RNA (rRNA) and ribosomal proteins. 18S rRNA is present in it. The 40S subunit plays a crucial role in the initiation of translation and the decoding of mRNA. 2.      Prokaryotic Small Ribosomal Subunit (30S): In prokaryotic cells, such as bacteria, the small ribosomal subunit is known as the "30S" subunit..Like its eukaryotic counterpart, it also contains both rRNA and ribosomal proteins. 16S rRNA is present in it. The 30S subunit is responsible for the initiation of protein synthesis and helps decode mRNA.

 

ii).   In the second step initiator tRNA binds to mRNA which has anticodon of mRNA. For example here AUG is present on mRNA so tRNA which has anticodon UAC joined with AUG.

 

 

iii). In the third step, binding of larger subunit takes place. Here the P site of larger subunit is attached to initiator tRNA. Here initiator tRNA has Methionine amino acid because mRNA has AUG which codes for it.

 

iv). Here one more tRNA attached with A site of larger subunit. Due to this a peptide bond is formed between the two amino acids. This bond is formed with the help of rRNA.

 

v).   In this step amino acid present on initiator tRNA transfers from P site to A site, so we can see two amino acids are present on A site in the fig. given below.

 

vi). As soon as the amino acid transfers to A site, the Ribosomal unit moves forward in such a way that now A site is empty as shown in the fig. given below. Here new tRNA with amino acid joins the A site and two amino acids again transfers from P site to A site and ribosomal unit moves to 3’ side again. This process continues unless and until ribosomal unit doesn’t reach to the stop codon.

 

12.If two such charged tRNAs are brought close enough, the formation of peptide bond between them would be favoured energetically.

13.The presence of a catalyst would enhance the rate of peptide bond formation.

 

C.     Termination

 

https://www.youtube.com/watch?v=5bLEDd-PSTQ (Translation Animation)

 

i).     In this step stop codon stops the further addition of tRNA at the A site, thereby a releasing factor is released which attaches to the A site due to which chain of amino acids separates from tRNA and released in the cytoplasm.

ii).   Or we can say that at the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.

 

 

14.The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond.

15.The UTRs are present at both 5' -end (before start codon) and at 3' -end (after stop codon).

16.They are required for efficient translation process.

17.The ribosome proceeds to the elongation phase of protein synthesis.

18.During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon.

19.The ribosome (ribosomal unit) moves from codon to codon along the mRNA.

 

5.9          REGULATION OF GENE EXPRESSION

 

1.     Regulation of gene expression refers to a very broad term that may occur at various levels. It can be defined as a mechanism of switching on/off of certain genes depending upon the requirement of the cell/organism.

 

Regulation of gene expression in prokaryotes

1.     Regulation of gene expression in prokaryotes is done by operon.

2.     Operons:-

i).     A cluster of functionally related genes having a common promoter and operator is known as operon.

a)     Promoter: This is like a "start sign" for the genes. It is a specific spot on the DNA where the process to read and use the genes begins. Think of it like a green light at a traffic signal that tells the cars to go. b)      Operator: This is like a "switch" for the genes. It is a spot on the DNA where certain proteins can attach and either turn the genes on or off. Imagine it like a switch on a toy that can either allow the toy to work or stop it from working.

 

ii).   The genes within an operon are transcribed together into a single mRNA molecule, which is then translated to produce multiple proteins.

 

2.     The operator region is adjacent to the promoter elements in most operons and in most cases the sequences of the operator bind with a repressor protein.

A repressor protein is a type of regulatory protein that can bind to specific DNA sequences known as operator regions. By binding to these operators, the repressor can inhibit the transcription of adjacent genes.

 

3.     Each operon has its specific operator and specific repressor.

i).     For example, lac operator is present only in the lac operon and it interacts specifically with lac repressor protein only.

5.8.1 The Lac operon

1.     The elucidation (clarification) of the lac operon was also a result of a close association between a geneticist, Francois Jacob and a biochemist, Jacque Monod.

2.     They were the first to elucidate (explains) a transcriptionally regulated system.

3.     In lac operon (here lac refers to lactose), a polycistronic structural gene is regulated by a common promoter and regulatory genes.

4.     Such arrangement is very common in bacteria and is referred to as operon.

i).     In case of Lac operon, lets take the example of E.coli. When it is on potato, it takes energy from the potato as it contains carbohydrate (which contain glucose).

ii).   2^nd case, if its present in milk, we know milk has lactose so in this case it requires breakdown of lactose into glucose and galactose which required certain genes, these genes are present in lac operon.

iii). To name few such examples, lac operon, trp operon, ara operon, his operon, val operon, etc.

a)     lac operon: Lactose operon b)     trp operon: Tryptophan operon c)     ara operon: Arabinose operon d)     his operon: Histidine operon e)     val operon: Valine operon

 

 

5.     As we studied in the transcriptional unit which have three parts:-

i)                   promoter

ii)                 structural gene

iii)               terminator

 

6.     Similarly, DNA of Lac operon divided into seven segments:- (pipo zya)

P

I

P

O

Z

Y

A

 

 

 

🔍 What is the Lac Operon?

The Lac Operon is a group of genes in E. coli that work together to help the bacteria digest lactose, a sugar found in milk.

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🧬 Main Parts of the Lac Operon

1.     Structural Genes:

o    lacZ – makes β-galactosidase, which breaks lactose into glucose and galactose.

o    lacY – makes permease, which helps lactose enter the cell.

o    lacA – makes transacetylase (its role is less clear).

2.     Regulatory Elements:

o    Promoter (P) – where RNA polymerase binds to start transcription.

o    Operator (O) – a DNA region where the repressor protein binds to block transcription.

o    Regulatory Gene (lac I) – makes the repressor protein.

Ø **Lactose operon **Inhibitor gene

Ø "lac" stands for lactose, because the operon is involved in breaking down lactose.

Ø "I" stands for inhibitor, because this gene produces a repressor protein that inhibits (blocks) the operon when lactose is not present.

Ø So, lacI = lactose inhibitor gene – it controls whether the operon is ON or OFF by making the repressor.

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🧪 How it Works

🚫 When Lactose is Absent or Glucose is present:

·         The lacI gene produces a repressor protein.

·         The repressor binds to the operator.

·         This blocks RNA polymerase, so the genes aren’t transcribed.

·         ❌ No enzymes are made.

✅ When Lactose is Present:

·         Lactose (or its form allolactose) binds to the repressor.

·         This makes the repressor protein inactive so it can’t bind to the operator.

·         RNA polymerase can now move forward and transcribe the genes.

·         Enzymes are made, and lactose can be digested.

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💡 Why is it Important?

The Lac Operon is a classic model in molecular biology that explains:

·         How cells save energy by only making proteins when needed.

·         How gene expression is regulated in response to environmental changes.

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📚 Summary Table

Condition	Repressor Status	Gene Expression
Lactose absent	Active (binds operator)	OFF (no enzymes made)
Lactose present	Inactive (can’t bind)	ON (enzymes made)

 

Lets take two conditions whether lactose is present or absent:-

a)     When lactose is absent (means glucose is present in the surrounding).

 

i).     RNA polymerase binds with promoter to start transcription.

ii).   After the transcription of I gene, it forms mRNA. This mRNA is known as repressor RNA which produces repressor protein with the process of translation.

iii). This repressor protein now binds with operator at the 4^th segment.

iv). RNA polymerase although binds with promoter at the 3^rd segment yet not able to initiate the process of transcription due to the binding of repressor protein with the operator.

v).   As the transcription not started, so formation of three enzymes is inhibited (not completely). Small amount of these enzymes formed in this condition also in case small amount of lactose is present in the medium.

vi). In this way, Lac operon switched off.

 

b)     When lactose is present.

 

i).     As we know that even though lac operon is switched off, it is still form some enzymes including permease.

ii).   In this case permease allows the permeability of lactose inside bacteria. After entering into the cell, lactose converts into allolactose with the help of Beta galactosidase.

iii). As the repressor protein is formed in the first case, repressor protein is formed in this case too. Allolactose binds with the repressor protein and makes it inactive.

iv). Now the repressor protein became inactive due to which it will not bind with the operator which will not effect the transcription of structural genes and thus large amount of enzymes will formed.

v).   As the large amount of enzyme formed, so enzyme permease will allow the permeability of lactose inside the bacteria where lactose breaks down into glucose and galactose.

 

 

5.10     HUMAN GENOME PROJECT

1.     In other words, genetic make-up of an organism or an individual lies in the DNA sequences.

2.     If two individuals differ, then their DNA sequences should also be different, at least at some places. These assumptions led to the quest of finding out the complete DNA sequence of human genome. Total genetic material present per cell is called its genome.

3.     With the establishment of genetic engineering techniques where it was possible to isolate and clone any piece of DNA and availability of simple and fast techniques for determining DNA sequences, a very ambitious project of sequencing human genome was launched in the year 1990.

4.     Human Genome Project (HGP) was called a mega project.

5.     The Human Genome Project was a 13-year project coordinated by the U.S. Department of Energy and the National Institute of Health. The project was completed in 2003.

6.     You can imagine the magnitude and the requirements for the project if we simply define the aims of the project as follows:

i).     Human genome is said to have approximately 3 x 10⁹ bp, and if the cost of sequencing required is US $ 3 per bp (the estimated cost in the beginning), the total estimated cost of the project would be approximately 9 billion US dollars.

ii).   Further, if the obtained sequences were to be stored in typed form in books, and if each page of the book contained 1000 letters and each book contained 1000 pages, then 3300 such books would be required to store the information of DNA sequence from a single human cell.

7.     The enormous amount of data expected to be generated also necessitated the use of high speed computational devices for data storage and retrieval, and analysis.

8.     HGP was closely associated with the rapid development of a new area in biology called Bioinformatics.

9.     Bioinformatics is a field that combines biology and computer science to analyze and interpret biological data, such as DNA sequences, to gain insights into biological processes and make discoveries in genetics and genomics.

10.Goals of HGP Some of the important goals of HGP were as follows:

i).     Identify all the approximately 20,000-25,000 genes in human DNA;

ii).   Determine the sequences of the 3 billion chemical base pairs that make up human DNA

iii). Store this information in databases

iv). Improve tools for data analysis

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

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

11.Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

12.Besides providing clues to understanding human biology, learning about non-human organisms DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation.

13.Many non-human model organisms, such as bacteria, yeast, Caenorhabditis elegans (a free living non-pathogenic nematode, a round worm), Drosophila (the fruit fly), plants (rice and Arabidopsis), etc., have also been sequenced.

14.Methodologies : The methods involved two major approaches.

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

ii).   The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).

15.For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller sizes (recall DNA is a very long polymer, and there are technical limitations in sequencing very long pieces of DNA) and cloned in suitable host using specialised vectors.

16.The host is a living organism, such as a bacterium (e.g., E. coli), where the cloned DNA fragments are inserted and replicated in order to get large amounts of DNA.

17.A  vector is a circular DNA of bacteria, yeast etc which is used to clone DNA fragments. Vectors are DNA molecules, often derived from viruses or plasmids, that can carry and replicate DNA fragments of interest. Scientists can insert their DNA of interest into these vectors, and the vectors are then introduced into a host organism where they can replicate and produce copies of the cloned DNA fragments.

18.We cannot directly insert the fragments in host as it’s a foreign DNA to yeast or bacteria and this fragment of DNA can be killed by host.

 

19.The commonly used hosts were bacteria and yeast, and the vectors were called as BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes) depends on the host used.

20.The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger. (Remember, Sanger is also credited for developing method for determination of amino acid sequences in proteins).

21.Alignment of these sequences was humanly not possible. Therefore, specialised computer based programs were developed (Figure 5.15).

 

22.These sequences were subsequently annotated and were assigned to each chromosome.

23.The sequence of chromosome 1 was completed only in May 2006 (this was the last of the 24 human chromosomes – 22 autosomes and X and Y – to be sequenced).

5.9.1 Salient Features of Human Genome

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

1.     The human genome contains 3164.7 million bp.

2.     The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin (helps in muscle function) at 2.4 million bases.

3.     The total number of genes is estimated at 30,000–much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (99.9 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 shed light on 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 single base DNA differences (SNPs – single nucleotide polymorphism, pronounced as ‘snips’) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease-associated sequences and tracing human history.

10.The single base pair change sometimes cause disease like sickle cell anemia and some times an organism becomes resistant to specific disease.

 

5.10.2 Applications and Future Challenges

1.     Deriving meaningful knowledge from the DNA sequences will define research through the coming decades leading to our understanding of biological systems.

2.     This enormous task will require the expertise and creativity of tens of thousands of scientists from varied disciplines in both the public and private sectors worldwide.

3.     One of the greatest impacts of having the HG sequence may well be enabling a radically new approach to biological research.

4.     In the past, researchers studied one or a few genes at a time.

5.     With whole-genome sequences and new high-throughput technologies, we can approach questions systematically and on a much broader scale.

6.     They can study all the genes in a genome, for example, all the transcripts in a particular tissue or organ or tumor, or how tens of thousands of genes and proteins work together in interconnected networks to orchestrate (arrange) the chemistry of life.

 

 

5.11     DNA FINGERPRINTING

1.     DNA fingerprinting is a very quick way to compare the DNA sequences of any two individuals.

2.     It is laboratory technique/chemical test that shows the genetic make up of a person.

3.     As stated in the preceding section, 99.9 per cent of base sequence among humans is the same. 0.1 % is different from each other which make every individual unique in their phenotypic appearance.

4.     This 0.1 % of DNA have repetitive sequences. This is also known as satellite DNA.

5.     DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. For ex. GACAGACAGACA etc.

6.     These repetitive DNA are separated from bulk genomic DNA as different peaks during density gradient centrifugation.

7.     The bulk DNA forms a major peak (99.9%) and the other small peaks are referred to as satellite DNA (0.1%).

8.     Depending on base composition (A : T rich or G:C rich), length of segment, and number of repetitive units, the satellite DNA is classified into many categories, such as micro-satellites (A : T rich) , mini-satellites (More G:C) etc.

micro-satellites	mini-satellites
A : T rich	G:C rich
2-6 bp present in one repetitive unit.	10-100 bp present in one repetitive unit.

 

9.     We use mini-satellites in finger printing which is called as Variable Number of Tandem Repeats (VNTR).

10.The technique of DNA Fingerprinting was initially developed by Alec Jeffreys.

i).     He used a satellite DNA as probe that shows very high degree of polymorphism.

ii).   It was called as Variable Number of Tandem Repeats (VNTR).

iii). The technique, as used earlier, involved Southern blot hybridisation using radiolabelled VNTR as a probe.

 

11.These sequences normally do not code for any proteins, but they form a large portion of human genome.

12.These sequence show high degree of polymorphism and form the basis of DNA fingerprinting. Since DNA from every tissue (such as blood, hair-follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications.

13. What is polymorphism?

i)       As polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA fingerprinting, it is essential that we understand what DNA polymorphism means in simple terms.

ii)     Polymorphism means many different forms of a specific DNA sequence among people in the population.

Example:

Ø One person may have a DNA segment like GATAGATA,

Ø Another may have GATAGATAGATA,

Ø Another might have GATA repeated 5 times.

iii)   Polymorphism (variation at genetic level) arises due to mutations.

 

iv)   For polymorphism, there should be at least 2 variants (in the above fig. there are 3 variants). It should be equal to or more than 1% of population.

v)     Another example :

Imagine a specific position in the DNA where most people have the base A (adenine), but some people have G (guanine).

In a sample population:

97% have A

3% have G

Here:

There are two variants: A and G ✅

G is found in more than 1% of people ✅

So, this is called a polymorphism.

❌ Counter Example (Not a polymorphism):

If only 0.01% people have G and 99.99% have A, then G is too rare

This would be called a rare mutation, not a polymorphism.

📌 Summary:

A polymorphism is a common DNA variation. It must have at least two forms, and the less common form should be found in at least 1% of people. This ensures it's a normal genetic difference, not a rare disease-causing mutation.

 

vi)   New mutations may arise in an individual either in somatic cells or in the germ cells (cells that produce gametes in sexually reproducing organisms).

vii) If a germ cell mutation does not seriously impair individual’s ability to have offspring who can transmit the mutation, it can spread to the other members of population (through sexual reproduction).

viii)                       In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.

ix)   The probability of such variation to be observed in noncoding DNA sequence would be higher as mutations in these sequences may not have any immediate effect/impact in an individual’s reproductive ability.

x)     These mutations keep on accumulating generation after generation, and form one of the basis of variability/polymorphism. There is a variety of different types of polymorphisms ranging from single nucleotide change to very large scale changes.

DNA polymorphism refers to genetic variations within a population that are inherited and exist at a relatively high frequency.

 

 

 

 

 

14.Process of DNA finger printing is as follows:-

1.     Isolation of DNA.

i).     The process of DNA fingerprinting starts with isolating DNA from any part of the body such as blood, semen, vaginal fluids, hair roots, teeth, bones, etc. 

ii).   Polymerase chain reaction (PCR) is the next step in the process. In many situations, there is only a small amount of DNA available for DNA fingerprinting.  Because of this, in a test tube, DNA replication is must occur to make more DNA. The DNA and the cells will undergo DNA replication in order to make more DNA to be tested. 

2.     After the DNA is isolated and more copies of the DNA have been made, the DNA will be tested.

i).     The scientist will treat DNA with restriction enzymes (an enzymes that cuts DNA near specific recognition nucleotide sequences known as restriction sites).

ii).   This will produce different sized fragments of DNA.

3.     Gel electrophoresis is the next step in this process of DNA fingerprinting.

i).     Here the principle of separation is based on the charge and size of the DNA. As we know that DNA is negatively charged so it starts moving towards Anode (+).

ii).   During gel electrophoresis, an electrical current is applied to a gel mixture, which includes the samples of the DNA. An alkaline solution is also added to this mixture which breaks the bond between the strands of DNA and convert it into ssDNA.

 

iii). The electric current causes the DNA strands to move through the gel. Smaller the fragments farther they move.

iv). This separates the molecules of different sizes. Still these fragments are not visible.

4.     Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.

i)       In this step a synthetic membrane is put on the gel which absorbs the DNA fragments.

5.     Hybridisation using labelled VNTR probe.

i).     These are radioactive complementary codes of DNA found at the crime scene.

ii).   As we know that the above DNA is ssDNA, so when we use these radioactive probes on the synthetic membrane it combines with their respective complementary DNA.

6.     Detection of hybridised DNA fragments by autoradiography

 

i)       Once the filter is exposed to the x-ray film, the radioactive DNA sequences are shown and can be seen with the naked eye. This creates a banding pattern or what we know as DNA fingerprints. 

A schematic representation of DNA fingerprinting is shown in Figure 5.16.

               

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