A1.1 Notes Ch10,11,13,14 (1) Biology Notes PDF

Notes by Sarah Lind Genetics 244 – Chapter 10 – DNA Structure and Analysis Note: The highlighted bits are some details that appeared in the 2023 A1.1. However, you need context in order to understand the questions and the details. Therefore, you should not solely focus on the highlighted bits...

Notes by Sarah Lind Genetics 244 – Chapter 10 – DNA Structure and Analysis Note: The highlighted bits are some details that appeared in the 2023 A1.1. However, you need context in order to understand the questions and the details. Therefore, you should not solely focus on the highlighted bits if you want to succeed in this module. Good luck !!!! Lecture 2: DNA Structure and Analysis 1 10.1 Properties of Genetic Material There are 4 crucial characteristics that a molecule must exhibit to serve as genetic material. 1. Replication: genetic material is replicated and partitioned through the process of mitosis and meiosis 2. Storage of information: the need for the genetic material to encode the gene products as well as having the capability of storing such diverse info. and transmitting it to progeny cells 3. Expression of stored information: The central dogma of molecular genetics (DNA makes RNA, which makes proteins) 4. Variation by mutation: Mutations can be passed to future generations and can be distributed within a population. 10.2 Why were proteins initially favoured? Proteins were diverse and abundant in cells Much more was known about proteins than nucleic acid chemistry Levene’s tetranucleotide hypothesis: early experiments suggested that the 4 base molecules ACGT occur in equal ratios – later disproved by Erwin Chargaff 10.3 Evidence Favoring DNA as the Genetic Material o 1944: Initial event that led to the acceptance of DNA as the genetic material => Publication by Avery et al. The publication proved that the “transforming principle” in bacteria that is responsible for hereditary was DNA and not protein. o 1927: The Transformation Principle & Griffith’s Critical Experiment Griffith provided the foundation for Avery’s work. He performed several experiments using strains of the bacterium Streptococcus pneumoniae Strains that were virulent caused pneumonia in vertebrates  Virulent strains have a polysacc capsule à cannot be readily engulfed and destroyed by phagocytic cells in the host animal’s system Griffith used types IIR and IIIS for his critical experiments Summary of his experiments: Critical experiment à he injected the mouse with both living IIR and heat-killed IIIS He expected the mouse to survive as those cells are individually ineffective, however, the mouse died He recovered live S-strain from the dead mouse tissue He concluded that something from the S strain was passed onto the R strain, making the R-strain virulent The R-strain “transformed” into the S- strain à named this the “Transforming principle” Did not view transformation as a genetic event, but rather as a physiological modification of some sort o Transformation: The Avery et al. Experiment (1944) What molecule served as the “transforming principle?” => DNA Also used the same strains of which could be transformed from non-virulent to virulent Avery isolated different cellular components from the virulent strain à polysaccharides, proteins, RNA & DNA à and used them to transform the non- virulent strain Only the DNA from the virulent strain was capable of transforming the non- virulent strains into virulent ones 2 o 1952: The Hershey-Chase Experiment Involved the study of Escherichia coli and one of its infecting viruses, bacteriophage T2. 3  Phages were used to demonstrate that DNA is the genetic material The phage consisted of a DNA molecule, surrounded by a protein coat When the phage infects the bacteria à they inject DNA into the cell The protein coat remains on the outside of the cell The first part of the experiment: - Phages were produced in a medium containing 𝑆 !" radioactively labelled amino acids - Resulted in a phage population with 𝑆 !" labelled proteins but no radioactive label in the DNA - The phages were then allowed to infect the bacteria and injected their DNA - The radioactively labelled protein coat remained on the outside of the cell - The phages that were produced in the cell contained no radioactivity The second part of the experiment: - Phages were produced in a medium containing 𝑃!# labelled DNA - No radioactive label on the protein - After infecting the bacteria, it was found that most of the DNA had been transferred into the bacterial cell - The progeny phages contained 𝑃!# and not 𝑆 !" It was concluded that DNA enters the host cell and directs phage reproduction Key Feature of the experiment: Hershey and Chase used radioisotopes 32-P and 35-S to follow the molecular components of phages during infection. Because DNA contains phosphorus (P) but not sulfur, 32-P effectively labels DNA; because proteins contain sulfur (S) but not phosphorus, 35-S labels protein. 4 o Transfection experiments: DNA was purified from bacteriophage ϕX174 When this DNA was added to E.coli protoplasts, it resulted in the complete production of ϕX174 bacteriophages They use protoplasts because purified DNA cannot penetrate the bacterium’s cell wal Proves that the bacteriophage’s DNA alone contains all the necessary information for the production of mature viruses 10.4 Indirect Evidence that Supports that DNA is the Genetic Material in Eukaryotes o Distribution of DNA DNA found in mitochondria + chloroplasts (known to perform genetic functions) à DNA found primarily where genetic functions occur Opposed to proteins, which are found everywhere in the cell The amount of DNA determines the no. of chromosome sets (ploidy) in a cell Diploid cells have twice as much DNA as haploid cells of the same species o Mutagenesis UV light is an agent capable of inducing mutations in the genetic material The molecule serving as the genetic material will absorb at the wavelength(s) found to be mutagenic UV light is most mutagenic at 260nm and both DNA + RNA absorb UV most strongly at 260nm Proteins absorb most strongly at 280nm à no significant mutagenic effects are observed at that wavelength 5 Direct Evidence that Supports that DNA is the Genetic Material in Eukaryotes o Recombinant DNA studies It was found that injecting human DNA encoding for the 𝛽-globin gene into a fertilized mouse egg, would later be present and expressed in adult mouse tissue This gene was also transmitted to and expressed in that mouse’s progeny These mouse are examples of transgenic animals This clearly demonstrates that DNA meets the requirement of expression of genetic information in eukaryotes 10.5 RNA as Genetic Material Some viruses contain an RNA core rather than DNA RNA serves as the genetic material in these viruses Called retroviruses Their RNA serves as a template for the synthesis of the complementary DNA (Reverse Transcription) o Tobacco Mosaic Virus (TMV) - TMV is a small virus composed of a single molecule of spring-like RNA encapsulated in a cylindrical protein coat - When purified RNA from TMV was spread on tobacco leaves à the characteristic lesions caused by the viral infection appeared - o Fraenkel-Conrat experimentally proved that in the absence of DNA, RNA acts as the genetic material In one experiment protein and RNA components of the TMV were separated and both were used to infect the tobacco leaf separately In the case of protein subunits, there were no symptoms on the leaf and no progeny viruses were obtained. RNA part caused the infection and symptoms appeared on the tobacco leaf In the other experiment, two strains of TMV (type A and type B) showing different symptoms were taken Protein and RNA components were separated to form hybrid viruses à viruses were synthesized using RNA of type A and protein of type B and vice-versa It was observed that symptoms on the leaf belonged to the virus strain from which RNA was taken These experiments proved that the genetic information of TMV is stored in the RNA and not in the protein. 6  MAIN POINTS: (don’t need to worry about textbook detail too much) § Properties of DNA § Griffith’s experiment § Avery et al. experiment § Hershey-Chase experiment § Fraenkel-Conrat experiment Lecture 3: DNA Structure and Analysis 2 Chemical Composition of Nucleic Acids Even though DNA replicates à it is not alive à it’s a polymer Nucleotides = building blocks of Nucleic Acid They consist of: - A nitrogenous base - Pentose sugar - Phosphate group 2 types of nitrogen bases are found: - Purines (double-ringed) - Pyrimidines (single-ringed) Purines: (gave a diagram and we had to be able to identify them) - Adenine - Guanine Pyrimidines: - Cytosine - Thymine - Uracil 7 o Nucleoside: Nitrogenous base (purine/pyrimidine) + Pentose sugar (ribose/deoxyribose) Nucleoside = base + sugar o Nucleotide: a Nucleoside mono/di/triphosphate (NMP/NDP/NTP) Nucleotide = Nucleoside + Phosphate group o Polynucleotide: 2 nucleotides are joined together by a phosphate group linking the 2 sugars This bond = phosphodiester bond Joining of 2 nucleotides = dinucleotides Oligonucleotides = less than 20 Polynucleotides = more than 20 8  (A): - C-3 very important!!! - A new nucleotide attaches to carbon 3 via a phosphodiester bond - Even though DNA is synthesized in a 5’à3’ direction, a new nucleotide attaches to a pre-existing 3’ end (B): - Short Hand - Top à Nitrogenous bases (C1 position) - Nearly vertical lines à pentose sugar - Phosphodiester bond à diagonal line with a P in the middle that links the 3’C of one sugar and the 5’C of a neighbouring sugar. Carbon 3' is very important because it has the OH group that DNA polymerase needs for forming the phosphodiester bond between the phosphate of the "old nucleotide" and the sugar of the "new nucleotide". Base Composition Studies 1. The amount of Adenine residues is proportional to the amount of Thymine residues in DNA (the same goes for C and G). 2. Therefore (A+T) = (C+G) 3. However, the ration of (A+T)/(C+G) does not necessarily equal 1 X-Ray diffraction allowed us to visualize the structure of DNA When fibres of a DNA molecule is subjected to X-ray bombardment à the X-rays scatter in a pattern that depends on the molecule’s atomic structure. The Watson-Crick Model The model has the following major features: § 2 right-handed helical polynucleotide chains, spiralling around a central axis § The 2 chains are antiparallel => 5’à3’ ends run in opposite directions § Nitrogen bases are paired by hydrogen bonds: A+T have 2 and C+G have 3 9  § Nitrogen bases are stacked flat perpendicular to the axis § Larger major grooves alternate with smaller minor grooves Right-handed helices are the most common as opposed to left Stabilizing factors: 1. Hydrogen bonds: - Weak electrostatic attraction btwn H and an atom w an unshared electron pair (N/O) - H becomes partially positive and N/O becomes partially negative 2. The arrangement of sugars and bases along the axis: - The hydrophobic nitrogenous bases are stacked horizontally on the interior of the axis - The hydrophilic sugar-phosphate backbones are on the outside of the axis B-DNA is the most important biological form of DNA RNA Structure o How it differs from DNA: Ribose replaces deoxyribose Uracil replaces Thymine Generally a single-stranded, linear molecule 3 Major Classes of RNA: - rRNA - mRNA - tRNA Their nucleotide sequence is complementary to the DNA sequence that served as a template for their synthesis 4 Methods for Analyses of Nucleic Acids (Need to be able to understand and describe) 1) Absorption of UV light § Nucleic acids absorb UV light most strongly at wavelengths of 254-260nm § Interaction btwn UV light and the ring systems of purines and pyrimidines § This characteristic can be used to determine the conc. of nucleic acids in a particular solution 2) De- and Renaturation of Nucleic Acids § Heat/other stresses can cause DNA to denature § H-bonds break and the helix unwinds § However, no covalent bonds break § The viscosity of DNA decreases § UV absorption increases à Hyperchromic shift à easiest change to measure 10  § Bc G-C base pairs have 1 more hydrogen bond than A-T pairs à they are more stable under heat treatment § DNA w more G-C pairs require higher temps to denature completely § Tm = melting temp à 50% of strands are unwound § At max. optical density à denaturation is complete à only single stands exist § Melting profiles are GC dependent § Renaturation? à if DNA is cooled slowly, random collisions btwn complementary strands will result in their reassociation 3) Molecular Hybridization § Based on the reassociation of complementary single strands of nucleic acids § Isolated DNA strands from 2 distinct organisms (w a reasonable degree of base complementary btwn them) can form double-stranded molecular hybrids under proper conditions § Hybridization can also occur btwn single strands of DNA and RNA § If the RNA has been transcribed from the DNA à molecular hybridization can occur à creating a DNA: RNA duplex § DNA molecules are heated à causing strand separation à The strands are cooled in the presence of single-stranded RNA § Applications: FISH, PCR & Southern Blotting 4) Gel electrophoresis (asked questions about Prac 1 as well) § Separation of different size fragments of DNA in a molecular matrix under the influence of an electric field § DNA, RNA (bc of phosphate groups of the nucleotides) and proteins are negatively charged 11  § If 2 molecules have roughly the same shape and mass à the one with the greatest net charge will migrate more rapidly to the electrode of opposite polarity § Smaller molecules migrate through the gel faster than larger molecules Mistakes in the picture: T shouldn’t have 2 rings and C shouldn’t have 2 rings à same goes for A and G (purines are switched with pyrimidines) Gave us this diagram and we had to identify the purines and pyridines and had to identify 3’ and 5’ ends 12 Chapter 11: DNA Replication and Recombination Lecture 4: The Mode of DNA Replication Semiconservative replication: - After 1 round of replication à the new DNA double helix would be a hybrid that consisted of one strand of old DNA bound to one strand of newly synthesized DNA - 2nd round à Hybrids would separate and each strand would pair with a newly synthesized strand - Every subsequent round would result in fewer hybrids and more completely new double helices The Conservative and Dispersive models were also proposed Conservative: The entire dsDNA molecule serves as a template for the new dsDNA molecule Dispersive: replication of a dsDNA molecule results in 2 dsDNA molecules that are mixtures/hybrids of “old” and “new” DNA à sections of “old” and “new” DNA on the same strand The Meselson and Stahl experiment: Proof for semiconservative replication 1) Producing Generation 0 E. coli is grown in a medium with 15-Nitrogen (“heavy nitrogen” – contains one extra neutron) for several generations until all the bacteria are labelled with 15-N 13  Ultracentrifugation (separation method according to density) shows that all DNA is heavy 2) Producing Generation I The 15-N cells were then transferred into a medium containing only 14-N The cells were allowed to replicate once It was found that each replicated molecule was composed of one new 14-N strand and one old 15-N strand – the expected result for semiconservative replication 3) Producing Generation II Generation I was allowed to replicate a second time in 14-N medium The ultracentrifugation showed 2 types of DNA: - 50/50 heavy & light - 100 light - The 2 types of DNA were in the same proportion ( ½ all light and ½ mixed) 4) Producing Generation III Generation II was allowed to replicate a 3rd time in 14-N medium 2 Types of DNA were evident but not in the same proportion - One quarter = 50/50 mixed - Three quarters = 100 light 14 Question: What would you expect to see across subsequent generations of E.coli grown in 14-N? Answer: The relative proportion of DNA labelled with 15-N will decrease as predicted by the semiconservative replication model Semiconservative Replication in Eukaryotes o Taylor et.al experiment: Used the root tip (great source of dividing cells) cells of Vicia faba Sister chromatids were labelled with 3-H thymidine The replication cycle was then followed using autoradiography Step 1: Replication I and labelling all DNA in the cells Unlabeled chromosomes were allowed to replicate in 3-H-thymidine In metaphase, both sister chromatids are labelled Step 2: Replication II and a second meiotic division is allowed to occur in unlabeled thymine In metaphase – 2 things can be observed a) When there is no sister chromatid exchange (SCE) à only one chromatid is labelled b) When there is SCE à reciprocal regions of both chromatids are labelled Together, the Meselson- Stahl and the Taylor et.al experiment soon led to the general acceptance of the semiconservative mode of replication. 15  Origin of Replication o Prokaryotes: Only a single circular chromosome is present One origin of replication – oriC Only one replicon (DNA molecule that replicates from a single origin of replication) One termination site – ter Replication starts at oriC and proceeds bidirectionally with 2 replication forks (asked what is a part of prokaryotic replication à NB that it is bidirectionally with 2 replication forks) The replication fork moves along the chromosome to produce a copy (replicon) Replication fork Replication fork DNA synthesis in microorganisms DNA polymerase 1 à enzyme that has the ability to synthesize DNA in vitro 3 Major requirements for DNA synthesis in vitro - All 4 deoxyribonucleoside triphosphates - A primer - Template DNA During chain elongation (5’à3’), 2 terminal phosphates from the 5’-C of each dNTP are removed by DNA poly I The remaining phosphate then binds to the 3’-OH group of the last dNMP (mono- phosphate) in the chain which it is added 16 DNA polymerase II, III, IV & V To date, 4 other DNA poly enzymes have been discovered Must know this table!!! (asked this table) The roles of the 5 polymerases in vivo – know the names and roles of DNA polymerases in E.coli!! o DNA polymerase III à very important!! § Responsible for DNA synthesis during replication 17  § 3’-5’ exonuclease activity serves as a proofreading mechanism (removes nucleotides in the 3’à5’ direction) (Exonuclease meaning à as they are adding nucleotides, it can remove incorrect nucleotides and then continue forward “Proofreaders”) o DNA polymerase I à also important! § Removes RNA primers and fills the gaps after their removal § 5’-3’ exonuclease activity serves as a proofreading mechanism à can go BACK and remove incorrect nt and then continue forward o DNA polymerase II, IV and V § Probably involved in the repair of DNA that has been damaged by external forces (e.g., UV light) DNA Poly III Holoenzyme DNA poly III is a large and complex enzyme Primary enzyme complex involved in prokaryotic DNA replication Functions: To add nucleotides to the RNA primer & to synthesize the leading and lagging strands of DNA Comprised of: - 2x Core enzyme - Sliding clamp loader - Sliding clamp (increases the processivity of the enzyme by holding it to the DNA strand) The holoenzyme & other proteins at the replication fork are known as replisome 18  Lecture 5: A Model for DNA Synthesis DNA replication proceeds in 2 directions physically (bidirectional) – BUT proceeds in one biochemical direction only (5’à3’) For this process to occur, some critical steps need to happen: 1. Single-stranded binding proteins bind to each strand to prevent H-bonds from reforming 2. Gyrase (enzyme belonging to DNA topoisomerases) relieves tension that develops as the helix unwinds 3. Primase must provide the necessary RNA fragments that provide the hydroxyl group so that DNA pol III can commence the polymerization process 4. Continuous synthesis of complementary DNA is possible on the leading strand (5’-3’) and synthesis of the lagging strand (3’-5’) is discontinuous 5. Primers must be removed and gaps must be filled prior to the 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. A proofreading mechanism must exist to correct errors. Leading strand = TOWARDS the replication fork Lagging strand = AWAY from replication fork à requires a lot more primers bc it is working away from fork and so it needs to catch up DNA polymerase is one enzyme that clamps 2 separate strands (not 2 separate entities) 19 Unwinding the Helix A protein, DnaA, initiates the unwinding of the helix by first binding 9mer and then 13mer regions DNA helicase then binds the ssDNA region (unwinds from 5’-3’) and recruits pol III holoenzymes to bind to the newly formed replication fork Ss binding proteins bind to ssDNA to prevent re-annealing As unwinding proceeds, tension is created ahead of the replication fork – gyrase relieves tension Initiation of DNA synthesis Initiation occurs as soon as a small portion of the helix is unwound DNA poly III requires the free 3’ end of a primer (3’ OH) in order to start elongation à but when the 2 strands are separated there is no 3’ OH ∴ RNA primers are synthesized on the DNA parental strand by primase à they provide the 3’-OH by DNA polymerase III Elongation is performed by DNA poly III DNA poly I (5’-3’ exonuclease activity) removes RNA primers and fills the gaps with dNTPs complementary to the template strand Ligase seals the strands à restores the phosphodiester bonds in the sugar- phosphate backbone Continuous and Discontinuous Synthesis Leading strand = continuous strand synthesis Various initiation points are needed on the other strand (lagging strand) – DNA synthesis is, therefore, discontinuous on this strand. 20  Discontinuous DNA fragments are called Okazaki fragments à are joined together at a later stage by an enzyme called ligase How does DNA polymerase III manage to concurrently synthesize DNA on both the leading and the lagging strands? Concurrent synthesis at the replication fork is achieved by forming a loop in the lagging strand. The orientation of the lagging strand is thus inverted - the biochemical direction is unaffected, however. β-subunit clamp keeps the core enzyme in position. Proofreading Noncomplementary nucleotides are sometimes incorporated during DNA synthesis To compensate for this, both DNA polymerase I and III have 3'-5' exonuclease activity à can detect and correct mismatched nucleotides 21  In the case of the DNA polymerase III holoenzyme, the proofreading function is attributed to the epsilon (ε) subunit. Lecture 6: Question: Why are primers made from RNA instead of DNA Answer: DNA poly cannot add nucleotides to ssDNA à it can only elongate from the 3’OH of an existing nucleotide Primase is activated by helicase and binds ssDNA – generating a small RNA complementary to the template that provides the 3’-OH needed for elongation by DNA poly 22 What happens at the replication bubble of E.coli? 23 24 DNA Synthesis in Eukaryotes Needs the same requirements as prokaryotic synthesis: dNPTs, template and primer But is more complex - More DNA - Complexed in nucleosomes - Linear chromosomes Prior, to replication à DNA dissociates from its histones but reassociates after it is completed During replication à histone count doubles ∴ DNA synthesis and histone production are coupled events Þ Nucleosome: DNA wound around eight histone proteins Þ Heterochromatin = densely packed à transcription cannot occur Þ Euchromatin = loosely packed à Transcription can occur Prokaryotic DNA replication Eukaryotic DNA replication Single origin of replication Multiple origins of replication Single expanding replication loop Multiple replication loopsà accelerates the speed of a complete replication process Single termination site on the other side of Termination involves merging of 2 Ori replication forks and replication of chromosome ends known as telomeres 25  DNA poly III is the major DNA polymerase DNA polymerase alpha, delta and epsilon as involved the major polymerization enzymes Proceeds at a much faster rate than Eukaryotes Ozaki fragments are a lot longer than in eukaryotes Eukaryotic DNA polymerase Pol α, pol δ, and pol ε are the major forms involved in initiation and elongation Pol α: - Subunit 1 synthesizes RNA primer - Subunit 2 adds 10-20 nucleotides - It then dissociates from the template and is replaced by pol δ and pol ε Pol δ synthesizes lagging strand Pol ε synthesizes leading strand Pol γ = replication of mitochondrial DNA NB table!!! (need to know name and function) Translesion synthesis: DNA synthesis through regions of the template that contain damage/distortions Replication in Eukaryotes (v similar to prokaryotes) 1. A helicase using the energy from ATP hydrolysis opens up the DNA helix. 2. Replication forks are formed at each replication origin as the DNA unwinds. 3. The opening of the double helix causes over-winding, or supercoiling, in the DNA ahead of the replication fork. These are resolved with the action of topoisomerases. 26  4. Primers are formed by the enzyme primase, and using the primer, DNA pol can start synthesis. 5. While the leading strand is continuously synthesized by the enzyme pol δ, the lagging strand is synthesized by pol ε. A sliding clamp protein holds the DNA pol in place so that it does not slide off the DNA. 6. RNase H removes the RNA primer, which is then replaced with DNA nucleotides. 7. The Okazaki fragments in the lagging strand are joined together after the replacement of the RNA primers with DNA. 8. The gaps that remain are sealed by DNA ligase, which forms the phosphodiester bond. DNA Synthesis at the End of Linear Chromosomes (Telomerase) What does telomeres do? à It creates inert chromosome ends that protect from improper fusion or degradation of the chromosome Unlike prokaryotic chromosomes, eukaryotic chromosomes are linear. On the lagging strand, DNA is synthesized in short stretches, each of which is initiated by a separate primer. When the replication fork reaches the end of the linear chromosome, there is no place for a primer to be made for the DNA fragment to be copied at the end of the chromosome. These ends thus remain unpaired, and over time these ends may get progressively shorter as cells continue to divide. The ends of the linear chromosomes are known as telomeres, which have repetitive sequences that code for no particular gene (G-rich) The telomerase enzyme contains a catalytic part and a built-in RNA template. It attaches to the end of the chromosome, and complementary bases to the RNA template are added on the 3ʹ end of the DNA strand. Once the 3ʹ ends of the lagging strand template is sufficiently elongated, DNA polymerase α can add the nucleotides complementary to the ends of the chromosomes. DNA ligase repairs the phosphodiester bonds Cell Aging Cells that undergo cell division continue to have their telomeres shortened because most somatic cells do not make telomerase. This means that telomere shortening is associated with ageing. Þ TERT gene, which encodes telomerase reverse transcriptase, an enzyme essential for the maintenance of telomeres during DNA replication Þ 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 27 DNA Recombination Reciprocal exchange of homologous chromosomes a) Two homologous dsDNA strands b) Endonuclease nicking à Single-stranded nick in identical positions by an endonuclease enzyme c) Strand displacement à the loose ends are displaced and pair to their complementary sequences on the opp duplex d) Ligation à a ligase joins the loose ends to their neighboring nucleotide primers to form hybrid heteroduplex DNA molecules 28  Steps a-d e) Strand displacement results in a cross-bridged configuration f) These cross-bridges move down the chromosome in a process called branch migration to create increased lengths of heteroduplex DNA on both homologs - When migration ceases à duplexes bend 29  Steps e & f g) Duplex separation àone half of duplex rotates 180 degrees to form an intermediate planar structure (chi-form/Holliday structure) h) Endonuclease nicking à 2 homologous strands that were previously uninvolved in the process are nicked by endonuclease 30  i) Ligation à completes the process of recombinant duplex formation Steps g-i The most important proteins in homologous recombination are RecA (in E. coli) and Rad51 (in eukaryotes) These proteins are loaded onto ssDNA ends after endonuclease nicking and then search for a homologous DNA sequence in another molecule Many other enzymes regulate the actions of RecA and Rad51 Gene conversion – non-reciprocal DNA exchange 31 Chapter 13 – The Genetic Code and Transcription Lecture 7: 32 Characteristics of the Genetic Code (Need to memorize) Linear, ribonucleotide bases -RNA 3 Ribonucleotides – triplet à codon Unambiguous – 1 codon corresponds to one amino acid Degenerate/redundant – 18 AAs out of 20 have more than one codon Start codon = AUG Stop codons = UAA, UAG, UGA (need to know stop + start codons) Non-overlapping – any single base is part of only one triplet Colinear – AA sequence makes up the protein Universal – used by almost all viruses, prokaryotes, archaea and eukaryotes Never have a case where a codon will translate into more than 1 AA But more than 1 codon can translate into the same AA 4x4x4 = 64 3-letter words BUT 20 AA => redundancy DNA Encodes Proteins Directly? The First Hypothesis Researchers initially thought that DNA itself directly encodes protein during their synthesis Ribosomes discovered à thought that info in DNA was transferred to RNA of the ribosome But evidence showed the presence of an unstable intermediate à mRNA And rRNA was extremely stable à concluded that mRNA was bringing info to the ribosome 33 o Triplet Nature of the Genetic Code § Experimental evidence supporting the triplet nature à Using phage T4 and proflavin, they studied frameshift mutations, which result from the addition or deletion (indel) of one or more nucleotides within a gene and subsequently the mRNA transcribed from it. o Frameshift Mutations: first proof of triplet nature of code § Deletion of a triplet à DNA can still replicate but will be “missing something” § Deletion of 1 or 2 nucleotides à whole message destroyed § If a triplet is inserted after a whole triplet à polypeptide will contain an extra AA à NOT a frameshift mutation Need to identify if a scenario will be a frameshift mutation or not 34 Lecture 8: Characterizing Specific Codons o Nirenberg & Matthaei Experiment The use of in vitro and polynucleotide phosphorylase Polynucleotide phosphorylase (isolated from bacteria) allowed them to make synthetic mRNA which they could use as templates for polypeptide synthesis in a cell-free system Polynucleotide phosphorylase degrades RNA but can be “forced” to make RNA if large amounts of rNDPs are present Does not require a DNA template Adds rNDPs in a random order à depends on the relative conc of each rNDPs à VERY IMPORTANT 35  In vitro protein synthesis à need a cell free extract Obtain this extract by lysing the cell In order to track protein synthesis à need one or more radioactively labelled AA The probability of the insertion of a specific ribonucleotide is proportional to the availability of that molecule relative to other available ribonucleotides à CRITICAL point in understanding N&M’s experiment N&M first synthesized RNA homopolymers à i.e they only supplied one type of nucleotide They used all 4 nucleotides in 4 separate rxns They repeated each experiment 20 times to make sure all 20 AA were used by using 1 diff radio labelled AA each time They tested each mRNA and looked at the AA that was incorporated into the proteins UUUUUU… + 14-C labelled AA à labelled homopolymer polyphenylalanine UUU = phenylalanine à first code cracked AAA = lysine CCC= proline GGG = Did not work à the mRNA folds back onto itself (seen something similar in telomeres à has G quartets) Translation Overview 36  Protein checked for radio activity Next phase in their experiments Used mathematical predictions for mixed copolymers (codons containing more than one nucleotide) What is the order of the bases? By examining the % of the diff AA in proteins made from synthetic mRNA If we know the relative proportion off each base in the synthetic DNA à can predict the frequency of each possible codon it contains and compare it to the % observed in protein Example: - A:C = 1:5 à 1/6 prob of A and 5/6 prob for C $ - AAA = (%)! = 0.5% 37  They translated the mRNA and then look at the % of the various AA in that protein They found that PROLINE occurred 69% of the time à ∴no. of codons must add up to 69% CCC was already known to code for proline à57.9% Missing 11.5% à concluded that it is also made up from 2C:1A NB to know that the order of the bases was still unknown Must know how to calculate % for Assessments à They could even throw in another nucleotide or switch up the ratios (asked in previous years) Triplet Binding Assay Nirenberg & Leder Helped them to determine the order of bases in each codon Ribosomes in vitro will bind to an RNA sequence as short as 3 nucleotides à exact length of a codon This complex is then able to attract the complementary tRNA tRNA contains an amino acid at one end and an anticodon on the other end Could create a complex that would attract the right AA to the particular codon 38  The AA they wanted to test was radio labelled with 14-C and then combined with its cognate tRNA to create a charged tRNA molecule Codons of known seq could be synthesized to act as mRNA If radioactivity is retained on the filter à then the charged tRNA had bound to the codon that was in the ribosome Allowed them to assign specific codon seq 50/64 triplets assigned to a specific AA Also confirmed that the code is degenerate (more than 1 codon for AA) and unambiguous (a specific codon cannot translate into more than 1 AA) Repeating Copolymers Khorana 1. mRNA polymers of repeats were synthesised 2. Synthetic mRNAs were added to a cell-free system 3. Predicted prop. of AA were found to be incorporated in the resulting polypeptides 4. Data was combined with copolymer and triplet binding experiments 5. Specific codon assignments were made Some codons did not result in protein synthesis à proposed the existence of stop codons 39 The Coding Dictionary 61 codons = AA 3 = termination signals (UAA, UGA, UAG) à DO NOT ENCODE FOR ANY AA Met = initiation signal Met & Trp = only AA that have 1 codon each Ordered Nature and the Wobble Hypothesis o “Ordered” degeneracy Codons from a class of AA often share a middle base - U/C = hydrophobic Determines function and conformation of - G/C = hydrophilic the protein Benefit à buffers the effect of mutations as the change in the codon may result in the exchange of another chemically similar AA à minimizes the effect of the mutation on the protein structure o Wobble Hypothesis Often only the 3rd base differs in the set of codons encoding an AA ∴ The first 2 bases are the most important in attracting tRNA More flexible base pairing at position 3 Reduces the no. of diff forms of tRNA present in the cell to cover all 20 AA à economy measure Nonsense Mutation If a mutation changes any of the other codons in the mRNA molecule into a stop codon à protein synthesis will be terminated early Resulting in only a partial polypeptide being made Overlapping Genes vs. Overlapping Code The genetic code is non-overlapping à each nucleotide belongs to only 1 codon BUT multiple initiation pts may exist on an mRNA à overlapping open reading frames 40  Initiation of 2 diff AUG out of frame with each other will generate diff AA sequences Pro: can pack a lot of genes into small genomes Con: single mutation can affect multiple peptides Lecture 9: Asked this exact question 41 Expression of Genetic Information Transcription = process of making RNA from a DNA template RNA synthesis occurs in the nucleus Protein synthesis occurs in the cytoplasm à ∴ RNA migrates to the cytoplasm Proportionally equal quantities of RNA and protein RNA Polymerase Weiss Found an enzyme that was capable of synthesising RNA from a DNA template RNA poly has similar substrate requirements as DNA poly HOWEVER - Nucleotides contained ribose sugar instead of deoxyribose - No primer is required to initiate synthesis It polymerizes in 5’à3’ Prokaryotic RNA Poly Eukaryotic RNA Poly Core Enzyme Subunits 5 subunits 3 types of RNA Poly: 𝛼 & , 𝛼 && , 𝛽, 𝛽 & , 𝜔 (alpha, RNA poly I, RNA poly II beta and omega) and RNA poly III Each has more subunits than prokaryotic RNA polymerase Holoenzyme 𝜎 Sigma factor is an - additional subunit of the holoenzyme It regulates initiation Diff sigma factors allow for recognition of diff promoters 42 Components Involved in the Transcription Process DNA Transcription factors RNA Poly ATP o Promoter § is a cis element = next to, on the same DNA strand § Consensus Seq = sequence that is conserved across a lot of species § In Prokaryote - -10 = Pribnow box (functions of these) à sequences recognises/signals the sigma factor - -35 = -35 Region § -10/-35 = Genetic co-ordinates à how far upstream they are from the transcription site 43 Transcription Process in Prokaryotes A bacterial transcription unit contains 3 critical regions - The promoter - RNA coding region - The terminator (hairpin loop and/or Rho protein) 44 45  Hairpin Loop à NB concept - Secondary structure that forms spontaneously by folding back onto itself - ssRNA forms H-bonds w complementary base pairs Polycistronic vs Monocistronic o Polycistronic One mRNA with multiple translation start sites mRNA encodes for several proteins In prokaryotes o Monocistronic One mRNA with one translation start site mRNA encodes for one protein In eukaryotes (but w many exceptions) 46 Transcription Process in Eukaryotes Set up: - Strand of DNA - Transcription unit (largest region the DNA strand) - TATA Box (Cis Element) - An enhancer region may also be involved 1) Initiation TFIID (TF2D) with a TBP component binds to the DNA using the TATA box to position TFIID near the transcription site TFIIA and TFIIB then attach to the complex These transcription factors help RNA Poly to bind Energy must be added to the system in order for transcription to begin Each of the types of RNA polymerase (I, II, III) recognizes a different promoter sequence and requires different transcription factors. 2) Elongation RNA Poly is released from the transcription factors and elongation proceeds in the 5’à3’ direction RNA Poly synthesizes an RNA template from the strand of DNA RNA polymerase II transcribes the major share of eukaryotic genes 3) Termination (don’t need to know much detail on termination compared to prokaryotes) Unlike bacteria, there is no specific seq that signals for the termination of transcription Due to enzyme instability, the transcript in enzymatically cleaved 10-35 bases further downstream in the 3’ direction Forms the pre-mature transcript § Transcription factors = trans elements § Enhancers and silencers = cis elements § RNA Poly I and RNA Poly III à involved in rRNA and tRNA transcription Must know this table !! 47 Processing of Eukaryotic RNA The development of the mature monocistronic eukaryotic transcript involves several different processing steps: o 5’ Capping 7-methylguanosine (7mG) cap added prior to the completion of transcription Protection of the 5’ end against the nucleases enzyme It also facilitates the transport of mature mRNA from the nucleus into the cytoplasm o Addition of Poly-A tail Found at the 3’ end Up to 250 A’s added by poly-A-polymerase Makes RNA molecule more stable Prevents degradation o Intron Splicing Some DA seq are not expressed in the mRNA or AA seq that they encode Specific sites for the excision of introns Removed by splicing Intron à internal seq that is not expresse Exon à internal seq that is expressed Alternative splicing à 1 gene can introduce diff proteins à source of variation 48 49 Chapter 14 – Protein and Translation Lecture 10: Overview Translation = the biological polymerisation of AA into polypeptide chains Process requires: - AA - mRNA - ribosomes - means of directing he aa into the correct position (tRNA) Ribosomes Macromolecules Abundant in cells Composed of a large and a small subunit Monosome = large and small subunits assembled Redundancy in rRNA genes à multiple gene copies Polycistronic transcription of rRNA à in Prokaryotes AND Eukaryotes *S= Svedberg coefficient – associated w density, mass and shape of the macromolecule (“S” ARE NOT ADDITIVE) Ribosomal Units in Prokaryotes and in Eukaryotes (note how the S units do not add up) - Must know the names of the subunits !! Prokaryotes Eukaryotes Overall 70S (50S + 30S) 80S (60S + 40S) Large Subunit 23S + 5S 28S + 5.8S + 5S Small Subunit 16S 18S tRNA Stable Practically the same in prokaryotes and eukaryotes Transcribed as a larger precursor à spliced into 4S tRNA molecules Undergoes post-transcriptional modification to produce unique bases These unique bases (inosinic acid, ribothymidylic acid pseudouridylic acid) are important for stability and mRNA-tRNA binding 50  tRNAs are never translated à they are what they are after transcription Has 3 hairpin loops à sequence has stretches that are complementary which allows H-bonds to spontaneously form 3D structure of tRNA formed from single-strands o 2D Structure of tRNA 2D cloverleaf model 4 large and 1 small stem Anticodon 3’-5’ Acceptor stem with AA binding site Lengths of stems and loops are very conserved; the anticodon loop is always found in same location on the cloverleaf structure AA attaches to CCA at the 3’end (don’t read it as ACC bc we read it in the 5’-3’ direction) Need to be able to label a tRNA diagram (acceptor arm, anticodon etc) Must be able to identify which codon a tRNA anticodon will code for Must know which aa it will code for as well o tRNA 3D structure L-shaped Anticodon loop and amino acid binding sites on opposite legs of L 51 o Charging tRNAs (aminoacylation) Adding the AA to the 3’ AA binding site tRNA in a free or charged conformation Aminoacyl tRNA synthases catalyze the formation of charged tRNA à a covalent bond is formed btwn the AA and the 3’ end of the tRNA = charged tRNA molecule Isoaccepting tRNAs = tRNAs that have diff anticodons but still carry the same AA 52  Lecture 11: Translation in Prokaryotes o Components: Large and small ribosomal subunits mRNA charged tRNAs GTP (not ATP) Initiation factors - IF1,2,3 Elongation factors - EF-Tu & EF-G Release Factors - RF1,2,3 (need to know these factors !!! especially if mentioned further) 1) Initiation rRNA & IFs bind to the Shine-Dalgarno seq in mRNA (upstream of AUG) The 3 IFs bind to the small subunit and attract mRNA fMet = start codon in bacteria 53  The initiation complex is formed The initiation comples binds w large subunit forming A and P site E site = Exit Site P site = Peptidyl site A site = Aminoacyl site 2) Elongation Step 1: Addition of second AA to 𝑡𝑅𝑁𝐴 '()* 54  Step 2: First cycle of Elongation is completed Step 3: New tRNA can enter the ribosome 55  The process repeats itself to form a peptide chain 30S subunit à decode codons 50S subunit à catalyses peptide bonding Elongation summary: 3) Termination Termination codon = signal for GTP dependent release factors which cleave polypeptide chain from tRNA - U Are Gone (UAG) - U Are Annoying (UAA) - U Go Away (UGA) Ribosomal complex dissociates into subunits Poly peptide folds into tertiary protein 56 o Polysomes (polyribosome): A single mRNA is translated by multiple ribosomes Translation in Eukaryotes § In prokaryotes à transcription and translation happen at the same time and in the same place § In eukaryotes à mature mRNA must migrate to the cytoplasm across nuclear pores to be translated § Ribosomes are larger and more complex than in prokaryotes § Transcription and translation spatially and temporally separated § mRNA has a 5’ 7-mG cap: - essential for effective initiation - serves as a ribosome binding site § Kozak consensus = analogue to Shine-Dalgarno seq of prokaryotes § Initiation tRNA = 𝑡𝑅𝑁𝐴()* o The Process Met-tRNAi binds GTP and eIF2 The complex binds to the small subunit Complex needs elF4 to bind to 5’-cap of mRNA Complex “scans” RNA sequence to find initiation codon (Kozak sequence A/GCCAUGG) Anticodon base pairs with the initiation codon Large subunit binds Initiation factors released; GTP hydrolyzed for energy Met-tRNAi in P site, 2nd codon in A Elongation and termination are very similar to prokaryotes (names and number of protein factors differ). Þ Translation factors that start w a small e = eukaryotic factors 57  Closes the loop 5’ and 3’ end of the transcript is close together à can translate w the same ribosome Must Know these details!! à especially names of the factors/proteins (asked us the diagram above and to give the functions of the labelled components) Function of small subunit: decodes codons Function of large subunit: catalyses peptide bonding One gene, One polypeptide Is the refinement of the previous hypotheses: "one gene, one enzyme" was not clear how it could work; and "One gene, one protein" was contested because not all proteins are enzymes. "One gene, one polypeptide" agrees with the fact that most proteins have multiple polypeptides encoded by different genes, a fact that is very clear to us now but took a series of experiments to clarify. 1. Based on pedigree studies in the early times, it was evident that sickle-cell anaemia follows a Mendelian mode of inheritance, with only two alleles present. Homozygotes for one of the alleles are unaffected, homozygous for the other allele are affected, and heterozygotes are largely unaffected carriers. 2. Two different types of haemoglobin were found in affected (HbS) vs unaffected individuals (HbA). This was based on electrophoresis - therefore, the charges of the two haemoglobins had to differ. Because it was known that haemoglobin has four heme groups (non-protein) and a globin (four peptide chains), the chemical change had to be in one of these components. 3. Later, it was proved that the difference resided in the globin, specifically in one of the polypeptides. This was determined using an old method called "protein fingerprinting" (we will talk about it in one of our lectures). 58  4. Later, it was proved that the difference between haemoglobin HbA and HbS was due to a single amino acid change - a substitution of Val for Glu in the sixth position of the beta globin chain. Lecture 12: Protein Structure Biological functionality depends on AA sequence and 3D structure Synthesized as a linear, non-branched molecule Comprised of: - NH2 group - COOH group - R group à gives the identity of the protein (eg. Polar/non-polar) 4 classes of proteins: polar, non-polar, negatively charged and positively charged Peptide bond covalently binds 2 AA to form di-, tri- polypeptides o Primary AA sequence of a linear polypeptide: Chemical characteristics of amino acids important for higher order structure o Secondary structure: Loose association of amino acids in certain areas of polypeptide 2 most common secondary structures: 1. Alpha-helix (spiral) 2. Beta-sheet (ribbons) 59 o Tertiary structure: 3D structure (whole polypeptide) Characteristics that stabilize tertiary structure: - Covalent disulfide bonds between closely aligned cysteine residues. - Nearly all hydrophilic R-groups on outer surface of molecule. - Nearly all hydrophobic R-groups on inside of molecule. o Quaternary structure: Only relevant where proteins are composed of more than one polypeptide chain Many enzymes have this structure (DNA & RNA Polymerases) Eg. Haemoglobin - Is a Heterotetramer = a macromolecule structure consisting of 4 noncovalently associated subunits, of which not all are identical - 2 alpha chains - 2 beta chains - Co-factors (Heme groups) 60 Summary: Posttranslational Modification: 1. The N-terminal amino acid is often removed or modified. Eg. either the formyl group or the entire formyl methionine residue in bacterial polypeptides is usually removed enzymatically. In eukaryotic polypeptide chains, the amino group of the initial methionine residue is often removed, and the amino group of the N-terminal residue may be modified (acetylated). 2. Individual amino acid residues are sometimes modified. Eg., phosphates may be added to the hydroxyl groups of certain amino acids, such as tyrosine. Creates negatively charged residues that may form an ionic bond with other molecules or change the local conformation of the protein. Phosphorylation is extremely important in regulating many cellular activities and is a result of the action of enzymes called kinases. Methyl groups or acetyl groups may be added enzymatically, which can affect the function of the modified polypeptide chain. 3. Carbohydrate side chains are sometimes attached. These are added covalently àproducing glycoproteins An important category of cell-surface molecules, such as those specifying the antigens in the ABO blood-type system in humans. 4. Polypeptide chains may be trimmed. Eg., the insulin gene is first translated into a longer molecule that is enzymatically trimmed to insulin’s final 51-amino-acid form. 5. Target peptides may be removed. 61  Proteins often function in specific locations of the cell such as the plasma membrane or a particular organelle. Proteins are directed to their appropriate destinations by short internal sequences (3–7 amino acids long) called target peptides, which function like postal codes for the cell often enzymatically cleaved after the protein has been delivered to its proper location. 6. Folded polypeptide chains are often complexed with prosthetic groups. The tertiary and quaternary levels of protein structure often include and are dependent on non-proteinaceous elements (prosthetic groups) are commonly vitamins, metals, or metal-containing molecules. Eg. The four iron-containing heme groups present in haemoglobin Protein folding and Misfolding (Need to understand and remember!!) Some proteins rely on chaperones for their correct folding Chaperones = proteins that mediate the folding process by - binding folding polypeptides to exclude the formation of alternative incorrect conformations. - Using ATP energy to ensure proper conformation NB to note à like enzymes, chaperones do not become part of the final product Even in the presence of chaperones à misfolding can still occur Systems of “quality control” exists o Quality control system: Misfolded proteins in prokaryotes: - Digested by ATP-dependent protease In eukaryotes: - Misfolded protein tagged by ubiquitin - Enzymes known as ubiquitin ligases recognize misfolded proteins and catalyze the attachment of ubiquitin molecules. - Once a protein is tagged with several of these residues, it becomes a substrate for the proteasome, a large protein complex with protease activity that releases the ubiquitins and degrades the misfolded protein o Consequences of Misfolding: Creutzfeld-Jakob disease(CJD), Scrapieand Bovine spongiform encephalopathy(BSE, mad cow disease) caused by prions (aggregates of a misfolded protein) Prion = Type of protein that can trigger normal proteins in the brain to fold abnormally PrP(C) is normal form of protein: a-helix PrP(Sc) is infectious form: pleated b-sheet Contact causes refolding àleads to chain reaction which devastates the brain Huntingtons, Alzheimersand Parkinsonsdiseases linked to abnormal protein aggregates in the brain. 62 Protein Domains Functional areas (50-300 AAs) Fold into unique, stable conformations independently of the rest of the molecule Often made up of a-helices and b-sheets The significance of the domain resides in the tertiary structures of proteins Eg. A domain may serve as a catalytic site or a ligand binding site 63 Exon Shuffling During evolution, exons may have been reshuffled btwn genes in eukaryotes, with the result that different genes have similar domains Different from alternative splicing In this case, it is building proteins by putting together exons that appear in different regions of the genome LDL receptor protein à Domain is homologous with a portion of the gene encoding an immune system protein = NB concept !! LDL has numerous functional domains Signal seq is removed and the 5 exons code for various functional domains of the protein and appear to have been recruited from other genes during evolution 64