Genetics 244 Summaries CHP10,11,13 PDF

lOMoARcPSD|34035292 Genetics 244 Summaries CHP10,11,13 Genetics (Universiteit Stellenbosch) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Asande Khuzwayo ([email protected]) ...

lOMoARcPSD|34035292 Genetics 244 Summaries CHP10,11,13 Genetics (Universiteit Stellenbosch) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 1 Genetics 244 Summaries Chapter 10: DNA Structure and Analysis ✓ Explain the general characteristics of genetic material ✓ Know the historical studies conducted to prove DNA as the genetic material in prokaryotes ✓ Understand and know the indirect and direct evidence to prove DNA as genetic material in eukaryotes ✓ Understand that RNA act as genetic material in many viruses ✓ Understand the chemical composition of nucleic acids ✓ Know the basic structures of DNA and RNA, and differences between them ✓ Be able to discuss the different types of analyses of nucleic acids Four General Characteristics of Genetic Materials For a molecule to serve as genetic material, it must exhibit 4 crucial characteristics: 1. Replication: to be able to replicate and reproduce through the process of Mitosis and Meiosis 2. Storage of information: molecule should act as repository of entire genome, despite the fact that at any point in time, these molecules only express part of this genetic potential (Gene Expression) 3. Expression of stored information: transcription and translation of DNA into protein forms the central dogma of molecular genetics. “DNA makes RNA, which makes protein” 4. Variation by mutation: a mutation is the change in chemical composition of DNA. Historical Timeline of the Discovery of DNA as Genetic Materials in Prokaryotes 1927- Griffith’s transformation of bacteria 1944- Avery et al 1952- Hershey-Chase Griffith’s Experiment: Griffith experimented on several strains of bacterium Diplococcus pneumoniae. Some strains were virulent (infectious strains which caused pneumonia), and some were avirulent ( non-infectious strain) Viral strains have polysaccharide capsules (making them more resistant to phagocytic cells) whereas avirulent strains do not this capsule (thus they are readily destroyed by phagocytic cells) Encapsulated, viral strains of bacteria form smooth, shiny surfaced colonies , S. Non-encapsulated, avirulent strains form rough colonies, R. In Griffith’s experiment, he injected Living IIR (avirulent) cells combined with heat-killed IIIS (virulent) cells into mice. Neither cell type would cause death in mice when injected alone. After 5 days all the mice that received both IIR and IIIS cells died. Analysis of their blood revealed a large number of living IIIS (virulent) bacteria. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 2 Griffith concluded that the heat-killed IIIS bacteria somehow converted live avirulent IIR cells into virulent IIIS cells- accrediting this phenomenon to a transforming principle. The Avery, MacLeod, and McCarty Experiment: Samples of Virulent IIIS cells were centrifuged, collected and heat killed until researchers obtained a soluble filtrate that retained the ability to induce transformation of type IIR avirulent cells Protein and polysaccharides were then removed via chloroform extraction and digesting enzymes respectively. Precipitation with ethanol yielded a fibrous mass that still retained the ability to induce transformation of type IIR avirulent cells. Finally, remaining RNA proteins were destroyed by ribonuclease (RNase). Nevertheless, transforming activity still remained. Chemical testing of the final product gave strong positive reactions for DNA. As the final test, the final sample was treated with Deoxyribonuclease (DNase). Digestion with this enzyme destroyed the transforming activity of the filtrate—thus Avery and his coworkers were certain that the active transforming principle in these experiments was DNA. Avery, MacLeod, and McCarty emphasized that, once transformation occurs, the capsular polysaccharide is produced in successive generations. Transformation is therefore heritable, and the process affects the genetic material. The Hershey-Chase Experiment: Evidence supporting DNA as the genetic material was provided during the study of the bacterium Escherichia coli and one of its infecting viruses, bacteriophage T2. Often referred to simply as a phage, the virus consists of a protein coat surrounding a core of DNA. These phage’s undergo the lytic cycle wherby phage’s inject their genetic components into a host cell. Following infection, the viral component “commandeers” the cellular machinery of the host and causes viral reproduction. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 3 H&C wanted to determine which molecular component of the phage— DNA or protein (or both)—entered the bacterial cell and directed viral reproduction H&C used radioisotopes 𝑃32 and 𝑆 35 to follow the molecular components of phages during infection. 𝑃32 labels DNA and 𝑆 35 labels protein. This is a key feature in the experiment labelled phages and unlabelled bacteria were mixed, an adsorption complex was formed as the phages attached their tail fibres to the bacterial wall. By tracing the radioisotopes, Hershey and Chase were able to demonstrate that most of the 32P-labeled DNA had been transferred into the bacterial cell following adsorption; on the other hand, almost all of the 35S- labeled protein remained outside the bacterial cell and was recovered in the phage “ghosts” (empty phage coats) after the blender treatment. the bacterial cells, which now contained viral DNA, were eventually lysed as new phages were produced. These progeny phages contained 𝑃32 , but not 𝑆 35. Hershey and Chase interpreted these results as indicating that the protein of the phage coat remains outside the host cell and is not involved in directing the production of new phages. On the other hand, and most important, phage DNA enters the host cell and directs phage reproduction. Indirect and direct evidence to prove DNA as genetic material in eukaryotes ✓ Indirect: Distribution of DNA ✓ Indirect: Mutagenesis ✓ Direct: Recombinant DNA studies Indirect Evidence: Distribution of DNA Genetic material should be found where it functions (in nucleus)- both DNA and protein fit this criterion. However: Protein also abundant in cytoplasm but DNA is not Both mitochondria and chloroplasts are known to perform genetic functions, and DNA is present inn both of these organelles Thus DNA is found only where primary genetic functions occur, whears protein is found everywhere in the cell. These observations favours the fact that DNA, instead of protein, is the genetic material In Addition, the amount of DNA and the number set of chromosomes is closely correlated. No Such consistent correlations can be observed between chromosomes and proteins. Indirect Evidence: Mutagenesis Ultraviolet (UV) light is able to induce mutations in the genetic material. The effects of ultraviolet light absorption in organisms can be measure by the number of mutations it induces. When the data is plotted, and action spectrum of UV light as a mutagenic agent is obtained. This is called the absorption spectrum Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 4 Molecules serving as the genetic material is expected to absorb at the wavelengths found to be mutagenic. UV light is most mutagenic at the wavelength of 260nm. Both DNA and RNA absorb UV light most strongly at 260nm, wheras proteins absorbs most strongly at 280nm with no observed mutagenic effects. This indirect evidence supports idea that nucleic acid rather than protein is genetic material. Direct Evidence: Recombinant DNA studies The strongest evidence is provided by molecular analysis utilizing recombinant DNA technology. In this procedure: Segments of eukaryotic DNA corresponding to specific genes are isolated and spliced into bacterial DNA. This is then inserted into a bacterial cell whose genetic expression is monitored. If a eukaryotic gene is introduced into a bacterial cell, the subsequent production of the eukaryotic protein by the bacterial cell demonstrates that the eukaryotic DNA is present and functional in the bacterial cell. RNA as Genetic Material Some viruses contain RNA core rather than DNA core, serving as the main genetic material. It was demonstrated that when purified RNA from Tobacco mosaic viruses (TMV) was spread onto tobacco leaves, the characteristic lesions caused by the viral infection subsequently appeared, thus concluding that RNA is the genetic material of the TM virus. Fraenkel-Conrat’s experiment using TMV hybrid viruses: Hybrid viruses were constructed by using the protein coat of one TMV strain to cover the RNA of another strain. Symptoms on leaves always corresponded to the strain that contributed the RNA component of the hybrid. Retroviruses: Viruses which perform reverse transcription. In this process, RNA is used as a template for the synthesis of the complementary DNA molecule under the instruction of reverser transcriptase enzyme. This DNA intermediate can be incorporated into the genome of the host cell, and when the host DNA is transcribed, copies of the original retroviral RNA chromosomes are produced. E.g. AIDS. Chemical Composition of Nucleic Acids Nucleotides: Nucleotides are the building blocks of nucleic acid monomers. These structural units consist of: ✓ Nitrogenous Base: ✓ Pentose sugar ✓ Phosphate group Two kinds of nitrogenous bases: the nine- member double-ring purines and the six-member single-ring pyrimidines. Two types of purines (Adenine and Guanine) and three types of pyrimidines (cytosine, thymine, and uracil). Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 5 Sugars found in Ribonucleic acids (RNA) contain ribose, while sugars in deoxyribonucleic acids (DNA) contain deoxyribose (Compared with ribose, deoxyribose has a hydrogen atom rather than a hydroxyl group at the C-2ʹ position). Building of Nucleotides: Bonding between nucleotides are very specific: C-1’ atom of sugar + N-1’ of pyrimidine or N-9’ of purine Phosphate group + C-2’, C-3’ or C-5’ of sugar (C-5’ bonding is the most biological prevalent) Nucleoside – nitrogen base (purine or pyrimidine) plus pentose sugar (ribose or deoxyribose). Nucleotide – nitrogen base plus pentose sugar plus a phosphate group. Depending on the amount of phosphate groups that are attached to the nucleoside (1,2 or 3 respectively), the molecule is called a nucleoside monophosphate (NMP), nucleoside diphosphate (NDP) or nucleoside triphosphate (NTP). Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 6 Structure of DNA The Watson-Crick Model Watson and Crick published their analysis of DNA structure in 1953. The model has the major features: 1. Two long polynucleotide chains are coiled around a central axis, forming a right-handed double helix. 2. The two chains are antiparallel; that is, their C-5′-to- C-3′ orientations run in opposite directions. 3. The bases of both chains are flat structures lying perpendicular to the axis; they are “stacked” on one another, 3.4 Å (0.34 nm) apart, on the inside of the double helix. 4. The nitrogenous bases of opposite chains are paired as the result of the formation of hydrogen bonds; in DNA, only A “ T and G ‚C pairs occur. (most significant feature in terms of explaining DNA’s genetic function). 5. Each complete turn of the helix is 34 Å (3.4 nm) long; thus, each turn of the helix is the length of a series of 10 base pairs 6. A larger major groove alternating with a smaller minor groove winds along the length of the molecule. 7. The double helix has a diameter of 20 Å (2.0 nm). The specific A =T and G ≡C base pairing is described as complementarity and results from the chemical affinity that produces the hydrogen bonds in each pair of bases. These Hydrogen bonds provide chemical stability necessary to hold the two chains together. Arrangement of the components in this way produces the major and minor grooves along the molecule’s length. Another stabilizing factor is the arrangement of sugars and bases along the axis. In the Watson– Crick model, the hydrophobic (“water-fearing”) nitrogenous bases are stacked almost horizontally on the interior of the axis and are thus shielded from the watery environment that surrounds the molecule within the cell. The hydrophilic (“water- loving”) sugar-phosphate backbones are on the outside of the axis, where both components may interact with water. These molecular arrangements provide significant chemical stabilization to the helix. Both right-handed and left-handed helixes exist, with the right-handed form being the most common. There are different types of DNA (from A-DNA to E-DNA, as well as Z-DNA); of these, B-DNA is the most important biological form. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 7 Differences between DNA and RNA Nucleic Acid Analytical Techniques ✓ Absorption of Ultraviolet Light ✓ Denaturation and Renaturation of Nucleic Acid ✓ Molecular Hybridization ✓ Fluorescent in situ Hybridization (FISH) ✓ Gel Electrophoresis Absorption of Ultraviolet Light Nucleic acids absorb ultraviolet (UV) light most strongly at wavelengths of 254 to 260 nm due to the interaction between UV light and the ring systems of the purines and pyrimidines. Thus, UV light can be used in the localization, isolation, and characterization of molecules that contain nitrogenous bases (i.e., nucleosides, nucleotides, and polynucleotides). Denaturation and Renaturation of Nucleic Acid Heat or other stresses can cause a complex molecule like DNA to denature; or lose its function due to the unfolding of its three-dimensional structure. During Denaturation, the Hydrogen bond of the duplex (double-stranded) structure break, the helix unwinds, and the strands separate. The viscosity of the DNA decreases, and both the UV absorption and the buoyant density increase. The increase in UV absorption of heated DNA in solution, called the hyperchromic shift Because G ‚ C base pairs have one more hydrogen bond than do A “ T pairs (see Figure 10–14), they are more stable under heat treatment. Thus, DNA with a greater proportion of G ‚ C pairs than A “ T pairs requires higher temperatures to denature completely When absorption at 260 nm is monitored and plotted against temperature during heating, a melting profile of the DNA is obtained. Midpoint of this profile, or curve, is called the melting temperature (Tm) and represents the point at which 50 percent of the strands are unwound, or denatured. When the curve plateaus at its maximum optical density, denaturation is complete, and only single strands exist. Analysis of melting profiles provides a characterization of DNA and an alternative method of estimating its base composition. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 8 Molecular Hybridization Based on renaturation (re-association) of complementary single strands of nucleic acids with nucleic acid from a different organism in order to form hybrid duplex structures. One of the most powerful analytic techniques in molecular biology. Applications include: FISH, PCR and Southern Blotting Fluorescent in situ Hybridization (FISH) In this procedure, mitotic or interphase cells are fixed to slides and subjected to hybridization conditions. Single stranded DNA or RNA is added, and hybridization is monitored. The added nucleic acid serves as a “probe,” since it will hybridize only with the specific chromosomal areas for which it is sufficiently complementary. Gel Electrophoresis Electrophoresis separates, or resolves, molecules in a mixture by causing them to migrate under the influence of an electric field. Samples are placed on a porous substance, like semisolid gels, which is in turn placed in a solution that can conduct electricity. If two molecules have approximately the same shape and mass, the one with the greatest net charge will migrate more rapidly toward the electrode of opposite polarity (DNA, RNA and proteins are negatively charged). Smaller molecules migrate at a faster rate through the gel than larger molecules. This is because the porous gel matrix restricts migration of larger molecules more than it restricts smaller molecules Once electrophoresis is complete, bands representing the variously sized molecules are currently identified by use of a fluorescent dye that binds to nucleic acids. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 9 Chapter 11: DNA replication and Recombination ✓ Be able to explain the different modes of DNA replication and know the experiments to prove these in both pro- and eukaryotes ✓ Describe DNA synthesis in microorganisms and know the enzymes involved in this process ✓ Understand and explain the model of DNA replication ✓ Understand the genetic control of replication ✓ Understand eukaryotic replication and know the differences from prokaryotes ✓ Understand and know the mechanism of DNA Recombination at the molecular level Different Modes of DNA Replication 1. Semi-conservative replication: Each replicated DNA molecule would consist of one “old” and one “new” strand 2. Conservative replication: Two newly created strands then come together and the parental strands reassociate. The original helix is thus “conserved.” 3. Dispersive replication: Parental strands are dispersed into two new double helices following replication. Hence, each strand consists of both old and new DNA. This mode would involve cleavage of the parental strands during replication. The Meselson-Stahl Experiment In 1958, Matthew Meselson and Franklin Stahl published the results of an experiment providing strong evidence that semiconservative replication is the mode used by bacterial cells to produce new DNA molecules. A “heavy” isotope of nitrogen, 𝑁 15 contains one more neutron than the naturally occurring 14N isotope; thus, molecules containing 𝑁 14 are more dense than those containing 𝑁 14. Unlike radioactive isotopes, 𝑁 15 is stable. Sedimentation equilibrium centrifugation allows 𝑁 15 to be distinguished from 𝑁 14. This method separates molecules such as DNA into bands by spinning them at high speeds in the presence of another molecule, such as cesium chloride, that forms a density gradient from the top to the bottom of the spinning tube. Density gradient centrifugation allows very small differences—like those between 𝑁 15 and 𝑁 14 DNA—to be detected. 𝑁 15 would be toward end of test tube as it is denser. 1. They began by growing E. coli in medium N15H4Cl (ammonium chloride) containing a "heavy" isotope of nitrogen, 𝑁 15. After many generations growing in the , 𝑁 15 medium, the nitrogenous bases of the bacteria's DNA were all labeled with heavy , 𝑁 15. 2. DNA isolated from cells at the start of the experiment (“generation 0,” just before the switch to 𝑁 14 medium) produced a single band after centrifugation. This result made sense because the DNA should have contained only heavy 𝑁 15 at this time. 3. Then, the bacteria were switched to medium containing a , 𝑁 14 isotope and allowed to grow for several generations. DNA made after the switch would have to be made up of 𝑁 14 as this would have been the only nitrogen available for DNA synthesis. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 10 4. DNA isolated after one generation (one round of DNA replication) also produced a single band when centrifuged. However, this band was higher, intermediate in density between the 𝑁 15 DNA and the light 𝑁 14 DNA. a. The intermediate band told Meselson and Stahl that the DNA molecules made in the first round of replication was a hybrid of light and heavy DNA. This result fit with the dispersive and semi-conservative models, but not with the conservative model. b. The conservative model would have predicted two distinct bands in this generation (a band for the heavy original molecule and a band for the light, newly made molecule). 5. Information from the second generation let Meselson and Stahl determine which of the remaining models (semi-conservative or dispersive) was actually correct. When second-generation DNA was centrifuged, it produced two bands. One was in the same position as the intermediate band from the first generation, while the second was higher (appeared to be labeled only with 𝑁 14 ). a. This result told Meselson and Stahl that the DNA was being replicated semi-conservatively. The pattern of two distinct bands—one at the position of a hybrid molecule and one at the position of a light molecule—is just what we'd expect for semi-conservative replication In contrast, in dispersive replication, all the molecules should have bits of old and new DNA, making it impossible to get a "purely light" molecule. Semiconservative Replication in Eukaryotes Taylor et al. were able to monitor the process of replication by labelling DNA with 𝐻3 -thymidine (a radioactive precursor of DNA) and performing autoradiography (a common technique that pinpoints the location of a radioisotope in a cell). After the first replication cycle in the presence of the isotope, both sister chromatids show radioactivity, indicating that each chromatid contains one new radioactive DNA strand and one old unlabelled strand. After the second replication cycle, which takes place in unlabelled medium, only one of the two sister chromatids of each chromosome should be radioactive because half of the parent strands are unlabelled. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 11 Origins, Forks, and Units of Replication Replication fork- the point along the chromosome where the strands of the helix are unwound during replication. o A fork will initially appear at the point of origin of synthesis and then move along the DNA duplex as replication proceeds. o If replication is bidirectional, two such forks will be present, migrating in opposite directions away from the origin. Replicon- the length of DNA that is replicated following one initiation event at a single origin For most prokaryotes (where only a single circular chromosome is present), only one origin of replication is found (in E. coli this specific region is called oriC) and therefore also only one replicon. Replication starts at oriC and proceeds bidirectionally with two replication forks. It terminates at a termination region (called ter in E. coli). In eukaryotes replication also proceeds bidirectionally, but from multiple origins and with various replication forks migrating along the chromosome. In the end all forks merge, resulting in a completed round of replication for the chromosome. Enzymes involved in DNA synthesis in Microorganisms DNA Polymerase I Studies of the enzymology of DNA replication were first reported by Arthur Kornberg and colleagues in 1957. They isolated an enzyme from E. coli that was able to direct DNA synthesis in a cell-free (in vitro) system. They were able to prove that DNA polymerase I is capable of synthesizing the complete DNA of a small phage in vitro, and that this DNA could be used to successfully transfect E. coli protoplasts. Kornberg determined that there were three major requirements for in vitro DNA synthesis under the direction of DNA polymerase I: 1. All four deoxyribonucleoside triphosphates (dNTPs) 2. Template DNA 3. A primer Without the above two requirements, either none or very little synthesis of DNA occurs. The way in which each nucleotide is added to the growing chain is a function of the specificity of DNA polymerase I. the precursor dNTP contains the three phosphate groups attached to the 5′-carbon of deoxyribose As the two terminal phosphates are cleaved during synthesis, the remaining phosphate attached to the 5′-carbon is covalently linked to the 3′-OH group of the deoxyribose to which it is added. Each step provides a newly exposed 3′-OH group that can participate in the next addition of a nucleotide as DNA synthesis proceeds Thus, Chain elongation occurs in the 5’ to 3’ direction by the addition of one nucleotide at a time to the growing 3’ end. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 12 DNA Polymerase II, III, IV, and V DeLucia and Cairns discovered a mutant strain of E.coli that was deficient in Polymerase I activity. This mutation was named polA1. Despite the absence of polymerase I, this mutant strain of E coli still duplicated its DNA and successfully reproduced. However, the cells were deficient in their ability to repair DNA, whereas nonmutant bacteria (which possessed) Polymerase I, are able to repair damage to DNA. These observations led to 2 conclusions: 1. At least one enzyme other than DNA Pol I is responsible for replicating DNA in vivo in E. coli cells 2. DNA Pol I serve a secondary function in vivo, and is now believed to be critical to the maintenance of fidelity of DNA synthesis. To date, five unique polymerases have been isolated (including Polymerase I): DNA polymerase III o Responsible for DNA synthesis during replication. Allows elongation. o 3'-5' exonuclease activity serves as a proofreading mechanism (so does DNA polymerase II and I) DNA polymerase I o Removes RNA primers and fills the gaps left after their removal. o Exonuclease activity serves as a proofreading mechanism here as well. DNA polymerase II, IV and V o Probably involved in repair of DNA that has been damaged by external forces (e.g. UV light). The DNA Pol III Holoenzyme The active form of DNA Pol III is referred to as the holoenzyme. It’s made up of 10 unique polypeptide subunits, each with a specific function during DNA synthesis. ✓ The largest subunit, 𝛼, along with subunits 𝜀 and 𝜃, form a complex called the core enzyme. Each holoenzyme contains 2-3 core enzyme complexes. o 𝛼 subunit is responsible for DNA synthesis along the template strands while the 𝜀 subunit possess 3’ to 5’ exonuclease capabilities essential for proofreading. ✓ A second group of five subunits are complexed to form a sliding clamp loader ✓ The sliding clamp loader and the core enzyme couple up to form the sliding DNA clamp, a critical component of the holoenzyme. o The clamp leads the way during synthesis, maintaining the binding of the core enzyme to the template during polymerization of nucleotides. o Thus, the length of the DNA that is replicated by the core enzyme before it detaches from the template ( a property referred to as processivity) is vastly increased The holoenzyme complexed with other proteins at the replication fork is called a replisome. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 13 A Model for DNA Synthesis (Prokaryotes) We have now established that replication is under the direction of DNA polymerase III; semiconservative in nature; and proceeds in two directions (physically in two directions, i.e. bidirectional – BUT proceeds in one biochemical direction only: 5' to 3’) Crucial steps for the successful replication of DNA 1. A mechanism must exist by which the helix can be unwound locally to form a replication bubble; it must then be kept in this “open” state in order for synthesis to proceed on both strands. 2. As the helix unwinds, tension develops further down the strand – this tension must be reduced. 3. A primer must be synthesized for DNA polymerase III to be able to commence with the polymerisation process – polymerase cannot initiate de novo synthesis. 4. Because DNA replication is always performed in the 5’→3’ direction, and the strands of the parental molecule are orientated antiparallel to each other, continuous synthesis of complementary DNA strands by DNA polymerase III is possible on one strand (leading strand) only. Synthesis on the other strand (lagging strand) is discontinuous. 5. Primers must be removed, and the gaps left after their removal filled up, prior to completion of replication. 6. The DNA synthesized to fill the gaps caused by primer removal, must be ligated to (joined with) their adjacent DNA strands. 7. Even though DNA polymerases are very accurate, occasional errors are made - a proofreading mechanism must exist to correct such errors. Unwinding of the DNA Helix Origin of replication in E. coli is called OriC. ✓ Consists of 245 DNA base pairs ✓ Characterized by 5 repeating sequences of 9 base pairs (9mers) and 3 repeating sequences of 13 base pairs (13mers) ✓ AT-rich thus they are relatively less stable than average double-helical sequences of DNA, enhancing helical unwinding. Stages of unwinding: A specific initiator protein (DnaA) initiates replication by binding to a region of 9mers. This newly formed complex then undergoes a slight conformation change and associates with the region of 13mers, causing the helix to destabilize, exposing single-stranded regions of DNA (ssDNA) DNA helicase then binds to the ssDNA region ad recruits the DNA pol III holoenzyme to bind to the newly formed replication fork to formally initiate replication. Once the helix has been opened, Single-stranded binding proteins (SSBs) bind to ssDNA to ensure that re-annealing (base pairing) does not occur As unwinding proceeds, tension is created ahead of the replication fork – the result is supercoiling of the helix. Tension is relieved by the action of gyrase, an enzyme that is part of a bigger family of proteins called topoisomerases Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 14 Initiation of DNA Synthesis Initiation occurs as soon as a small portion of the helix is unwound. DNA Pol III requires a primer with a free 3’-hydroxyl group in order to elongate a polynucleotide. RNA serves as the primer that initiates DNA synthesis. A short segment of RNA, complementary to DNA, is first synthesized on the DNA template under the direction of Primase (a form of RNA polymerase enzyme). Primase is recruited to the replication fork by DNA helicase, which does not require a free 3’- hydroxyl end to initiate synthesis Elongation is performed by DNA Pol III Synthesis continues until another RNA primer is encountered. DNA polymerase I (5’-3’ exonuclease activity) removes RNA primers and fills the gap with dNTP’s complementary to the parental strand. Continuous and Discontinuous DNA Synthesis Because DNA Pol III synthesizes DNA in only the 5’–3’direction, synthesis along an advancing replication fork occurs in one direction on one strand and in the opposite direction on the other- bearing in mind that the two strands of DNA in a double helix run anti-parallel to each other. Continuous DNA synthesis: Only one strand can serve as template for continuous DNA synthesis Newly synthesized DNA is called the leading strand Discontinuous DNA synthesis: As many points of initiation is necessary on the opposite DNA template, it results in discontinues DNA synthesis of the lagging strand Discontinuous DNA fragments (1000-2000 nucleotides) are called Okazaki fragments. RNA primers are part of each such fragments. At a later stage, RNA primers are removed by Pol I and replaced by nucleotides. Okazaki fragments are then joined together by DNA ligase enzyme which catalyse the formation of phosphodiester bonds, sealing the nick between the fragments. Concurrent Synthesis Leading and lagging strands are replicated simultaneously, with each strand acted upon by one of the two core enzymes which are part of the DNA Pol III holoenzyme. This is possible because when the lagging strand is spooled out to form a loop, the looping inverts the orientation of the template, but not the direction of the actual synthesis on the lagging strand (which is always in the 5’ to 3’ direction). Each parental strand can now be replicated by one of the monomers of the DNA Pol III holoenzyme. Another important feature of the holoenzyme, besides facilitating synthesis at the replication fork, is the donut shaped sliding DNA clamp (linked to the advancing core enzyme) that surrounds the unreplicated double helix. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 15 The 𝛽-subunit clamp prevents the core enzyme from dissociating from the template as polymerization proceeds. By doing so, the clamp is responsible for vastly increasing the processivity of the core enzyme by enabling more nucleotides to be continually added prior to dissociation from the template. This function is critical to the rapid in vivo rate of DNA synthesis during replication. Proofreading and Error Correction Although the action of DNA polymerases is very accurate, synthesis is not perfect and a noncomplementary nucleotide is occasionally inserted erroneously. To compensate for such inaccuracies, the DNA polymerases all possess 3’ to 5’ exonuclease activity. This property imparts the potential for them to detect and excise a mismatched nucleotide (in the 3’ to 5’ direction). Once the mismatched nucleotide is removed, 5’ to 3’ synthesis can again proceed. This process is known as proofreading This process increases the fidelity of synthesis by a factor of around 100x 𝜀 subunits of the core enzyme , of Pol III holoenzyme, is directly involved in the proofreading step. Summary of DNA replication At the advancing fork, a helicase is unwinding the double helix. Once unwound, single- stranded binding proteins associate with the strands, preventing the reformation of the helix. In advance of the replication fork, DNA gyrase functions to diminish the tension created as the helix supercoils. Each of the core enzymes of DNA Pol III holoenzyme is bound to one of the template strands by a sliding DNA clamp. Continuous synthesis occurs on the leading strand, while the lagging strand must loop around in order for simultaneous (concurrent) synthesis to occur on both strands. Not shown in the figure, but essential to replication on the lagging strand, is the action of DNA polymerase I and DNA ligase, which together replace the RNA primers with DNA and join the Okazaki fragments, respectively. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 16 DNA Synthesis in Eukaryotes Eukaryotic DNA replication shares many features with replication in bacteria. In both systems, double- stranded DNA is unwound at replication origins, replication forks are formed, and bidirectional DNA synthesis creates leading and lagging strands from single-stranded DNA templates under the direction of DNA polymerase. Eukaryotic polymerases have the same fundamental requirements for DNA synthesis as do bacterial polymerases: four deoxyribonucleoside triphosphates, a template, and a primer. However, eukaryotic DNA replication is more complex. This is because eukaryotic cells contain much more DNA, this DNA is complexed with nucleosomes, and eukaryotic chromosomes are linear rather than circular. Eukaryote vs Prokaryote Prokaryotes Eukaryotes Site of DNA replication Cytoplasm Nucleus Number of Origin replications One Many (~25000 replicons) (replicon) Number of nucleotides in the 100-200 nucleotides Approximately 150 nucleotides replicon Number of replication forks Two Many. Visible as “replication bubbles” under electron microscope. Types of DNA polymerase Leading and lagging strands Leading strand synthesized by synthesised by DNA Pol III. DNA Pol delta (𝛿) More copies of polymerases Lagging strand synthesized by found (50K vs 15) DNA Pol alpha (𝛼) Size of Okazaki fragments Large, approx. 1500 nucleotides Short, approx. 150 nucleotides Rate of DNA synthesis 25x faster than eukaryotes ~2000 nucleotides per min Initiation at Multiple Replication Origins Origins in yeast are called Autonomously replicating sequences (ARSs). They consist of approximately 120 base pairs and a consensus sequence composed of 11 base pairs. Eukaryotic replication origins not only act as sites of replication initiation, but also control the timing of DNA replication. These regulatory functions are carried out by Prereplication complexes (pre-RC)- a complex of more than 20 proteins which assemble at replication origins. Replication process: In the early G1 phase of the cell cycle, replication origins are recognized by a six-protein complex known as an Origin Recognition Complex (ORC), which tags the origin as a site of initiation of replication. Throughout the G1 phase of the cell cycle, other proteins associate with the ORC to form the pre- RC. The presence of a pre-RC at an origin “licenses” that origin for replication. Once DNA polymerases initiate synthesis at the origin, the pre-RC is disrupted and does not reassemble again until the G1 phase of the next cell cycle. o This is an important mechanism because it distinguishes segments of DNA that have completed replication from segments of unreplicated DNA, thus maintaining orderly and efficient replication Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 17 The initiation of DNA replication is also regulated at the pre-RC. o A number of cell-cycle kinases that phosphorylate replication proteins, along with helicases that unwind DNA, associate with the pre-RC and are essential for initiation. (The kinases are activated in S phase, at which time they phosphorylate other proteins that trigger the initiation of DNA replication). The end result is the unwinding of DNA at the replication forks, the stabilization of single-stranded DNA, the association of DNA polymerases with the origins, and the initiation of DNA synthesis. Eukaryotic DNA Polymerases To accommodate the increased number of replicons, eukaryotic cells contain many more DNA polymerase molecules than do bacterial cells. Pol 𝛼, 𝛿, and 𝜀 are the major forms of the enzyme involved in initiation and elongation during eukaryotic nuclear DNA synthesis. Pol 𝛼: o Two of the four subunits of the Pol α enzyme synthesize RNA primers on both the leading and lagging strands. o After the RNA primer reaches a length of about 10 ribonucleotides, another one of the four subunit adds 10–20 complementary deoxyribonucleotides. o Pol 𝛼 is said to possess low processivity, a term that refers to the strength of the association between the enzyme and its substrate, and thus the length of DNA that is synthesized before the enzyme dissociates from the template o Once the primer is in place, an event known as polymerase switching occurs, whereby Pol a dissociates from the template and is replaced by Pol 𝛿 or 𝜀. Pol δ and Pol ε o Pol δ synthesizes lagging strand. o Pol ε synthesizes leading strand. Pol γ o Found exclusively in mitochondrial DNA thus is responsible for replication of mitochondrial DNA DNA synthesis at end of Linear chromosomes Unlike the closed, circular DNA of bacteria and most bacteriophages, eukaryotic chromosomes are linear. During replication, two special problems arise at the ends of these linear double-stranded DNA molecules 1. Firstly, the double-stranded “ends” of DNA molecules at the termini of linear chromosomes potentially resemble the Double-Stranded Breaks (DSBs) that can occur when a chromosome becomes broken internally as a result of DNA damage. In such cases, either: a. The double-stranded loose ends can fuse, resulting in chromosome fusions and translocations b. The ends do not fuse and thus they become vulnerable to degradation by nucleases 2. Secondly, problems occur during DNA replication, because DNA polymerases cannot synthesize new DNA at the tips of single-stranded 5’ ends. Because no free 3'-OH group is available for the polymerase enzyme to attach a new nucleotide to, the gap cannot be filled by DNA polymerases Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 18 To deal with these two problems, linear eukaryotic chromosomes end in distinctive sequences called telomeres, which help preserve the integrity and stability of the chromosomes. Telomeres create inert chromosome ends, protecting intact eukaryotic chromosomes from improper fusion or degradation, as well as the 5’ replication problem. Thus, they solve both problems mentioned above. Telomere structure (Solves first problem) All telomeres in vertebras contain short tandem repeating TTAGGG sequences at the ends of the 3’-ending G-rich strands which repeat several thousand times in somatic cells. The 3′-ending G-rich strand extends as an overhang, lacking a complement, and thus forms a single- stranded tail at the terminus of each telomere. Though not considered complementary in the same way as A-T and G-C base pairs are, G-containing nucleotides are nevertheless capable of base pairing with one another when several are aligned opposite another G-rich sequence. Thus, the G-rich single-stranded tails are capable of looping back on themselves, forming multiple G-G hydrogen bonds to create what are referred to as G-quartets (stabilized by telomerases). These resulting loops at the chromosome ends are called called t-loops, and they essentially close off the ends of chromosomes making them resistant to nuclease digestion and DNA fusions. Replication at the Telomere (Solves second problem) It has already been established that DNA replication initiates from short RNA primers, synthesized on both leading and lagging strands. Primers are necessary because DNA polymerase requires a free 3′-OH on which to initiate synthesis. After replication is completed, these RNA primers are removed. The resulting gaps within the new daughter strands are filled by DNA polymerase and sealed by ligase. The problem arises at the gaps left at the 5′ ends of the newly synthesized DNA [gaps (b) and (c) in of diagram) These gaps cannot be filled by DNA polymerase because no free 3’-OH groups are available for the initiation of synthesis. Each gap which remains on the newly synthesized DNA strands at each successive round of synthesis causes the shortening of the double stranded ends of the chromosome by the length of the RNA primer. With each round of replication, the shortening becomes more severe in each daughter cell, eventually extending beyond the telomere and potentially deleting gene-coding regions. The solution to this so-called end-replication problem is provided by a unique eukaryotic enzyme called telomerase (a type of ribonucleoprotein). Telomerases are capable of adding several more repeats of this six-nucleotide sequence TTGGGG to the 3′end of the G-rich strand (using 5′–3′ synthesis) The telomerase RNA component (TERC) serves as both a “guide” to proper attachment of the enzyme to the telomere and a “template” for synthesis of its DNA complement The telomerase reverse transcriptase, called TERT, is the catalytic subunit of the telomerase enzyme. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 19 Part of the RNA sequence (TERC) of the enzyme (shown in green) base-pairs with the ending sequence of the single-stranded overhanging DNA, while the remainder of the RNA extends beyond the overhang. Next, reverse transcription (TERT) of this extending RNA sequence—synthesizing DNA on an RNA template—extends the length of the G-rich lagging strand. It is believed that the enzyme telomerase is then translocated toward the (newly formed) end of the strand, and the same events are repeated, continuing the extension process. Once the telomere has been lengthened by telomerase, conventional DNA synthesis ensues. Primase lays down a primer near the end of the telomere, then DNA polymerase and ligase fill most of the gap. DNA ligase repairs the phosphodiester bonds. In most eukaryotic somatic cells, telomerase is not active, and thus, with each cell division, the telomeres of each chromosome do shorten. After many divisions, the telomere may be seriously eroded, causing the cell to lose the capacity for further division. Most stem cells and malignant cells, on the other hand, maintain telomerase activity and this may contribute to their immortality DNA Recombination Models of Homologous Recombination Genetic exchange at equivalent positions along two chromosomes with substantial DNA sequence homology is referred to as general, or homologous recombination. Homologous Recombination occurs in the following steps: a) A single-stranded nick is introduced into two identical positions of two homologous DNA duplexes by an endonuclease enzyme. b) The loose ends of strands created in this manner are displaced and subsequently pair to their complementary sequences on the opposite duplex c) A ligase then joins the loose ends to their neighbouring nucleotide polymers, thus creating hybrid heteroduplex DNA molecules. d) A strand displacement results in a cross-bridge configuration (the heteroduplex DNA molecule is held together by a cross-bridge structure). e) These cross-bridges move down the chromosome by a process referred to as branch migration, resulting in an increased length of heteroduplex DNA on both homologs. f) Separation of duplexes is initiated by a 180⁰ rotation on one half of the duplex, creating an intermediate planar structure called a chi-form, or Holiday structure. g) The two homologous strands that were previously uninvolved in the process are now nicked by endonuclease h) Ligation occurs as in step 3, and two recombinant duplexes are created. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 20 Enzymes and Proteins Involved in Homologous Recombination Perhaps the most important proteins involved in homologous recombination are: ✓ RecA (in E. coli) ✓ Rad51 (in eukaryotic cells). These proteins are loaded onto the single-stranded DNA ends formed after the endonuclease nicking step. RecA and Rad51 then search for a homologous sequence of DNA in another DNA molecule and bring about strand invasion and displacement Gene Conversion Gene conversion is characterized by a nonreciprocal genetic exchange between two DNA molecules. It is thought to be a consequence of homologous recombination. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 21 Chapter 13: The Genetic Code and transcription ✓ Know the characteristics of the genetic code ✓ Describe the early studies on the mechanism and the deciphering of the genetic code ✓ Interpret the coding dictionary ✓ Understand the concept of overlapping genes ✓ Describe the process of transcription in prokaryotes and eukaryotes in detail and know the differences between the two ✓ Explain the role of introns and know the types of splicing ✓ Describe RNA editing Characteristics of the Genetic Code 1. The genetic code is written in linear form- using ribonucleotide bases, composed of mRNA molecules, as letters. The ribonucleotide sequence is derived from the complementary nucleotide bases in DNA. 2. Each “word” within the mRNA consists of three ribonucleotide letters referred to as a triplet code. Each group of triplet codes is called a codon. 3. The code is unambiguous- each triplet specifies only a single amino acid. 4. The code is degenerate, meaning that a given amino acid can be specified by more than one triplet codon. 5. The code contains one “start” and three “stop” signals that initiate and terminate translation, respectively. 6. The code is commaless- there is no internal punctuation. Once translation of mRNA begins, the codons are read one after another with no breaks between them until a terminal signal is reached. 7. The code is nonoverlapping. After translation commences, any single ribonucleotide within the mRNA is part of only one triplet. 8. The sequence of codons in a gene is colinear, with the sequence of amino acids making up the encoded protein. 9. The code is nearly universal, with only minor exception. Early studies Establishing Transcription Mechanism ✓ Triple Nature of Code ✓ Nonoverlapping Nature of Code ✓ Commaless and Degenerate Nature The Triplet Nature of the Code Experimental work done by Crick et al. provided the first solid evidence for the triplet code. These researchers induced insertion and deletion mutation in the rII locus of phage T4. Wild-type phage cause lysis and plaque formation in strain K12, but the rII mutations prevents this infection process in strain K12. Insertion or deletion of a single nucleotide causes the reading frame to shift- such mutations are called frameshift mutations. Upon translation, the amino aid sequence of the encoded protein will be altered radically starting at the point where the mutation occurred. Crick and his colleagues reasoned that a second frameshift mutation might result in a revertant phage, which would display wild-type behaviour and successfully infect E coli K12. For example, if Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 22 the original mutant contained an insertion (+), a second event causing a deletion (−) close to the insertion would restore the original reading frame They found that various combinations of one plus (+) and one minus (−) caused reversion to wild- type behaviour. When two pluses or two minuses occurred in the same sequence, the correct reading frame was not re-established. This argued against a doublet (two letter) code. However, when three pluses three minuses were present together, the original frame was re- established. These observations strongly supported the triplet nature of the code. Nonoverlapping Nature of the Code Brenner et al., TMV experiments and Crick all separately established that the code could not be overlapping for any one transcript Brenner et al. deduced that if the codes were overlapping, the compositions of tripeptide sequences within proteins should be somewhat limited. However, Brenner failed to find such restrictions in tripeptide sequences. Thus he concluded that the code is not overlapping. With an overlapping code, two adjacent amino acids would be affected by a point mutation. However, mutations in the genes coding for the protein coat of tobacco mosaic virus (TMV) revealed only single amino acid changes. Thus concluding nonoverlapping nature of code. Crick predicted that DNA would not serve as a direct template for the formation of proteins. Crick reasoned that any affinity between nucleotides and an amino acid would require hydrogen bonding. Chemically, however, such specific affinities seemed unlikely. Instead, Crick proposed that there must be an “adaptor molecule” that could covalently bind to the amino acid, yet also be capable of hydrogen bonding to a nucleotide sequence. o With an overlapping code, various adaptors would somehow have to overlap one another at nucleotide sites during translation, making the translation process overly complex, in Crick’s opinion, and possibly inefficient o Crick’s prediction was correct—transfer RNA (tRNA) serves as the adaptor in protein synthesis, and the ribosome accommodates two tRNA molecules at a time. These three arguments strongly suggest that during translation, the genetic code is nonoverlapping. This concept has been upheld without exception. The Commaless and Degenerate Nature of the Code Crick hypothesized, on the basis of genetic evidence, that the code would be commaless—that is, he believed no internal punctuation would occur along the reading frame. Crick also speculated that only 20 of the 64 possible codons would specify an amino acid and that the remaining 44 would carry no coding assignment. However, Crick’s frameshift studies suggested that the code is degenerate, thus disproved his 20/44 hypothesis. In addition, If 44 of the 64 possible codons were blank, referred to as nonsense codons, and did not specify an amino acid, at least one blank codon would very likely occur in the string of nucleotides still out of frame. If such a nonsense codon was encountered during protein synthesis, the process would probably stop or be terminated at that point. If so, the product of the rII locus would not be made, and restoration would not occur. But because the various mutant combinations were able to reproduce on E. coli K12, Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 23 Crick and his colleagues concluded that, in all likelihood, most, if not all, of the remaining 44 triplets were not blank. It followed that the genetic code was degenerate. Early studies Leading to Deciphering of the Code Nirenberg & Matthaei became the first to characterize specific coding sequences, laying a cornerstone for the complete analysis of the genetic code. Their success was dependant on the use of two experimental tools which allowed the production of synthetic mRNAs: an in vitro (cell-free in a test tube) protein- synthesizing system and the enzyme polynucleotide phosphorylase. These synthesized mRNAs served as templates for polypeptide synthesis in the cell-free system. Synthesizing Polypeptides in a Cell-Free System In the cell-free protein-synthesizing system, amino acids are incorporated into polypeptide chains. The process begins with an in vitro mixture containing all the essential factors for protein synthesis in the cell: Ribosomes, tRNAs, amino acids and other molecules essential to translation. To allow scientists to follow (or trace) the progress of protein synthesis, one or more of the amino acids must be radioactive. Finally, an mRNA must be added, to serve as the template to be translated. However in 1961 mRNA had yet to be isolated. Therefore the enzyme polynucleotide phosphorylase was used as it allowed the artificial synthesis of RNA templates which could be added to the cell-free system. In contrast to RNA polymerase, polynucleotide phosphorylase does not require a DNA template. As a result, the order in which ribonucleotides are added is random- depending on the relative concentrations of the four ribonucleotide diphosphates present in the reaction mixture. NB! The probability of the insertion of a specific ribonucleotide is proportional to the availability of that molecule relative to other available ribonucleotides. Together, the cell-free system for protein synthesis and the availability of synthetic mRNAs provided a means of deciphering the ribonucleotide composition of various codons encoding specific amino acids. Homopolymer Codes For their initial experiments, Nirenberg and Matthaei synthesized RNA homopolymers, RNA molecules containing only one type of ribonucleotide and used them for synthesizing polypeptides in vitro. In other words, the mRNA used in the cell-free protein-synthesizing system was either UUUUUU… AAAAA… etc. They tested each of these types of mRNA individually to see which, if any, amino acids were consequently incorporated into newly synthesized proteins They always made all 20 amino acids available, but for each experiment they attached a unique radioactive label to each different amino acid thus enabling them to tell when that amino acid had been incorporated into the resulting polypeptide. o UUUUUUUU= poly-phenylalanine o AAA= lysine o CCC= proline o Since Poly G molecules fold back on one another, it was not a functional template. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 24 These specific triplet codon assignments were possible only because homopolymers were used. This is because in this method, one can only conclude the composition of the template, but one cannot conclude the order of the nucleotides in each triplet, but since three identical letters have only one possible sequence, the actual codons for phenylalanine, lysine and proline could be identified. Mixed Copolymers With the initial success of these techniques, Nirenberg and Matthaei, and Ochoa and colleagues turned to the use of RNA heteropolymers. In this experiment, two or more different ribonucleoside diphosphates were used in combination to form the artificial codons. The reasoning behind this experiment was that if they knew the relative proportion of each type of ribonucleoside diphosphate in their synthetic mRNA, they could predict the frequency of each of the possible triplet codons it contained. If they then added the mRNA to the cell-free system and ascertained the percentage of each amino acid present in the resulting polypeptide, they could then analyse the results and predict the composition of the triplets that had specified those particular amino acids. Suppose that only A and C are used for synthesizing mRNA. - In a 1A:5C ratio, there is a 1/6 possibility for an A and a 5/6 chance for a C to occupy each position. On this basis, we can calculate the frequency of any given triplet appearing in the codon 1 - For AAA, the frequency is ( )3 or about 0.4 percent. 6 1 2 5 - For AAC, ACA, and CAA, the frequencies are identical— that is, ( ) ( )or about 2.3 percent for 6 6 each triplet - Together, all three 2A:1C triplets account for 6.9 percent 1 5 - In the same way, each of three 1A:2C triplets accounts for ( )( )2 or 11.6 percent (or a total of 6 6 34.8 percent) 5 - CCC is represented by ( )3 , or 57.9 percent of the triplets. 6 By examining the percentages of the different amino acids incorporated into the polypeptide synthesized under the direction of this message, we can propose probable base compositions for each of those amino acids - Because proline appears 69 percent of the time, we could propose that proline is encoded by CCC (57.9 percent) and also by one of the codons consisting of 2C:1A (11.6 percent) - Histidine, at 14 percent, is probably coded by one 2C:1A codon (11.6 percent) and one 1C:2A codon (2.3 percent). - Threonine, at 12 percent, is likely coded by only one 2C:1A codon. This experiment determines the composition of the triplet code words corresponding to the 20 amino acids, but not the specific sequence of the triplets. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 25 The Triple-Binding Assay Nirenberg et al. developed the triplet-binding assay, leading to specific assignments of triplet codons. This technique took advantage of the observation that ribosomes, when presented in vitro with an RNA sequence as short as three ribonucleotides, will bind to it and form a complex similar to what is found in vivo. The triplet RNA sequence acts like a codon in mRNA, attracting a tRNA molecule containing a complementary sequence called an anticodon. Specific triplet sequences could be synthesized in the laboratory to serve as templates. All that was needed was a method to determine which tRNA– amino acid was bound to the triplet RNA– ribosome complex. The amino acid to be tested was made radioactive and combined with its cognate tRNA, creating a “charged” tRNA. Because codon compositions (though not exact sequences) were known, it was possible to narrow the decision as to which amino acids should be tested for each specific triplet The radioactively charged tRNA, the RNA triplet, and ribosomes were incubated together on a nitrocellulose filter, which retains ribosomes.. If radioactivity is not retained on the filter, an incorrect amino acid has been tested. If radioactivity remains on the filter, it does so because the charged tRNA has bound to the RNA triplet associated with the ribosome, which itself remains on the filter. In such a case, a specific codon assignment can be made. In the end, about 50 of the 64 triplets were assigned. These specific assignments led to two major conclusions. 1. The genetic code is degenerate. 2. The genetic code is unambiguous Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 26 Repeating Copolymers Khorana was able to chemically synthesize long RNA molecules consisting of short sequences repeated many times. First he created individual short sequences (di-, tri- and tetranucleotides) , then replicated them many times and finally joined them enzymatically to form long polynucleotides A dinucleotide is converted into an mRNA with two repeating triplet codons (depending on point at which initiation occurs), trinucleotide is converted into mRNA with 3 repeating nucleotides and so on. When these synthetic mRNAs were added to a cell-free system, the predicted proportions of amino acids were found to be incorporated in the resulting polypeptides. Data from this experiment, incorporated with data drawn from mixed copolymer and triplet binding experiment, assignment was possible Suppose that the repeating trinucleotide sequence UUCUUCUUC… is used - Three possible repeating triplets, depending on point of initiation, can be read- UUC, UCU and CUU. - When placed in a cell-free translation system, three different polypeptide homopolymers— containing phenylalanine (phe), serine (ser), or leucine (leu)—are produced. Thus, we know that each of the three triplets encodes one of the three amino acids, but we do not know which codes which. On the other hand, the repeating dinucleotide sequence UCUCUCUC… is used - Two possible repeating triplets- UCU and CUC - When placed in a cell-free translation system, two different polypeptide homopolymers— containing serine (ser), or leucine (leu)—are produced. - Considering both experiments, , we can conclude that UCU, which is common to both experiments, must encode either leucine or serine but not phenylalanine. Thus, either CUU or UUC encodes leucine or serine, while the other encodes phenylalanine. To derive more specific information, we can examine the results of using the repeating tetranucleotide sequence UUAC - Four possible repeating triplets produced - UUA, UAC, ACU, and CUU. CUU is the one we are interested in - Three amino acids are incorporated by this experiment: leucine, threonine, and tyrosine - Because CUU must specify only serine or leucine, and because, of these two, only leucine appears in the resulting polypeptide, we may conclude that CUU specifies leucine Khorana also reached the conclusions that there are presence of termination codes, since neither of the codes GAUA and GUAA directed the incorporation of more than a few amino acids, too few to detect. It was later shown that UAG is the termination code. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 27 The Coding Dictionary The various techniques applied to decipher the genetic code have yielded a dictionary of 61 triplet codons assigned to amino acids. The remaining three codons are termination signals, not specifying any amino acid Degeneracy and the Wobble Hypothesis Looking at the code, it is evident that it is degenerate- almost all amino acids are specified by more than one codes. The pattern of degeneracy is also evident- sets of codons specifying the same amino acid are grouped such that the first two letters are the same with only the third differing. Crick observed this pattern in degeneracy and postulated the Wobble Hypothesis He predicted that the initial two ribonucleotides of triplet codes are often more critical than the third member in attracting the correct tRNA. He postulated that hydrogen bonding at the third position of the codon-anticodon interaction would be less spatially constrained and need not adhere as strictly to the established base-pairing rules. The wobble hypothesis proposes a more flexible set of base-pairing rules at the third position of the code. This relaxed base-pairing requirement, or “wobble”, allows the anticodon of a single form of tRNA to pair with more than one triplet in mRNA. It also buffers the severe impact of mutations on the fidelity of translation. Initiation, Termination, and Suppression In bacteria, the initial amino acid inserted into all polypeptide chains is a modified form of methionine— N-formylmethionine (fmet). The only codon coding for methionine, AUG, is called the initiator codon. ( however, when AUG appears internally in mRNA, rather than the initiating position, unformylated methionine is inserted) UAG, UAA, and UGA serve as termination codons, punctuation signals that do not code for any amino acids. They are not recognized by tRNA molecule, and translation terminates when they are encountered. Mutation that produce any of the three codons internally in a gene also result in termination. In this case, only a partial polypeptide is synthesized, since it is prematurely released from the ribosome. When such a change occurs in DNA, its called a nonsense mutation. If a second mutation suppresses a nonsense mutation its called a suppression mutation. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 28 Confirmation of Gene Code in Phage MS2 Studies MS2 is a bacteriophage that infects E. coli. When the constitutions of this gene and its encoded protein were compared, they found it exhibited collinearity with one another. Meaning the linear sequence of triplet codons formed by the nucleotides corresponds precisely with the linear sequence of amino acids in protein. Upon analysis, it was clear that the genetic code for the MS2 virus, was identical to that established in bacterial systems. Other evidence suggests that the code is also identical in eukaryotes, thus providing confirmation of what seemed to be a universal code. However, several studies have reported that the coding properties of DNA derived from mitochondrial DNA (mtDNA) undermines the hypothesis of the universality of the genetic code. Overlapping Genes Genetic code is nonoverlapping, meaning that each ribonucleotide in the code for a given polypeptide is part of only one codon. However this doe not rule out the possibility that a single mRNA may have multiple initiation points for translation- an overlapping gene. If a mRNA has multiple initiation points, it could theoretically create several different reading frames within the same mRNA, thus specifying more than one polypeptide. This concept is called over-lapping genes The gene is thus referred to as an Open Reading Frame (ORF). ORFs are defined as a DNA sequence that produces an RNA that has a start and stop codon, between which is a series of triplet codons specifying the amino acids making up a polypeptide. Advantage- overlapping reading frames optimizes the limited amount of DNA present. Disadvantage- a single mutation may affect more than one protein and thus increase the chances that the change will be deleterious or lethal. Transcription of RNA on DNA template The process by which RNA molecules are synthesized on a DNA template is called transcription. Transcription is very significant as it is the initial step of information flow within the cell. The idea that RNA is involved as an intermediate molecule in the process of information flow between DNA and protein is suggested by the following observations: 1. DNA is chromosome (nucleus) bound, however protein synthesis occurs in association with ribosomes located outside the nucleus. Thus DNA isn’t directly involved in protein synthesis. 2. RNA is synthesized in the nucleus of eukaryotic cells, in which DNA is found, and is chemically similar to DNA. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 29 3. Following its synthesis, most RNA migrates to the cytoplasm, in which protein synthesis (translation occurs) 4. The amount of RNA is generally proportional to the amount of protein in a cell. Collectively these observations suggest that genetic information, stored in DNA, is transferred to an RNA intermediate, which directs the synthesis of its proteins. Studies with Bacteria and Phages proving Existence of mRNA ✓ Volkin et al. ✓ Brenner et al.- Role of ribosomes ✓ Spiegelman et al.- RNA hybridisation Volkin: Volkin and his colleagues reported their analysis of RNA produced immediately after bacteriophage infection of E. coli Using the isotope 32P to follow newly synthesized RNA, they found that its base composition closely resembled that of the phage DNA, but was different from that of bacterial RNA This newly synthesized RNA was unstable (short lived); however, its production was shown to precede the synthesis of new phage proteins. Thus, Volkin and his co-workers considered the possibility that synthesis of RNA is a preliminary step in the process of protein synthesis. Brenner- The Role of Ribosomes: The role of the ribosome was not clear- either ribosomes were specific for the protein synthesized in association with it, or ribosomes were unspecific “workbenches” for protein synthesis and that specific genetic information rests with a mRNA. In their experiment- they labelled uninfected E coli ribosomes with heavy isotopes and then allowed phage infection to occur in the presence of radioactive precursors They were able to demonstrate that the synthesis of phage proteins (under the direction of newly synthesized RNA) occurred on bacterial ribosomes that were present prior to infection. The ribosomes appeared to be nonspecific, strengthening the case that another type of RNA serves as an intermediary in the process of protein synthesis. Spiegelman- RNA hybridisation: Spiegelman et al. isolated 𝑃32 -labelled RNA following the infection of bacteria and used it I molecular hybridization studies. They tried hybridizing this RNA to the DNA of both phages and bacteria in separate experiments The RNA hybridized only with the phage DNA, showing that it was complementary in base sequence to the viral genetic information. These experiments agree with the concept of a mRNA being made on a DNA template and then directing the synthesis of specific proteins in association with ribosomes. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 30 RNA Polymerase To prove that RNA can be synthesized on a DNA template, it was necessary to demonstrate that there is an enzyme capable of directing this synthesis. RNA polymerase was discovered- it has the same general substrate requirement as DNA Polymerase. The only two differences being that it contained ribose rather than deoxyribose sugar, and that unlike DNA polymerase, no primer was required to initiate synthesis. The overall reaction summarizing the synthesis of RNA on DNA template can be expressed as: Where nucleoside triphosphates (NTPs) serve as substrates for the enzyme which catalyses the polymerisation of nucleoside monophosphates (nucleotides) into polynucleotide chains (NMP)n by 5’-3’ phosphodiester bonds. The sequential equation summarizes the sequential addition of each ribonucleotide (NMP) to the growing polyribonucleotide chain (NMP)n+1, using a nucleoside triphosphate (NTP) as the precursor. The Active form of the RNA polymerase is called the Holoenzyme - Prokaryotes (E. Coli) RNA polymerase has 5 subunits: α,β,β’,ω,σ - Eukaryotes have 3 types: RNA Polymerase I, II and III Transcription in Prokaryotes Promoters, Template Binding, and the σ Subunit Transcription results in the synthesis of a single-stranded RNA molecule complementary to a region along only one of the two strands of the DNA double helix. The DNA strand being transcribed is the template strand The other partner strand is called the partner strand The initial step in prokaryotic gene transcription is referred to as Template Binding. - The site of initial binding is established when the RNA Polymerase σ subunit recognizes specific DNA sequences called Promoters which are located in the 5’ region, upstream from the point of initial transcription of a gene. - Once this occurs, the helix is denatured (unwound), making the template strand of the DNA accessible to the action of the enzyme. - The point at which transcription actually begins is called the Transcription Start Site. Promoter sequences are incredibly important as the interacting of promoters with RNA polymerase governs the efficiency of transcription by regulating the initiation of transcription. The following points are examples of interactions regulating the initiation of transcription: Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 31 ✓ Two types of consensus sequences (sequences which are similar in different genes of the same organism) have been found within bacterial promoters: 1. TATAAT l located in the -10 region, or Pribnow box, 10 nucleotides upstream from Transcription start site 2. TTGACA located on the -35 region, 35 nucleotide upstream from the sire of transcription ✓ The degree of RNA polymerase binding to different promoters varies greatly, causing variable gene expression. Mutations in promoter sequences may severely reduce the initiation of gene expression. In molecular genetics, then, cis-elements are those that are located on the same DNA molecule (such as all promoter sequences). In contrast, trans-acting factors are molecules that bind to these DNA elements. The σ70 form of the σ subunit tis the most important form. Each form recognizes different promoter sequences, which in turn provides specificity to the initiation of transcription. Initiation, Elongation, and Termination of RNA Synthesis Once RNA polymerase has recognized and bound to the promoter, DNA is converted from its double-stranded form to an open structure, exposing the template strand. - The enzyme then proceeds to initiate RNA synthesis. Whereby the first 5’-ribonucleoside triphosphate, which is complementary to the first nucleotide, is inserted at the start site. No primer is requires - Subsequent ribonucleotide complements are inserted and linked together by phosphodiester bonds as RNA polymerization proceeds. o This process continues in a 5’ to 3’ direction, creating a temporary 8-bp DNA/RNA duplex whose chain runs anti-parallel to one another - After these ribonucleotides have been added to the growing RNA chain, the s subunit dissociates from the holoenzyme, and chain elongation proceeds under the direction of the core enzyme. o Chain elongation in E. coli occurs at a rate of 50 nucleotides/second at 37⁰ - Like DNA Polymerase, RNA polymerase can perform Proofreading - The enzyme traverses the entire gene until eventually it encounters a specific nucleotide sequence that acts as a termination signal o Such termination sequences, about 40 bp in length, are very important in prokaryotes because of the close proximity of the end of one gene to the upstream sequences of the adjacent gene. o The termination sequence is unique in that this termination region causes the newly formed transcript to fold back on itself, forming a hairpin secondary structure, held together by H- bonds. o Termination is also dependent on the termination factor, rho (ρ) - When the transcribed RNA molecule is released from the DNA template, termination is achieved and the core polymerase enzyme dissociates Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 32 In bacteria, groups of genes whose protein products are involved in the same metabolic pathways often cluster together along the chromosome. The genes are contagious, and all but the last gene lack the termination sequence. - This results in one large mRNA, encoding more than one protein. - These genes are referred to as complementation groups and are called cistrons. The RNA is called a polycistronic mRNA The products of cistrons are usually all needed by the cell at the same time, so this is an efficient way to transcribe and subsequently translate the needed genetic information. Transcription in Eukaryotes Differences in Transcription Prokaryotes vs Eukaryotes Most of the general aspects of the mechanics of these processes are similar in eukaryotes, but there are several notable differences: 1. Transcription in eukaryotes occurs within the nucleus under the direction of three separate forms of RNA polymerase. Unlike prokaryotic transcription, in eukaryotes the RNA transcript is not free to associate with ribosomes prior to the completion of transcription. For the mRNA to be translated, it must move out of the nucleus into the cytoplasm. 2. Initiation of eukaryotic genes requires the compact chromatin fiber, characterized by nucleosome coiling, to be uncoiled and the DNA made accessible to the RNA polymerase and other regulatory proteins. This is called chromatin remodeling 3. Initiation and regulation of transcription entail a more extensive interaction between cis-acting DNA sequences and trans-acting protein factors involved in stimulating and initiating transcription. a. For example, eukaryotic RNA polymerases rely on transcription factors (TFs) to scan and bind to DNA b. Enhances and silences are also located in the regulatory region upstream from initiation point. 4. Eukaryotes’ primary RNA transcript undergo processing to produce mature mRNA. The primary transcript, called pre-mRNAs, are most often much larger than mature mRNAs. a. These pre-mRNAs are found only in the nucleus and referred to collectively as heterogenous nuclear RNA (hnRNA). Only about 25% of hnRNA molecules are converted into mRNA b. From those that are converted, many sequences are excised and the remaining segments are spliced back together. This concept is called splicing. c. The initial processing step involves the addition of a 5’ cap and a 3’ tail Initiation of Transcription in eukaryotes RNA Polymerase II is responsible for transcription of eukaryotic mRNA. It contains 2 large subunits and 10- 15 smaller subunits. The activity of RNAP-II is dependent on both cis-acting elements surrounding the gene itself and a number of trans-acting transcription factors that bind to these DNA elements. At least four cis-acting DNA elements regulate the initiation of transcription by RNAP-II 1. Core Promoter- determines where RNAP-II binds to DNA and where it begins copying the DNA into RNA 2. Proximal Promoter Elements 3. Enhancers- increase transcription levels → Each eukaryotic gene has its own unique arrangement of 4. Silencers- decreases transcription levels proximal promoter, enhancer, and silencer elements. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 33 In prokaryotes, the DNA sequence recognized by RNA polymerase is also called the promoter. However, in eukaryotes, transcriptional initiation is controlled by regulatory proteins that bind to larger number of cis-acting DNA elements. In many eukaryotic genes, a cis-acting core-promoter element is the TATA Box, which is located about 30 nucleotides upstream in the -30 region. o The sequence and function of TATA boxes are analogous to those found in the −10 promoter region of prokaryotic genes (except that RNAP binds directly to -10 region in prokaryotes) o A wide range of core-promoter and proximal-promoter elements are also found within eukaryotic gene-regulatory regions, and each can have an effect on the efficiency of transcription Complementing the cis-acting regulatory sequences are various trans-acting factors the facilitate RNAP II binding and, therefore, the initiation of transcription. These proteins are referred to as Transcription Factors. The two general TF categories: o General Transcription Factors (GTFs)- these are absolutely required for all RNAP II-mediated transcription. The general transcription factors are essential because RNAP II cannot bind directly to eukaryotic core-promoter sites and initiate transcription without their presence ▪ TFIID binds directly to TATA-box sequence and other TFs, along with RNAP II, bind sequentially to TFIID, forming an pre-initiation complex o Transcriptional activators and repressors- influence the efficiency of the rate of RNAP II transcription initiation. ▪ These bind to enhancer and silencer elements and regulate transcription initiation by aiding or preventing assembly of pre-initiation complexes and the release of RNAP II from pre-initiation into full transcription elongation. ▪ Seem to supplant the role of the σ factor in prokaryotic enzymes and are important in eukaryotic gene regulation. Processing Eukaryotic RNA RNA transcripts require significant alteration before they are transported to the cytoplasm and then translated. mRNA is initially transcribed as a precursor molecule much larger than that which is translated into a protein. Initial transcript must be processed in the nucleus before it appears as a mature mRNA molecule. Caps and tails An important posttranscriptional modification of eukaryotic RNA transcripts destined to become mRNAs occurs at the 5′ end of these molecules, where a 7-methylguanosine (7-mG) cap is added. - The cap is added even before synthesis of the initial transcript is complete and appears to be important to subsequent processing within the nucleus. - The cap stabilizes the mRNA by protecting the 5’ end of the molecule from nuclease attack - Subsequently, yt is thought to facilitate the transport of mature mRNAs across the nuclear membrane into the cytoplasm and in the initiation of translation of the mRNA into protein. Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 34 Both pre-RNAs and mRNAs contain at their 3’ end a stretch of as many as 250 adenylic acid residues. This Poly-A sequence is added after the 3’ end of the initial transcript is cleaved enzymatically at the position some 10-35 ribonucleotides from a highly conserved AAUAAA polyadenylation sequence. - Poly A has now been found at the 3’ end of almost all mRNAs studied in a variety of eukaryotes. - In the absence of this tail, these RNA transcripts are rapidly degraded. Both the 5′ cap and the 3′ poly-A tail are critical if an mRNA transcript is to be transported to the cytoplasm and translated Removal of Introns Genes of animal viruses contain internal nucleotide sequences that are not expressed in the amino acid sequence of the proteins they encode. These internal DNA sequences are represented in initial RNA transcripts, but they are removed before the mature mRNA is translated - DNA sequences that are not represented in the final mRNA product are also called introns - Those nucleotide sequences retained and expressed are called exons The presence of introns are evident in the heteroduplexes which form after molecular hybridization of purified, functionally mature mRNAs with DNA containing the genes from which the RNA was originally transcribed - Heteroduplexes form when non perfect complementation occurs - This is because introns are present in the DNA but absent in the mature mRNA and thus the introns loop out and remain unpaired. Direct comparison of nucleotide sequences of DNA with those of mRNA and their correlating amino acid sequence also illustrates the presence of introns. The first introns were found in the β-globin gene in mice and rabbits. Splicing of Introns In the process of splicing, endonucleolytic “cuts” are made at each end of an intron, the intron is removed, and the terminal ends of the adjacent exons are ligated by an enzyme Group I and II are self-splicing RNAs - The introns that are part of the primary transcript of rRNAs, require no additional components for intron excision; the intron itself is the source of the enzymatic activity necessary for removal. - RNAs capable of such catalytic activity are referred to as ribozymes Splicing of Introns which do not fall under Group I and II are mediated by a huge molecular complex called a spliceosomes Downloaded by Asande Khuzwayo ([email protected])  lOMoARcPSD|34035292 35 RNA Editing RNA editing, the nucleotide sequence of a pre-mRNA is actually changed prior to translation. As a result, the ribonucleotide sequence of the mature RNA differs from the sequence encoded in the exons of the DNA from which the RNA was transcribed. Although other variations exist, there are two main types of RNA editing: 1. Substitution editing: identities of individual nucleotide bases are altered 2. insertion/deletion editing: nucleotides are added to or subtracted from the total number of bases Downloaded by Asande Khuzwayo ([email protected])