SCIE1106 - All Lectures Summarised PDF

lOMoARcPSD|40737841 SCIE1106 - all lectures summarised Molecular Biology Of The Cell (University of Western Australia) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Harshdeep K...

lOMoARcPSD|40737841 SCIE1106 - all lectures summarised Molecular Biology Of The Cell (University of Western Australia) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 1 – MACROMOLECULES  explain what are atoms, molecules, covalent bonds and macromolecules,  describe the composi琀椀on, classi昀椀ca琀椀on, forma琀椀on and breakdown of carbohydrates,  explain alpha and beta monosaccharides and the structural implica琀椀ons for the forma琀椀on of di- and polysaccharides  describe examples of cellular roles of carbohydrates,  composi琀椀on and characteris琀椀cs of lipids,  explain the biological roles of lipids  Molecule = elements = cant be broken or converted by chemical means  Atom = smallest par琀椀cle of element – dis琀椀nct chemical proper琀椀es o un昀椀lled outer electron shell = reac琀椀ve o complete outer shell = stabilised  Electrons = determine chemical behaviour  Covalent bond = sharing of electrons o Single – 2 electrons shared and allows free rota琀椀on o Double – 4 electrons shared and is rigid  Macromolecules = constructed by linking small molecules together by covalent bonds o DNA, RNA, Proteins, Polysaccharides and Lipids  Building blocks of these = macromolecules & energy source o Carbon = 4 covalent bonds = small organic molecules (central role)  form highly stable chains and rings POLYSACCHARIDES = MANY CARBOHYDRATES  Monosaccharides = simplest carbohydrates = 1 sugar O Hydrate of carbon (CH2O)n O Glucose = 6 C ring (5 Cs and 1 O) O Aldose sugar (glucose) or ketose sugar (fructose)  Disaccharide = 2 monosaccharides = sucrose O Linked via glycosidic linkage  Oligosaccharide = 3-50 monosaccharides  Polysaccharides = 100-1000 O Branched or linear Forma琀椀on/breakdown of polysaccharides  Condensa琀椀on reac琀椀on = covalent bond forms and release water o Forms nucleic acids  phosphoanhydride bonds between phosphates in nucleo琀椀des  phosphodiester bonds link nucleo琀椀des = nucleic acids o Forms proteins – pep琀椀de bonds link amino acids o Forms polysaccharides - glycosidic linkage between 2 monosaccharides = C-O-C Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Hydrolysis reac琀椀on = cleaves bond and requires water o Catalysed by enzymes Glucose cyclisa琀椀on = interconver琀椀ble glucose – both exist in equilibrium  Alpha (1st Carbon OH below ring) and Beta (OH above ring) glucose Structural role of carbohydrates  Cellulose = polysaccharide of repea琀椀ng B 1-4 linked D-glucose units = linear Energy source of carbohydrates  Glycogen and starch = polysaccharides of glucose o A 1-4 linked = linear o A 1-6 linked = branched LIPIDS = FATTY ACIDS  Fa琀琀y acids = amphipathic = hydrophilic carboxylic head (chemically reac琀椀ve) AND hydrophobic hydrocarbon tails  insoluble in water  soluble in fat & organic solvents o Saturated FAs  no double bonds = 琀椀ght packing = solid o Unsaturated FAs  double bond(s) = link carbon atoms in hydrophobic tail  kinks formed = no close packing = liquid at room temperature  FAs Biological roles = energy (6x usable energy as glucose/weight) o stored as triacyl-glyceride molecules = 3 FAs on 1 glycerol  glycerol ester linkage to carboxylic acid head of fa琀琀y acid  Deriva琀椀ves = phospholipids, glycolipids, steroids Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Phospholipids  glycerol linked to 2 FAs and a phosphate group o 2 OH in glycerol linked to FAs but 3rd OH linked to phosphoric acid (phosphate linked to small polar/hydrophilic group)  strongly amphipathic and component of cellular membranes o bilayer = energe琀椀cally favourable  unsaturated tails increase membrane 昀氀uidity Glycolipids  glycerol linked to 2 FAs and a sugar group o amphipathic = OH of sugar o non-cytosolic surface of membrane o cell to cell recogni琀椀on Steroids  cholesterol = part of membrane  non-membrane = hormones o signalling molecules and pass through membrane LECTURE 2 - MACROMOLECULES  describe the building blocks, composi琀椀on, structure, forma琀椀on and breakdown of nucleic acids  biological roles of nucleo琀椀des and deriva琀椀ves  describe the building blocks, composi琀椀on, structure, forma琀椀on and breakdown of proteins  noncovalent interac琀椀ons and impact on stability and interac琀椀on of macromolecules  describe origin of non-covalent forces NUCLEIC ACIDS = NUCLEOTIDES linear polymers of nucleo琀椀des  Phosphodiester bonds link adjacent nucleo琀椀des in nucleic acids o OH on 3'C of pentose covalently linked to phosphate group a琀琀ached to 5' C of adjacent pentose = sugar phosphate backbone  informa琀椀on storage and retrieval o RNA = transient carrier of informa琀椀on o DNA - long-term storage of hereditary material DNA  2 DNA strands = an琀椀-parallel = direc琀椀onality = 5’ end  3’ end o sugar-phosphate backbone = outside & bases = inward o H-bonds between bases hold strands together  G≡C (3 H-bonds) & A=T (2 H-bonds)  ra琀椀o of purines to pyrimidines is 1:1 = complementary strands  Nucleo琀椀des O Nitrogen bases = pyrimidines (C & T/U = single ring) and purines (A & G = 2 rings) O Pentose sugar – deoxy ribose sugar = DNA and ribose = RNA  Nucleoside = N + P O Phosphate = phosphoanhydride bonds between phosphates (condensa琀椀on rxn)  Joined to C5 OH of sugar  Nega琀椀vely charged = DNA and RNA are nega琀椀vely charged  AMP/ADP/ATP  Carriers of chemical energy o Phosphoanhydride bond hydrolysed ➔ATP Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 o cleavage of bond releases high energy ➔ used by prokaryo琀椀c and eukaryo琀椀c cells  Signalling molecules o ac琀椀vate enzymes and turns on genes - cAMP  Coenzymes - combined with other groups o Ac琀椀vated = transfer groups between molecules (acetyl CoA) PROTEINS = AMINO ACIDS  alpha carbon covalently a琀琀ached to carboxylic acid, amino group & side chain (R group)  pep琀椀de bonds = link amino acids  carboxylic acid group reacts with amino group of another amino acid = condensa琀椀on reac琀椀on  Direc琀椀onality = N-terminus to C-terminus Noncovalent interac琀椀ons O Mediates molecular interac琀椀ons O Stabilise macromolecule structures  Ionic interac琀椀ons = opposite charges a琀琀ract o Absence of water = strong  Hydrogen bonds = polarized molecule with H between 2 electron a琀琀rac琀椀ng atoms  Hydrophobic bonds = hydrophobic molecules cluster together to exclude water molecules  Van der Waals = a琀琀rac琀椀on & repulsions o shi昀琀 in electron density around nucleus = charge  nearby atom a琀琀racted or repelled. o Weak interac琀椀ons LECTURE 3 – PROKARYOTES AND EUKARYOTES  cells = basic unit of life sharing a basic chemistry; store gene琀椀c material as DNA - replicated & passed on by cell division + same 20 amino acids  iden琀椀fy and contrast characteris琀椀cs of prokaryo琀椀c and eukaryo琀椀c cells,  describe the composi琀椀on and roles of cellular membranes,  explain the roles of cellular organelles and compartments,  explain the origin of mitochondria and chloroplasts - endosymbiosis theory  describe in detail mitochondrial and chloroplas琀椀c structures,  describe the genera琀椀on of cellular energy,  explain the role of proton gradients in ATP produc琀椀on,  describe the produc琀椀on of carbohydrates in chloroplasts Prokaryotes E.coli – (eubacteria) = gut bacterium = familiar environments Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Beggiatoa – (archea) = sulphurous (hos琀椀le) environments o Division based on molecular biological characterisa琀椀ons  Diverse = di昀昀erent environments + di昀昀erent growth forms o Organotrophic = organic molecule as energy source o Phototrophic = light as energy source o lithotrophic = inorganic molecules as energy source  simple cells – few micrometres (μm) long  tough, protec琀椀ve cell wall  no membrane-bound organelles = ribosomes + 昀氀agellum o ribosome = 70S = smaller complex of proteins and rRNAs  no nucleus = circular DNA free in cytosol  reproduce quickly; divide every 20 min Ribosomes  sites of protein synthesis  large complexes of proteins & rRNAs  prokaryo琀椀c = 70S  eukaryotes = 80S o cytosolic (free/a琀琀ached to ER) = 80S o mitochondria & chloroplasts = 70S = bacterial size Eukaryotes  Unicellular = pro琀椀sts + yeast  Mul琀椀cellular = animals, plants (mul琀椀cellular algae) & fungi Membranes  Bilayer of phospholipids - asymmetrical arrangement  Proteins = integral & peripheral  Selec琀椀vely permeable o Small hydrophobic & uncharged molecules cross freely. o Larger uncharged polar molecules & charged solutes interact with transmembrane proteins (transporters) to cross phospholipid bilayer  Compartmentalise cells o separate cells from environments and organelles from each other & cytosol  Double membranes surround; nucleus, mitochondria & chloroplasts Plasma membrane  cell signalling  transport of solutes  cell growth & mo琀椀lity o Carbohydrate groups a琀琀ached to lipids = glycolipids o Carbohydrate groups a琀琀ached to proteins = glycoproteins  external (non-cytosolic) side of plasma membrane  cell-to-cell communica琀椀on, protec琀椀on + adhesion  SINGLE membrane organelles Endoplasmic re琀椀culum Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  sacs & tubules (cisternae)  cytosol  synthesis of cell membrane components & molecules exported from a cell Rough ER = ribosomes a琀琀ached  protein synthesis  Carbohydrate addi琀椀on to proteins  con琀椀nuous with nuclear membrane Smooth ER  no ribosomes = lipid synthesis Golgi apparatus  昀氀a琀琀ened sacs (cisternae) o cis face = adjacent to ER, where vesicles arrive from ER o trans face = toward plasma membrane  vesicles pinch o昀昀 & fuse with cisternae = carry proteins modi昀椀ed (added sugar)  correla琀椀on of enzyme loca琀椀on & catalysis in sugar-modi昀椀ca琀椀on pathway  synthesis and packaging of molecules to be secreted from cell  rou琀椀ng of newly synthesised proteins to correct cellular loca琀椀ons Endomembrane system  Exocy琀椀c pathway = membrane growth + secre琀椀on  Proteins synthesised on rough ER & glycosylated = glycolproteins  Vesicles with glycoproteins bud o昀昀 ER & fuse with cis Golgi cisternae  Glycoproteins glycosylated through Golgi cisternae by vesicle budding & fusion  trans face of the Golgi = vesicles directed to plasma membrane or lysosome/vacuole  Endocy琀椀c pathway = inges琀椀on and degrada琀椀on  Plasma membrane region  vesicles bud inward to form vesicles  fuse with early endosomes  molecules degraded in lysosome/vacuole  products reused by cell Peroxisomes  contain oxida琀椀ve enzymes  animals = sites of detoxi昀椀ca琀椀on  plants = sites of detoxi昀椀ca琀椀on + photorespira琀椀on (carbon recycling) o conversion of stored fats into sucrose during germina琀椀on of seeds (= glyoxysomes) Vacuoles  sites of degrada琀椀on  storage organs  detoxi昀椀ca琀椀on sites  pigment deposi琀椀on LECTURE 4 – PROKARYOTES AND EUKARYOTES  DOUBLE membrane organelles Nucleus  double membrane = con琀椀nuous with ER + interrupted by pores = pass selected molecules  cellular DNA o heterochroma琀椀n = DNA + proteins = highly condensed o euchroma琀椀n = DNA + proteins = not condensed un琀椀l mitosis  contains nucleolus = ribosomal RNA (rRNA) synthesis & ribosomal subunit assembly Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Mitochondria  Cellular respira琀椀on and ATP produc琀椀on – oxida琀椀ve phosphoryla琀椀on  smooth outer membrane = permeable to ions & small molecules  inner membrane = highly folded (cristae) & impermeable o transport proteins = movement across inner membrane o electron transport chain & ATP synthase Mitochondrial matrix o mitochondrial genome (DNA) = mitochondria; tRNA, rRNA and mRNA  proteins for DNA synthesis and oxida琀椀ve reac琀椀ons o mitochondrial ribosomes = 70S o enzymes for TCA (tricarboxylic acid) cycle Chloroplasts o photosynthesis = light harves琀椀ng and carbohydrate produc琀椀on  outer membrane = permeable (ions and small molecules)  inner membrane = impermeable (transport proteins) o Thylakoids = folded inner membrane  Pigments, ETC and ATP synthase Stroma = similar to matrix  outside thylakoid membrane o DNA = tRNA, rRNA and mRNA  proteins for DNA synthesis and photosynthesis o Ribosomes – 70S o Enzymes for carbohydrate produc琀椀on Endosymbiosis theory  Eukaryo琀椀c cell ingested; aerobic bacterium undigested  mitochondrion ➔ eukaryo琀椀c cells  Eukaryo琀椀c cells ingested photosynthe琀椀c bacterium; but undigested  chloroplast ➔ plant cells Energy produc琀椀on  Diges琀椀ve enzymes = breaks down macromolecules to amino acids, glucose & FAs/glycerol  Products enter cytosol  gradual oxida琀椀on (glucose  pyruvate acetyl CoA) + ATP produc琀椀on and NADH (reducing molecules) o Glucose  pyruvate through glycolysis = ATP and NADH  Pyruvate (+ aa and FAs) enter mitochondrion = oxidised to acetyl CoA = full oxida琀椀on + ATP o Acetyl CoA oxidised by TCA cycle = CO2 + NADH & FADH2 (reducing power – high energy electrons)  ETC (electron transport chain) = NADH & FADH2 donate electrons to ETC in inner mitochondrial membrane  through mul琀椀protein complexes = loses energy (=H2O)  Energy = pump protons to space between outer and inner membrane = proton gradient  higher conc of protons in inner membrane than matrix  Protons move through the ATP synthase back into matrix = ATP produc琀椀on (ADP + phosphate)  ETC = oxida琀椀on of NADH & FADH2 + reduc琀椀on of O2  H2O & ATP produc琀椀on. Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Light harves琀椀ng reac琀椀ons – chloroplasts  Light (pigments in thylakoid membranes)  reducing power = NADPH and chemical energy ATP o H2O = electron donor and NADPH = electron acceptor  ETC = protons move across thylakoid membranes (stroma  lumen) = proton gradient = ATP synthesis (proteins move from lumen back to stroma through chloroplas琀椀c ATP synthase)  Calvin cycle = NADPH and ATP used for carbohydrate synthesis from CO2 in stroma o Rubisco enzyme catalyses reac琀椀on LECTURE 5 – DNA STRUCTURE  hypotheses on origin of life on earth.  main events which led to the discovery of DNA.  describe the main features of DNA.  how the gene琀椀c code works.  “The Central Dogma”  terminology for bases, nucleo琀椀des, nucleosides, deoxy, ribo.  De昀椀ne; an琀椀parallel, complementary base pairing, coding strand, codon, right-handed helix, major groove Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 4 hypotheses on origin of life Organic chemical synthesis in reducing atmosphere  Early earth had a reducing atmosphere – high H and CH3 (+NH3)  Prebio琀椀c soup – amino acids and nucleo琀椀des Carriage by meteorites/comets  Panspermia – disappearance of life and uniformity  Organic compounds in space; glycine in comet o Complex chemicals on Titan + meteorites carry microbes Synthesis on metal sulphides in deep sea vents  Vents – biological ac琀椀vity independent of solar energy; chemical source = prebio琀椀c soup  Prebio琀椀c soup self-organises into life suppor琀椀ng networks on metal sulphide surfaces; incorporate into membranes RNA world  First self-replica琀椀ng en琀椀ty  RNA store informa琀椀on and catalyse reac琀椀ons (ribozymes) Discovery of DNA structure  J. Watson - X ray di昀昀rac琀椀on image of DNA by M. Wilkins  F. Crick works on helical di昀昀rac琀椀on in proteins  X ray data from R. Franklin  Watson and Crick = 3-stranded DNA model.  model’s inconsistencies with Franklin’s data  E. Charga昀昀 = A/T and C/G ra琀椀o  Watson & Crick double helical, base-paired model of DNA structure together with X ray di昀昀rac琀椀on papers from Wilkins and Franklin groups Features of DNA  Gene琀椀c material  Base paired, an琀椀parallel & right-handed double helix o Complementary strands = one strand sequenxe de昀椀nes other due to base pairing o Coding and non-coding strands  Code = triplets of ACGT (gene  proteins)  3 bases = 1 codon (codes for amino acid  proteins)  DNA code 5’  3’ o Control regions = sequences not organised into triplets unlike DNA code  = Upstream (5’) and Downstream (3’) controls not in code ≠ part of gene/protein  DNA interac琀椀on with Proteins o Proteins interact with bases in “major groove”  recognize speci昀椀c base sequences  Nucleo琀椀des: DNA; sugar = deoxyribose = pre昀椀x deoxy (Deoxyadenosine triphosphate (dATP) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 o RNA = ribose = pre昀椀x ribo o Nucleo琀椀des linked via phosphodiester bonds; 5’ phosphate reacts with 3’ OH of another LECTURE 6 – DNA REPLICATION  De昀椀ne; semiconserva琀椀ve, origin, bidirec琀椀onal, replica琀椀on fork, Okazaki fragment.  mechanism of leading and lagging strand replica琀椀on and role of the RNA primer.  func琀椀ons of the proteins at the DNA replica琀椀on fork.  List major DNA polymerases of prokaryotes and eukaryotes and their func琀椀ons. Chromosome replica琀椀on  Complementary base pairing = semiconserva琀椀ve DNA replica琀椀on  DNA synthesis ini琀椀ates at origins o Synthesis moves bidirec琀椀onally away from origin via 2 replica琀椀on forks = replica琀椀on bubble o 5’  3’ o Requires primer Complementary base pairing = accurate DNA replica琀椀on  Each strand of dsDNA molecule = template for synthesis of new complementary strand Semi conserva琀椀ve DNA replica琀椀on  Daughter molecule = parental and new strand  1000 nucleo琀椀des per second without error DNA synthesis ini琀椀ates at Origins  dsDNA pried apart at replica琀椀on origin by DNA helicase; posi琀椀on iden琀椀昀椀ed by ori DNA sequence Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  protein machine moves along replica琀椀on fork  DNA polymerase adds deoxyribonucleo琀椀des to 3’ end of new (primer) strand a琀琀ached to template (parent) strand o creates phosphodiester bonds o Proofreads = reduce error and catalyses synthesis Bidirec琀椀onal synthesis from Origins  Circular/short chromosomes of prokaryotes = single origin of replica琀椀on  Long linear chromosomes of eukaryotes = mul琀椀ple origins of replica琀椀on Replica琀椀on fork structure  daughter strands polymerised 5’  3’  Leading strand synthesised con琀椀nuously  Lagging strand synthesised discon琀椀nuously = short Okazaki fragments o DNA polymerase can only a琀琀ach to 3’ end Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 DNA primase = primer for DNA synthesis  DNA Primase = enzyme = synthesises short strand of RNA on a DNA template  Okazaki fragments primed by RNA primer at 3’ of lagging strand (synthesised by DNA primase)  DNA ligase joins Okazaki fragments by sugar-phosphate backbone Replica琀椀on machine  Proteins at a replica琀椀on fork cooperate = machine  Single-strand DNA-binding proteins stabilise ssDNA and aid the helicase  Helicase unwinds double helix to form ssDNA for replica琀椀on  Sliding clamp moves behind DNA polymerase across DNA template strands to ensure a琀琀achment of polymerase.  Clamp loader assembles the clamp on the DNA using ATP energy  Lagging strand DNA = folded = brings DNA polymerase into a complex with leading strand DNA polymerase. DNA polymerase in prokaryotes  Polymerase 1 o DNA repair o Matura琀椀on of Okazaki fragments (removes RNA primer and 昀椀lls gaps)  Polymerase 2 o DNA repair  Polymerase 3 o DNA replica琀椀on enzyme (both strands) DNA polymerase in eukaryotes  Alpha α = primase and elongates primer with short length of DNA (mul琀椀 subunit enzyme)  Beta β = DNA repair  Gamma γ = replica琀椀on of mitochondrial DNA  Delta δ = synthesis of lagging strand  Epsilon ε = synthesis of leading strand Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 7 – MUTATION AND REPAIR OF DNA  Explain muta琀椀on of DNA  Describe the processes that result in the muta琀椀on of DNA  Describe the consequences of depurina琀椀on, deanima琀椀on, thymine dimer forma琀椀on and double stranded breaks on DNA replica琀椀on  What transposable DNA elements and infec琀椀ous agents introduce muta琀椀ons into DNA  Explain the 2 mechanisms for DNA repair: O Mismatch repair system O homologous recombina琀椀on Muta琀椀on  permanent and heritable change in the DNA sequence o damaged DNA = replica琀椀on problems o change caused by replica琀椀on errors (rare)  DNA replica琀椀on = High Fidelity (error by polymerase = one in a billion) o base paired structure of DNA o primer requirements of DNA polymerases + proofreading of polymerase (diagram)  3’5’ exonuclease ac琀椀vity removes incorrect nucleo琀椀de  restora琀椀on of correct nucleo琀椀de sequence  possible incorrect N base incorporated Chemical factors Nucleo琀椀de instability = DNA polymerase randomly assigns nucleo琀椀des to match damaged nucleo琀椀de = change nucleo琀椀de sequence  昀椀xed & inherited  Depurina琀椀on = loss of A/G from DNA backbone  Deamina琀椀on = loss of NH2 from A/G/C (= U) Mutagenic chemical  Alkyla琀椀on = electrophiles (carcinogens) add alkyl groups to N bases = stall replica琀椀on  Intercala琀椀on = compound inserts into ds helix = distor琀椀on but no change in bases o Ethidium bromide Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Radia琀椀on UV radia琀椀on  Forms thymine dimers Gamma Rays  A琀琀ack DNA bonds (single and double stranded breaks) o Produce free electrons = a琀琀ack backbone o Generate hydroxide free radicals Mobile DNA Infec琀椀ous agents (viruses and bacteriophages) = Insert/recombine into a target DNA molecule  Recombina琀椀on = breaking & rejoining of DNA molecules = new combina琀椀ons o Non-homologous = NO similarity between DNA molecules (bacterial DNA and phage DNA)  targeted recombina琀椀on catalysed by integrases/transposases (enzymes)  phage encoded integrase = promotes recombina琀椀on between a琀琀achment sites o Homologous = both donor and acceptor DNA have similar sequences  Retroviruses (HIV) integrate into host DNA = use host replica琀椀on machinery o Ly琀椀c (enter and lyse/kill host) o Lysogenic (enter and integrate into host chromosome)  Inser琀椀on of foreign DNA = disrupts coding region/gene Transposable elements  Transposons = linear DNA molecules o Move within/between chromosomes o Insert into di昀昀erent DNA sequences = disrupt gene o Excision of transposon = small duplica琀椀on of DNA = disrupt gene  Transposase monomers (enzymes) encoded by transposon = a琀琀ach at short inverted repeat sequences = allow break away from chromosome by forming transpososome loop o Loop integrated into other chromosome at short direct repeats of target DNA sequences  Cut and paste non replica琀椀ve transposi琀椀on o Transposase encoded by transposon cut out transposon and insert into new DNA  Replica琀椀ve transposi琀椀on o Transposon replicated by DNA replica琀椀on  into target DNA = both donor and target DNA have transposon Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Repair Mismatch (mis-paired nucleo琀椀des)  repairs muta琀椀ons in newly synthesised DNA strand = restore to original sequence of template strand  repair proteins recognise and excise strand of DNA containing mismatch o bind to mismatch sequence and excised by nucleases = ssDNA patch o 2nd strand synthesised by repair DNA polymerase using free 3’OH group as primer o Liga琀椀on of DNA backbone by DNA ligase Homologous recombina琀椀on  repairs double stranded breaks in phosphodiester backbone of DNA o = restora琀椀on of original sequence + rearrangements of local regions of DNA sequence  Similar sequence regions align  double strands break = cross over = new combina琀椀ons of DNA a昀琀er repair  Exonuclease removes nucleo琀椀des from 5’  3’ = single stranded 3’ overhang = migrates to other chromosome of homologous sequences  DNA polymerase synthesises complementary strand  Holiday Junc琀椀on = crossed strands = rotate  sec琀椀on of strand joined to another strand = exchange DNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 8 – TRANSCRIPTION MECHANISM  Describe the structure of RNA and the di昀昀erences to DNA  De昀椀ne; transcrip琀椀on, understand how DNA is transcribed to RNA  di昀昀erences in the transcrip琀椀on processes between eukaryotes and prokaryotes  Describe and understand the func琀椀on of RNA polymerase  Describe the transcrip琀椀on ini琀椀a琀椀on and termina琀椀on signals in DNA  Describe the process of capping and polyadenyla琀椀on  List the di昀昀erent classes of RNA molecules RNA Structure  Ribose sugar  A, C, G, U  Single stranded Transcrip琀椀on = RNA from DNA template Ini琀椀a琀椀on  elonga琀椀on  termina琀椀on Conven琀椀ons  Coding/sense strand = 5’  3’  RNA polymerase = transcribes complementary mRNA strand of an琀椀-parallel/template strand from P to T = same sequence of coding strand (T replaced with U) o Template strand = 3’  5’ from promotor to terminator Requirements of enzyma琀椀c synthesis of RNA  4 ribonucleo琀椀de 5’ triphosphates (5’ATP, 5’GTP, 5’CTP, 5’UTP)  Mg2+  DNA template (no primer)  RNA polymerase (RNAP) Transcrip琀椀on in prokaryotes  RNA polymerase (holoenzyme) binds to promoter on dsDNA  Sigma factor leaves = core enzyme = begins RNA synthesis (elonga琀椀on) = mRNA strand is complementary to template strand from P to T  RNA polymerase reaches terminator = stop = complete mRNA synthesis & (sigma factor rebinds) o Ini琀椀a琀椀on signal = promoter  Holoenzyme binds to TGTTGA sequence = upstream/promoter region  TATAATG box shows where consensus sequence in coding strand ends and where coding region begins o Termina琀椀on signal = Hair-pin structure Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Inverted repeat followed by a stretch of T bases (coding strand of DNA sequence)  protein factor (Rho); makes polymerase fall o昀昀 hair loop RNA Polymerase  Forms phosphodiester bonds between ribonucleosides  Uses energy stored in ribonucleoside triphosphates for polymerisa琀椀on  Unlike DNA polymerase; RNA synthesis starts without a primer o Error rate higher than DNA polymerase o Unwinding of DNA template doesn’t require a helicase or ATP. E. coli RNA polymerase  5 SUBUNITS  complex of α2ββ’ ω = core enzyme  σ = promoter recogni琀椀on (pyridine-rich DNA sequence of >10 bases). o σ + core enzyme = holoenzyme = ini琀椀a琀椀on  correct ini琀椀a琀椀on achieved = σ dissociates from holoenzyme = core enzyme con琀椀nues elonga琀椀on of RNA chain. Messenger RNA Prokaryo琀椀c messenger RNA  mRNA = copy of coding strand (DNA) = determine amino acid sequence of proteins  DNA segment (coding sequence corresponding to protein) + start and stop signals = cistron. o A single mRNA encoding 1 polypep琀椀de = monocistronic mRNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 o Polycistronic mRNAs = code for several di昀昀erent polypep琀椀de chains  mRNA’s have 5’ leaders, 3’ termini and intercistronic regions called spacers (polycistronic)  Prokaryo琀椀c mRNA’s have short half-lives = minutes  Primary transcript not processed Eukaryo琀椀c messenger RNA  Gene contains coding (exons) and non-coding (introns) regions  RNA is processed (Iden琀椀昀椀es RNA molecule as an mRNA) o RNA capping (G with methyl group added to 5’ end) o Polyadenyla琀椀on ( “A”s added to 3’ end)  Introns removed by a spliceosome = recognises boundaries between introns and exons  Exons s琀椀tched back together and translated  mRNA molecules eventually degraded LECTURE 9 – TRANSCRIPTION PRODUCTS  Describe the major classes of RNA molecules  Describe the di昀昀erences between prokaryo琀椀c and eukaryo琀椀c mRNA  Describe the structure and func琀椀on of tRNA and func琀椀on of aminoacyl-tRNA synthetases  Describe di昀昀erences between prokaryo琀椀c and eukaryo琀椀c ribosomes  Understand the roles of ribosome binding sites in transla琀椀on of mRNA to protein  De昀椀ne; polycistronic mRNA, polyribosome, mRNA cap, poly-A tail, codon, an琀椀codon, amino acyl tRNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Major classes of RNA Messenger = mRNA  Copy of DNA – codes for protein  Decoded in sets of 3 nucleo琀椀des = codons Transfer = tRNA  Small adaptor molecules which align speci昀椀c amino acids opposite their triplet codon in the mRNA molecule during transla琀椀on of mRNA into protein o Amino acids can’t recognise mRNA codon = requires adaptor (tRNA) Ribosomal = rRNA  Integral part of ribosomal structures – part of protein synthesising machinery RNA transcripts Eukaryotes = Transcrip琀椀on/transla琀椀on = separate – di昀昀erent compartments = control Prokaryotes = Transcrip琀椀on/transla琀椀on = coupled – simultaneous Eukaryotes:  dsDNA = gene encoded by exons – separated by introns  transcrip琀椀on ini琀椀ated = RNA polymerase binds to template strand (3’  5’) of DNA and synthesises mRNA (copy of coding strand; 5’  3’)  post transcrip琀椀on modi昀椀ca琀椀ons = o splicing = removes introns by spliceosomes o 5’ cap methylated G recognises stop = polymerase drops o昀昀  polyadenyla琀椀on = poly-A-tail at end of gene = mature mRNA  Mature mRNA released into cytoplasm o No cap or poly-A-tail = mRNA degrada琀椀on  Ribosome recognises mRNA start codon (AUG) and protein synthesis beings eukaryo琀椀c mRNA = monocistronic Roles of 5’ end cap and 3’ end polyadenyla琀椀on o assist the export of mRNA molecules from the nucleus o Protect mRNA from degrada琀椀on in cytoplasm and increase its half-life o Allows binding of ribosomes = promote transla琀椀on  CAP = 7-methylguanosine  Poly-A tail = many adenosines linked to 3’ end of mRNA (not in DNA sequence) o = binding site for proteins Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 prokaryo琀椀c mRNA = polycistronic single prokaryo琀椀c mRNA molecule can encode several di昀昀erent proteins prokaryo琀椀c ribosomes ini琀椀ate transla琀椀on at ribosome-binding sites, located in the interior of an mRNA molecule = prokaryotes synthesize >1 type of protein from 1 polycistronic mRNA tRNA  Small adaptor molecules = align amino acids opposite triplet codon in mRNA molecule during transla琀椀on of mRNA into protein o Amino acids can’t recognise mRNA codon = requires adaptor (tRNA)  Bases other than A, G, C and U are present in tRNA due to post-transcrip琀椀onal modi昀椀ca琀椀on o inosine (modi昀椀ed A), pseudouridine & dihydrouridine Structure  Internal complementary base pairing of RNA gives cloverleaf structure  An琀椀codon sequence of 3 bases determines mRNA codon binding  Amino acid a琀琀achment site is at 3’ end of tRNA  Modi昀椀ed bases o D = dihydrouracil o Ψ = pseudouridine  3D shape determines a琀琀achment of correct amino acid (matching codon/an琀椀codon pair) by aminoacyl-tRNA synthetases Adaptors Gene琀椀c code translated by 2 adaptors  Aminoacyl-tRNA synthetase couples amino acid to corresponding tRNA o For every amino acid = 1 aminoacyl-tRNA synthetase - links amino acid to several tRNAs o Enzymes responsible for a琀琀achment of correct amino acid to tRNA:  Aa speci昀椀ed by an琀椀codon of tRNA and aa a琀琀achment sequence on the 3’ end o correct amino acid a琀琀ached to tRNA = tRNA is charged/acylated  ATP hydrolysis required  mischarging = incorrect amino acid a琀琀achment  tRNA = an琀椀codon forms base pairs with codon of mRNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 rRNA  integral part of ribosomal structures + important parts of the protein synthesising machinery o mRNA message decoded on ribosomes Ribosome = large mul琀椀-domain complex  scans along mRNA and captures correct aminoacyl tRNA o Correct base-pairing = covalent linking of aa onto growing polypep琀椀de chain  Small subunit matches tRNA to the codon on the mRNA  Large subunit catalyses the forma琀椀on of the pep琀椀de bonds that link aa together = polypep琀椀de  Each ribosome has 1 binding site for mRNA and 3 binding sites for tRNA o tRNA binding sites:  A = new tRNA enters ribosomal complex  P = tRNA a琀琀ached to the polypep琀椀de chain  E = empty tRNA exits ribosomal complex Prokaryo琀椀c Ribosome structure Large ribosomal subunit = 5S, 23S rRNA Small ribosomal subunit = 16S rRNA Eukaryo琀椀c Ribosome structure Large ribosomal subunit = 5S, 5.8S & 28S rRNA Small ribosomal subunit = 18S rRNA Elonga琀椀on of polypep琀椀de 1. Aminoacyl-tRNA binds to A-site, spent tRNA leaves from E-site. 2. New pep琀椀de bond formed 3. Large subunit moves to next codon 4. Small subunit moves to next codon, A- site now empty, ready for next aminoacyl-tRNA 5. Cycle 1 to 4 repeats Polyribosomes large assembly of ribosomes - 80 nts apart on a single mRNA translate proteins o ribosomes can simultaneously translate the same mRNA molecule = increases produc琀椀on capacity of a protein Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 10 – TRANSLATION Describe the gene琀椀c code Explain the history and experimental design associated with the discovery of the gene琀椀c code Explain the redundancy in the gene琀椀c code at the mRNA level Explain the redundancy in the recogni琀椀on of the gene琀椀c code by tRNAs using wobble hypothesis Gene琀椀c code = redundant/degenerate Transcrip琀椀on = DNA to RNA 4 bases matching 4 bases, copy Transla琀椀on = RNA  Protein 4 bases and 20 amino acids, translate DNA/RNA language = 4 bases (GATC/ GAUC). Max number of codons in groups of three: 64 possible codons Code = triplet - each codon consists of a unique combina琀椀on of 3 nucleo琀椀des code has punctua琀椀ons at end of the coding region in mRNA: start and Stop codons codon code = redundant; amino acids speci昀椀ed by more than one codon o methionine and tryptophan are speci昀椀ed by a single codon code = consistent - each codon speci昀椀es a single amino acid code = universal - viruses, bacteria, plants, and animals use the same code o some minor varia琀椀ons in mitochondria and a few fungi Unique codons AUG = Methionine = Ini琀椀a琀椀on signal for transla琀椀on Prokaryotes = formyl-methionine Eukaryotes = methionine 3 codons not recognised by tRNAs STOP codons =Termina琀椀on signal for transla琀椀on UAA, UAG & UGA History 4 bases + codon + 20 aa ar琀椀昀椀cial RNA composed of uracil, poly(U) = synthesize protein of poly-phenylalanine. cell-free system o E. coli cytosol frac琀椀on o destroyed na琀椀ve mRNA o added radiolabelled amino acids Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Repeated with poly(A) mRNA = poly-Lysine & poly(C) mRNA to create a poly-Proline RNA strands of repea琀椀ng di- tri- and tetra-nucleo琀椀de sequences made through transcrip琀椀on of DNA into RNA with RNA polymerase. Poly UG repeated and poly AG trap trinucleo琀椀des (size of a codon) with the ribosome = trap corresponding charged aminoacyl-tRNA (radiolabelled) decipher all the remaining codon sequences = what amino acid they code for. Wobble Hypothesis Degeneracy in tRNA recogni琀椀on of the gene琀椀c code o Some tRNA's can bind at >1 codon = "wobble”. o inosine (modi昀椀ed adenosine) – 1st posi琀椀on in the an琀椀codon of the tRNA = bind to U, A, or C in the 3rd posi琀椀on of the codon = 1 tRNA recognizes 3 di昀昀erent codons. codon and an琀椀-codon = an琀椀-parallel. o 1st posi琀椀on of the codon = 5' end of codon (mRNA) and binds to the 3 rd posi琀椀on (3‘ end) of the an琀椀-codon (tRNA) LECTURE 11 – TRANSLATION PROCESS Describe the 3 stages of protein synthesis Understand transla琀椀onal reading frames Describe and know the di昀昀erences in ini琀椀a琀椀on of transla琀椀on in eukaryotes and prokaryotes Describe the ini琀椀a琀椀on signals in eukaryotes and prokaryotes Describe the di昀昀erence between the ini琀椀a琀椀on methionine in eukaryotes and prokaryotes Describe the processes of elonga琀椀on and termina琀椀on of transla琀椀on Describe why certain an琀椀bio琀椀cs can inhibit bacterial transla琀椀on and not eukaryo琀椀c transla琀椀on Stages of protein synthesis Ini琀椀a琀椀on = mRNA + ribosomes + ini琀椀a琀椀ng tRNA Elonga琀椀on = pep琀椀de bonds + movement Termina琀椀on = dissocia琀椀on of ribosomes and pep琀椀des Transla琀椀onal reading frames Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Transla琀椀on ini琀椀a琀椀on signals transla琀椀on machinery picks the right AUG as the ini琀椀a琀椀on signal Needs other ini琀椀a琀椀on signals; ribosome binding site mo琀椀f in the mRNA Eukaryotes = ribosomes - scanning ability to 昀椀nd the right site Prokaryotes = small subunit of ribosome binds to Shine Dalgarno sequence (5’AGGAGGU3’) located upstream of AUG codon Ini琀椀a琀椀on in eukaryotes Ini琀椀a琀椀on site = de昀椀nes the correct open reading frame. 1. Ini琀椀a琀椀on Met-tRNAi loaded into “P” site of small ribosomal subunit with ini琀椀a琀椀on factors (eIF2-GTP) 2. loaded small ribosomal subunit a琀琀aches to 5’end of mRNA (assisted by 5’end cap) 3. scans along mRNA (5’ to 3’ direc琀椀on) - iden琀椀昀椀es 1st AUG codon, surrounded by a long consensus sequence (Kozac sequence) Kozac sequence = degenerate = ribosome ini琀椀ate transla琀椀on from mul琀椀ple start codons = short polypep琀椀des 4. ini琀椀a琀椀on factors detach and large subunit binds = complete ribosomal complex 1st amino acid in ‘P’ site = ready for chain elonga琀椀on Ini琀椀a琀椀on in prokaryotes 1. 30S (small subunit) interacts with ini琀椀a琀椀on factor (IF) - complex binds Shine Dalgarno sequence upstream of AUG start codon 2. Ini琀椀ator tRNA binds (fMet-tRNAi) aligns with start codon (AUG) in mRNA 3. 50S associates with 30S, releasing IF = 70S complex tRNA occupies P-site of 50S subunit (A site = empty) Ini琀椀a琀椀on tRNA-Met Prokaryotes: Ini琀椀a琀椀on of transla琀椀on requires ini琀椀ator fMet-tRNAi. ini琀椀ator tRNA in bacteria = formylated methionine (N-formylmethionine tRNA or f Met-tRNAi) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Eukaryotes - Ini琀椀a琀椀on of transla琀椀on requires ini琀椀ator Met-tRNAi ini琀椀ator tRNA = Met-tRNAi (not formylated) Elonga琀椀on tRNA-Met Prokaryotes AND eukaryotes = Met codon inside coding region = elonga琀椀on Met-tRNA used. Met-tRNA has a di昀昀erent stem loop structure = preferen琀椀ally binds elonga琀椀on co-factors Met appears inside an open reading frame = Met-tRNA used  binds elonga琀椀on co-factors. Elonga琀椀on Use ATP to move ribosome 5’  3’, inser琀椀ng tRNAs above codons and catalysing pep琀椀de bond forma琀椀on between pep琀椀des 1. aminoacyl-tRNA molecule binds to vacant A-site on the ribosome 2. new pep琀椀de bond formed between amino acids a琀琀ached to the tRNAs in the P and A sites = tRNA-pep琀椀de in the A-site 3. mRNA moves a distance of 3 nucleo琀椀des (codon) through small subunit (uses ATP  ADP+PPi) o de-acylated tRNA molecule ejected from E site o A-site-tRNA-pep琀椀de  P site o A-site = empty and located over the downstream codon = ready to receive another aminoacyl-tRNA molecule = base pairs with codon posi琀椀on at which the growing pep琀椀de chain is a琀琀ached to a tRNA does not change during elonga琀椀on: always linked to tRNA present at P-site of the large subunit Ribosomes = rRNA and polypep琀椀des Polypep琀椀des = structural rRNA forms structured pocket of H networks + cataly琀椀c proper琀椀es ribosomes = ribozymes o rRNA performs condensa琀椀on of carboxy-terminus of amino acid on tRNA in P site with the amino-terminus of the amino acid in tRNA in the A-site = pep琀椀de bond o pep琀椀de-tRNA in A-site  P-site as ribosome moves down mRNA in the 3’ direc琀椀on. Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Termina琀椀on Termina琀椀on ini琀椀ated by stop codons: UAA, UAG, UGA = don’t match tRNA Releasing factors bind to a stop codon that reaches ‘A’ on ribosome = alters ac琀椀vity of pep琀椀dyl transferase (which forms pep琀椀de bond in large subunit) o adds water molecule to pep琀椀de and releases it ribosome releases mRNA  2 subunits dissociate and can bind to same/another mRNA at 5’end Termina琀椀on = ribosome encounters stop codon in A-site = hydrolysis reac琀椀on  release pep琀椀de from 昀椀nal tRNA = dissocia琀椀on of ribosome from mRNA. An琀椀bacterial ac琀椀on of an琀椀bio琀椀cs = inhibi琀椀ng transla琀椀on. Streptomycin/neomycin bind 30S subunit (small) = prevents transi琀椀on from ini琀椀a琀椀on to elonga琀椀on Tetracyclines/puromycin mimic structure of charged tRNA’s and block ‘A’ site of prokaryo琀椀c ribosome Chloramphenicol blocks pep琀椀dyl transferase reac琀椀on Erythromycin blocks transloca琀椀on reac琀椀on Eukaryo琀椀c ribosomes usually not inhibited (except for ribosomes in mitochondria - like prokaryotes) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 15 – PROTEIN STRUCTURE describe the composi琀椀on and characteris琀椀cs of amino acids classify amino acids according to their side chains characteris琀椀cs (basic, acidic, uncharged polar, nonpolar)  determine net charge of a polypep琀椀de memorise and use the three-le琀琀er amino acid code describe the forma琀椀on and characteris琀椀cs of pep琀椀de bonds describe noncovalent and covalent interac琀椀ons that play a role in determining the conforma琀椀on of a polypep琀椀de describe primary, secondary, ter琀椀ary and quaternary protein structures, -helices, -sheets and random coils, polypep琀椀de domains, homo- and heteromeric protein Amino acids α-carbon = asymmetric = 2 mirror images = stereoisomers L and D Proteins contain L-amino acids o Glycine = not a stereoisomer = neither L/D pH7.0 = amino and carboxyl groups ionised charge = pH dependent amino and carboxy groups = uncharged if part of the pep琀椀de bond of a protein. o only N-terminal amino group and C-terminal carboxy group = charged at neutral pH. Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Polypep琀椀des Pep琀椀de bonds form backbone of polypep琀椀de chain Condensa琀椀on reac琀椀on of a carboxylic acid and amino group oligopep琀椀de ≈ ten amino acids polypep琀椀de ≥ ten amino acids Flexible Molecules Resonance of amide structure = bond between the carbonyl carbon and nitrogen to have par琀椀al double bond character o = restricts rota琀椀on around pep琀椀de bond ➔ planar, rigid structure o Impacts 3D pep琀椀de structure Non-covalent interac琀椀ons restrict conforma琀椀on of polypep琀椀des  Ionic/electrosta琀椀c interac琀椀ons  Hydrophobic bonds  Hydrogen bonds  Van der Waals Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Covalent interac琀椀ons Disulphide bonds = stabilise conforma琀椀on (cys-cys) o Formed in oxida琀椀ng condi琀椀ons Denatura琀椀on and Renatura琀椀on of Proteins Denature highly concentrated urea = loses structural conforma琀椀on by ‘breaking’ non-covalent interac琀椀ons ➔ protein = inac琀椀ve reducing environment = break disul昀椀de bond Renature Remove urea/protein in oxidizing (cooling) environment = protein refold into na琀椀ve con昀椀gura琀椀on ➔ amino acid sequence = ‘memory’ o Not every protein can do this Protein structure Primary (1) = order (sequence) of amino acids Secondary (2) = local folded structures (folding mo琀椀fs in proteins) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 β-sheets rigid cores formed by β-sheets carbonyl oxygens on polypep琀椀de backbone in -strand form hydrogen bonds with hydrogen on nitrogen group of another - strand to form a –sheet. o H-bonds keep -strands together R-group s琀椀ck outwards from sheets = not involved in holding sheets together β-sheets can form with β-strands in same protein/polypep琀椀de or between β-strands in di昀昀erent polypep琀椀de chains An琀椀-parallel β-sheet neighbouring β-strands run in opposite orienta琀椀on (NC terminus & CN terminus) Parallel β-sheet neighbouring β-strands run in same orienta琀椀on (both NC terminus/CN terminus) α -sheets H-bonds between carbonyl oxygen of pep琀椀de bond and amide hydrogen of amino acid 4 residues away = stabilise helical structure o R group s琀椀ck out Transmembrane proteins contain α-helical regions o transmembrane domains = formed by α-helices o hydrophobic side chains s琀椀ck out into hydrophobic lipid bilayer membrane Random coil unstructured units DON’T form regular secondary structure and NOT characterized by regular H-bonding pa琀琀ern found in 2 loca琀椀ons in proteins: Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 o Terminal ends - both N-terminus and C-terminus o Loops - between regular secondary structure elements (α– helices & β–sheets) Protein domain/mo琀椀f region of a polypep琀椀de that can fold independently into a compact, stable structure o domain of 4 alpha helices o domain of alpha helices and beta strands o domain of beta strands = beta sheet occur in di昀昀erent proteins with di昀昀erent amino acids similar domains in proteins with similar func琀椀ons in evolu琀椀onarily distant organisms o increased organism complexity increases number of domains Ter琀椀ary (3) = full 3-dimensional conforma琀椀on ALL -helices, -sheets, random coils, loops of a polypep琀椀de chain Quaternary (4) = 3-dimensional rela琀椀onship of polypep琀椀des in a protein made up of >1 protein. Mul琀椀subunit proteins (Each protein = a subunit) Homodimer: made up of two iden琀椀cal protein subunits. Heterodimer: made up of two di昀昀erent protein subunits. o Haemoglobin tetramer = heterotetramer LECTURE 16 – PROTEIN FUNCTION amino acid sequences show evolu琀椀onary rela琀椀onships of proteins and determine, shape, 昀氀exibility and func琀椀on of proteins Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 explain that all proteins bind other molecules named ligands via non-covalent interac琀椀ons de昀椀ne di昀昀erent types of proteins and describe their features explain how enzymes work and how they are regulated explain how motor proteins move cargos in cells Amino acid sequence = determines shape, 昀氀exibility and func琀椀on of proteins Proteins Bind Other Molecules – ligands Ligand = Ion (Mg2+), Small Molecules (ATP) or Macromolecules (proteins, DNA) Binding to binding sites is speci昀椀c due to shape complementarity o Binding = via non-covalent interac琀椀ons Binding of ligand = protein folds to provide close 昀椀t o Non 昀椀琀 ng ligands fall out of binding; sum of non-covalent interac琀椀ons = weak ➔ unwanted associa琀椀ons prevented o Correct ligand binds ➔ many non-covalent interac琀椀ons; 琀椀ght 昀椀t into binding site An琀椀bodies Bind Ligands – An琀椀gens An琀椀bodies: produced by immune system against foreign molecules/an琀椀gens = ligands 2 heavy and 2 light chains, variable and constant domains bind an琀椀gens with an琀椀gen binding site = an琀椀gen destruc琀椀on Variable domains have variable loops ➔ to changing length & amino acid sequence ➔ speci昀椀city of an琀椀body to an琀椀gen ➔ poten琀椀ally billions produced by an immune system Structural proteins = Fibrous Kera琀椀n protein = α-helical regions Kera琀椀n monomers assemble into dimers  Dimers form staggered tetramers  8 tetramers form intermediate 昀椀laments Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Gene琀椀c disorders characterised by blistering of skin from muta琀椀ons in kera琀椀n genes (clumping & disrup琀椀on of kera琀椀n network in basal cells) Rela琀椀onships and func琀椀onality Rela琀椀onships = similar/di昀昀erent amino acid sequence → similar sequences cluster (subfamily of proteins) Func琀椀onality re昀氀ected in clustering Enzymes Proteins that speed up reac琀椀ons by reducing ac琀椀va琀椀on energy required for reac琀椀on to occur o Enzyme (E) binds speci昀椀c ligand/substrate (S) = enzyme substrate complex (ES) o Enzyme catalyses a change (cleavage) of substrate = enzyme product complex (EP) o EP dissociates = release product (P) from enzyme (E)  Binding = noncovalent interac琀椀ons in ac琀椀ve site  Organiza琀椀on of atoms in the ac琀椀ve site is op琀椀mized for catalysis Encourage reac琀椀on by; o Bind to 2 S molecules and orients them for reac琀椀on o Bind S to E = rearranges electrons in S = delta charges = favour reac琀椀on o Enzyme deforms bound S  transi琀椀on state (S breaks and released) Not consumed during reac琀椀on = available to bind new substrates and catalyse reac琀椀on repeatedly Substrate speci昀椀c: o pep琀椀dase break pep琀椀dase-speci昀椀c pep琀椀de bonds in target proteins Regulated: o by binding other molecules/by phosphoryla琀椀on o gene expression and compartmentalise enzymes o regulate enzyme degrada琀椀on Posi琀椀ve regula琀椀on = increase number of ac琀椀ve enzymes by binding a di昀昀erent molecule to the substrate. Posi琀椀ve e昀昀ector molecule = X = binds to regulatory site of protein Frequency of enzyme molecule binding to glucose increase = higher metabolism of glucose Nega琀椀ve regula琀椀on = feedback inhibi琀椀on Nega琀椀ve e昀昀ector molecule = binds to regulatory site of protein o Product made late in pathway = X o Inhibits enzyme catalysing a reac琀椀on early in pathway Regulates connected metabolic pathways o P of 1 enzyme-catalysed reac琀椀on = S for another → metabolic pathways o Feedback inhibi琀椀on at mul琀椀ple points regulates connected metabolic pathways Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Allosteric regula琀椀on = Enzymes regulated by e昀昀ector molecules = Allosteric Proteins o >2 di昀昀erent conforma琀椀ons; ac琀椀vity regulated by switching conforma琀椀ons Regulatory and ac琀椀ve sites “communicate” o e昀昀ector bound to regulatory site = changes shape of enzyme o Binding of posi琀椀ve regulator = allosteric ac琀椀vator changes shape of ac琀椀ve site = S binds be琀琀er o Binding of nega琀椀ve regulator = allosteric inhibitor changes shape of ac琀椀ve site = S doesn’t bind Phosphoryla琀椀on regula琀椀on Protein kinases = catalyse phosphoryla琀椀on of amino acids with OH (serine, threonine and tyrosine) o Proteins = ac琀椀vated/inac琀椀vated by phosphoryla琀椀on Protein phosphatase removes phosphate groups from amino acid Motor proteins Kinesin & dynein o move cytoplasmic components (cargo) along microtubules (cytoskeleton) in opposite direc琀椀ons ATP hydrolysis (ATP ADP) occurs at head regions of motor proteins Cargo bound at tail regions of motor proteins Move by conforma琀椀onal changes Movement of motor proteins = coupled with hydrolysis of ATP o ATP bound = conforma琀椀onal changes in motor protein o Hydrolysis of ATP (ATP ADP)= another conforma琀椀onal change conforma琀椀onal changes = motor protein to move; through cell/along other proteins LECTURE 12 – CLONING Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 de昀椀ne; recombinant DNA molecule, DNA cloning, restric琀椀on enzyme, palindromic sequence, blunt and s琀椀cky (cohesive) ends, DNA ligase, plasmid vector, agarose gel electrophoresis, large- scale protein produc琀椀on, protein expression, pep琀椀de tag understand the enzymes, DNA sequences, DNA products and experimental steps used in DNA cloning recognise that restric琀椀on enzymes are part of a bacterial defence system describe the characteris琀椀cs of plasmid vectors and how they can be used to clone DNA fragments and to produce large-scale amounts of proteins Recombinant DNA molecule = molecule of DNA originated from 2 or more DNA fragments that aren’t found together in nature DNA Cloning = Produc琀椀on of iden琀椀cal copies of a par琀椀cular DNA molecule Isola琀椀on of a par琀椀cular piece of DNA from the rest of a cell’s DNA o = copies of DNA fragments  produce proteins/study gene muta琀椀ons Use restric琀椀on enzymes = enzyma琀椀cally insert DNA fragment/plasmids into plasmid vector by cu琀 ng (DNA/plasmid) at sequence speci昀椀c sites = recombinant plasmid Restric琀椀on enzymes = Endonucleases Digest dsDNA at internal phosphodiester bonds Cut DNA at speci昀椀c sites = complimentary ends O de昀椀ned by nucleo琀椀de (recogni琀椀on) sequences = restric琀椀on sites/sequences Palindromes (Palindromic DNA sequences) = same sequence forward & backwards (5’ 3’ & 3’5’) Restric琀椀on enzymes recognise and bind to speci昀椀c palindromic sequences and cut them O 4-8 nucleo琀椀des Leave 3’-OH & 5’-phosphate groups on both cut strands DNA fragments with complementary s琀椀cky ends can anneal Blunt (DNA) ends Restric琀椀on enzyme cut both strands at same posi琀椀on S琀椀cky (cohesive) ends cut makes 5’ or 3’ overhang Name of restric琀椀on enzyme re昀氀ects organism of origin 1st le琀琀er from Genus (Escherichia) and the next 2 from the Species (coli) = EcoRI Func琀椀on of restric琀椀on enzymes in nature = Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 O Protect cells from invaders (virus = not methylated) O Acts together with a methylase = protects own (methylated) DNA ligase Covalently links 3’-hydroxyl and 5’- phosphate groups Uses ATP o E.g; DNA ligases join Okazaki fragments during DNA replica琀椀on Plasmids Small, extrachromosomal, double-stranded, circular DNA molecules (found in bacterial cells) Carry genes bene昀椀cial for survival of the bacterium under certain condi琀椀ons Dis琀椀nct from the bacterial chromosome - replicates independently from chromosomal DNA “Vectors” = move DNA from one place to another Promoter: expression of a gene cloned downstream of the promoter start of making RNA (ac琀椀vate transcrip琀椀on) Restric琀椀on enzyme sites: cloning of DNA of interest. (inserted cDNA) many unique restric琀椀on enzyme sites = mul琀椀ple cloning site or polylinker Origin of replica琀椀on: Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 replica琀椀on of the plasmid = ensure it’s passed on to daughter cells during bacterial division. = several copies of a plasmid are found in a bacterial cell. Selec琀椀on marker: select bacteria that have the plasmid. (grows = contains vector) o marker = an琀椀bio琀椀c resistance gene, then bacteria without marker will die in the presence of the an琀椀bio琀椀c. Plasmid (vector) map 1. mul琀椀ple cloning site or polylinker ➔ insert DNA of interest 2. origin of replica琀椀on ➔ star琀椀ng point of plasmid DNA replica琀椀on in bacteria 3. an琀椀bio琀椀c resistance gene – selectable marker  Makes bacteria resistant to ampicillin Transforma琀椀on of bacterial cells Heat shock = 42 degrees o Opens up cell wall = looser membrane = DNA into cell Electropora琀椀on o Current applied to cells = bacterial cell wall/membrane permeable to DNA Agarose Gell red algal carbohydrate boil in bu昀昀er = dissolve  into tray containing comb = solid gel with loading pockets (cooled down) Pull out combs, transfer tray with gel into an electrophoresis tank and overlay with a bu昀昀er Agarose Gel Electrophoresis Agarose gel placed in an electrophoresis chamber with a posi琀椀ve and nega琀椀ve electrode voltage ➔ DNA (nega琀椀vely charged) migrates to posi琀椀ve electrode (smallest move fastest) o DNA fragments separated by size Ethidium bromide (昀氀uorescent dyes) binds DNA (& RNA) and 昀氀uoresces when viewed with ultraviolet light Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Large scale produc琀椀on of proteins Host cells = bacteria/yeast cells/mammalian cells/insect cells Expression vector encodes o highly ac琀椀ve promoter region Pep琀椀de tag for isola琀椀on of expressed protein o nucleo琀椀de sequence coding for a cluster of his琀椀dine residues (His-tagged) cluster of His residues can bind to metal ions o His-tagged, overexpressed proteins o eluted from column by reducing the pH of the solvent LECTURE 13 – COPYING DNA AND RNA IN VITRO list the components of a PCR and explain their roles in a PCR reaction. describe the steps in a PCR cycle and relate this to amplification of DNA. explain the binding of a PCR primer pair to a complementary double-stranded template and extension of the primers by DNA polymerases, including Taq DNA polymerase. describe cDNA synthesis, including the properties of the enzymes used. explain using examples the differences between PCR, RT-PCR and qRT-PCR and when each is used Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 DNA fragments all/parts of DNA sequence = known – related sequence synthesise DNA fragment using PCR o use DNA polymerases = replicate DNA in vitro DNA polymerase replicate DNA in vivo and in vitro DNA polymerase release a pyrophosphate (2 phosphates) from dNTPs and adds the resul琀椀ng dNMP to the 3’OH end of the primer strand via a nucleophilic a琀琀ack mechanism. o Pyrophosphate cleaved by pyrophosphatases (irreversible rxn) o Deoxynucleo琀椀de monophosphate added to 3’ end of newly synthesised strand o Template strand read 3’ 5’ = new strand synthesised 5’3’ Use a heat stable DNA polymerase = Taq polymerase o from thermophilic bacteria Components of PCR Template DNA = double stranded DNA to be ampli昀椀ed 2 oligonucleo琀椀de primers = short, single stranded DNA oligonucleo琀椀des o bind to either the upper/lower template strand Taq DNA polymerase = enzyme that synthesises copies of the template DNA dNTPs = building blocks for newly synthesised DNAs o dATP, dCTP, dGTP and dTTP Bu昀昀er = ensures reac琀椀on condi琀椀ons (pH and Mg ions) are suitable for ampli昀椀ca琀椀on Magnesium ions = cofactor for Taq DNA polymerase. o enzyme cannot work without it. PCR machine = correct temperatures and 琀椀ming for individual PCR steps o Thermocycler =  2 primers; 1 to hybridize to each strand of a target DNA  DNA polymerase I fragment (Klenow fragment) from E. coli for exponen琀椀al DNA replica琀椀on.  Now thermostable DNA polymerases. Oligonucleo琀椀de primers base pair with template DNA strand 20-30 nucleo琀椀des in length DNA sequence info required to design primers Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 PCR – Polymerase Chain Reac琀椀on in vitro method for genera琀椀ng large copy numbers of a speci昀椀c DNA fragment repeats (cycles) of 3 steps  large ampli昀椀ca琀椀on of the template DNA o dsDNA  2 ssDNA (high temperatures) o 2 primers for each ssDNA  DNA synthesis begins with Taq polymerase = 2 ds DNA molecules synthesised = 1st cycle # of DNA molecules generated from 1 template molecule = 2n, (n = cycle number) (PCR) – amplifying DNA in vitro 1. 95℃ = separate 2 DNA strands ➔ denaturing dsDNA into ssDNA (template DNA) by disrup琀椀ng IMFs 2. 50℃ - 60℃ = primers to hybridise (anneal) to template via base pairing 3. 72℃ = Working temperature for Taq DNA polymerase. o Deoxynucleo琀椀des incorporated into new DNA strand by polymerase using old DNA strand as a template. Applica琀椀ons of PCR Cloning a DNA fragment Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Parental studies  Molecular marker = Any site (locus) in the genome of an organism at which the DNA base sequence varies among the di昀昀erent individuals of a popula琀椀on. Environmental studies  molecular PCR markers - determine microbial diversity o = assess environmental impacts and rehabilita琀椀on  Rhizobium leguminosarum = bacterium o PCR tests for markers = diversity of bacteria o Contaminated = low diversity/uniform size vs non-contaminated = high diversity/variable size RT-PCR = Reverse Transcrip琀椀on-Polymerase Chain Reac琀椀on  produce a DNA copy of an mRNA Reverse Transcrip琀椀on (RT) of RNA  eukaryo琀椀c mRNAs = 3’ poly-A tail, hybridize with poly(T) primer o exon sequences -mRNA  Reverse transcriptase: Makes RNA:DNA hybrid o RNA-dependent DNA polymerase found in retroviruses o requires primer = poly T primer (complements poly A tail of mRNA)  poly A and poly T primer = hybrid recognised by transcriptase o Synthesizes DNA in 5’3’ direc琀椀on by adding dNTPs (DNA copy of mRNA)  uses RNA strand as template strand  RNaseH: degrades RNA in RNA:DNA hybrid o = ssDNA molecule used in PCR to make DNA o some RNA NOT degraded ➔ primer for complementary DNA (=cDNA) strand synthesis Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  DNA polymerase: synthesis complementary DNA (cDNA) strand o cDNA ampli昀椀ed (cloned) through vectors & bacteria or using PCR Applica琀椀on of RT-PCR – Gene expression studies  Study evolu琀椀onary origins of mammalian immunoglobulin heavy chain isotypes in di昀昀erent an琀椀body types (IgM, IgD, IgG, IgE, and IgA). o inves琀椀gated 琀椀ssue speci昀椀c mRNA expression of the di昀昀erent immunoglobulins  IgO only expressed in spleen  RT-PCR detec琀椀on of the platypus Ig (immunoglobulin) gene expression in di昀昀erent 琀椀ssues. o NO Northern blo琀 ng to detect 琀椀ssue expression of the platypus Ig genes - unavailability of high-quality RNA = performed RT-PCR Summary: Synthesising a DNA strand in vitro requires:  primer that base pairs with the DNA template to be sequenced  DNA polymerase to extends the primer, synthesises a new DNA strand  dNTPs as building blocks LECTURE 14 – ANALYSING DNA AND RNA  Describe the workings of Sanger DNA sequencing  describe the condi琀椀ons double-stranded nucleic acid molecules are denatured and renatured  hybridisa琀椀on can occur between complementary DNA and/or RNA molecules  describe the di昀昀erence between heterologous and homologous probes  describe the process of random priming and the use of labelled nucleo琀椀des  describe Southern, northern, and in situ hybridisa琀椀on techniques and give examples of the uses of these techniques. Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Di-deoxyribonucleo琀椀de triphosphates (ddNTPs) cause chain termina琀椀on during DNA synthesis  lack 3’OH on the sugar – prevents nucleo琀椀des from adding Sanger DNA sequencing ddNTPs  determine at which nucleo琀椀de chain termina琀椀on occurs  dNTP:ddNTP ra琀椀o determines how long synthesised DNA is Labelled DNA Primer  label with radioac琀椀ve phosphate  DNA detected/visualised through X-ray 昀椀lm Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Sanger sequencing done in 1 reac琀椀on tube using 昀氀uorescently labelled ddNTPs  4 ddNTPs labelled with 4 di昀昀erent 昀氀uorescent labels  DNA synthesis done in the presence of all 4 dNTPs and all 4 labelled ddNTPs in 1 tube  ddNTPs = chain termina琀椀on  synthesised products separated by size using capillary gel electrophoresis ➔ shortest products run fastest  laser excites 昀氀uorescent labels = 昀氀uorescent labels emit a label speci昀椀c light = detected  computer records the label speci昀椀c signal over 琀椀me and produces a chromatogram. Iden琀椀fying DNA fragments in SIMPLE DNA samples/genomes  restric琀椀on digest + agarose gel electrophoresis = map of simple DNA genomes  Lambda genome = speci昀椀c DNA fragment - easily iden琀椀昀椀ed Iden琀椀fying DNA fragments in COMPLEX DNA samples/genomes  Diges琀椀on of complex genomes with restric琀椀on enzymes = many fragments with di昀昀erent size  Smear visualised in agrose gel of digested DNA = DNA fragments with small size di昀昀erences migra琀椀ng close together Iden琀椀fying a speci昀椀c DNA fragment in complex mixture = Hybridising a labelled probe to DNA fragment Hybridisa琀椀on Denatura琀椀on and renatura琀椀on of DNA  Renatura琀椀on = restores DNA double helices by reforming nucleo琀椀de pairs o Cool or lower pH  Denatura琀椀on = H-bonds between base pairs are disrupted = denature dsDNA  ssDNA o High pH or temperature Hybridisa琀椀on also occurs between:  complementary RNA strands → double-stranded RNA molecules  complementary DNA and RNA strands → RNA/DNA hybrids Hybrid of 2 nucleic acid molecules Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Hybrid = probe to a ssDNA strand Probes = Single stranded, usually DNA (15-1000’s nucleo琀椀des) o Hybridize/anneal to single stranded DNA molecules o Hybridiza琀椀on temperature impacts outcome  Homologous probes: detect iden琀椀cal nucleic acid molecules o 100% complementa琀椀on at higher temperatures  Heterologous probes: detect related nucleic acid molecules o lower temperature – imperfect base pairing but form stable double helices Formamide = lowers mel琀椀ng temperature of nucleic acid duplexes  = facilitates forma琀椀on of complementary strands Hybridisa琀椀on techniques – detect speci昀椀c DNA and RNA molecules DNA:  Southern blo琀 ng = gel based, uses a blot and hybridisa琀椀on  Fluorescent in situ hybridiza琀椀on (FISH) = cellular based, hybridisa琀椀on done using 琀椀ssue sec琀椀ons RNA:  northern blo琀 ng = gel based, uses a blot and hybridisa琀椀on  In situ hybridisa琀椀on = cellular based, hybridisa琀椀on done using 琀椀ssue sec琀椀ons Southern/northern blo琀 ng 1. Nucleic acids separated by size by agarose gel electrophoresis  Separa琀椀on depends on agarose concentra琀椀on  Unlabelled RNA/DNA  labelled RNA/DNA = known sizes 2. Separated nucleic acids blo琀琀ed on to nitrocellulose paper/nylon membrane by suc琀椀on of bu昀昀er through gel and paper  membrane permeable to bu昀昀er but DNA/RNA gets transferred to membrane (stuck) 3. Membrane + labelled probe in bu昀昀er  labelled probe hybridised  Remove nitrocellulose/nylon membrane with adhered DNA or RNA  Labelled, denatured probe added in a bu昀昀er to the membrane  Labelled probe hybridises to complementary DNA (Southern) or RNA (northern) Southern blo琀 ng Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Separa琀椀ng nucleic acids = digest DNA with restric琀椀on enzyme  0.5-2% agarose gel = separates nucleic acids  migrates as dsDNA  denature DNA in alkaline solu琀椀on ➔ separates dsDNA to ssDNA  Transfer using alkaline solu琀椀on or bu昀昀er - Bu昀昀er carries DNA from gel to nitrocellulose/nylon membrane  DNA adheres to membrane whilst bu昀昀er goes through to paper towels Used for DNA analysis:  iden琀椀fy related genes, number and sizes  construc琀椀on of restric琀椀on maps Northern blo琀 ng  Separa琀椀ng using 1-1.5% agrose gel with denaturing agent  removes 2o structures from RNA = RNA migrates as single stranded molecule  Transfer using salt solu琀椀on (bu昀昀er) - carries RNA from gel to nitrocellulose paper/nylon membrane  RNA adheres to membrane whilst bu昀昀er goes through to paper towels Used for RNA analysis:  iden琀椀fy related transcripts, number and size(s) of related transcripts  determine pa琀琀erns of gene expression o 琀椀ssue/cell types, development In situ hybridisa琀椀on (ISH)  uses a labelled complementary DNA or RNA probe to localize a speci昀椀c DNA/RNA sequence in a por琀椀on/sec琀椀on of 琀椀ssue or cell (in situ)  RNA detec琀椀on – using an琀椀sense probes o An琀椀sense probes = labelled sense probe with a complementary sequence to transcript = localise transcripts o Sense probe = same sequence as transcripts = nega琀椀ve control/no hybridisa琀椀on RNA analysis:  localize transcripts within 琀椀ssues/cells  determine expression pa琀琀erns of speci昀椀c genes in 琀椀ssues Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Fluorescent in situ hybridisa琀椀on (FISH)  approximate mapping of genes directly on chromosomes 1. Cells & nuclei broken open 2. Chromosomes spread on a microscope slide 3. Chromosomes denatured 4. Fluorescent labelled DNA probe hybridized to chromosomes 5. Visualized under a 昀氀uorescent microscope - probe hybridized to complementary sequence on a chromosome  Each sister chroma琀椀d has a copy of the gene of interested Random priming – labelling DNA molecules  making a probe  DNA denatured = ssDNA – act as templates for DNA synthesis  Hexanucleo琀椀des = DNA primer with 6 nucleo琀椀des o mixture of varia琀椀ons of hexanucleo琀椀des increases chances of any DNA molecule being labelled  DNA polymerase adds nucleo琀椀des to primer = synthesise complementary DNA strands. o Labelled nucleo琀椀des used = popula琀椀on of DNA molecules that contain labelled examples of all sequences of both DNA strands Label DNA probes – making a probe chemically labelled dNTP  DNA labelled with digoxygenin (DIG) labelled dNTP - detected via an琀椀-digoxygenin an琀椀bodies radioac琀椀ve labelled dNTP - dATP labelled with radioac琀椀ve isotope 32P  DNA labelled with radioac琀椀ve labelled dNTP is detected via exposure to X-ray 昀椀lm LECTURE 17 – REGULATION OF GENE ACTIVITY IN PROKARYOTES  di昀昀erences between gene regula琀椀on in prokaryotes and eukaryotes  Understand the e昀昀ect of and reasons for gene regula琀椀on  4 ways that a cell can control the proteins it makes  Describe the basic principles of coordinate regula琀椀on, catabolic vs. anabolic pathways, and posi琀椀ve vs. nega琀椀ve regula琀椀on  Describe the tryptophan operon of E. coli and its nega琀椀ve regula琀椀on, using diagrams  De昀椀ne the terms operon, promoter, operator, repressor and polycistronic mRNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Genes = DNA sequence that codes for a polypep琀椀de, tRNA or rRNA  Cons琀椀tu琀椀ve genes: always expressed o essen琀椀al for basics of life; DNA synthesis, replica琀椀on and repair, RNA & protein synthesis  Con琀椀ngency genes: encode products that confer an advantage under special condi琀椀ons O heat stress, pH stress, starva琀椀on, carbon source availability Gene regula琀椀on Prokaryotes  purpose of gene control = provide max growth  challenges = con琀椀nual environmental changes  range of gene expression = low basal level – quick response o Total switch-o昀昀 is rare: “o昀昀”  produc琀椀on of mRNA = transcrip琀椀on and transla琀椀on in cytoplasm. Eukaryotes  purpose of gene control = regulate development and di昀昀eren琀椀a琀椀on  challenges = constant environmental  range of gene expression = Total switch-o昀昀 is common  produc琀椀on of mRNA = o transcrip琀椀on in nucleosome o RNA splicing in nucleus o transla琀椀on in cytoplasm Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Regula琀椀on of gene ac琀椀vity in Prokaryotes  number of protein molecules produced by ac琀椀ve genes varies from gene to gene and varies in response to the environment.  = sa琀椀sfy needs + avoid wasteful synthesis o Molecules needed = synthesized only when needed o enzyme that consumes the substrate of a 2nd enzyme = inhibited if end-product of 2nd enzyme is required o 2 pathways for energy produc琀椀on - cell ‘chooses’ 1 that yields the most energy  Mul琀椀ple mechanisms to regulate gene expression o both transcrip琀椀on and transla琀椀on can be regulated. o on/o昀昀 regula琀椀on: system turned on when needed, o昀昀 when not needed  ini琀椀a琀椀on of transcrip琀椀on = most important point of control 4 ways cell controls amount of proteins made 1. Transcrip琀椀onal control: o control when and frequency gene is transcribed 2. RNA processing control: o control how the RNA transcript is spliced/processed  correct 3’ poly A tail or 5’ G methylated cap – eukaryotes 3. Transla琀椀onal control: o selec琀椀ng which mRNAs in the cytoplasm are translated by ribosomes 4. Post-transla琀椀onal control: (protein ac琀椀vity control) o selec琀椀vely ac琀椀va琀椀ng or inac琀椀va琀椀ng proteins a昀琀er synthesis 3 principles of regula琀椀on Coordinate regula琀椀on  Metabolic pathway to convert compound A to compound D  ALL or NO enzymes are present  Prokaryotes = Polycistronic mRNA encodes all 3 enzymes = all 3 genes translated Degrada琀椀ve vs biosynthe琀椀c pathways  Degrada琀椀ve = catabolic pathway o Large complex substrate molecule = regulatory molecule  Availability determines if enzymes are synthesised  present = broken down to smaller simpler end product = genes on o release heat/energy  biosynthe琀椀c = anabolic o larger complex end product = regulatory molecule o uses energy Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Nega琀椀ve vs Posi琀椀ve regula琀椀on Nega琀椀ve:  Repressor protein prevents transcrip琀椀on. o repressor bound to the DNA = prevents transcrip琀椀on of the gene. o Binds to operator on DNA = genes switched o昀昀  Inducer = antagonist of repressor = needed to remove repressor from DNA and allow ini琀椀a琀椀on of transcrip琀椀on  Corepressor binds to repressor and allows repressor to block transcrip琀椀on. Posi琀椀ve:  Ac琀椀vator binds to DNA = allows transcrip琀椀on. o ac琀椀vator binds to the DNA = ac琀椀va琀椀on of transcrip琀椀on of the gene.  Ligand (small molecule; sugars, amino acids, salts, metals) binds the ac琀椀vator. o determines whether the ac琀椀vator binds the DNA. De昀椀ni琀椀ons  Operon: group of genes adjacent to each other on the chromosome that are transcribed from a single promoter into a single mRNA molecule. Operons are not found in eukaryotes.  Promoter: DNA sequence that RNA polymerase binds to  opens DNA double helix and begin mRNA synthesis (transcrip琀椀on) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Operator: short region of DNA which the repressor protein binds to o controls expression of the genes adjacent to it in the operon.  Repressor: Protein that binds to an operator sequence to prevent transcrip琀椀on of the adjacent genes in the operon.  Polycistronic mRNA: RNA with >1 coding region = translated into many di昀昀erent proteins; formed when an operon is transcribed. Tryptophan Operon  biosynthesis operon - manufactures the amino acid tryptophan (Trp)  5 genes encode the enzymes needed to make Trp.  In the Trp operon; o operator region is within the promoter sequence - 15 bp long Nega琀椀ve regula琀椀on of the tryptophan operon  NO (Low) tryptophan present in surroundings = genes switched ON (transcribed) o Inac琀椀ve repressor (TrpR) = no tryptophan present  polymerase binds = genes on  High Trp present  enters bacterial cell = enzymes no longer needed = genes switched OFF o Tryptophan present binds to repressor – changes conforma琀椀on so repressor can bind to operator = makes repressor ac琀椀ve and polymerase cant bind to promoter = genes o昀昀 o Tryptophan = corepressor  trpR encodes TrpR repressor - synthesized in inac琀椀ve form  NO tryptophan = repressor unable to bind to the operator  tryptophan (corepressor) binds to repressor protein TrpR = change in 3-D structure of TrpR o = repressor TrpR to bind to operator o tryptophan repressor binds to operator = blocks access of RNA polymerase to promoter  = prevents transcrip琀椀on of operon and produc琀椀on of the tryptophan-producing enzymes  enzymes encoded by trp operon = biosynthe琀椀c pathway = wasteful to make if tryptophan is available  Tryptophan = “expensive” for the cell to make: o cell consumes many metabolites and lots of energy to make tryptophan = trp operon func琀椀ons only when tryptophan levels are low, and tryptophan must be made from precursor molecules in the cell Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 18 – LACTOSE OPERON AND NEGATIVE REGULATION  usage of glucose and lactose in E. coli  List the enzymes of the lactose operon and their func琀椀ons  Describe the lactose operon and its nega琀椀ve regula琀椀on, using diagrams  Explain the terms inducer, inducible, on-o昀昀 regula琀椀on and diauxic growth Lactose Operon  operon - transport and metabolism of lactose in E. coli.  Lactose = a carbon and energy source o Glucose = preferred C source for E. coli o supply E. coli with glucose and lactose = use glucose un琀椀l exhausted  stop growing brie昀氀y  start growing again using lactose  operator = adjacent to promoter  lacI gene = upstream (5ꞌ) of the operon. o transcribed from its own promoter and translated separately = repressor protein Growth of E. coli with glucose and lactose provided  diauxic growth = cellular growth in 2 phases  2nd lag phase = cells adjusted by turning on lac operon = produce enzymes to break down lactose Enzymes - lactose metabolism in E. coli lactose permease  encoded by lacY gene  Transports lactose into the cell β-galactosidase  encoded by lacZ gene Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841  Breaks lactose down to its component sugars: glucose and galactose (catabolic/degrada琀椀ve) = Catalyses hydrolysis of lactose o Catalyses lactose  allolactose = inducer of β-galactosidase synthesis Regula琀椀on of lactose operon Inducible system  (lactose) Allolactose = inducer of the produc琀椀on of the 2 enzymes o inducer = s琀椀mulates synthesis of an inducible protein NO lactose = genes OFF and enzymes at basal level Lactose present = genes ON Nega琀椀ve regula琀椀on  NO lactose = enzymes not needed = switched OFF  inducer lactose (allolactose) present = enzymes needed = switched ON o Inducer binds to repressor protein, alters repressor conforma琀椀on = prevents repressor binding to operator site on DNA Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Nega琀椀ve regula琀椀on of the lactose operon summary  Lactose metabolism - 2 proteins/enzymes = β-galactosidase and lactose permease o genes (Z and Y + A) and transacetylase = clustered together and transcribed from 1 promoter = polycistronic mRNA = form an operon (lacZ, lacY, lacA)  Nega琀椀ve control of lac operon = o repressor protein binds to the operator = prevent RNA polymerase from binding to the promoter = no transcrip琀椀on o lactose present – allolactose (deriva琀椀ve) acts as an inducer  binds to the repressor = dissociate from the operator: transcrip琀椀on of structural genes occurs = genes ON Lac repressor  tetramer with 2 iden琀椀cal binding sites  lac operator = 3 sites: O1, O2, O3 o repressor binds O1 (main operator) and either O2 or O3 = forms DNA loop o loop contains the -35 and -10 binding sites recognised by RNA polymerase o sites = inaccessible to RNA polymerase when repressor is bound Lac Operon vs Tryptophan operon – Nega琀椀ve regula琀椀on  lac operon = catabolic (degrada琀椀ve) operon o NO lactose = switched OFF o Allolactose (inducer) PRESENT = enzymes needed = switched ON  trp operon = anabolic (biosynthe琀椀c) operon o NO tryptophan = genes switched ON o Tryptophan PRESENT = enzymes no longer needed =switched OFF Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 LECTURE 19 – LACTOSE OPERON AND POSITIVE REGULATION  components in posi琀椀ve regula琀椀on of the Lac operon  Describe the posi琀椀ve regula琀椀on of the Lac operon, using diagrams  3 regulatory DNA-binding proteins in bacteria  Understand how proteins can bind to speci昀椀c regions of DNA, and the roles of some conforma琀椀onal changes caused by this binding Components in posi琀椀ve regula琀椀on of the Lac operon  Lac promoter: “weak” promoter o RNA polymerase recognises it poorly  ac琀椀vator protein = CAP: help RNA polymerase bind to the promoter. o binds to DNA near the promoter o CAP = catabolite ac琀椀vator protein/ CRP = cAMP receptor protein  Cyclic AMP (cAMP): intracellular signalling/messenger molecule. o Responds to changes in glucose concentra琀椀on o CAP must bind to cAMP before CAP can bind to DNA = acts as an inducer  CAP-cAMP dimer binds to DNA Posi琀椀ve regula琀椀on  Glucose = preferred sugar  Lac operon = inac琀椀ve if glucose is present  Cells sense the lack of glucose  ac琀椀vate transcrip琀椀on of the Lac operon (lactose present) o Lac operon on = accumulate enzymes that break down lactose  Inverse rela琀椀onship between glucose & cAMP concentra琀椀on o ↑ glucose = ↓ cAMP  = low cAMP = CAP can’t bind to form dimer = operon OFF  cAMP binds to CAP protein  conforma琀椀onal change = Allows CAP to bind to CAP site  bound CAP-cAMP dimer interacts with RNA polymerase = s琀椀mulates transcrip琀椀on ini琀椀a琀椀on.  CAP-cAMP dimer binds to DNA and ac琀椀vates transcrip琀椀on = Lac operon ac琀椀vated when glucose concentra琀椀on is low  metabolise alterna琀椀ve energy source = lactose  ↑ cAMP levels = ↑ CAP-cAMP dimer levels o more cAMP available to bind to CAP  operon = highly expressed only when lactose is present, and glucose is absent Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 o ↑ lactose = allolactose present = repressor cant bind = operon ON o ↑ glucose = ↓ cAMP = CAP can’t bind to form dimer = operon OFF (no CAP = polymerase cant bind well) Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 Signalling molecules = cAMP and tryptophan  Can’t bind to DNA to ac琀椀vate/inac琀椀vate DNA binding proteins  Transcrip琀椀on = controlled by proteins binding to regulatory DNA sequences  Regulatory DNA-binding proteins o Trp repressor TrpR – binds to trp operon with trp present o Lac repressor LacI – binds to lac operon with lactose absent  Repressor binds to operator DNA  Tetrameric – each dimeric repressor binds to 21 base pair sequence o CAP ac琀椀vator – binds to lac operon with glucose absent  outside of DNA helix read by proteins o Repressors recognise operator DNA o CAP recognises CAP-binding site DNA  surface of the protein 昀椀ts 琀椀ghtly against the surface of the speci昀椀c DNA region it recognises o protein inserts into the major groove of the DNA double helix  protein forms bonds with bases – H-bonds, hydrophobic or ionic o NOT covalent bonds o 20 contacts between protein and DNA = very strong & speci昀椀c binding Binding of a regulatory protein to DNA  protein-DNA interface consist of 10-20 contacts involving di昀昀erent amino acids  DNA-binding helix-turn-helix mo琀椀f = DNA-binding mo琀椀fs. o C terminal of protein = recogni琀椀on helix = recognises DNA major groove. helix-turn-helix DNA-binding proteins.  bind as dimers = 2 copies of the recogni琀椀on helix separated by 1 turn of the DNA helix. Conforma琀椀onal changes in DNA CAP structure and DNA binding:  CAP binds to major groove in DNA adjacent to promoter of the lac operon.  CAP-cAMP bends the DNA by 90o = s琀椀mulates transcrip琀椀on of operon  enhances binding of RNA polymerase to DNA o Binding of CAP-cAMP dimer to CAP-binding site near the promoter = DNA to bend ~ 90°  = RNA polymerase binds to the promoter more e昀케ciently, s琀椀mulates transcrip琀椀on Downloaded by Harshdeep Kaur ([email protected])  lOMoARcPSD|40737841 DNA looping stabilises protein-DNA interac琀椀ons:  lac repressor binds to 2 operators (O1 and O2/O3)  reac琀椀on to the double-bound form is favoured o lower a昀케nity site Oa (O2) = increases binding of repressor to high a昀케nity site Om (O1)  higher levels of repression of the lac operon  Binding of lactose to lac repressor changes repressor conforma琀椀on = Prevents repressor binding to operator DNA  Binding of lac repressor to operator = DNA loops = Prevents RNA polymerase from transcribing DNA Binding of tryptophan to TrpR (repressor protein) changes the conforma琀椀on of the repressor.  conforma琀椀onal change enables repressor to bind 琀椀ghtly to operator = blocks transcrip琀椀on of genes encoding the enzymes that produce tryptophan (trp operon).  helix-turn-helix protein  Tryptophan binding increases the distance between the 2 recogni琀椀on helices in the homodimer, = repressor to 昀椀t 琀椀ghtly on the operator o Binding of tryptophan to TrpR changes repressor conforma琀椀on = repressor binds 琀椀ghtly to operator DNA LECTURE 20 – REGULATION OF EUKARYOTIC GENE EXPRESSION – PACKING & UNPACKING OF DNA  explain experimental evidence that shows that almost all cell have the same genome  func琀椀onal eukaryo琀椀c organism.  relate development and di昀昀eren琀椀a琀椀on of a mul琀椀cellular eukaryote to the regula琀椀on of gene expression.  discuss why eukaryo琀椀c control of gene expression