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Molecular Biology II

by: Mrs. Carol Pagac

Molecular Biology II PCB 4524

Mrs. Carol Pagac
University of Central Florida
GPA 3.63

Debopam Chakrabarti

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Debopam Chakrabarti
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This 80 page Class Notes was uploaded by Mrs. Carol Pagac on Thursday October 22, 2015. The Class Notes belongs to PCB 4524 at University of Central Florida taught by Debopam Chakrabarti in Fall. Since its upload, it has received 111 views. For similar materials see /class/227686/pcb-4524-university-of-central-florida in Biology Molecular Cell & Dev at University of Central Florida.

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Date Created: 10/22/15
l 0quot 0 Models for Homologous Recombination a Key steps of homologous recombination i Alignment of two homologous DNA molecules 1 Homologous identical or near identical for at least lOObp 2 Alleles small regions of sequence and may carry dilTerent sequence variants ii Introduction of breaks in the DNA iii Formation of initial short regions of base pairing between the two recombining DNA molecules 1 Strand invasion occurs a Forms a heteroduplex DNA i Strands of DNA o en containing mismatched base pairs generated from strand invasion iv Formation of a Holliday junction 1 Two DNA molecules that are connected by a crossing of the strands 2 The junction can move by branch migration a Occurs when the lnction moves by melting and forming base pairs along the DNA v Cleavage of the Holliday junction 1 Regenerates two separate duplex DNA molecules 2 Known as resolution Strand Invasion is a Key Step in Homologous Recombination i Recombination is initiated by the presence of a doublestranded break in one of the DNA molecules ii Strand invasion is the central step in homologous recombination iii Strand invasion generates a Holliday junction that can then move along the DNA by branch migration 1 This migration increases the length of DNA exchanged iv Repair of mismatches in the heteroduplex DNA can have important genetic consequences Resolving Holliday Junctions is a Key Step to Finishing Genetic Exchange i Finishing recombination requires resolution of the Holliday junction by cutting the DNA strands near the site of the cross ii Vertical cut sites of rotated Holliday junctions occur in the two DNA strands that are composed entirely of DNA from one of the two parental DNA molecules 1 Product are quotsplicedquot a Recombination products contain the two original duplexes and are now spliced together such that regions from the parental DNA molecules are covalently joined together by a region of hybrid duplex b Also called crossover product iii Horizontal cuts of rotated Holliday junctions occur inthe two DN strands that contain regions of sequences from both paternal molecules 1 A er resolution and covalent joining of the strands at these sties the resulting DNA molecules contain a region of quotpatchquot or hybrid DNA a Known as patch product 2 In these products recombination does NOT result in reassortment of the genes anking the site of the initial cleavage these are known as noncrossover products d The DoubleStranded BreakRepair Model Describes Many Recombination Events 1 iii Since homologous recombination is initiated by a doublestranded break the model describes the type of genetic exchange is the doublestranded break repair pathway 1 Starts with the introduction of a DSB in one of two homologous duplex DNA molecules A er introduction of the DSB a DNA cleaving enzyme sequentially degrades the broken DNA molecule to generate regions of ssDNA 1 This process creates singlestranded extensions known as ssDNA tails on the broken DNA molecules these ssDNA tails terminate with 339 ends The ssDNA tails then invade the unbroken homologous DNA duplex l The invading strand basepairs with its complementary strand in the other DNA molecule Invading strands end with 3 termini therefore they are primers for new DNA synthesis a Elongation from these DNA ends using complementary strands in the homologous duplex as atemplate serves to regenerate the regions of DNA that were destroyed during the process of the strands at the break site N 2 Homologous Recombinatino Protein Machines a RecBCD Pathway 1 ii Is the E coli major DSBrepair pathway Requires a DSB on one of the recombining two DNA molecules 1 No protein has been found in Bacteria that induces DSBs 2 Breaks generated as a result of DNA damage or failure of a replication fork are the major sources of the initiating events b RecBCD HelicaseNuclease Processes Broken DNA Molecules for Recombination i iii iv DNA molecules with ssDNA tails are the preferred substrate for initiating strand exchange between regions of homologous sequence RecBCD helps load the RecA strand exchange protein onto the ssDNA ends RecBCD is composed ofthree subunits l RecB 2 RecC 3 RecD a And has both DNA helicase and nuclease activity b Binds to DNA molecules at the site of a DSB and tracks along DNA using the energy of ATP hydrolysis i As a result the DNA is unwound with or without the accompanying nucleolytic destruction of one or both strands c The activities of RecBCD are controlled by specific DNA sequence elements known as Chi Sites Crossover Hotspot Instigator Mechanism V N L RecBCD enters the DNA at the site of the DSB and moves along the DNA unwinding the strands RecB and RecD subunits are both DNA helicases a RecB is a 3 gt5 helicase slow and has a multifunctional DNA nuclease domain b RecD is a 539 gt3 helicase processes faster than RecB RecC functions to recognize the Chi sites a Upon encountering the Chi sequence the nuclease activity of RecBCD is altered 1 RecBCD moves into the sequence distal to the Chi site and no longer cleaves the DNA strand with 3 gt5 polarity After the encounter the other DNA strand 539 gt3 polarity is cleaved even more frequently than prior to the Chi site 1 As a result of this change a duplex DNA molecule is converted into one with a 3 single strand extension terminating with the Chi sequence at the 3 end 1 Structure is ideal for assemble of RecA and initiation of stand exchange Molecular basis of the change in RecBCD enzyme activity after the encounter with a Chi site is due to inactivator of the RecD subunit and a change in the way the DNA travels through the multisubunit RecBCD complex N Structure of RecBCD Protein complex has an overall triangular shape with the Duple DNA entering the protein from the top point of the triangle a Here the DNA encounters a quotpinquot structure protruding from the RecC subunit that splits the duplex and guides the two individual strands of DNA to the two motors within the enzyme 1 i iii The RecC subunit channels the 3 strand to the RecB motor and the 539 strand do the RecD motor 1 RecC although not a helicase contributes to the overall efficiency of the helicase activity of the complex The organization of the channels within the enzyme causes the 3 tail to be fed along a groove that emerges at the nuclease active site on the RecB subunit 1 Results in the strand is now effectively and processively degraded prior to the enzyme encountering a Chi site The 539 tail also moves by the nuclease active site upon leaving the RecD motor but it is digested less frequently than the 3 trail because it must compete with the more favorable positioned 3 strand b Upon entering the Chi site the situation changes 1 RecC recognizes and binds tightly to this DNA site and once the 3 end is bound it is prevented from entering the nuclease 1 Binding both prevents further digestion of the 3 tail and promotes digestion of the 539 tail by removing its competitor ii The ssDNAtail generated by RecBCD must be coated by the RecA protein before recombination can occur 1 Cells also contain ssDNA binding proteins that can bind to the DNA 1 To ensure that RecA and NOT SSBs bind to DNA RecBCD interacts directly with RecA and promotes its assembly 1 Loading activity involves a direct protein protein interaction between the nuclease domain of the RecB subunit and the RecA protein and serves to load RecA on the DNA with the 3 tail c Chi Sites Control RecBCD i Chi sites increase frequency of recombination approximately 10 fold 1 Elevated recombination frequencies are observed for approximately lOkb distal to the chi site and drop off gradually over a distance a This is because the DNA between the DSB Where RecBCD enters and the chi site is cut into small pieces by the enzyme and is therefore not available for recombination ii The ability of chi sites to control the nuclease activity of RecBCD also helps bacterial cells protect themselves from foreign DNA that may enter via phage infection or conjugation l Eightnucleotide chi site GCTGGTGG is highly overrepresented in the E coli genome which occurs about 80 times which is equivalent to approximately 1009 chi sites 2 The E coli DNA that enters an E 001139 cell is likely to be processed by RecBCD in a manner that generates the 3 ssDNA tails and thus to be activated for recombination a DNA from another species will lack frequent chi sites i RecBCD action on this DNA will lead to its extensive degradation rather than activation for recombination iii The DNAdegradation activity of RecBCD has multiple consequences I Needed to process DNA at a break site for the subsequent steps ofRecA assembly and strand invasion a In this manner RecBCD promotes recombination b Because RecBCD degrades DNA to activate it the overall process of homologous recombination must also involve DNA synthesis to regenerated the degraded strands 2 Functions simply to destroy DNA as it does when foreign DNA lacking frequent chi sites enters cell a Protects cells from the potentially deleterious consequences of taking up foreign sequences d RecA Protein Assembles on SingleStranded DNA and Promotes Strand Invasion iii RecA is the central protein in homologous recombination l Founder member of a family of enzymes called strandexchange protein a Catalyze the pairing of homolo gous DNA molecules i Pairing involves both the search for sequence matches between the two molecules and the generation of regions of base pairing between these molecules The important features of the DNA molecules are 1 DNA sequencing complementarity between the two partner molecules 2 A region of ssDNA on at least one molecule to allow RecA assembly 3 Presence of a DNA end within the region of complementarity enabling the DNA strands in the newly formed duplex to intertwine The active form of RecA is a proteinDNA lament l Filament is huge and variable in size laments that contain approximately 100 subunits of RecA and 300 nucleotides of DNA are common a The lament can accommodate one two three or four strands of DNA The structure of DNA within the lament is highly extended compared to either uncoated ssDNA or a standard Bform helix 1 The distance between adjacent bases is 5 A in contrast to the 34A spacing 2 Upon RecA binding to the DNA molecule it will be extended by approximately 15 fold To form a lament subunits of RecA bind cooperatively to DNA 1 RecA binding and assembly are much more rapid on ssDNA than on dsDNA a Filament grows by the addition of RecA subunits in the 5 gt3 direction such that a DNA strand that reminates in 3 ends is most likely to be coated by RecA e Newly BasePaired Partners are Established Within the RecA Filament 1 RecAcatalyzed strand exchange can be divided into distinct reaction stages 1 RecA lament must assemble on one of the participating DNA molecules a Assembly occurs on a molecule containing a region of ssDNA ssDNA tail b This RecAssDNA complex is the active form that participates in the search for a homology 2 Search for homology a RecA must look for basepair complementarity between the DNA within the lament and a new DNA molecule b This homology search is promoted by RecA because the lament structure has two distinct DNA binding sites i Primary site 1 Bound by the rst DNA molecule ii Secondary site 1 Can be occupied by a doublestranded DNA 2 Binding is rapid weak transient and independent of DNA sequence c How does the RecA lament sense homology 9 i Not yet clear ii DNA in the secondary binding site is transiently opened and tested for complementarity with the ssDNA in the primary site 1 This testing is presumable via base pairing interactions 2 Experiments suggest that the initial alignment may involve base ipping of some of the bases in the DNA duplex In vitro experiments indicate that a sequence match of just 15bp provides a sulTicient signal to the RecA lament that a match has been found 3 Once a region of basepair complementarity has been located RecA promotes the formation of a stable complex between these two DNA molecules a The RecAbound three stranded structure is called a joint molecule and usually contains several hundred base pairs of hybrid DNA i It is within the joint molecule that the actual exchange of DNA strands occurs 1 The DNA strand in the primary binding site becomes base paired with its complement in the DNA duplex bound in the secondary site Strand exchange require the breaking of one set of base pairs and the formation of a new set of identical base pairs b RecA binds preferentially to the DNA product a er strand exchange has occurred and it is this binding energy that actually drives the exchange reaction toward the new DNA con guration RecA Homologs are Present in All Organisms i Strandexchange proteins of the RecA family are present in all forms of life 1 Bestcharacterized list a RecA in Eubacteria b RadA in Archaea c Rad51 and Dmcl in Eukaryota d UvsX protein in Bacteriophage T4 The RuvAB Complex Specifically Recognizes Holliday Junctions and Promotes Branch Migration i After the strand invasion step of recombination is complete the two 1 quot39 D A 39 39 are beaDNAbranchknownasa Holliday junction 1 Movement of the site of this branch requires exchange of DNA base pairs between the two homologous DNA duplexes ii RuvA protein is the Holliday junctionspecific DNAbinding protein that recognizes the structure of the DNA junction regardless of its specific DNA sequence 1 Recognizes and binds Holliday junctions and recruits the RuvB protein to this site E N iii Pquot RuvB is a hexameric ATPase similar to the hexameric helicases involved in DNA replication 1 Provides the energy to drive the exchange of base pairs that move the DNA branch a This energy is needed to move the branch rapidly and in one direction Structural models of RuvAB complexes at a Holliday junction show how a tetramer of RuvA together with two hexamers of RuvB work together to power this DNAexchange process Rqu Cleaves Speci c DNA Strands at the Holliday Junction to Finish Recombination i iii Completion of recombination requires that the Holliday junction between the two recombining DNA molecules be resolved 1 In bacteria the major Holliday junction resolving endonuclease is Rqu a Discovered and purified based on its ability to cut DNA junctions made by RecA in vitro Resolution by Rqu occurs when Rqu recognizes the Holliday junctionlikely in a complex the RuvA and RuvBand speci cally nicks two of the homologous DNA strands that have the same polarity l Cleavage results in DNA ends that terminate with 5 phosphates and 339 OH groups that can be directly joined by DNA ligase Rqu cleaves DNA with modest sequence specificity l Cleavage takes place only at sites conforming to the consensus 539ATTT GC3 a Cleavage occurs after the second T in the sequence this sequence is found frequently in DNA averaging one every 64 nucleotides i This ensures at least some branch migration occurs before resolution 3 Homologous Recombination in Eukaryotes a Homologous Recombination has Additional Functions in Eukaryotes i iii iv Homologous recombination in bacteria is required to 1 Repair DBSs in DNA 2 Restart collapsed replication forks 3 Allow a cell s chromosomal DNA to recombine with DNA that enters via phage infection of conjugation Homologous recombination is also required for DNA repair and the restarting of collapsed replication forks in eukaryotic cells Animals with mutations that interfere with homologous recombination are predisposed to certain types ofcancer In eukaryotes homologous recombination is required for proper chromosome pairing and thus for maintaining the integrity of the genome in meiosis b Homologous Recombination is Required for Chromosome Segregation During Meiosis i Before division the cell has two copies of each chromosome one each that was inherited from its two parents and during S phase the chromosomes are replicated to give a total DNA content of 4N C ii In preparation of the rst nuclear division these duplicated homologous chromosomes must pair and align at the center of the cell iii Recombination must be completed before the rst nuclear division to allow the homologs to properly align and then separate 1 During this process the sister chromatids remain paired 2 In the second nuclear division it is the sister chromatids that separate iv In the absence of recombination chromosomes o en fail to align properly for the rst meiotic division and as a result there is a high incidence of chromosome loss 1 This improper segregation of chromosomes nondisjunction leads to a large number of gametes without the correct chromosome complement V The homologous recombination events that occur during meiosis are called meiotic recombination 1 Frequently gives rise to crossing over between genes on the two homologous parental chromosomes The genetic exchange can be observed cytologically Programmed Generation of DoubleStranded DNA Breaks Occurs During Meiosis i The developmental program needed for cells to successfully complete meiosis involves turning on the expression of many genes that are NOT needed during normal growth 1 SP011 Encodes a protein that introduces DBSs in chromosomal DNA to initiate meiotic recombination b Protein cuts the DNA at many chromosomal locations with little sequence selectivity but at a very speci c time during meiosis P c Spoll mediated DNA cleavage occurs right around the time when the replicated homologous chromosomes start to repair i Spo 11 cut sites are not randomly distributed along the DNA rather are located most commonly in chromosomal regions that are not tightly packed with nucleosomes such as promoters controlling gene transcription d Mechanism i Speci c tyrosine side chain in the Spoll protein attacks the phosphodiester backbone to cut the DNA and generate a covalent compleX between the protein and the severed DNA strand 1 Two subunits of Spo ll cleave the DNA two nucleotides apart on the two DNA strands to make a staggered DSB l Spo ll shares this DNA cleavage mechanism with the DNA topoisomerases and the sitespeci c recombinases 2 The fact that Spo ll cleavage involves a covalent protein DNA compleX has two consequences 1 The 539 ends ofthe DNA at the site of Spoll cleavage are covalently bound to the enzyme 1 These Spo l llinked 5 DNA ends that are the initial sites of DNA processing to create the ssDNA tails required for assembly of RecA like proteins and initiation of strand invasion ii The energy of the cleaved DNA phosphodiester bond is stored in the bound proteinDNA linkage and so the DNA strands can be resealed by a simple reversal of the cleavage reaction 1 Resealing can occur when cells receive a signal to stop proceeding with meiosis d MRX Protein Processes the Cleaved DNA Ends for Assembly of the RecALike FD StrandExchange Proteins i During meiotic recombination the MRXenzyme is responsible for this DNA processing event 1 This complex although NOT homologous to RecBCD is a multisubunit DNA nuclease 2 Composed of a protein subunits called a Mre l l b Rad50 c er2 ii Processing of the DNA at the break site occurs exclusively on the DNA strand that terminates with a 539 end the strands covalently attached to the Spoll protein 1 The strands terminating with 3 ends are NOT degraded a This reaction is therefore called a 5 gt3 resection i Generates the long ssDNA tails with 3 ends that are o en lkb or longer iii The MRX complex is thought to remove the DNAlinked Spoll Dmcl Is a RecALike Protein That Speci cally Functions in Meiotic Recombination i Eukaryotes encode two wellcharacterized homolo gs of the bacterial RecA protein 1 Rad51 a Widely expressed in cells dividing mitotically and meiotically 2 Dmcl a Expressed only as cells enter meiosis i Both proteins lnction in meiotic recombination ii Strand exchange during meiosis occurs between a particular type of homologous DNA partner 1 Recall meiotic recombination occurs at atime when there are four complete dsDNA molecules and the two homologs each of which have been copied to generate two sister chromatids a Although the two homolo gs likely contain small sequence differences and carry distinct alleles for various genes the majority of the DNA sequence among these four copies of the chromosome will be identical i Dmcldependent recombination is preferentially between the nonsister homologous chromatids rather than between the sisters 1 Rationale meiotic recombination promotes interhomolog connections to assist alignment of the chromosomes for division f Many Proteins Function Together to Promote Meiotic Recombination 1 iii iv Protein involved in the critical stages of DSB formation DNA processing to generate 339 ssDNA tails and strange exchange during meiotic recombination have been identi ed and characterized Many proteins appear to interact with the known recombination enzymes and it seems likely that these proteins lnction in the context of a large multicomponent complex 1 These large proteinDNA complexes are known as recombination factories and can be visualized in cells Various other proteins have been shown to be involved with RadSl to help promote recombination and DSB repair 1 Rad52 is another essential recombination protein that interacts with RadSl a Functions to promote assembly of RadSl DNA laments the active form of RadSl i It accomplishes this by antagonizing the action of RPA the major ssDNAbinding protein present in eukaryotic cells 1 In hindsight Rad52 shares an activity with the E coli RecBCD protein which helps RecA load onto ssDNA that would otherwise have been bound to SSB ii Rad52 protein also promotes the annealing and base pairing of complementary ssDNA molecules 2 The product of the BRCA2 gene also participates in RadSlmediated DSB repair By analogy with Bacteria we expect that eukaryotic cells encode proteins that promote the branch migration and Holliday junction resolution steps of recombination 1 In fact enzymes promoting these reactions are being identified a Complex containing a Rad5 llike protein called RadSlC and XRCC3 which both have been found to contain Holliday junction resolvase activity 4 MatingType Switch a In addition to promoting DNA pairing DNA repair and genetic exchange homologous recombination also serves to change the DNA sequence at a specific chromosomal location 1 O en used to regulate gene expression b S cerevisiae is a singlecell eukaryote that can exist as any of three different cell WP i es Haploid Sc cells can be either oftwo mating types 1 a 2 0L 9 2391 0 a In addition when an a or or cell come into close proximity they can lse to form an aot diploid cell i The cell may then go through meiosis to form two haploid a cells and two or cells The mating type genes encoding transcriptional regulators i Regulators control expression of target genes whose products de ne each cell type ii Matingtype genes expressed in a given cell are those found at the mating type locus MAT locus in that cell 1 This in a cells the al gene is present at the MAT locus whereas in or cells the 11 and 12 genes are present at the MAT locus Cells can switch their mating type by recombination i In addition to the a or or present at the MAT locus in each cell there is an additional copy of both the a and or genes present but NOT expressed elsewhere in the genome 1 These silent copies are found at loci called HMR and HML a Known as silent cassettes b Their lnction is to provide a storehouse of genetic information that can be used to switch a cells39 mating type c The switch requires the transfer of genetic information from the HM sites to the MAT locus via homologous recombination MatingType Switching is Initiated by a SiteSpecific DoubleStrand Break i Matingtype switching is initiated by the introduction of the DSB at the MAT locus 1 This reaction is performed by a specialized DNAcleaving enzyme called HO endonuclease a Expression of the H0 gene is tightly regulated to ensure that switching occurs only when it should b Sequencespecific endonuclease the only sites inthe yeast chromosome that carry HO recognition sequences are the mating type loci HO cutting introduces a staggered break in the chromosome and in contrast to Spoll cleavage HO simply hydrolyzes the DNA and does not remain covalently linked to the cut strand ii 539 gt3 resection of the DNA at the site of the HOinduced break occurs by the same mechanism used during meiotic recombination Resection depends on the MRX protein complex and is specific for the strands that terminate with 539 ends In contrast the 3 terminating ends are very stable and not subjected to nuclease digestion a Once the 3 ssDNA tails have been generated the DNA becomes coated by the RadSl and Rad52 proteins i These RadSl protein coated strands then search for homologous chromosomal regions to initiate strand invasion and genetic exchange 9 N iii Matingtype switching is unidirectional sequence information is quotmovedquot from the MAT locus from HMR and HML but information never quotgoesquot in the other direction 1 The cut MAT locus is always the recipient partner during recombination and the HMR and HML sites remain unchanged by the recombination process a This stems from the fact that HO endonuclease cannot cleave its recognition sequence at either HML or HMR because the chromatin structure renders these sites inaccessible for this enzyme iv The Rad51coated 339 ssDNA tails from the MAT locus choose the DNA at either the HMR or HML locus for strand invasion 1 If the DNA sequence at MAT is a then invasion occurs with HML which carries the quotstoragequot copy of the or sequences 2 If the DNA sequence at MAT is or the invasion occurs with HMR which carries the storage copy of the a sequences a A er recombination the genetic information that was at the chosen HM loci is present at the MAT loci as well i This genetic change occurs without a reciprocal swap of information from MAT to the HR locus f MatingType Switching is a Gene Conversion Event and Not Associated with Crossing Over i Evidence indicates that alter the strand invasion step this recombination pathway diverges from the DSBrepair mechanism 1 One hint that the mechanism is distinct is that the crossover class of recombination products is never observed during matingtype switch Recall in the DSBrepair pathway resolution of the Holliday junction intermediates gives two classes of products the splice crossover or the patch noncrossover a Theories state that these resolutions occur at equal frequencies however crossover products are rarely observed in matingtype switching ii Gene conversion without crossing over a new recombination model termed synthesis dependent strand annealing SDSA l The initiating even is the introduction of a DSB at the recombination site 2 A er strand discharge the invading 3 end serves as the primer to initiate new DNA synthesis a In contrast to what occurs during the DSBrepair pathway a complete replication fork is assembled at this site N 3 Both leadingstrand and laggingstrand DNA synthesis occurs a In contrast to normal DNA replication the newly synthesized strands are displaced from the template 4 As a result anew doublestranded DNA segment is synthesized joined to the DNA site that was originally cut by HO and resected by MRX 5 Completing recombination requires that the other quotoldquot DNA strand present at MAT the 339 ending strand not cleaved by MRX be removed 5 Genetic C a 0 2391 a The newly synthesized DNAan exact copy of the information in the partner DNA moleculereplaces the information that was originally present 1 ofthe 39 ofIY 39 Recombination A central feature of homologous recombination is that it can occur between any two regions of DNA regardless of the sequence provided these regions are sulTiciently similar i NONE of the steps in homologous recombination require recognition of a highly specific DNA sequence Distortions in genetic maps compared to physical maps occur when a region of DNA does not have the average probability of participating in recombination i Regions with a higher than average probability are quothot spotsquot whereas regions that participate less commonly than average segment are quotcold spotsquot 1 Therefore two genes that have a hot spot between them appear in a genetic map to be farther apart than is true in a physical map of the same region 2 In contrast genes separated by a quotcoldquot interval appear by genetic mapping to be closer together than is true from their physical distance Encounter two examples of hot and cold spots i Regions near chi spots and Spoll cleavage sites have a higher than average probability of initiating recombination whereas regions having few such sites are correspondingly quotcoldquot One Cause of Gene Conversion is DNA Repair During Recombination i Gene conversion is commonly observed during normal homologous recombination events such as those responsible for genetic exchange in bacteria and for pairing chromosomes during meiosis matingtype switching in yeast ii Illustration of Gene Conversion 1 Consider a cell undergoing meiosis has the A allele on one homolo g and the a allele on the other a A er DNA replication four copies of this gene are present and the genotype would be A A a a b In the absence of gene conversion two gametes carrying the A allele and two gametes carrying the a allele would be generated i Ifinstead the gametes with genotypes A a a a orA A A a are formed then a gene conversion even has occurred in which one copy of the A gene has been converted into a or vice versa ii How does this arise Two ways that gene conversion can occur during the DSBrepair pathway iii 1 Consider what could happen if the A gene was very close to the site of the a In this case when the 3 ssDNA tails invade the homologous duplexes and are elongated they may copy the a information which could replace the A information in the product chromosome upon completion of recombination 2 Involves the repair of basepair mismatches that occur in the recombination intermediates a If either strand invasion or branch migration includes the Aa gene a segment of heterodupleX DNA carrying the A sequence on one strand would be formed i iii This region of DNA carrying basepair mismatches could be recognized and acted upon by the cellular mismatch repair enzymes These enzymes are specialized for xing basepair mismatches in DNA 1 When they detect a mismatched base pair these enzymes exist a short stretch of DNA from one of the two strands 2 A repair DNA polymerase then lls in the gap now with the properly basepaired sequence When working 0 recombination intermediates the mismatch repair enzymes will likely choose randomly which strand will repair 1 Therefore alter their action both strands will carry the sequence coding either the A information or the a information depending on which strand was quotfixedquot by the repair enzymes and gene conversion will be observed 1 Chapter 11 SiteSpecific Recombination and Transposition ofDNA 2 Introduction a Two classes of genetic recombination are responsible for many important DNA rearrangements 1 ii Conservative SiteSpeci c Recombination CSSR l Recombination between two de ned sequence elements Transpositional Recombination Transposition 1 Recombination between specific sequences and nonspeci c DNA sites b In many organisms transposition is the major source of spontaneous mutation and nearly halfthe human genome consists of sequences derived from transposable elements 0 Conservative sitespeci c recombination and transposition share key mechanistic features 1 Recombinases Proteins that recognize speci c sequences where recombination will occur within a DNA molecule Bing these speci c sites together to form a proteinDNA complex bridging the DNA sites a Bridges are known as synaptic complexes i The recombinase catalyzes the cleavage and rejoining of the DNA molecules either to invert a DNA segment or to move a segment to a new site N 3 Conservative SiteSpeci c Recombination a SiteSpeci c Recombination occurs at Speci c DNA Sequences in the Target DNA 1 iii CSSR is responsible for many reactions in which a de ned segment of DNA is rearranged Key feature of these reactions is that the segment of DNA that will be moved carries specific short sequence elements called recombination sites 1 Sites where DNA exchange occurs 2 Example integration of the bacteriophage I genome into the bacterial chromosome a During 7 integration recombination always occurs at exactly the same nucleotide sequence within two recombination sites i One on the phage DNA ii On the bacterial DNA Recombination sites carry two classes of sequence elements 1 Sequences specifically bound by the recombinases 2 Sequences where DNA cleavage and rejoining occur Recombination sites are often short approximately 20bp CSSR can generate three different types of DNA rearrangements Insertion of a segment of DNA into a speci c site as occurs during bacteriophage 7 DNA integration Deletion of a DNA segment Inversion of a DNA segment a Whether recombination results in DNA insertion deletion or inversion depends on the organization of the recombinase W Vi viii recognition sites on the DNA molecule or molecules that participate in recombination Each recombination site is organized as a pair or recombinase recognition sequences that are positioned symmetrically 1 These sequences ank a central short asymmetric sequence crossover region where DNA cleavage and rejoining occurs Due to crossover regions being asymmetric a given recombination site always has a de ned polarity l The orientation of two sites present on a single DNA molecule will be related to each other in either a Inverted repeat i Recombination will invelt the DNA segment between the two sites b Direct repeat i Recombination deletes the DNA segment between the two sites Insertion speci cally occurs when recombination sites on two different molecules are brought together for DNA exchange b SiteSpeci c Recombinases Cleave and Rejoin DNA Using a Covalent ProteinDNA Intermediate i iii Two families of conservativesite speci c recombinase l Serine recombinases 2 Tyrosine recombinases a Fundamental mechanism used by both families is that upon cleavage of the DNA a covalent proteinDNA intermediate is generated Serine recombinases the side chain of a serine residue within the protein39s active site attacks a speci c phosphodiester bond in the recombination site i This reaction introduces a singlestrand break in the DNA and simultaneously generates a covalent linkage between the serine and a phosphate at this DNA cleavage site c Tyrosine recombinases it is the side chain of the activesite tyrosine that attacks and then becomes joined to the DNA The covalent proteinDNA intermediate conserves the energy of the cleaved phosphodiester bond within the proteinDNA linkage 1 As a result the DNA strands can be rejoined by reversal of the cleavage process Reversal an OH group from the cleaved DNA attacks the covalent bond that links the protein to the DNA This covalently seals the DNA break and regenerates the free nonDNA bound recombinase The mechanism is conservative because every DNA bond that is broken during the reaction is resealed by the recombinase NO external energy such as energy released by ATP hydrolysis is needed for DNA cleavage and joining by these proteins 0quot N C 3 1 FD Serine Recombinases Introduce DoubleStrand Breaks in DNA and Then Swap Strands to Promote Recombination i CSSR always occurs between two recombination sites 1 Sites may be on the same DNA molecule for inversion or deletion 2 Sites may be on two different molecules for integration ii Serine recombinases cleave the four strands prior to strand exchange four single strands of DNA two from each duplex 1 One molecule of the recombinase protein promotes each of these cleavage reactions therefore a minimum of four subunits of the recombinase is required For recombination to occur a requirement of 4 serine recombinases are required R1 R2 R3 and R4 1 For recombination to occur R2 segment of the top DNA molecule must recombine with the R3 segment of the bottom DNA molecule iii 2 Similarly the R1 segment of the top molecule must recombine with the R4 segment of the bottom DNA molecule 3 Once this DNA swap has occurred the 339OH ends of each of the cleaved DNA strands can attack the recombinaseDNA bond in their new partnered segment 4 This reaction liberates the recombinase and covalently seals the DNA Structure f the Serine RecombinaseDNA Complex Indicates that Subunits Rotate to Achieve Strand Exchange i The most dramatic feature of the structure is the large at interface between the quottopquot and quotbottomquot recombinase dimers 1 This structure is largely hydrophobic and slippers providing little barrier to impede rotation of the top and bottom halves of the complex around each other ii Some regions of complementary positive and negative charge can serve to stabilize the structure speci cally in the initial and the 1800 rotated orientation Mechanism 1 DNA cleavage to form the covalent enzymeDNA intermediate 2 An 180O rotation of the dimers in the proteinDNA complex 3 Attack of the 339OH DNA ends on the resolvaseDNA linkages to join the strands in the new recombined configuration Tyrosine Recombinases Break and Rejoin One Pair of DNA Strands at a Time i In contrast to the serine recombinases the tyrosine recombinases cleave and rejoin two DNA strands first and only then cleave and rejoin the other two strands ii Four molecules of the recombinase is needed one to cleave each of the four individual DNA strands To start recombination 1 The subunits of recombinase bound to the le recombinasebinding sites R1 and R3 each cleave the top strand of the DNA molecule to which they are bound a This cleavage occurs at the rst nucleotide of the crossover region iii 0quot The right top strand from the top DNA molecule and the right top strand from the bottom DNA molecule swap partners c These two DNA strands are then joined now in the recombined con gurations d The firststrand exchange reaction generates a branched DNA intermediate called a Holliday Junction 2 Once the first strand exchange is complete two more recombinase subunits R2 and R4 cleave the bottom strands of each DNA molecule a These strands switch partners and then are joined by the reversal of the cleavage reaction b The secondstrand exchange reaction undoes the Holliday junction to yield the rearranged DNA products f Structure of Tyrosine Recombinases Bound to DNA Reveal the Mechanism of DNA Exchange i Mechanism of sitespecific recombination is best understood for tyrosine recombinases ii Class example is the structure of the Cre Recombinase bound to two dilTerent con gurations of the recombining DNA 1 Cre is an enzyme encoded by Phage Pl which lnctions to circularize the linear phage genome during infection 2 The recombination sites on the DNA where Cre acts are called 10x sites a Cre 10x is a simple example of recombination by the tyrosine recombinase family Only Cre protein and the lox sites are needed for complete recombination The Crelox structures reveal that recombination requires four subunits of Cre each molecule bound to one binding site on the substrate DNA molecules 5 The conformation of the DNA is generally a square planar fourway junction with each arm of this junction bound by one subunit of Cre 6 Cre exists in two distinct conformations a Only one of these conformations can Cre cleave and rejoin DNA iii Mechanism for Cre l Firststrand cleavage of DNA forms the CreDNA intermediate I a Two free 339OH and the 539 end linking to Tyrosine by phosphodiester bond 2 Firststrand exchange forms the Holliday junction intermediate 3 Secondstrand cleavage forms the CreDNA intermediate 11 a Two 3 OH and the 539 end linking to Tyrosine by phosphodiester bond 4 Secondstrand exchange yields recombined DNA 4 Biological Roles ofSiteSpeciflc Recombination a Site specific recombination is also widely used to help maintain the structural integrity of circular DNA molecules during cycles of DNA replication homologous recombination and cell division b All reactions depend critically on the assembly of the recombinase protein on the DNA and the bringing together of the two recombination sites 5 0 i For some recombination reactions this assembly is simple requiring only the recombinase and its DNA recognition sequences Cre ii Other reactions require accessory proteins 1 These proteins include architectural proteins that bind speci c DNA sequences and bend the DNA a They organize DNA into a speci c shape and thereby stimulate recombination Can also control the direction of a recombination reaction IE to ensure that integration of a DNA segment occurs while preventing the reverse reaction DNA excision 7t Integrase Promotes the Integration and Excision of a Viral Genome into the Host Cell Chromosome i When bacteriophage 7t infects a host bacterium a series of regulatory events result either in establishment of the quiescent lysogenic state or in phage multiplication lytic growth 1 Establishment of a lysogen requires the integration of the phage DNA from the host chromosome ii To integrate the 7 integrase protein Mnt catalyzes recombination between two speci c attachment sites att l The attP site is on the Phage DNA P for phage 2 The attB site is on the bacterial chromosome B for Bacteria iii Mnt is a tyrosine recombinase and the mechanism of strand exchange follows the pathway for Cre protein 1 Mnt requires accessory proteins to help the required proteinDNA complex to assemble which contrasts Cre a These proteins controlthe reaction to ensure that DNA integration and DNA excision occur at the right time in the page life cycle iv Important to the 7 integration is the highly asymmetrical organization of the attP and attB sites 1 Both sites carry a central core segment of approximately 30bp 2 These core recombination sites each consist of two Mntbinding sites and a crossover region where strand exchange occurs attB consists only of the central region attP is much longer 240bp and carries several additional protein binding sites a Flanking each side of the core region of attP are DNA regions known as the arms i Arms carry a variety of proteinbinding sites including additional sites bound by Mnt V Mnt is unusual because it has two domains involved in sequencespeci c DNA binding 1 One domain binds to the arm recombinase recognition sites 2 One domain binds to the core recognition site Vi Integration requires attB attP Hut and an architectural protein called integration host factor IHF 1 Sequence dependent DNAbinding protein that introduces large bends gt160quot in DNA 0quot LA LAN Arms of attP carry three IHF binding sites The lnction of IHF is to bring together the Mnt sites on the DNA arms where Mnt binds strongly with the sites present at the central fore where Mnt binds weakly a vii Bending of the DNA mediated by IHF allows Mnt to nd the weak core sites and to catalyze recombination When recombination is complete the circular phage genome is stably integrated into the host chromosome 1 As a result two new hybrid sites are generated at the junctions between the phage and the host DNA 2 These sites are called a b attL le attR right i Both of these sites contain the core region but the two arm regions are now separated from each other ii Thus neither of the two core regions in this new arrangement is competent to assemble an active Mnt recombinase complex via the mechanism during integration 1 The DNA sites important for assemble are simply not in the right place d Bacteriophage 7t Excision Requires a New DNABending Protein i How does 7 excise 1 An additional architectural protein this one phageencoded is essential for excisive recombination a 09quot 2391 FD This protein is called Xis excise it binds to speci c DNA sequences and introduces bends in the DNA Xis is similar to IHF Xis recognizes two sequence motifs present in one arm of attR i Binding these sites introduces a large bend gt140 and together Xis Mnt and IHF stimulate excision by assembling an active proteinDNA complex at attR 1 This complex then interacts productively with proteins assembled at attL and recombination occurs In addition to stimulation excision recombination between attL and attR DNA binding by Xis also inhibits integration recombination between attP and attB The DNA structure created upon Xis binding to attP is compatible with proper assembly of Mnt and IHF at this site i Xis is a phageencoded protein and is only made when the phage is triggered to enter lytic growth ii Xis dual action as a stimulatory cofactor for excision and an inhibitor of integration ensures that the phage genome will be free remain free from the host chromosome when Xis is present e The Hin Recombinase Inverts a Segment of DNA Allowing Expression of Alternative Genes iii The Salmonella Hin recombinase inverts a segment of the bacterial chromosome to allow expression of two alternative sets of genes Hin recombination is an example of a class of recombination reactions known as programmed rearrangements relatively common in bacteria 1 These reactions o en function to quotpreadaptquot a portion of a population to a sudden change in the environment In the case of Hin inversion recombination is used to help the bacteria evade the host immune system 1 The genes that are controlled by the inversion process encode two alternative forms of agellin H1 and H2 forms the protein component of the agella filament By using Hin to switch between these alternative forms some of the individuals in the bacterial population can avoid recognition of his surface structure by the immune system The chromosomal region inverted by Hin is approximately 1000bp and is anked by specific recombination sites called hixL left and hixR right 1 These sequences are in inverted orientation with respect to one another 2 Hin is a serine recombinase promoting inversion using the basic mechanism described above for this enzyme family The invertible segment carries the gene Encoding Hin as well as a promoter which in one orientation is positioned to express the genes located outside the invertible segment directly adjacent to the hixR site a When the invertible regions is in the quotonquot orientation these adjacent genes are expressed whereas when the segment is ipped into quotoffquot position these adjacent genes are silenced N L 4 The two genes under control of this quot ippingquot promoter are a jB i Encodes the H2 agellin b jA i Encodes atranscriptional repressor of the gene for the H1 agellin 5 The H1 agellin gene is located at a distant site this in the quoton orientation H2 agellin and the H1 repressor are expressed a These cells have exclusively H2type agella on their surface 6 In the quotoffquot orientation however neither H2 nor the H1 repressor is synthesized and the H1type agella are present f Hin Recombination Requires a DNA Enhancer 1 iii Hin recombination requires a sequence in addition to the hix sites 1 This short approximately 60bp sequence is an enhancer that stimulates the rate of recombination approximately 1000 fold Enhancer functions requires the bacterial Fis protein factor for inversion stimulation 1 Like IHF Fis is a sitespecific DNA bending protein 2 It makes proteinprotein contacts with Hin that are important for recombination The Fisenhancer complex activates the catalytic steps of recombination Pquot l Hin can actually assemble and pair the hix recombination sites to form a synaptic complex in the absence of the Fisenhancer complex a This contrasts with the role of IHF in 7 integration where the accessory protein is essential for assembly of the recombinaseDNA complex For Fis activation of Hin the three DNA sites hixL hixR and enhancer need to come together 1 Formation of this threeway complex is greatly facilitated by negative DNA supercoiling which stabilizes the association of the distant DNA sites V Another bacterial architectural protein HU also facilitates assembly of this invertasome complex 1 HU is a close structural homolog of IHF yet in contrast to IHF it binds DNA in a sequenceindependent manner vi What is the biological rationale for control of Hin inversion by the Fis enhancer complex 1 The principal function is to ensure that recombination occurs only between hix sites that are present on the same DNA molecule a This selectivity ensures that the invertible segment is ipped frequently but also that intermolecular DNA rearrangements which could disrupt the integrity of the bacterial chromosome are avoided vii In contrast to integration and excision of bacteriophage 7t Hin catalyzed inversion is not highly regulated inversion occurs stochastically such that within a population of cells there will always be come cells that carry the invertible segment in each orientation Recombinases Convert Multimeric DNA Molecules into Monomers i Sitespeci c recombination is critical to the maintenance of circular DNA molecules within cells ii A problem with circular DNA molecules is that they sometimes form dimers and even higher multimeric forms during the process of homologous recombination l Sitespecific recombination can be used to convert these DNA multimers back into monomers A single homologous recombination event can generate one large circular chromosome with two copies of all the genes dimeric chromosome 1 At the time of cell division the dimer poses a major problem as there will be only one rather than two DNA molecules to be segregated into the two daughter cells Due to the multimerizaation many circular DNA molecules carry sequences recognized by sitespecific recombinases v Proteins that lnction at the sequences are called resolvases as they quotresolvequot dimers into monomers 1 These proteins speci cally catalyze resolution DNA deletion reaction but not the reverse reaction conversion of monomers into dimers There Are Other Mechanisms to Direct Recombination to Speci c Segments of DNA iii ii Matingtype occur in yeast 1 Occur by a targeted geneconversion event VDJ Recombination 5 Transposition Some Genetic Elements Move to new Chromosomal Locations by Transposition a i iii Specific form of genetic recombination that moves certain genetic elements from one DNA site to another 1 Mobile genetic elements are called transposable elements or transposons a Movement occurs through recombination between the DNA sequences at the very ends of the transposable element and a sequence in the DNA of the host cell movement can occur with or without duplication of the element Transposable elements show little sequence selectivity in their choice of insertion sites 1 Transposons can insert within genes o en completely disrupting gene function 2 Can insert within the regulatory sequences of a gene where their presence may lead to changes in how that gene is expressed Transposable elements are the most common source of new mutations in many organisms these elements are an important cause of mutations leading to genetic disease in humans Transposable elements are present in the genomes of all lifeforms Comparative analysis of genome sequences showed two observations 1 Transposonrelated sequences can make up huge fractions of the genome of an organism a More than 50 of both the human and maize genomes are r J of r r lat d b Contribution is in contrast to the small percentage of the sequence that actually encodes cellular proteins gt2 2 Transposon content in different genomes is highly variable a Compared to humans or maize the y and yeast genomes are quotgenerichquot and quottransposonpoorquot b There Are Three Principal Classes of Transposable Elements C i ii DNA Transposons Virus Like Retrotransposons 1 This class includes the retroviruses 2 These elements are also called long terminal repeat LTR retrotransposons Poly A Retrotransposons 1 These elements are also called nonviral retrotransposons DNA Transposons Carry a Transposase Gene Flanked by Recombination Sites i DNA transposons carry both DNA sequences that lnction as recombination sites and genes encoding proteins that participate in recombination l The recombination sties are at the two ends of the element and are organized as invertedrepeat sequences 3 1 Fquot f a These terminal inverted repeats vary in length from approximately 25 to a few hundred base pairs are not exact sequence repeats and carry the recombinase recognition sequences i The recombinase responsible for transpositions are usually called transposases or integrases ii DNA transposons carry a gene encoding their own transposase 1 May carry a few additional genes sometimes encoding proteins that regulate transposition or provide a lnction useful to the element or its host a Many bacterial DNA transposons carry genes encoding proteins that promote resistance to one or more antibiotics i The presence of the transposon therefore causes the host cell to be resistant to that antibiotic The DNA sequences immediately anking the transposon have a short 2 20bp segment of duplicated sequence 1 Segment are organized as direct repeats called target site duplications and are generated during the process of recombination Transposons Exist as Both Autonomous and Nonautonomous Elements i Autonomous Elements 1 DNA transposons that carry a pair of terminal inverted repeats and a transposase gene have everything they need to promote their own transposition ii Nonautonomous Elements 1 Genomes also contain many simpler mobile DNA segments 2 These elements carry only the terminal inverted repeats a The cisacting sequences needed for transposition In a cell that also carries an autonomous transposon encoding a transposase that will recognize these terminal inverted repeats the nonautonomous element will be able to transpose 1 In the absence of the quothelperquot transposon to donate the transposase nonautonomous elements will remain unable to move VirusLike Retrotransposons and Retroviruses Carry Terminal Repeat Sequences and Two Genes Important for Recombination i Terminal inverted repeats are embedded within longer repeated sequences these sequences are organized on the two ends of the element as direct repeats and are called LTRs ii Viruslike retrotransposons encode two proteins needed for their mobility l Integrase transposase 2 Reverse Transcriptase a Special type of DNA polymerase that can use an RNA template to synthesize DNA b This enzyme is needed for transposition because an RNA intermediate is required for the transposition reaction c Retrotransposons can move only to new DNA sites within a cell but can never leave that cell PolyA Retrotransposons Look Like Genes iii iii iii iv Do not have the terminal inverted repeats present in their other transposon classes 1 Instead the two ends of the element have distinct sequences a One end is called the 539 UTR whereas the other end has a region called the 3 UTR followed by a stretch of AT base pairs called the poly A sequence i These elements are also anked by short targetsite duplications Retrotransposons carry two genes Fl a Encodes an RNAbinding protein 2 ORF2 a Encodes a protein with both reverse transcriptase activity and an endonuclease activity b Essential roles during recombination Commonly in both autonomous and nonautonomous forms Genome sequence analysis reveals that there are many truncated elements that do not have a complete 5 UTR sequence and have lost their ability to transpose g DNA Transposition by a CutandPaste Mechanisms 1 Simplest transposition reaction The movement of a DNA transposon by a nonreplicative mechanism 1 Involves the excision of the transposon from its initial location in the host DNA 2 Followed by integration of this existed transposon into a new DNA site a This process is called the cut and paste mechanism 3 To initiate recombination the transposase binds to the terminal inverted repeats at the end of the transposon 4 Once the t1 r 1 39 these 1 it brings the two ends of the transposon DNA together to generate a stable proteinDNA complex a This complex is called the synaptic complex or transposome b It contains a multimer of transposaseusually two or four subunits and the two DNA ends c This complex functions to ensure that the DNA cleavage and joining reactions needed to move the transposon occur simultaneously on the two ends of the elements DNA d It also protects the DNA ends from cellular enzymes during recombination 5 Next step is excision of the transposon DNA from its original location in the genome a The transposase subunits within the transposome rst cleave one DNA strand at each end of the transposon exactly at the junction between the transposon DNA and the host sequence in which it is inserted region called the anking host DNA 6 The transposase cleaves the DNA such that the transposon sequence terminates with free 339OH groups at each end of the elements DNA 7 To nish the reaction the other DNA strand at each end of the element must also be cleaved Di erent transposons use different mechanisms to cleave these second DNA strands with the 539 ends A er excision of the transposon the 3 OH ends of the transposon DNA the ends first liberated by the r atta l the DNA r39 r39 J39 bonds at the site of the new insertion a This DNA segment is called the target DNA 2 As a result of this attack the transposon DNA is covalently joined to the DNA at the target site 3 During each DNAjoining reaction a nick is introduced into the target DNA a DNAjoining reaction occurs by a onestep transesteri cation reaction called DNA strand transfer b The transposome ensures that the two ends of the transposon DNA attack the two DNA strands of the same target site together iii Once DNA strand transfer is complete the job of the transposome is complete h The Intermediate in CutandPaste Transposition is Finished by Gap Repair i The structure of the DNA intermediate generated a er DNA strand transfer has the 339 ends of the transposon DNA attached to the target DNA 1 This also carries the two nicks of the target DNA that were generated during the process of DNA strand transfer The gaps are lled by a DNArepair polymerase encoded by the host cell 1 Target DNA is cleaved during the DNA strand transfer step to generate 3 OH ends that can serve as the primers for this repair synthesis a Filling in the gaps gives rise to the targetsite duplications that ank transposons The length of the targetsite duplication reveals the distance between the sites attacked on the two strands of the target DNA during DNA strand transfer c A er gaprepair synthesis DNA ligase is needed to seal the DNA strands Cutandpaste transposition also leaves a doublestrand break in the DNA at the site of the quotoldquot insertion and must be repaired to maintain the integrity of the host cell39s genome There Are Multiple Mechanisms for Cleaving the Nontransferred Strand during DNA Replication i Transposase cleaves the 339 ends of the element DNA and promotes DNA strand transfer to catalyze cutandpaste transposition l Transposons that move by this mechanisms also need to cleave the 5 terminating strands at the junctions between the transposon and the anking host DNA a These DNA strands are called the nontransferred strands as their 539 ends are not directly linked to the target DNA during the DNA strand transfer reaction ii 0quot iii i v iii vi An enzyme other than the transposase can be used to cleave the nontransferred strand 1 The bacterial transposon Tn7 encodes a specific protein TnsA that does this job a TnsA has a structure similar to that of a restriction endonuclease b Assembles with the Tn7encoded transposase TnsB protein i By working alltogether the transposase and TnsA excise the transposon from its original target site Other way of cleaving the nontransferred strand is promoted by the transposase itself using a DNA transesteri cation mechanism that is similar to DNA strand transfer 1 Transposons Tn5 and TnlO cleave the nontransferred strand by generating a structure known as a quotDNA hairpinquot a To form the hairpin the transposase uses the initially cleaved 339 OH end of the transposon DNA to attack a phosphodiester bond directly across the DNA duplex on the opposite strand i This reaction both cleaves the attacked DNA strand and covalently joins the 3 end of the transposon DNA to one side of the break 1 Results the two DNA strands are covalently joined by a looped end in the shape of a hairpin 2 The hairpin DNA end is then cleaved by the transposase to generate a standard doublestrand break in the DNA a The opening reaction occurs on both ends of the transposon DNA b Once the steps are complete the 3 OH ends of the element DNA are ready to be joined to a new target DNA by the DNA strand transfer reaction The Hermes transposon is a member of the hAT family of elements using a DNAhairpin intermediate to excise the transposon from the old DNA insertion site 1 In this case the order of the strand cleavage and transesteri cation reactions is different such that the DNA hairpins are formed in the host cell39s DNA rather than onthe end of the transposable element DNA cleavage via a transesteri cation reaction can also occur between the two ends of the transposon 1 In this case one cleaved 339OH end attacks the DNA strand at the opposite end of the element s DNA and the resulting DNA intermediate is further processed to generate the excised transposon 2 IS3 family uses this mechanism Why might transposases use transesterification as a cleavage mechanism I Probably an economic solution 2 Transposases have the ability to promote a Sitespecific hydrolysis of the 339 ends of the transposon DNA b Transesterification of this end into nonspecific DNA site DNA Transposition by a Replicative Mechanism i Some DNA transposons move using a mechanism called replicative transposition in which the element DNA is duplicated during each round of transposition ii Mechanism is similar to cutandpaste 1 Assembly of the transposase protein on the two ends of the transposon DNA to generate a transpososome a Transpososome formation is essential to coordinate the DNA cleavage and joining reactions on the two ends of the transposons DNA 2 DNA cleavage at the ends of the transposon DNA a Reaction is catalyzed by the transposase within the transpososome b Transposase introduces a nick into the DNA at each of the two junctions between the transposon sequence and the anking host DNA i This cleavage liberates two 339OH DNA ends on the transposon sequence ii In contrast to the cutandpaste transposition the transposon DNA is not excised from the host sequences at this stage this is the major difference between replicative and cutandpaste transposition 3 The 339 OH ends of the transposon DNA are then joined to the target DNA site by the DNA strand transfer reaction a The mechanism is the same as the cutandpaste mechanism i The only difference the intermediate generated by DNA strand transfer is in this case a double branched DNA molecule ii In this intermediate the 3 ends of the transposon are covalently joined to the new target site whereas the 539 ends of the transposon sequence remain joined to the old anking DNA 4 The two DNA branches within this intermediate have the structure of a replication fork a A er DNA strand transfer the DNA replication proteins from the host cell can assemble at these forks iii Replicative transposition frequently causes chromosomal inversions and deletions that can be highly detrimental to the host cell k VirusLike Retrotransposons and Retroviruses Move Using an RNA Intermediate i Viruslike retrotransposons and retroviruses insert into new sites in the genome of the host cell using the same steps of DNA cleavage and DNA strand transfer 1 In contrast to the DNA transposons recombination for these retroelements involves an RNA intermediate ii A cycle of transposition starts with transcription of the retrotransposon retroviral DNA sequence into RNA by a cellular RNA polymerase l N Transcription initiates at a promoter sequence within one of the LTRs and continues across the element to generate a nearly fulllength RNA copy of the element39s DNA The RNA is then reversetranscribed to generate a doublestranded DNA molecule this DNA molecule is called cDNA and is free from any anking host DNA sequences iii It is the cDNA that is recognized by the integrase protein for recombination with a new target DNA site 1 N L Integrase assembles on the ends of this cDNA and then cleaves a few nucleotides off the 3 end of each strand a This cleavage reaction is identical to the DNA cleavage step of DNA transposition b A mechanism to cleave the second strand is unnecessary for the elements i As the direct precursor DNA for interaction is generated form the RNA template by reverse transcription it is already in the form of an excised transposon Integrase then catalyzes the insertion of these cleaved 3 ends into a DNA target site in the hostcell genome using the DNA strand transfer reaction a This target site can have essentially any DNA sequence Host cell DNA repair proteins ll the gaps at the target site generated during DNA strand transfer to complete recombination a This gap repair reaction generates the target site duplications iv Because transcription to generate the RNA intermediate initiates within one of the LTRs this RNA does not carry the entire LTR sequence the sequence between the transcription start site and the end of the element is missing 1 A special mechanism is needed to regenerate the fulllength element sequence during reverse transcription v The pathway of reverse transcription involves two internal priming events and two strand switches 1 These switching events result in the duplication of sequences at the ends of the cDNA a cDNA has completely reconstructed LTR sequences to compensate for regions of sequence lost during transcription 1 DNA Transposases and Retroviral Integrases are Members of a Protein Superfamily i DNA cleavage of the 3 ends of the transposon DNA cDNA and DNA strand transfer are common steps used for DNA transposition and the movement of viruslike retrotransposons and retroviruses l N This conserved recombination mechanism is re ected in the structure of the transposaseintegrase proteins The catalytic domain contains three evolutionary invariant acidic amino acids a Two aspartates D b Glutamate E i Therefore recombinases of this class are referred to as DDE motif transpo sase inte grase proteins 0 ii These acidic amino acids form part of the active site and coordinate divalent metal ions that are required for activity iii An unusual feature of the transposaseintegrase proteins is that they use this same active site to catalyze both the DNA cleavage and DNA strand transfer rather than having two active sites ii In contrast to the highly conserved structure of the catalytic domains the remaining regions of proteins in this family are not conserved iii Transposases and integrases are only active when assembled into a synaptic complex called a transpososome on DNA 1 The transposase subunit that is bond to the recombinase recognition sequences on one of these DNA fragments donates the catalytic domain that promotes the DNA cleavage and DNA strand transfer reactions on the other end of the transposon m PolyA Retrotransposons Move by a quotReverse Splicing Mechanism i The polyA retrotransposons move using an RNA intermediate but use a mechanism different from that used by the viruslike elements ii Mechanism is called target site primed reverse transcription 1 Transcription of the DNA of an integrated element by a cellular RNA polymerase a Although the promoter is embedded in the 539UTR it can in this case direct RNA synthesis to begin at the first nucleotide of the element39s sequence The newly synthesized RNA is exported to the cytoplasm and translated to generate the ORFl and ORF2 proteins a These proteins remain associated With the RNA that encoded them b An element promotes its own transposition and does not donate proteins to competing elements The proteinRNA complex then reenters the nucleus and associates with the cellular DNA a ORF2 protein has both a DNA endonuclease activity and a reverse transcriptase activity i The endonuclease initiates the integration reaction by introducing a nick in the chromosomal DNA ii Trich sequences are preferred cleavage sites 1 Presence of these Ts at the cleavage site permits the DNA to basepair with the polyA tail sequence of the element RNA b The 3 OH DNA end generated by the nicking reaction then serves as the primer for reverse transcription of the element RNA i The ORF2 protein also catalyzes this DNA synthesis N E Table Structural Features Mechanism of Movement Examples Type DNA Mediated Transposition I Copying of element DNA Bacterial Terminal inverted repeats Phage Mu replicative that ank antibiotic accompanying each round tr r 39 and tr r of insertion into a new genes target site Bacterial cutand Terminal inverted repeats Excision of DNA from old Tn5 TnlO paste transpo sons that ank antibiotic target site and insertion into resistance and transposase new site genes Eukaryotic Inverted repeats that ank Excision of DNA from old P Transposons coding region with introns target site and insertion into elements new site hAT RNAMediated 39 Transposition VirusLike Approx 250600bp direct Transcription into RNA Ty Retrotransposons terminal repeats LTRs from promoter in le LTR elements anking genes for reverse by RNA polymerase II transcriptase integrase followed by reverse and retroviruslike Gag transcription and insertion protein at target site PolyA 339 AT rich sequence and Transcription into RNA Alu Retrotransposons 539UTR ank genes from internal promoter sequences encoding an RNAbinding protein and reverse transcriptase targetprimed reverse transcription initiated by endonuclease cleavage 7 Examples of Transposable Elements and Their Regulation a Two types of regulation appear as recurring themes i Transposons control the number of their copies present in a given cell By regulating copy number these elements limit their deleterious impact on the genome of the host cell ii Transposons control targetsite choice Two general types 1 Some elements preferentially insert into regions of the chromosome that tend not to be harm ll to the host cell a Safe havens is the name for them 2 Some transposons speci cally avoid transposing into their own DNA a Transposition target immunit b Is4 Family Transposons are Compact Elements with Multiple Mechanisms for Copy Number Control i The bacterial transposon TnlO is a wellcharacterized representation of the 184 family which includes Tn5 l TnlO is a compact element of 9kb and encodes a gene for its own transposase and genes imparting resistance to the antibiotic tetracycline P Transposes via the cutandpaste mechanism using the DNA hairpin strategy to cleave the nontransferred strands The sequence also has a site for IHF binding i Helps In the assembly of the roper transpososome complex needed for recombination as it does during bacteriophage 7t 0quot integration c Organized into three lnctional modules called composite transposon i Two outermost modules are ISlOL and IS 10R 1 Minitransposons 2 IS stands for insertion sequence 1 ISlOR encodes the gene for the transposase that recognizes the terminal inverted repeat sequences ofISlOR IS 10L and tnlO 2 ISlOL does not encode a lnctional transposase although very similar in sequence to IS 10R 1 IS 10R and TnlO are autonomous ISlOL is nonautonomous d Limits its copy number in any given cell by strategies that restrict its transposition frequency i Antisense RNA is one mechanism 1 Controls the expression of the transposase gene Near the end of IS 10R are two promoters that direct the synthesis of RNA by the host cell39s RNA polymerase i PIN directs the RNA synthesis inward l Responsible for the expression of the transposase gene ii POUT directs transcription outward l Regulate transposase expression by making an antisense FD l The RNAs synthesized from PIN and POUT overlap by 36bp and therefore can pair by hydrogen bonding between these complementary regions 1 This pairing prevents the binding of ribosomes to the PIN transcript and thus synthesis of the transposase protein iii More copies of TnlO will transcribe more of the antisense RNA which in turn will limit expression of the transposase ene c TnlO Transposition is Coupled to Cellular DNA Replication i When the TnlO DNA is hemimethylated transposition is most likely to occur 1 This coupling of transcription to the methylation state is due to the presence of two critical GATC sites in the transposon sequence a One site is in the promoter for the transposase gene b The other is in the binding site for the transposase within one of the inverted terminal repeats N Both RNApolymerase an transposase bind more tightly to the hemimethylated sequences than to their llly methylated versions a As a result when the DNA is hemimethylated the transposase gene is most efficiently expressed and the transposase protein binds most efficiently to the DNA i Therefore transposition of Tn0 occurs at its highest frequency during the brief phase of the cell cycle just after its DNA has been replicated d Phage Mu is an Extremely Robust Transposon Fquot 1 ii iii Lysogenic bacteriophage like bacteriophage 7t Mu is also a large DNA transposon 1 This phage uses transposition to insert its DNA into the genome of the host ell during infection and in this way is similar to retroviruses 2 Uses multiple rounds of replicative transposition to amplify its DNA during lytic growth a During the lytic cycle Mu completes approximately 100 rounds of transposition an hour most efficient transposon known Mu is short for mutator and stems from this ability to transpose promiscuously cells carrying an inverted copy of the Mu DNA Frequently accumulate new mutations due to insertion of the phage DNA into cellular genes Mu genome is approx 40kb and carries more than 35 genes but only two encode proteins with dedicated roles in transcription 1 A gene encodes protein MuA a Transposase and a member of the DDE protein super family 2 B gene encodes protein MuB a ATPasae that stimulates MuA activity and controls the choice of the DNA target site Mu uses Target immunity to Avoid Transposing Into its Own DNA 1 ii iii Mu shows very little sequence preference at its target site How does Mu avoid transposing into its own DNA a situation that would likely result in serious disruptions of the phage s genes 1 Mu transposition is regulated by a process known as transposition target immunity a DNA sites surrounding a copy of the Mu element including the elements own DNA are rendered very poor targets for a new transposition even Observed for a number of different transposable elements and can work over very long distances Mu sequences within approx 15kb of an existing Mu insertion are immune to new insertions l Tn3 and Tn7 target immunity occurs over distances greaterthan lOOkb 2 Target immunity protects an element from transposing into itself or from having another new copy of the same type of element insert into its genome f Tclmariner Elements Are Extremely Successful DNA Elements in Eukaryotes i Recognizable elements of the Tcl mariner family are widespread in both invertebrate and vertebrate organisms ii Elements in this family are the most common DNA transposons present in eukaryotes iii Elements are the simplest autonomous transposons known 1 1525kb long and carry only a pair of terminal inverted repeat sequences and a gene encoding a transposase protein of the DDE transposase superfamily iv No accessory proteins are required which contrasts most transposons although the final steps of recombination do require cellular DNArepair proteins v Elements move by cutandpaste transposition mechanism 1 Transposon DNA is cleaved out of the old anking host DNA using pairs of cleavages that are staggered by two base pairs 2 These elements prefer to insert into DNA sites with the sequence 5 TA vi What happens to the quotemptyquot site in the host chromosome when a transposon excises 1 DNA sequence analysis of some sites that once carried a transposome reveals that sometimes the broken ends are lled in and then directly joined NHEJ 2 Repair reactions result in the incorporation of a few extra base pairs of DNA at the old insertion sites a These small DNA insertions are known as quotfootprintsquot as they are the traces le by a transposon that has traveled through a site in the genome vii Transposition of the elements is not well regulate contrasts many transposons 1 As a result many elements found by genome sequencing are dead unable to transpose g Yeast Ty Elements Transpose into Safe Havens in Genome i Viruslike retrotransposons ii Ty RNA is found in cells packaged into viruslike particles 1 These elements seem to be viruses that cannot escape one cell and infect new cells iii S cerevisiae carries members of the Tyl Ty2 Ty3 Ty4 and Ty5 families 1 Each class promotes its own mobility but does not mobilize elements of another class iv Preferentially integrate into specific chromosomal regions Tyl elements nearly always transpose into DNA within 200bp upstream of a start site for transcription by the host RNA polymerase III enzyme 2 Ty3 integration is also tightly linked to Pol III promoters integration is precisely targeted to the start site of transcription 2bp 3 Ty5 9In contrast preferentially integrates into regions of the genomes that are in a silenced transcriptionally quiescent state a Silenced regions targeted by Ty5 include the telomeres and the silent copies of the matingtype loci 4 The mechanism of regional targettype selection involves the formation of speci c proteinprotein complexes between the elements integrase bound in a complex to the cDNAand host speci c proteins bound to the chromosomal sites V Why the regional targetsite preference 1 Proposed that this target speci city enables the transposons to persist in a host organism by focusing most of their insertions away from important regions of the genome that are involved directly in coding for proteins h LINEs Promote Their Own Transposition and Even Transpose Cellular RNAs i Autonomous polyA retrotransposons known as LINEs are abundant in the genomes of vertebrate organisms l About 20 of the human genome ii Elements were rst recognized as a family of repeat sequences iii Name derived from this initial identi cation LINE is the acronym for long intersperse nuclear element 1 L1 is one ofthe best understood iv In addition to promoting their own mobility they also donate the proteins needed to reversetranscribe and integrate another related class of repeat sequences the nonautonomous polyA retrotransposons known as short interspersed nuclear elements SINEs l Alu sequence is an example of a widespread SINE sequence in the human genome v How do they avoid transposing cellular mRNA molecules for LINEs 1 All genes have a promoter and most are transcribed into an mRNA that will carry a polyA sequence at the 339 end ofthe molecule thus any mRNA should he an attractive quotsubstratequot for transposition of cellular RNA via the targetprimed reverse transcription mechanism vi For many cellular genes there are additional copies of a highly related sequence in the genome 1 Copies appear to have lost their promoter and their introns and o en carry truncations near their 539 ends a Known as processed pseudogenes and usually are not expressed by the cell i O en anked by short repeats in the target DNA 8 VDJ Recombination a Immune system of vertebrates has the job of recognizing and fending off invading organisms b Two specialized celltypes i B cells 1 Produce antibodies that circulate in the bloodstream ii T Cells 1 Produce cell surfacebound receptor proteins T Cell receptors c Recognition of a foreign molecule by either of these classes of proteins starts a cascade of events focused on destruction of the invader d The principal mechanism cells use to generate antibodies and Tcell receptors with such diversity relies on a specialized set of DNA rearrangement reactions known as VDJ Recombination Antibody and Tcell receipt genes are composed of gene segments that are assembled by a series of sequencespecific DNA rearrangements TCell Structure Fquot quot1 0 1 ii iii Genomic region encoding an antibody molecule is on page 366 Antibodies are constricted of two copies each of a light chain and a heavy chain The part of the protein that interacts with the foreign molecules Is called the antigen binding site 1 This binding region is constructed from VL and VH domains The light chain is the kappa locus 1 Region carries approx 300 gene segments coding for different versions of the lightchain VL protein region 2 There are also four gene segments encoding a short region of protein sequence called the J region followed by a single coding region for the CL region VDJ Recombination can lse the DNA between and pair of C and J segments therefore 1200 variants of the antibody light chain can be produced form the single genomic region 1 This segment is then brought together with the CL coding region via RNA splicing Early Events in VDJ Recombination Occur by a Mechanism Similar to Transposon Excision 1 Recombination sequences called recombination signal sequences ank the gene segments that are assembled by VDJ recombination 1 These signals all have two highly conserved sequence motifs a 7mer b 9mer i These motifs are bound by the recombinase l The recombination signal sequences come in two class 1 One class has the 7mer and 9mer mer motifs spaced by 12 bp of sequence 2 Second class has these motifs spaced by 23 bp 1 Recombination always have occurs between a pair of recombination signal sequences in which one partner has the 12bp spacer and the other partner has the 23bp spacer The recombinase responsible for recognizing and cleaving the recombination signal sequences is composed of two protein subunits l RAGl RecombinationActivating Gene a Recognize the recombination signal sequences and pair the two sites to form a proteinDNA synaptic compleX b Introduce singlestrand breaks in the DNA at each of the junctions between the recombination signal sequence and the gene segment that will be rearranged The site of cleavage s such that the proteincoding segment now has a free 3 OH DNA end 2 RAG2 a Recognize the recombination signal sequences and pair the two sites to form a proteinDNA synaptic complex iii Once the two DNA sequences in the synaptic complex have been nicked and quothairpinnedquot by the RAG recombinase cellular DNArepair proteins take over to nish the recombination repair 1 NHEJ proteins participate in the opening of the DNA hairpin 0 l Mechanisms of Transcription 2 Introduction a Transcription is chemically and enzymatically very similar to DNA replication i Both involve enzymes that synthesize a new strand of nucleic acid complementary to a DNA template strand ii Di erences New strand is made from ribonucleotides rather than deoxyribonucleotides 2 RNA polymerase does not need a primer rather it can initiate transcription de novo 3 The RNA product does not remain basepaired to the template DNA strand rather the enzyme displaces the growing chain only a few nucleotides behind where each ribonucleotide is added 4 Transcription although very accurate is less accurate than DNA replication 3 RNA Polymerases and the Transcription Cycle a RNA Polymerases Come in Different Forms but Share Many Features i RNA polymerase performs essentially the same reaction in all cells ii Bacteria have only a single RNA polymerase whereas eukaryotic cells have three 1 RNA Polymerase I RNA Pol I a Transcribes the large RNA precursor gene 2 RNA Polymerase 11 RNA Pol II a Focuses on eukaryotic transcription b Responsible for transcribing most genes 3 RNA Polymerase III RNA Pol III a Transcribes tRNA genes some small nuclear RNA genes and the 5S rRNA gene iii The bacterial polymerase core enzyme alone is capable of synthesizing RNA and comprises 1 Two copies of the or subunit a Homologous to RPB3 and RPBll 2 B subunit a Homologous to the large subunit found in RNA Pol II RPBl 3 339 subunit a Homologous to the large subunit found in RNA Pol II RPB2 4 m subunit a Homologous to RPB6 1 This enzyme is closely related to the eukaryotic polymerases iv The bacterial and yeast enzymes share an overall shape and organization indeed they are more alike than the comparison of the subunit sequences v The overall shape of each enzyme resembles a crab claw l The two pincers of the crab claw are made up predominantly of the two largest subunits of each enzyme B and B39 for the bacterial case and RPBl and RPB2 for the eukaryotic enzyme N The active site which is made up of regions from both these subunits is found at the base of the pincers within a region called the quotactive center cle quot a Works according to the twometal ion catalytic mechanism for nucleotide addition proposed for all types of polymerase i Active site contains one tightly bound Mg2 metal ion and the second Mg2 is brought in with each new nucleotide in the addition cycle released with the pyrophosphate Transcription by RNA Polymerase Proceeds in a Series of Steps i Three Phases 1 Initiation a A promoter is the DNA sequence that initially binds the RNA polymerase b Once formed the promoterpolymerase complex undergoes structural changes required for initiation to proceed a transcription bubble is formed c Transcription always occurs in a 5 gt3 direction d Unlike replication only one of the DNA strands acts as atemplate on which the RNA strand is built i As RNA polymerase binds promoters in a de ned orientation the same strand is always transcribed from a given promoter e The choice of a promoter determines which stretch of DNA is transcribed and is the main step at which regulation is imposed 2 Elongation a Once the RNA polymerase has synthesized a short stretch of RNA approximately lObp it shifts into the elongation phase b The enzyme performs a range of tasks including the catalysis of RNA synthesis i It unwinds the DNA in front and anneals it behind it ii It dissociates the growing RNA chain from the template as it moves along iii Performs proofreading functions 3 Termination a Once the polymerase has transcribed the length of the gene genes it must stop and release the RNA product b In some cells specific wellcharacterized sequences trigger termination and in others it is less clear on what goes on ii Transcription Initiation Involves Three De ned Steps 1 The first step is the initial binding of polymerase to a promoter to form a closed complex a In this form the DNA remains doublestranded and the enzyme is bound to one face of the helix 2 The second step is when the closed complex undergoes a transition to the open complex in which DNA strands separate over a distance of approximately l4bp around the start site to form the transcription bubble LA The nal step is when polymerase enters the phase of initial transcription followed by promoter escape a When DNA opens up the DNA template strand is now free and the rst two ribonucleotides are brought in to the active site aligned on the template strand and joined together i Incorporation of the rst ten ribonucleotides is rather inefficient process and at that stage the enzyme o en releases short transcripts and then begins synthesis again 1 Inthis phase the polymerasepromoter complex is called the initial transcribing complex ii Once an enzyme makes a transcription longer than 10 nucleotides it is said to have escaped the promoter 1 At this point it has formed a stable ternary complex containing 1 Enzyme 2 DNA 3 RNA 1 This is the transition to the elongation phase 4 The Transcription Cycle in Bacteria a Bacterial Promoters Vary in Strength and Sequence but Have Certain De ning Features i The bacterial core RNA polymerase can initiate transcription at any point on a DNA molecule and this can be shown in vitro using puri ed core enzyme ii In cells polymerase initiates transcription only at promoters 1 It is the addition of an initiation factor called 6 that converts core enzyme 5303 into the form that initiates only at promoters a This form of the enzyme is called the RNA polymerase holoenzyme iii In the case of E coll the predominant 6 factor is called 570 l Promoters 39 J this I 39J 39 39 570 share the following characteristic structure a Two conserved sequences each of six nucleotides are separated by a nonspecific stretch of 1719 nucleotides b The two de ned sequences are centered at approximately lObp and at approximately 35bp upstream of the site where RNA synthesis starts i The sequences are thus called the 35 and 10 regions or elements according to the numbering scheme 2 Although the case majority of G70 promoters contain recognizable 35 and 10 regions the sequences are not identical a By comparing many different promoters a consensus sequence can be derived i The consensus sequence re ects preferred 10 and 35 regions separated by the optimum spacing l7bp ii Promoters with sequences closer to the consensus are generally quotstrongerquot than those that match less well 1 Strength of a promoter is a term describing how many transcripts it initiates in a given time b An additional DNA element that binds RNA polymerase is found in some strong promoters those directing expression of the ribosomal RNA rRNA genes i Called UP element and increases polymerase binding by providing an additional speci c interaction between the enzyme and the DNA Another class of G70 promoters lacks a 35 region and instead has an quotextended 10quot element i Comprises a standard 10 region with an additional short sequence element at its upstream end ii Extra contacts made between polymerase and this additional sequence element compensate for the absence ofa 35 region 1 The E coli gal genes use this promoter iv An additional DNA element that binds RNA polymerase has been found just downstream from the 10 element 1 Called the discriminator 2 The strength of the interaction between the discriminator and polymerase in uences the stability of the complex between the enzyme and the promoter b The 6 Factor Mediates Binding of Polymerase to the Promoter i The 670 region can be divided into four regions called 6 region 1 through 6 region 4 l The regions that recognize the 10 and 35 elements of the promoter are regions 2 and 4 ii Two helices within region 4 form a common DNAbinding motif called helix tum helix motif 1 One of these helices inserts into the major groove and interacts with bases in the 35 region 2 The other helix lies across the top of the groove making contacts with the DNA backbone iii The 10 region is also recognized by an or helix 1 The interaction is less wellcharacterized and is more complicated a Rationale whereas the 35 region simply provides binding energy to secure polymerase to the promoter the 10 region has a more elaborate role in transcription initiation because it is within that element that DNA melting is initiated in the transition from the close to open complex 2 The 6 region that interacts with the 10 is doing more than binding DNA a The at helix involved in recognition of the 10 region contains several essential aromatic amino acids that can interact with bases on the nontemplate strand in a manner that stabilizes the melted DNA b The extended 10 element is recognized by an or helix in 6 region 3 i This helix makes contact with the two specific base pairs that constitute that element 0 C Unlike other elements in the promoter the UPelements are NOT recognized by 6 but it is recognized by a carboxyterminal domain of the or subunit called the xCTD l The 0LCTD is connected to the xNTD by a exible linker a Although xNTD is embedded in the body of the enzyme the 0LCTD can reach the upstream element and can do so even when that element is not located immediately adjacent to the 35 region but further upstream The 6 subunit is positioned within the holoenzyme structure in such a way as to make feasible the recognition of various promoter elements DNAbinding regions point away from the body of the enzyme rather than being embedded The spacing between those regions consistent with the distance between the DNA elements they recognize a 6 regions 2 and 4 are separated by about 75A when 6 is bound in the holoenzyme and this is about the same distance as that between the centers of the 10 and 35 elements of a typical 670 promoter This large spacing of the protein domains is accompanied by the region between 6 regions 2 and 4 by region 3especially region 32 also called the 63 4 linker N Transition to the Open Complex Involves Structural Changes in RNA Polymerase and in the i iii Promoter DNA The initial binding of RNA polymerase to the promoter DNA in the closed complex leaves the DNA in a doublestranded form The next stage in initiation requires the enzyme to become more intimately engaged with the promoter in the open complex The transition from closed to open complex involves structural changes in the enzyme and the opening of the DNA double helix to reveal the template and nontemplate strands 1 This melting occurs between positions 11 and 3 In the case of the bacterial enzyme bearing G70 isomerization does not require energy derived from ATP hydrolysis and is instead the result of a spontaneous conformational change in the DNAenzyme complex to a more energetically favorable form 1 Isomerization is irreversible and once completed typically guarantees that transcription will subsequently initiate Formation of the closed complex is readily reversible polymerase can as easily dissociate from the promoter as make the transition to the open complex Structure of the Holoenzyme 1 Channel runes between the pincers of the clawshaped enzyme 2 There are five channels into the enzyme a The NTPuptake channel i Allows ribonucleotides to enter the active center b The RNAexit channel 2 vi i Allows the growing RNA chain to leave the enzyme as it is synthesized during elongation c The remaining three channels allows the DNA entry and exit i The downstream DNA enters the active center cleft in double stranded form through the downstream channel between the pincers 1 Within the active center cle the DNA strands separate from position 3 ii The nontemplate strand exits the active center cle through the nontemplate strand NT channel and travels across the surface of the enzyme iii The template strand follows a path through the active center cle and exits through the templatestrand T channel Two structural changes are seen in the enzyme upon isomerization from the closed to open complex 1 The pincers at the front of the enzyme clamp down tightly on the downstream DNA 2 Major shi in the position of the aminoterminal region of o a When not bound to DNA 6 region ll lies within the active center cle of the holoenzyme locking the path that in the open complex is followed by the template DNA strand b In the open complex region ll shi s approximately 50A and is now found outside the enzyme allowing the DNA access to the cle i Region 11 is highly negatively charged thus inthe holoenzyme region ll acts as a molecular mimic l The space in the active center cle which may be occupied either by region 11 r by DNA is highly positively charged d Transcription is Initiated by RNA Polymerase Without the Need for a Primer FD i RNA polymerase can initiate a new RNA chain on a DNA template and thus does not need a primer 1 Requires that the initiating ribonucleotide be brought into the active site and held stably on the template while the next NTP is present with correct geometry for the chemistry of polymerization to occur a Difficult because RNA polymerase starts most transcripts with an A and that ribonucleotide binds the template nucleotide T with only two hydrogens bonds Enzyme has to make specific interactions with one or both of the initiating ribonucleotide and the second ribonucleotide holding one or both rigid in the correct orientation to allow chemical attack on the incoming NTP During Initial Transcription RNA Polymerase Remains Stationary and Pulls Downstream DNA Into Itself 1 RNA polymerase produces and releases short RNA transcripts of less than ten nucleotides abortive synthesis before escaping the promoter entering the elongation phase and synthesizing the proper transcript Three general models that were proposed N U The quottransient excursionquot model proposes transient cycles of forward and reverse translocation of RNA polymerase a Polymerase is thought to leave the promoter and translocate a short way along the DNA template synthesizing a short transcript before aborting transcription releasing the transcript and returning to its original location on the promoter quotInchwormingquot invokes a exible element within the polymerase that allows a module at the front of the enzyme containing the active site to move downstream synthesizing a short transcript before aborting and retracting the body of the enzyme still at the promoter quotScrunchingquot proposes that DNA downstream from the stationary promoterbound polymerase is pulled into the enzyme iii quotScrunchingquot is the proposed method that has been generally accepted 1 Based on experiments using singlemolecule analysis that allow the positions of different parts of polymerase to be measured relative to each other and to the template DNA during initial transcription f Promoter Escape Involves Breaking PolymerasePromote Interactions and Polymerase Coreo Interactions i It is not clear why RNA polymerase must undergo this period of abortive initiation before achieving escape but once again a region of the 6 factor appears to be invoked acting as a molecular mimic l 2 3 In this case it is the region 34 linker and it mimics RNA This region of 6 lies in the middle of the RNA exit channel in the open complex and for an RNA chain to be made longer than about ten nucleotides this region ofo must be ejected from that location a process that can take the enzyme several attempt The ejection of 6 region 34 linker probably accounts for 6 being more weakly associated with the elongating enzyme than it is with the open complex ii Scrunching is reversed upon escape l 2 The DNA unwound during scrunching is rewound with concomitant collapse of the transcription bubble from a size of 2224 nucleotides back down to 1214 nucleotides a This process provide the energy required by polymerase to break the polymerasepromoter and coreo interactions associated with escape Scrunching is a way to store and mobilize energy during transcription initiation and its release upon escape is what enables polymerase to break free of the promoter and dislodge 6 factor from core g The Elongating Polymerase Is a Processive Machine That Synthesizes and Proofreads i DNA passes through the elongating enzyme in a manner very similar to its passage through the open complex 1 Doublestranded DNA enters the front of the enzyme between the pincers a At the opening of the catalytic cle the strands separate to follow different paths through the enzyme before exiting via their respective ii iii h RNA i iii channels and reforming a double helix behind the elongating polymerase 2 Only eight or nine nucleotides of the growing RNA chain remain base paired to the DNA template at any given time the remainder of the RNA chain is peeled off and directed out of the enzyme through the RNA exit channel During elongation the enzyme adds one nucleotide at a time to the growing RNA transcript l The size of the bubble or the length of DNA that is not doublehelical remains constant throughout elongation as lbp is separated ahead of the processing enzyme lbp is formed behind it RNA polymerase carries out two proofreading functions as well 1 Pyrophosphorolytic Editing a The enzyme uses its active site in a simple back reaction to catalyze the removal of an incorrectly inserted ribonucleotide by reincorporation of PPi b The enzyme can then incorporate another ribonucleotide in its place in the growing RNA chain Note The enzyme can remove either correct or incorrect bases in this manner but spends longer hovering over mismatches that matches 2 Hydrolytic Editing a Polymerase backtracks b one or more nucleotides and cleaves the RNA product removing the errorcontaining sequence b Stimulated by Gre factors which both enhance hydrolytic editing lnction and serve as elongation stimulating factors they insure that polymerase elongates efficiently and helps overcome quotarrestquot at sequences that are difficult to transcribe i Combination of functions is comparable to those imposed on eukaryotic RNA polymerase II by the transcription factor TFIIS C c Nus proteinspoints polymerase in the elongation phase and promotes the process of elongation and termination Polymerase Can Become Arrested and Need Removing Under certain circumstances an elongating RNA polymerase can become arrested and cease transcribing One common cause of arrest is a damaged DNA strand The consequence s of arrest can be no products being made by the arrested polymerase and that same enzyme will cause a roadblock to other polymerases attempting to transcribe the same gene 1 To deal the cell has machinery that removes the arrested polymerase and at the same time recruits repair enzymes in particular the endonuclease UvrABC the repair mechanism is the transcription coupled repair 2 Both polymerase removal and repair enzyme recruitment are carried out by the TRCF protein a Has an ATPase activity L b Binds dsDNA upstream of the polymerase and uses the ATPase motor to translocate along the DNA until it encounters the stalled RNA polymerase The collision pushes polymerase forward either allowing it to restart elongation or cause dissociation of the ternary compleX of RNA polymerase template DNA and RNA transcript most common Transcription is Terminated by Signals Within the RNA Sequence i When RNA polymerase arrests during elongation it can be knocked off DNA by the action of the translocator TRCF 1 This termination is triggered by damaged DNA or by other unanticipated hindrances ii Termination is a normal and important lnction at the ends of genes 1 Sequences called terminators trigger the elongating polymerase to dissociate from the DNA and release the RNA chain it has made 2 Rho Proteins a Ringshaped protein with siX identical subunits binds to single stranded RNA as it exits the polymerase b Protein has an ATPase activity i Once attached to the transcript Rho uses the energy derived from ATP hydrolysis to induce termination Rho pushes polymerase forward relative to the DNA and RNA resulting in termination in a manner analogous to termination by RRCF 0 0 i Rho pulls RNA out of the polymerase resulting in termination ii Or Rho induces a conformational change in polymerase causing the enzyme to terminate How is Rho directed to a particular RNA molecule i There is some specificity in the sites it binds rut sites 1 These sites consist of stretches of approximately 40 nucleotides that do not fold into a secondary structure and are rich in C residues Rho fails to bind any transcript that is being translated l Transcription and translation are tightly coupled in Bacteriatranslation initiates on growing RNA transcripts as soon as they start exiting polymerase while they are still being synthesized l Rho typically terminates only those transcripts still being transcribed beyond the end of a gene or operon e Bacterial Terminator Rho i Rho Dependent 1 Ill de ned RNA elements called rut sites and for them to work require the action of the Rho factor 2 To work the require the action of the Rho factor ii Rho Independent 2391 N Also called intrinsic terminators because they need no other factors to work 2 Consist of two short sequence elements 1 Short inverted repeat of about 20 nucleotides 2 Stretch of approximately eight AT base pair 1 These elements do not affect the polymerase until they have been transcribed When polymerase transcribes an inverted repeat sequence the resulting RNA can form a stemloop structure by basepairing with itself 1 Formation of the hairpin causes termination by disrupting the elongation complex a The hairpin words as an efficient terminator only when it is followed by a stretch ofAU base pairs i Under those circumstances at the time the hairpin forms the growing RNA chain will be held on the template at the active site by only AU base pairs iii 5 Transcription in Eukaryotes a b 0 2391 The process of transcription is identical with RNA polymerase Eukaryotes have three different polymerases P01 1 Pol II and Pol 111 whereas bacteria have only one i Bacteria require only one additional initiation factor 6 whereas several initiation factors are required for ef cient and promoterspecific initiation in eukaryotes called general transcription factors GTFs In vivo the DNA template in eukaryotic ells is incorporated into nucleosomes i General transcription factors are not along suf cient to bind promoter sequences and elicit significant expression additional factors are required 1 DNAbinding regulatory proteins 2 Mediatorcomplex 3 Chromatinmodifying enzymes RNA Polymerase 11 Core Promoters are Made Up of Combinations of Four Different Sequence Elements i The eukaryotic core promoter refers to the minimal set of sequence elements required for accurate transcription initiation by the Pol II machinery 1 Typically 4060 nucleotides extending either upstream or downstream from the transcription start site ii Elements found in Pol 11 core promoters consist of l TFIIB recognition element BRE 2 TATA element box 3 Initiator Inr 4 Downstream promoter elements DPE DCE and MTE a Example Promoter typically have either a TATA element or a DPE element but NOT BOTH i O en a TATAcontaining promoter also contains a DCE b The Inr is the most common element found in combination with both TATA and DPEs iii Typically upstream of the core promoter there are other sequence elements required for efficient transcription 1 These elements constitute the regulatory sequences and can be groups into various categories re ecting their location and the organism in question a These elements include i Promoter proximal elements ii Upstream Activator Sequences UASs iii Enhancers iv Silencers boundary elements and insulators e RNA Polymerase 11 Forms a Preinitiation Complex with General Transcription Factors at the Promoter i iii The general transcription factors collectively perform the lnctions performed by 6 in bacterial transcription and some of the factors share regions of weak homology with different regions of o l The general transcription factors help polymerase bind to the promoter and melt the DNA 2 Also help polymerase escape from the promoter and initiate the elongation hase The complete set of general transcription factors and polymerase bound together at the promoter and poised for initiation is called the preinitiation complex 1 Many Pol II promoters contain a TATA element approximately 30 bp upstream of the transcription start site a This is Where preinitiation complex formation begins i The TAAT element is recognized by the general transcription factor called TFIID TFII denotes a transcription factor for Pol l TFIID is a multisubunit complex 1 Component of TFIID that binds to the TATA DNA sequence is called TBP TATAbinding protein 2 The other subunit in this complex are called TAFs for TBPassociated factors 1 Some TAFs recognize other core promoter elements such as Inr DPE and DCE Upon DNA binding TBP distorts the TATA sequence 1 The resulting TBPDNA complex provides a platform to recruit other general transcription factors and polymerase itself to the promoter a In vitro the order of formation is TFIIA TFIIB TFIIF together with polymerase and then TFIIE and TFIIH i Formation of the preinitiation complex sustaining these components is followed by promoter melting which requires hydrolysis of ATP and is mediated by TFIIH f Promoter Escape Requires Phosphorylation of the Polymerase quotTailquot i Recall that during abortive initiation the polymerase synthesizes a series of short transcripts 9 h ii In eukaryotes promoter escape involves two steps not seen in bacteria 1 ATP hydrolysis in addition to the ATP hydrolysis needed for melting 2 Phosphorylation of the polymerase Large subunit of Pol II has a carboxyterminal domain CTD which is referred to as the quottailquot 1 The CTD contains a series of repeats of the heptapeptide sequence YSPTSPS a There are 27 of these repeats inthe yeast Pol II CTD 32 inthe worm Caenorhabditis elegans 45 inDrosphz39la and 52 in humans The form of Pol II recruited to the promoter initially contains a largely unphosphorylated tail but the species found in the elongation complex bears multiple phosphoryl group on its tail 1 Addition of these phosphates help polymerase shed most of the general transcription factors used for initiation and which the enzyme leaves behind as it escapes the promoter TBP Binds to and Distorts DNA Using a B Sheet Inserted Into the Minor Groove i TBP uses an extensive region of B sheet to recognize the minor groove of the iii TATA element 1 More typically proteins recognize DNA using or helices inserted into the major groove ofDNA 2 The reasons for TBP being dilTerent in the recognition mechanism is linked to the need for that protein to distort the lo cal DNA structure ii How is specificity achieved 1 To select the TATA sequence TBP relies on the ability of that sequence to undergo a specific structural distortion 2 When TBP binds to DNA it cause the minor groove to be widened to an almost at conformation and bends the DNA by an angle of approximately a The interaction between TBP and DNA involves only a limited number of hydrogen bonds between the protein and the edges of the base pairs in the minor groove instead much of the specificity is imposed by two pairs of phenylalanine side chains that intercalate between the base pairs at either end of the recognition sequence and drive the strong bend in the DNA 3 AT base pairs are favored because they are more readily distorted to allow the initial opening of the minor groove a There are also extensive interactions between the phosphate backbone and basic residues in the B sheet adding to the overall binding energy of the interaction The Other General Transcription Factors Also Have Speci c Roles in Initiation i TAFs l TBP is associated with about ten TAFs a Two of the TAFs bind DNA elements at the promoter 2 Several of the TAFs have structural homology with histone proteins and it has been proposed that they might bind DNA in a similar manner L a TAF 42 and TAF62 fromDrosphz39la have been shown to form a structure similar to that of the H3 H4 tetramer i Found not only in the TFIID complex but also associated with some histone modi cation enzymes such as the yeast SAGA complex 3 Another TAF appears to regulate the binding of TBP to DNA a It uses an inhibitory ap that binds to the DNAbinding surface of TBP which is an example of molecular mimicry ii TFIIB l Singlepolypeptide chain protein 2 Enters the preinitiation complex a er TBP 3 The structure of the ternary complex of TFIIBTBPDNA shows specific TFIIB TBP and TFIIB DNA contacts a These include i Basespecific interactions with the major groove upstream to the BRE ii Minor groove downstream from the TATA element b Asymmetric binding of TFIIB to the TBPTATA complex accounts for the asymmetry in the rest of the assembly of the preinitiation complex and the unidirectional transcription that results c Contacts Pol II in the preinitiation complex i Appears to bridge the TATAbound TBP and polymerase d Segments of TFIIB insert into the RNAexit channel and active center cle of Pol II in a manner analogous to 6 region 34 linker in the bacterial case iii TFIIF l Twosubunit in humans factor associates with Pol II and is recruited to the promoter together with that enzyme 2 Binding of Pol IITFIIF stabilizes the DNATBPTFIIB complex is required before TFIIE and TFIIH are recruited to the preinitiation complex iv TFIIE and TFIIH l TFIIE consists of two subunits binds next and has roles in the recruitment and regulation of TFIIH 2 TFIIH controls the ATPdependent transition of the preinitiation complex to the open complex a It is also the largest and most complex of the general transcription factors it has two subunits and a molecular mass comparable to that of the polymerase b Within TFIIH are two subunits that lnction as ATPases and another that is a protein kinase with roles in promoter melting and escape In Vivo Transcription Initiation Requires Additional Proteins Including the Mediator Complexes i High regulated levels of transcription in vivo require additionally transcriptional regulatory proteins the Mediator complex and nucleosome modifying enzymes v F ii One reason for additional requirements is that the DNA template is packed into chromatin 1 This complicates binding to the promoter of polymerase and its associated factors a Transcriptional regulatory proteins called activators help recruit polymerase to the promoter stabilizing its binding i This recruitment is mediated through interactions between DNAbound activators chromatin modifying and remodeling factors and parts of the transcription machinery 1 One interaction is with the Mediator complex 1 Mediator is associated with the CTD tail of the large polymerase subunit through one surface while presenting other surfaces for interaction with DNA bound activators Deletion of individual subunits of Mediator often leads to loss of expression of only a small subset of genes different for each subunit 1 Re ects that fact that different activators are believed to interact with different Mediator subunits to bring polymerase to different genes Mediator aids initiation by regulating the CTD kinase in TFIIH v The need for nucleosome modi ers and remodelers also differs at different promoters or even at the same promoter under different circumstances Mediate Consists of Many Subunits Some Served from Yeast to Human i The yeast and human Mediator each include more than 20 subunits of which 7 show significant sequence homology between the two organisms 1 Very few of these subunits have any identi ed function ii Srb4Medl7 1 Essential for transcription of essentially all Pol 11 genes in vivo 2 Have a similar shape are very large even bigger than RNA polymerase iii The Mediator from both yeast and humans is organized in modules New Set of Factors Stimulate Pol II Elongation and RNA Proofreading i One polymerase has escaped the promoter and initiated transcription it shifts into the elongation phase 1 This transition involves the Pol II enzyme shedding most of its initiation factors a In their place another set of factors is recruited some of theses such as TFIIS and hSPTS are elongation factors ii The polymerase CTD lies directly adjacent to the channel through which the newly synthesized RNA exits the enzyme Various proteins are thought to stimulate elongation by Pol II 1 Kinase PTEFb is recruited to polymerase by transcriptional activators a Once bound to Pol II this protein phosphorylates the serine residue at position 2 of the CT repeats b That phosphorylation event correlates with elongation c Phosphorylates and thereby activates another protein hSPTS i Elongation factor 2 TATSFl iii iii a Elongation factor recruited by PTEFb iv EEL Family 1 Class of elongation factors 2 Bind to elongating polymerase and suppress transient pausing by the enzyme such pausing otherwise occurs at many sites along the DNA V TFIIS l Stimulates elongation but does not affect initiation 2 Stimulates elongation by limiting the length of time that polymerase pauses when it encounters sequences that would otherwise tend to slow the enzyme s progress 3 The length of time that polymerase pauses at any given site is reduced when TFIIS is present 4 It contributes to proofreading by polymerase 5 Stimulates an inherent RNase activity in polymerase allowing an alternative approach to removing misincorporated bases through local limited RNA degradation 1 Elongating RNA Polymerase Must Deal with Histones in its Path i Elongation take p lace in the presence of histones because the DNA template is incorporated into nucleosomes ii How does RNA polymerase transcribe through these potential barriers l Chromatin greatly impedes transcription 2 A factor called FACT facilitates chromatin transcription was identi ed in human cell extracts a This factor makes transcription on chromatin templates much more efficient b Heterodimer of two wellconserved proteins i Sptl6 ii S SRPl iii How does FACT work 1 Sptl6 binds to the H2NH2B dimer 2 SSRPl binds to the H3H4 tetramer 3 Fact can both dismantle histones by removing one H2AH2B dimer and reassemble them by restoring that dimer iv Ahead of a transcribing RNA polymerase FACT removes one H2NH2B dimer 1 This allows a polymerase to pass that nucleosome 2 FACT also has a histone chaperone activity this allows it to restore the H2AH2B dimer to the histone hexamer immediately behind the processing polymerase FACT allows polymerase to elongate and at the same time maintains the integrity of the chromatin through which the enzyme is transcribing m Elongating Polymerase is Associated With a new Set of Protein Factors Required for Various Types of RNA Processing i Once transcribed eukaryotic RNA has to be processed in various ways before being exported from the nucleus where it can be translated ii These processing events include the following 1 Capping of the 539 end of the RNA 2 Splicing 3 Polyadenylation of the 339 end of the RNA a The most complicated is splicingthe process whereby noncoding introns are removed from RN Ato generate the mature mRNA iii There is an overlap in proteins involved in elongation and those required for RNA processing 1 An elongation factor hSPTS also helps to recruit the 539capping enzyme to the CTD tail of polymerase phosphorylated at serine position 5 a hSPTS stimulates the 539 capping enzyme activity 2 Elongation factor TATSFl recruits components of the splicing machinery to polymerase with Ser2 phosphorylated tail a Elongation termination or transcription and RNA processing are interconnected presumably to ensure their proper coordination iv The first RNA processing event is capped 1 This involves the addition of a modi ed guanine base to the 539 end of the RNA a It is a methylated guanine and it is joined to the RNA transcript by an unusual 539 539 linkage involving three phosphates 2 The 539 cap is created in three enzymatic steps a A phosphate groups is removed from the 539 end of the transcript b The GMP moiety is added c That nucleotide is modi ed by the addition of a methyl group 3 The RNA is capped as soon as it emerges from the RNAexit channel of polymerase a This happens when the transcription cycle has progressed only to the transition between the initiation and elongation phases 4 A er capping dephosphorylation of Ser5 within the tail repeats may be responsible for dissociation of the capping machinery and lrther phosphorylation causes recruitment of the machinery needed for RNA splicing v The final RNA processing event polyadenylation of the 3 end of the mRNA is intimately linked with the termination of transcription 1 The polymerase CTD tail is involved in recruiting some of the enzymes necessary for polyadenylation Once polymerase has reached the end of a gene it encounters specific sequences that after being transcribed into RNA trigger the transfer of the polyadenylation enzymes to that RNA leading to four events a Cleavage of the message b Addition of many adenine residues to its 3 end c Degradation of the RNA remaining associated with RNA polymerase by a 539 gt3 ribonuclease d Termination of transcription vi Two protein complexes are carried by the CTD of polymerase as it approaches the end of the gene 1 CPSF cleavage and polyadenylation specificity factor 2 CstF cleavage stimulation factor N l The sequences that once transcribed into RNA trigger transfer of these factors to the RNA are called polyA signals a Once CPSF and CstF are bound to the RNA other proteins are recruited as well leading initially to RNA cleavage and then polyadenylation Polyadenylation is mediated by an enzyme called polyA polymerase which adds approximately 200 adenines to the RNA39s 3 end produced by the cleavage a This enzyme uses ATP as a precursor and adds the nucleotides using the same chemistry as RNA polymerase but does not require a template i The long tail of A39s is found in the RNA but not the DNA The mature mRNA is then transported from the nucleus 3 But what terminates transcription by polymerase a The enzyme does not terminate immediately alter the RNA is cleaved and polyadentylated n Transcription Termination is Linked to RNA Destruction by a Highly Processive N RNase i Polyadenylation is linked to termination although exactly how is still not quite clear ii Torpedo Termination l The free end of the second RNA is uncapped and thus can be distinguished from genuine transcripts a This new RNA is recognized by an RNase called Ratl in yeast and Xm2 in humans that is located onto the end of the RNA by another protein Rtt103 that binds the CTD of RNA polymerase i Ratl enzyme is very processive and quickly degrades the RNA in a 539 gt3 direction until it catches up to the stilltranscribing polymerase from which the RNA is being spewed ii Termination may not require any very specific interaction between Ratl and polymerase and might in fact be triggered in a manner rather similar to the Rhodependent termination in bacteria iii Although the torpedo termination is the favored one alternative method called the allosteric model 1 Termination depends on a conformational change in the elongating polymerase that reduces the processivity of the enzyme leading to spontaneous termination soon a erward a This change would be linked to polyadenylation and could be triggered by the transfer of the 339 processing enzymes from the CTD tail of polymerase to the RNA or by the subsequent binding to the CTD tail of other factors that induce a conformational change 6 Transcription by RNA Polymerases I and III a RNA Pol I and Pol III Recognize Promoters Using Distinct Sets of Transcription Factors but Still Require TBP i These enzymes are related to Pol II and even share several subunits but they initiate transcription from distinct promoters and transcribe distinct genes b Pol I is required for the expression of only one gene that encoding the rRNA precursor 1 There are many copies of that gene in each cell and it is expressed at far higher levels than any other gene perhaps explaining why it has its own dedicated polymerase Comprises two parts 1 Core element a Located around the start site of transcription 2 UCF upstream control element a Between 100 and 150bp upstream in humans c In addition to Pol I initiation requires two other factors called SLl and UBF 1 SLl comprises TBP and three TAFs speci c for Pol Itranscription 1 This complex binds to the core element a SLl binds DNA only in the presence of UBF this factor binds to UCE bringing in SLl and stimulation transcription from the core promoter by recruiting Pol I d Pol III Promoters are Found Downstream of Transcription Start Site 1 iii Pol III promoters come in various forms and the case majority have the unusual feature of being located downstream from the transcription start site 1 Some promoters those for the tRNA genes consists of two regions a Box A b Box B i Separated by a short element others contain BoxA and BoxC the 5S rRNA gene and still others contain a TATA element like those ofPol II Just as with Pol II and Pol I transcription by Pol III requires transcription factors in addition to polymerase in this case the factors are called TFIIIB and TFIIIC for the tRNA genes and those plus TFIIIA for the 5S rRNA gene The TFIIIC complex binds to the promoter region 1 This complex recruits TFIIIB to the DNA just upstream of the start site where it in turns recruits Pol III to the start site of transcription 2 The enzyme then initiates presumable displacing TFIIIC from the DNA template as it goes a Pol III uses TBP l Transcriptional Regulation in Prokaryotes 2 Principles of Transcription Regulation a 0quot 0 Gene Expression is Controlled by Regulatory Proteins i Genes are very often controlled by extracellular signals in the case of bacteria this means molecules present in the growth medium ii These signals are communicated to genes by regulatory proteins 1 Positive regulators activators 2 Negative regulators repressors Most Activators and Repressors Act at the Level of Transcription Initiation i Transcription initiation is the most energetically efficient step to regulate l Deciding on whether or not to express a gene at the rst step ensures that no energy or resources are wasted making part or all of an mRNA that will not then be sued ii Regulation at this first step is easier to do well 1 There is a single copy of each gene and so typically only a single promoter on a single DNA molecule must be regulated to control expression of a given gene Why is not all regulation focused on the step of transcription initiation 1 Later steps have two advantages a It allows for more inputs i If a gene is regulated at more than one step more signal can modulate its express or if the same signals can do so even more effectively b Regulation at steps later than transcription initiation can reduce the response time Many Promoters are Regulated by Activators that Help RNA Polymerase Bind DNA and by Repressors that Block that Binding i At many promoters in the absence of regulatory proteins RNA polymerase binds only weakly 1 This is due to one or more of the promoter elements discussed is absent or imperfect ii When polymerase does occasionally bind it spontaneously undergoes a transition to the open complex and initiates transcription 1 This gives a low level of constitutive expression called the basal level a Binding of the RNA polymerase is the rate limiting step To control such a promoter the repressor need only bind to a site overlapping the region bound by polymerase I The site on DNA where arepressor binds is called an operator To activate transcription from this promoter an activator can just help the polymerase bind the promoter 1 Mechanism a The activator uses one surface to bind to a site on the DNA near the promoter b With another surface the activator simultaneously interacts with RNA polymerase brinjng the enzyme to the promoter iii iii d FD quot1 i This is called recruitment and is an example of cooperative binding of proteins to DNA ii The interactions between the activator and the polymerase and between the activator and DNA serve merely adhesive roles the enzyme is active and the activator simply brings it to the nearby promoter v The lac genes of E 001139 are transcribed from a promoter that is regulated by an activator and a repress Some Activators and Repressors Work by Allostery and Regulate Steps in Transcriptional Initiation A er RNA Polymerase Binding i Not all promoters are limited the same way ii Activators that stimulate the kind of promoter work polymerase binds unaided by triggering a conformational change in either RNA polymerase or DNA they interact with the stable closed complex and induce a conformational change that causes transition to the open complex 1 This is called alloster Two examples of transcriptional activators working with allostery l The activator NtrC at the glnA promoter interacts with the RNA polymerase bound in a closed complex at the promoter transition to the open complex iii 2 The activator MerR at the merT promoter achieves the same effect but does so by inducing a conformational change in the promoter DNA 3 In the malT gene the absence an activator causes it to undergo abortive initiation and only the presence of an activator will it efficiently escape into elongation Action at a Distance and DNA Looping i Some proteins interact with each other even when bound to sites well separated on the DNA 1 To accommodate this interaction the DNA between the sties loops out bring in the sites into proximity to each other ii Example 1 NtrC activates from a distance its binding sites are normally located about 150bp upstream of the promoter and the activator works even when these sites are placed further away Distant DNA sties can be brought closer together to help loop formation 1 There are cases in which a protein binds between an activatorbinding sites and the promoter and helps the activator invert with polymerase by bending the DNA Cooperative Binding and Allostery Have Many Roles in Gene Regulation i Gene activation can be mediated by simple cooperative binding the activator interacts simultaneously with DNA and with polymerase and so recruits the enzyme to the promoter ii Groups of regulators o en bind DNA cooperatively two or more activators andor repressors interact with each other and with DNA and thereby help each other bind near a gene they all regulate l Cooperative binding of activators can also serve to integrate signals iii 2 Bacteriophage Lambda is an example g Antitermination and Beyond Not All Gene Regulation Targets Transcription Initiation i The bulk of gene regulation takes place at the initiation of transcription true in eukaryotes and bacteria ii Regulation is certainly not restricted 3 Regulation of Transcription Initiation Examples from Prokaryotes a An Activator and a Repressor Together Control the lac genes i The three lac genes lacZ lacY and lacA are arranged adjacently on the E coli genome and together are known as the lac operon l The lac promoter located at the 539 end of lacZ directs transcription of all three genes as a single mRNA polycistronic mRNA a This mRNA is transcribed to give three protein produces b The lacZ gene encodes the enzyme Bgalactosidase cleaves the sugar lactose into galactose and glucose c The lacY gene encodes the lactose permease protein that inserts into the cell membrane and transports lactose into the cell d The lacA gene encodes thiogalactoside transacetylase rids the cell of toxic thiogalactosidases that also get transported in by lacY ii These genes are expressed at high levels on when lactose is available and glucose is not 1 Two regulatory proteins are involved a Activator CAP i CAP stands for catabolite activator protein but also known as CRP for cAMP receptor protein ii The gene encoding CAP is located elsewhere on the bacterial chromosome not linked to the lac genes iii Mediates the effect of glucose b Repressor Lac repressor i Encoded by the lac gene which is located near the other lac genes but transcribed form its own promoter ii Mediates the lactose signal iii Lac repressor can bind to DNA and repress transcription only in the absence of lactose 1 In the presence of lactose the repressor is inactive and the genes derepressed iv CAP can bind DNA and activate the lac genes only in the absence of glucose b CAP and Lac Repressor Have Opposing Effects on RNA Polymerase Binding to the lac Promoter i The site bound by the Lac repressor is called the lac operator 1 21bp sequence is twofold symmetric and is recognized by two subunits of Lac repressor one binding to each halfsite ii The lac operator overlaps the promoter and so the repressor bound to the operator physically prevents RNA polymerase from binding to the promoter and thus initiating RNA synthesis iii RNA polymerase binds the lac promoter poorly in the absence of Cap even when there is no active repressor present 1 This is because the sequence of the approximately 35 region of the lac promoter is not optimal for its binding and the promoter lacks an UP element iv CAP binds as a dimer to a site similar in length to that of the lac operator but different in sequence 1 This site is located some 60bp upstream of the start site of transcription 2 When Cap binds to that site the activator helps polymerase binds to the promoter by interacting with the enzyme and recruiting it to the promoter c CAP Has Separate Activating and DNABinding Surfaces i Various experiments support the view that CAP activates the lac genes by simple recruitment of RNA polymerase I Mutant versions of CAP have been isolated that bind DNA but do not activate transcription a The existence of these positive control pc mutants demonstrates that to activate transcription the activator must be more than simply bind DNA near the promoter i Activation is NOT caused by the activator changing local DNA structure ii The amino acid substitutions in the positive control mutants identify the region of CAP that touches polymerase called the activating region ii What does the activating region of CAP touch RNA polymerase when activating the lac genes 1 This site is revealed by mutant forms of polymerase that can transcribe most genes normally but cannot be activated by CAP at the lac genes a These mutants have amino acid substitutions in the carboxy terminal domain CTD of the at subunit of RNA polymerase i This domain is attached to the aminoterminal domain NTD of at by a exible linker ii The xNTD is embedded in the body of the enzyme but the 0LCTD extends out from it and binds the UPelement of the prompter iii At the lac promoter Where there is no UPelement 0LCTD Binds to Cap and adjacent DNA instead d CAP and Lac Repressor Bind DNA Using a Common Structural Motif i In a typical case the protein binds as a homodimer to a site that is an inverted repeat 1 One monomer binds each halfsite with the axis of symmetry of the dimer lying over that of the binding site a Recognition of speci c DNA sequences is achieved using a conserved region of secondary structure called a helix tum helix i This domain is composed of two at helices one of whichthe recognition helix ts into the major groove of the DNA ii An 0L helix is just the right size to fit into the major groove allowing amino acid residues on its outer face to interact with chemical groups on the edges of base pairs ii The contacts made between the amino acid side chains protruding from the recognition helix and the edges of the bases can be mediated by direct Hbonds indirect Hbonds or van der Waals forces iii The second helix of the helixturnhelix domain sits across the major groove and makes contact with the DNA backbone ensuring proper presentation of the recognition helix and at the same time adding binding energy to the overall proteinDNA interaction iv Di erences l Lac repressor binds as at tetramer not a dimer a Each operator is contacted by only two of these subunits b The different oligomeric form does not alter the mechanism of DNA recognition The other two monomers within the tetramer can bind one of two other lac operators located 400bp downstream and 90bp upstream of the primary operator In some cases other regions of the protein outside the helixturnhelix domain also interact with the DNA a The 7 repressor makes additional contacts using aminoterminal arms i These reach around the DNA and interact with the minor groove on the back face of the helix In many cases binding of the protein does not alter the structure of the DNA a In some cases various distortions are seen in the proteinDNA complex i CAP induces a dramatic bend in the DNA partially wrapping it around the protein 1 This is caused by other regions of the protein outside the helixturnhelix domain interacting with sequences outside the operator v Not all repressors bind using a helixturnhelix e The Activities of Lac Repressor and CAP Are Controlled Allosterically by Their Signals i When lactose enters the cell it is converted to allolactose 1 It is allolactose that controls the Lac repressor ii The conversion of lactose to allolactose is catalyzed by Bgalactosidase itself encoded by one of the lac genes how is this possible 1 Even when lac genes are repressed an occasional transcript is made a This happens when RNA polymerase will manage to bind the promoter in place of the Lac repressor b This leakiness ensure that there is a low level of Bgalactosidase in the cell even in the absence of lactose and so there is enzyme poised to catalyze the conversion of lactose to allolactose 0 N U iii quot1 iii Allolactose binds to the Lac repressor and triggers a change in the shape conformation of that protein 1 In the absence of allolactose the repressor is present in a form that binds its site on DNA 2 Once allolactose has altered the shape of the repressor the protein can no longer bind DNA and so the lac genes are no longer repressed CAP activity is regulated in a similar manner 1 Glucose lowers the intracellular concentration of a small molecule cAMP a This molecule is the allosteric effector for CAP only when CAP is complexed with cAMP does the protein adopt a conformation that binds DNA b Only when glucose levels are low does CAP bind DNA and activate the lac genes The lac operon of E coli is one of the two systems used by French biologists Francois Jacob and Jacques Monod in formulating the early ideas about gene regulation Combinatorial Control CAP Controls Other Genes as Well i The lac genes provide an example of signal integration their expression is controlled by two signals each of which is communicated to the genes via the single regulatorthe Lac repressor and Cap respectively The gal genes are another set of E coli genes These genes encode enzymes involved in galactose metabolism 2 Only expressed when their substrate sugar in case galactose is present and the preferred energy source glucose is absent 3 The two signals are communicated to the genes via two regulators a Activator and repressor i The repressor encoded by the galR gene mediates the effects of the inducer galactose but the activator of the gal genes is again CAP l A regulator CAP works together with different repressors at different genes known as a control combinatorial control 1 In fact CAP acts at more than 100 genes in E coli Combinatorial control is a characteristic feature of gene regulation 1 When the same signal controls multiple genes by the same regulatory protein g Alternative 6 Factors Direct RNA Polymerase to Alternative Sets of Promoters 1 ii Recall that it is the 6 subunit of RNA polymerase that recognizes the promoter sequence 1 The lac promoter is recognized by the 570 subunit of RNA polymerase One alternative is the heat shock 6 factor 632 1 When E Call is subject to heat shock the amount of this new 6 factor increases in the cell it displaces the G70 from a proportion of RNA polymerase and it directs those enzymes to transcribe genes whose products protect the cell from the effects of heat shock 2 The level of G32 is increased by two mechanisms a Its translation is stimulated its mRNA is translated with greater efficiency a er heat shock than it was before b The protein is transiently stabilized iii Alternative factor 6 is factor 6 is considered in the next section is associated with a small fraction of the polymerase molecules in the cell and directs that enzyme to genes involved in nitrogen metabolism iv A series of alternative o s directs a particular program of gene expression 1 Two examples are found in the bacterium Bacillus subtilis a Bacteriophage SPOl infects B subtilis where it grows lytically to produce progeny phage i This process requires that the phage expresses its gene sin a carefully controlled order that control is imposed on polymerase by a series of alternative 6 factors ii Upon infection the bacterial RNA polymerase recognizes so called quotearlyquot phage promoters which direct transcription of genes that encode proteins needed early in infection 1 One of these genes gene 28 encodes an alternative 6 1 This displaces the bacterial 6 factor and directs the polymerase to a second set of promoters in the phage genome those associated with the socalled quotmiddle genesquot 1 One of these genes encodes the ofactor for the page quotlatequot genes h NtrC and MerR Transcriptional Activators that Work by Allostery Rather Than by Recruitment i Although a majority of activators work by recruitment there are exceptions ii Two examples of activators that work not be recruitment but by allosteric mechanisms are NtrC and MerR l NtrC controls expression of genes involved in nitrogen metabolism such as the glnA gene a At the glnA gene RNA polymerase is prebound to the promoter in a stale closed complex b The activator NtrC induces a conformational change in the enzyme triggering transition to the open complex 2 MerR controls a gene called merT which encodes an enzyme that makes cells resistant to the toxic effects of mercury a MerR also acts on an inactive RNA polymerasepromoter complex b MerR induces a conformational change that triggers open complex formation i NtrC has ATPase Activity and Works from DNA Sites Far Form the Gene i As with CAP NtrC has separate activating and DNAbinding domains and binds DNA only in the presence of a speci c signal 1 In the case of NtrC this signal is low nitrogen levels a Under these conditions NtrC is phosphorylated by a kinase NtrB and as a result undergoes a conformational change that reveals the activator39s DNAbinding domain v iii 2 Once active NtrC binds four sites approximately 150bp upstream of the promoter a NtrC binds to each of its sites as a dimer and through proteinprotein interactions between the dimers binds to the four sites in a highly cooperative manner The form of RNA polymerase that transcribes the anA gene contains the 654 subunit 1 This enzyme binds to the glnA promoter in a stable closed complex in the absence of NtrC 2 Once active NtrC bound to its sites upstream interacts directly with 554 a Requires that the DNA between the activatorbinding sites and the promoter form a loop to accommodate the interaction NtrC itself has an enzymatic activity it is an ATPase 1 Hydrolyzes the energy needed to induce a conformational change in polymerase 2 This change triggers polymerase to initiate transcription At some genes controlled by NtrC there is a binding site for another protein called IHF located between the NtrCbinding sites and the promoter 1 Upon binding IHF bends DNA when the IHFbinding site and hence the DNA bend are in the correct order the total conformation increases the activation by NtrC MerR Activates Transcription by Twisting Promoter DNA 1 When bound to a single DNAbinding site in the presence of Mercury MerR activates the merT gene 1 MerR binds the sequence between the 10 and 35 regions of the merT promoter gene transcribed by 670 containing polymerase 2 Binds to the opposite face of the DNA helix from that bound by RNA polymerase so polymerase binds to the promoter at the same times as MerR MerT promoter is unusual and distance between the 10 and 35 elements is l9bp and not 15l7bp 1 As a result these two sequence elements recognized by 6 are neither optimally separated or aligned somewhat rotated around the face of the helix with respect to each other 2 The binding of MerR in the presence of Hg locks the promoter in this conproitious conformation polymerase can bind but not in a manner that allows it to initiate transcription a THIS IS NOT BASAL TRANSCRIPTION 3 When MerR brings mercury the protein undergoes a conformational change that causes the DNA in the center of the promoter to twist a This structural distortion restores the disposition of the 10 and 35 regions to something close to that found of a strong 570 promoter b Note the activator does not interact with RNA Polymerase to activate transcription but instead alters the conformation of the DNA in the vicinity of the prebound enzyme k Some Repressors hold RNA Polymerase at the Promoter Site Rather Than Excluding It i The Lac repressor works in the simples steps possible by binding to a site overlapping the promoter it blocks RNA polymerase binding 1 In MerR the protein holds the promoter in a conformation incompatible with transcription initiation 2 Some repress work from binding sites that do not overlap the promoter AraC and Control of the araBAD Operon by Antiactivation i The promoter of the araBAD operon from E coli is activated in the presence of arabinose and the absence of glucose and directs expression of genes encoding enzymes required for arabinose metabolism ii AraC and CAP work together as activators 1 When arabinose is present AraC binds that sugar and adopts a configuration hat allows it to bind DNA as a dimer to the adjacent half sites am and aralz a Just upstream from these is a CAP site in the absence of glucose CAP binds here to help activation iii In the absence of arabinose the araBAD genes are not expressed 1 When not bound to arabinose AraC adopts a different conformation and binds DNA in a different way one monomer still binds the am site but the other monomer binds a distant halfsite called ara02 a As these two halfsites are l94bp apart when AraC binds in this fashion the DNA between the two sites forms a loop i When bound there is no monomer of AraC at math and as this is the position of which activation of araBAD promoter is mediated there is no activation in this configuration iv The magnitude of induction of the araBAD promoter by arabinose is very large and the promoter is often used in expression vectors Expression vectors are DNA constructs in which efficient synthesis of any protein can be ensured by fusing its gene to a strong promoter Fusing a gene to the araBAD promoter allows expression of the gene to be controlled by arabinose alone the gene can be kept off when its expression is undesirable and then both quotderepressedquot and quotinducedquot when its product is wanted by the addition of arabinose a This allows expression of genes with products that are toxic to the cells 4 The Case of Bacteriophage 7t Layers of Regulation a Bacteriophage 7 is a virus that infects E 001139 b Upon infection the phage can propagate in one of two ways i Lytically l Requires replication of the phage DNA and synthesis of new coat proteins 2 These components combine to form new phage particles that are released by lysis ofthe host cell ii Lysogenically l Involves the integration of the phage DNA into the bacterial chromosome where it is passively replicated at each cell division N C 2391 FD A lysogen is stable under normal circumstances but the phage dormant within it prophage can efficiently switch to lytic growth if the cell is exposed to agents that damage DNA i The switch from lysogenic growth to lytic growth is called lysogenic induction Alternative Patterns of Gene Expression Control Lytic and Lysogenic Growth i Bacteriophage 7 has a 50kb genome and approximately 50 genes 1 Most ofthese genes encode coat proteins proteins involved in DNA replication recombination and lysis 2 The products of the genes are important in making new phage particles during the lytic cycle ii Within the DNA there are two genes c1 and cm and three promoters PR PL and PRM 1 All of the other phage genes are outside the regions and are transcribed directly from PR rightward promoter and PL leftward promoter PRM promoter for repressor maintenance transcribes only the cI gene 1 Weak promoter and only directs efficient transcription when an activator is bound upstream similar to the lac promoter PR and PL are strong promoters they bind RNA polymerase efficiently and direct transcription without help from an activator v Lytic growth occurs when PR and PL are kept on and PRM is kept off for lysogenic growth it is vice versa Regulatory Proteins and Their Binding Sites i The cI gene encodes 7t repressor a protein of two domains joined by a exible linker region 1 The aminoterminal domain contains the DNAbinding region helixtum helix domain ii 7 repressor binds DNA as a dimer the main dimerization contacts are made between the carboxyterminal domains 1 A single dimer recognizes a l7bp sequence of DNA each monomer recognizing one halfsite again the same as in lac system The 9 repressor can both activate and repress transcription 1 When functioning as a repressor it works in the same way as the Lac repressor it binds to sites that overlap the promoter and excludes RNA polymerase 2 As an activator 9t repressor works like CAP by recruitment a A repressor39s activating region is in the aminoterminal domain of the protein b Its target on polymerase is a region of the 6 subunit adjacent to the part of 6 that recognizes the 35 region of the promoter Cro which stands for control of repressor and other things only represses transcription 1 It is a singledomain protein and again binds as a dimer to l7bp DNA sequences using a helixturnhelix motif v The 7 repressor and Cro can each bind to any one of six operators 1 These sites are recognized with different affinities by each of the proteins 2 Three of each are on the left and right iii iii iv f 9 Pquot a The three binding sites in the right operator are called 0R1 0R2 and 016 these sites are similar in sequence but not identical i Each can bind either a dimer of repressor or a dimer of Cro ii Repressor binds ten fold better to 0R1 than 0R2 iii The 0R3 binds the same a inity as that of OR iv Cro binds to 0R3 with the highest af nity and will bind to the other two if they are at higher concentrations 7 Repressor Binds to Operator Sites Cooperatively i This is critical to its lnction and occurs as follows 1 In addition to providing the dimerization contacts the carboxyterminal domain of 7 repressor mediates interactions between the dimers 2 Example a The repressor 0R1 helps the repressor bind to the loweraffmity site 0R2 by cooperative binding b Repressor thus binds both sites simultaneously and does so at a concentration that would be sufficient to bind only 0R1 were the two sites tested separately OM is not bounded repressor bound cooperatively at 0R1 and OR cannot make contact with a third dimer at that adjacent site simultaneously Repressor and Cro Binding in Different Patterns to Control lytic and Lysogenic Growth i How do repressor and Cro control the different patterns of gene expression associated with the different ways 7 can replicate 1 For lytic growth a single Cro dimer is bound to 0R3 this site overlaps PM and so Cro represses that promoter 2 As neither repressor is bound to 0R1 and 0R2 PR binds RNA polymerase and directs transcription of lytic genes PL does likewise ii During lysogeny PRM is on while PR and PL are o er l Repressor bound cooperatively at 0R1 and OM blocks RNA polymerase binding at PR repressing transcription from that promoter a Repressor bound at 0R2 ACTIVATES transcription from PRM Lysogenic Induction Requires Proteolytic Cleavage of 7 Repressor i The E coli sense and responds to DNA damage 1 Does this by activating the lnction of RecA a This enzyme is involved in recombination but it has another lnction it stimulates the proteolytic autocleavage of certain proteins 2 The primary substrate for this activity is a bacterial repressor called LeXA that represses genes encoding ENArepair enzymes a Activated RecA stimulates autocleavage of LeXA releasing repression of those genes called the SOS Response ii If the cell is a lysogen it is in the best interests of the prophage to escape under these threatening circumstances 1 To this end 7 repressor has evolved to resemble LeXA ensuring that 7 repressor too undergoes autocleavage in response to activated RecA L a The cleavage reaction removes the carboxyterminal domain of repressor and so dimerization and cooperatively are immediately lost b These functions are critical for repressor binding to 0R1 and 0R2 so the loss of cooperativity ensures that the repressor dissociates from those sites as well as from 0L1 and 0L2 i Loss of repression triggers transcription from PR and PL leading to lytic growth 1 Transcription from PR quickly produces Cro which binds 0m and blocks any further synthesis of repressor for PRM iii For induction to work effectively and efficiently the level of repressor in a lyso gen must be tightly regulated 1 It levels were to drop low under normal conditions the lysogen might spontaneously induce 2 If levels rose too high appropriate induction would be inef cient a The reason for the inefficient induction is because more repressor would have to be inactivated by RecA for the concentration to drop enough to vacate 0R1 and OR b Recall how repressors ensure that its level never drops too low it activates its own expression a process called positive autoregulation iv How does it ensure levels never get too high 1 The negative autoregulation works by the following mechanism a PRM is activated by the repressor Cro at OR to make more repressor b If the concentration gets too high repressor will bind to 0R3 as well and repress PRM i This prevents synthesis of new repressor until its concentration falls to a level at which it vacates 0R3 v quotInductionquot is used to describe both the switch from lysogenic growth to lytic growth in 7 and the switching on the lac genes in response to lactose 1 Just as lactose induces a conformational change in Lac repressor to relieve repression ofthe lac genes so too the inducing signals of work by causing a structural change proteolytic cleavage in 7 repressor Negative Autoregulation of Repressor Requires LongDistance Interactions and a Large DNA Loop i There is another level of cooperative binding seen in the prophage of a lyso gen one critical to negative autoregulation l Repressor dimers at 0R1 and OR interact with repressor dimers bound cooperatively at 0L1 and 0L2 a These interactions produce an octamer of repressor each dimer within the octamer is bound to a separate operator ii To accommodate the longdistance interaction between repressors and OR and OL the DNA between the operator regions about 35kb including the cI gene must form a loop I 1 When the loop is formed the OR subunits match up to the corresponding 0L subunits 2 This allows another two dimers of repressor to bind cooperatively to these two sites a This cooperativity means 0R3 binds repressor at a lower concentration than it otherwise wouldat a concentration only just a little higher than that required to bind 0R1 and OM i Repressor concentration is tightly controlled small decreases are compensated for by increased expression of its gene and increases by switching the gene off iii The structure of the carboxyterminal domain of 7 repressor reveals that basis of dimer formation but it also shows how two dimers interact to form the tetrameric form 1 The structure reveals the basis for the octamer form and shows that this is the highestorder oligomer repressor can form Anther Activator XCH Controls the Decision Between Lytic and Lysogenic Growth Upon Infection ofa New Host i Critical to the choice of grth are the products of two other 7 genes CH and cIII ii Like the 9 repressor the C11 protein is a transcriptional activator 1 It binds to a site upstream of a promoter called PRE repressor establishment and stimulates the transcription of the cI gene repressor gene from that promoter a The repressor gene can be transcribed from two different promoters PR1 and Pm iii The PRE promoter is weak 1 Has a poor 35 sequence 2 The CII protein binds to a site that overlaps the 35 region but is located on the opposite face of the DNA helix by directly interacting with polymerase CII helps polymerase bind to the promoter iv Only when sufficient repressor has been made from PRE can repressor bind to 0R1 and OR and direct its own synthesis from PRM l The repressor synthesis is established by transcription from one promoter stimulated by one activator and then maintained by transcription from another promoter under its own control positive autoregulation v CII Mechanism 1 Upon infection transcription is immediately initiated from the two constitutive promoters PR and PL a PR directs synthesis of both Cre and C11 i Cre expression favors lytic development once Cre reaches a certain level it will bind OM and block PRM b CII expression factors lysogenic development by directing transcription of the repressor gene i For successful lysogeny repressor must then bind to 0R1 and OM and activate PRM vi The efficiency with which CII direct transcription of the CI gene and hence the rate at which repressor is made is the critical stop in deciding how 7 will develop vii What determines how efficiently CII works in any given infection k The Number of Phage Particles Infecting a Given Cell Affects Whether the Infection Proceeds Lytically or Lysogenically i Multiplicity of infection moi is a measure of how many phage particles infect a given bacterial cell within a population 1 If the average number is one or fewer phage particles per cell the infection is more likely to result in lysis If the number of phage particles is two or more it is more likely to produce lysogenic a The more phage genomes that enter the cell and start transcribing from PR and PL the more C11 and C111 gets made and the greater the chance that at least one of those phage genomes will establish repressor synthesis and integrate into the bacterial chromosome 1 Growth Conditions of E 001139 Control the Stability of C11 Protein and Thus the yticLysogenic Choice When the phage infects a population of bacterial cells that are healthy and growing vigorously it tends to propagate lytically releasing progeny into an environmental rich in fresh host cells ii When conditions are poor for bacterial growth the phage is more likely to form lyso gens and wait there will likely be few host cells in the vicinity for any progeny phage to infect 1 C11 is a unstable protein in E coli it is degraded by a speci c protease known as Tsz H B encoded by the h gene a The speed with which CII can direct synthesis of repressor is determined by how quickly it is being degraded by FtsH i Cells lacking the h gene almost always form lyso gens upon infection by 7 in the absence of the protease C11 is stable and direct synthesis of ample repressor 2 If growth is good Tsz is very active C11 is destroyed efficiently repressor is not made and the phage tend to grow lytically Under poor growth conditions Tsz is not highly active a slow in degradation of C11 and a repressor is accumulated and a tendency toward lysogenic development is favored iii Levels of C11 are also modulated by the phage protein C111 1 C111 stabilizes CII probable because it acts as an alternative competing substrate for FtsH iv A second CIIdependent promoter P1 has a sequence similar to that of PRE and is located in front of the phage gene int this gene encodes the integrase enzyme that catalyzes sitespeci c recombination of 7 DNA into the bacterial chromosome to form the prophage v A third CHdependent promoter PAQ located in the middle of gene Q acts to retard lytic development and thus to promoter lysogenic development N 1 LA 1 This is because the FAQ RNA acts as an antisense message binding to the Q message and promoting its degradation In Transcriptional Antitermination in 7 Development i Two examples of transcriptional regulation a er initiation are found in 7 development 1 Positive transcriptional regulation called antitermination a The transcripts controlled by the 7 N and Q proteins are initiated perfectly well in the absence of those regulators i The transcripts terminate a few hundred to a few thousand nucleotides downstream from the promoter unless RNA polymerase has been modified by the regulator 7 N and Q proteins are therefore called antiterminators b N protein regulates early gene expression by acting at tree terminators i One to the le of the N gene ii One to the right of cm gene iii One between genes P and Q 1 Q protein has one target a terminator 200 nucleotides downstream from the late gene promoter PRv located between the Q and S genes 1 The late gene operon of 7 is transcribed from PR and is large for a prokaryotic transcription unit about 26kb a distance that takes about 10 minutes for RNA polymerase to transverse 2 N and Q work on genes that carry sequences specify for each regulator a N protein prevents termination in the early operon of 7 but not I other bacterial or phage operons i The specific recognition sequences for antiterminators are not found in the terminators where they act but instead occur somewhere between the promoter and the terminator 1 For N those sites are called nut forN utilization sites which are about 60 to 200 nucleotides downstream from PL and PR N does not bind to these sequences within DNA rather N binds to RNA transcribed from DNA containing a nut sequence Once RNA polymerase has passed a nut site N binds to the RNA and from there is loaded onto the polymerase itself 1 In this state the polymerase is resistant to the terminators found just beyond the N and cm genes b The 7 N works together with the products of the bacterial genes nusA nusB nusE and nusG i NusA protein is an important cellular transcription factor ii NusE is the small ribosomal subunit protein S10 but its role in N protein lnction is unknown N L iii No cellular lnction of NusB protein is known 1 These proteins form a complex with N at the nut site but N can work int her absence if present at high concentrations suggesting that it is N itself that promotes antitermination 3 The 7 Q protein recognizes DNA sequences QBE between the 10 and 35 regions of the late gene promoter a In the absence of Q polymerase binds PR and initiates transcription only to pause a er 16 or 17 nucleotides it then continues but terminates when it reaches the terminator th about 200bp upstream b If Q is present it binds to QBE once the polymerase has le the promoter and transfers from there to the nearby paused polymerase c With Q present the polymerase it hen able to transcribe through th ii It has become clear that the 6 factor of polymerase is involved in Q function 1 The reason polymerase pauses just a er initiation at PR is because it encounters a sequence resembling the 10 element of a promoter 2 Region 2 ofo typically recognizes that sequence binding to base pairs in the template strand n Retroregulation An Interplay of Controls on RNA Synthesis and Stability Determines int Gene Expression i The CII protein activates the promoter PI that directs expression of the int gene as well as the promoter PRE responsible for repressor synthesis 1 The Int protein is the enzyme that integrates the phage genome into that of the host cell during formation of a lysogen ii The int gene is transcribed fromPL as well as from P1 so one would have thought that integrase should be made even in the absence of C11 protein false 1 The reason is that int mRNA initiated at PL is degraded by cellular nucleases whereas mRNA initiated at P1 is stable and can be translated into integrase protein this occurs because the two messages have different structures at their 3 ends iii RNA initiated at P1 stops at a terminator about 300 nucleotides alter the end of the int gene and has atypical stemandloop structure followed by six uridine nucleotides 1 When RNA synthesis is initiated at PL RNA polymerase is modified by the N protein and thus goes through an beyond the terminator a This longer mRNA can form a stem that is a substrate for nucleases because the site responsible for this negative regulation is downstream from the gene it a ects and because degradation proceeds backwards through the gene the process is called retroregulation l The Genetic Code 2 The Code is Degenerate First Third Position 1 Position 5 U A G 3 End 1 endi 39 l UUU F UCU S UAU Y UGU C U U UUC39 F UCC S UAC Y UGC C C UUA L UCA S l A UUG L UCG S UGG W G CUU L CCU P CAU H CGU R U C CUC L CCC P CAC H CGC R C CUA L CCA P CAA Q CGA R A CUG L CCG P CAG Q CGG R G AUU I ACU T AAU N AGU S U A AUC 1 ACC T AAC N AGC S C AUA A ACA T AAA K AGA R A ACG T AAG K AGG R G GUU V GCU A GAU D GGU G U G GUC V GCC A GAC D GGC G C GUA V GCA A GAA E GGA G A GUG V GCG A GAG E GGG G G a One of the most striking features of the code is that 61 of the 64 possible triplets specify an amino acid with the other three being chainterminating signals i This means that many amino acids are speci ed by more than one codon known as degeneracy b Codons specifying the same amino acid are synonyms i When the first two nucleotides are identical the third nucleotide can either be Cytosine or Uracil and the codon will still code for the same amino acid c Perceiving Order in the Makeup of the Code i Inspection of the distribution of codons in the genetic code suggests that the code evolved in such a way as to minimize the deleterious effects f mutations l Mutations in the first position of the codon will o en give similar if not the same amino acid Codons with pyrimidines in the second position specify mostly N hydrophobic amino acids whereas those with purines in the second position correspond mostly to polar amino acids a Because transitions are the most common type of point mutations a change in the second position ofa codon will usually replace one amino acid with a very similar one E If a codon suffers a transition mutation in the third position rarely will a dilTerent amino acid be specified even if a transversion mutation in this position has occurred ii Whenever the first two positions of a codon are both occupied by G or C each of the four nucleotides in the third position speci es the same amino acid P A R or GD 1 Whenever the first two positions of the codon are both occupied by A or U the identity of the third nucleotide does make a difference a Since GC base pairs are stronger than AU base pairs mismatches in pairing the third codon base are often tolerated if the first two positions make strong GC base pairs thus having all four nucleotides inthe third position specify the same amino acid may have evolved as a safety mechanism to minimize errors in the reading of codons d Wobble inthe Anticodon i It was first proposed that a specific tRNA anticodon would exist for every codon 1 Evidence began to appear that highly puri ed tRNA species of known sequence could recognize several dilTerent codons ii In 1966 Francis Crick devised the wobble concept to explain the observations 1 0quot 0 2391 It states that the base at the 539 end of the anticodon is not as spatially confined as the other two allowing it to form hydrogen bonds with any of several bases located at the 339 end ofa codon Base in Base in Codon Anticodon G U or C l C G l i A i U l U A or G l I A U or C i The pairings permitted by the wobble rules are those that give ribose ribose distances close to that ofa standard AU or GC base pair i Purinepurine exception of IA or pyrimidinepyrimidine pairs would give riboseribose distances that are too long purinepurine or too short pyrimidinepyrimidine The wobble rules do not permit any single tRNA molecule to recognize four dilTerent codons The concept correctly predicted that at least three tRNAs exist for the six Serine codons UCU UCC UCA UCG AGU and AGC i Two other amino acids Leucine and Arginine that are encoded by six codons also have different tRNAs for the sets of codons that differ in the first or second position c In the 3D structure of tRNA the threeanticodon basesas well as the following 339 bases in the anticodon loopall point in roughly the same direction with FD 9 their exact conformations largely determined by stacking interactions between the at surfaces ofthe bases Three Codons Direct Chain Termination i Three codons do not correspond to any amino acid they signify chain termination l UAA UAG and UGA are read not be special tRNAs but by specific proteins known as release factors RFl and RF2 in bacteria eRFl in eukaryotes a RFs enter the A site of the ribosome and trigger hydrolysis of the peptidyltRNA occupying the P site resulting in the release of the newly synthesized protein How The Code Was Cracked Rev Chapter 2 Nucleic Acids Convey Genetic Information i By 1960 the general outline of how mRNA participates in protein synthesis had been established 1 There was little optimism that we would soon have a detailed understanding of the genetic code It was believed that identification of the codons for a given amino acid would require exact knowledge of both the nucleotide sequences of a gene and the corresponding amino acid in order in its protein product ii In 1961 the use of arti cial mRNA and the availability of cellfree systems for carrying out protein synthesis began to make it possible to crack the code Stimulation of Amino Acid Incorporation by Synthetic mRNA i Biochemists found that extracts prepared from cells of E 001139 that were actively engaged in protein synthesis were capable of incorporating radioactively labeled amino acids into proteins 1 Protein synthesis in these extracts proceeded rapidly for several minutes and then gradually came to a stop ii The dependence of cell free extracts on externally added mRNA provided an opportunity to elucidate the nature of the genetic code using synthetic polynucleotides l The synthetic templates were created using the enzyme polynucleotide phosphorylase that catalyzes the reaction XMPn XDP lt gt XMP n1 P where X represents the base and XMPn represents RNA of length n polynucleotides c Polynucleotide phosphorylase is normally responsible for breaking down RNA and under physiological conditions favors the degradation of RNA into nucleoside diphosphates i By use of high nucleoside diphosphate concentrations this enzyme can be made to catalyze the formation of irrtemucleotide 339gt5 phosphodiester bonds and thus make RNA molecules NO template DNA or RNA is required for RNA synthesis with this enzyme the base composition of the synthetic product depends entirely on the ratio of the various ribonucleoside diphosphates added to the reaction mixture N a When only adenosine diphosphate is used the resulting RNA contains only adenylic acid and is called polyadenylic acid or poly A PolyU Codes for Polyphenylalanine i High concentrations of magnesium circumvents the need for initiation factors and the special initiator fMettRNA allowing chain initiation to take place without the proper signals in the mRNA ii PolyU was the rst synthetic polyribonucleotide discovered to have mRNA activity 1 It selects r39 39 39 39 tRNA 39 39 39 39 39J thereby forming a polypeptide chain containing only phenylalanine 2 Codon for phenylalanine is composed of a group of three uridylic acid residues UUU Mixed Copolymers Allowed Additional Codon Assignments i PolyAC molecules can contain either different codons l Pquot 1 CCA CAC ACC CAA ACA AAC AAA ii When AC copolymers attach to ribosomes they cause the incorporation of N Q H and T in addition to Proline previously assigned to CCC codons and the Lysine previously assigned to AAA codons iii Since AC copolymer containing much more A than C promotes the incorporation of many more N than H residues we conclude that N is coded by two As and one C and that H is coded by two Cs and one A j Transfer RNA Binding to De ned Trinucleotide Codons i A direct way of ordering the nucleotides within some of the codons was developed in 1964 1 This method utilized the fact that even in the absence of all the factors required for protein synthesis speci c aminoacyltRNA molecules can bind to ribosomemRNA complexes a IE When polyU is mixed with ribosomes only phenylalanyl tRNA will attach i This speci c binding does not demand the presence of long mRNA molecules the binding of a trinucleotide to a ribosome is suf cient k Codon Assignments from Repeating Copolymers i Organic chemical and enzymatic techniques were being used to prepare synthetic polyribonucleotides with known repeating sequences 1 Ribosomes start protein synthesis at random points along these regular copolymers yet they incorporate speci c amino acids into polypeptides wsgwww Copolymer Codons Amino Acids Incorporated or Codon Recognized Polypeptide Made Assignment CUn CUClUCUlCUC Leucine 5 CUC3 Serine 5 UCU3 UGn UGUl GUGlUGU Cysteine 5 UGU3 Valine 5 GUG3 ACn ACAlCAClACA Threonine 5 ACA3 Histidine 5 CAC3 AGn AGAl GAGl AGA Arginine 5 AGA3 Glutamjne 5 GAG339 AUCn AUCl AUCl AUC Polyisoleucine 5 AUG339 UCAlUCAlUCA Polyserine 5 UCA3 CAUlCAUlCAU Polyhistidine 5 CAU3 3 Three Rules Govern the Genetic Code a The genetic code is subject to three rules i Holds that codons are read in a 539 3 direction 1 In principal the coding sequence for the dipeptide NHzThrArgCOOH could be written as 539 ACGCGA3 or as 339GCAAGC5 the rst is correct ii Codons are nonoverlapping and the message contains no gaps Means that successive codons are represented by adjacent trinucleotides in register The coding sequence for the tripeptide NHzThrArgSerCOOH is represented by three continuous and nonoverlapping triplets in the sequence 539 ACTCGAUCU3 iii Message is translated in a xed reading frame which is set by the initiation codon b Three Kinds of Point Mutations Alter the Genetic Code i Recall from Chapter 9 Mutability and Repair of DNA ii An alteration that changes a codon speci c for one amino acid to a codon specific for another amino acid is called a missense mutation l A gene bearing a missense mutation produces a protein product in which a single amino acid has been substituted for another 2 Sicklecell anemia Q6Bglobulin subunit is changed to a V FleCQlQG STOP N UUU UCU UGU UGU CAA CAG GGU UAA llilllilllr i F slc C QlV G STOP UUU UCU UGU UGU CAA GUU GGU UAA c An alteration that changes a codon specific for one amino acid to a stop codon is called a nonsensestop mutation i Arises in the middle of a genetic message an incomplete polypeptide is released from the ribosome owing to premature chain termination ii Causes truncations F s H C C Q N Q G STOP UUU UCU UGU UGU CAA CAG GGU UAA l i i i i l i F S C C STOP UUU UCU UGU UGU UAG dThird point mutation is a frameshift mutation which are insertions or deletions of one or a small number of base pairs that alter the reading frame AAHAAAHAAA GCU GCU GCU GCU GCU GCU GCU GCU i y i 39 i l y 39 i l i i A A s C C C C C GCU GCU AGC UGC UGC UGC UGC UGC c Genetic Proof That the Code is Read in Units of Three i Genetic crosses were carried out to create a mutant phage harboring three inferred singlebasepair insertion mutations at nearby positions in a single base The tree insertions would have scrambled a short stretch of codons but the protein encoded by the gene in question was able to tolerate the local alteration to its amino acid sequence Because the gene could tolerate three but not one or two the genetic code must be red in three units 4 Suppressor Mutations can Reside in the Same or a Different Gene a O en the effects of harm ll mutations can be reversed by a second genetic change i Some of these mutations are easy to understand being simple reverse back mutations which change an altered nucleotide sequence back to its original arrangement ii Mutations occuring at different locations on the chromosome that suppress the change due to a mutation at site A by producing an additional genetic change at site B are more difficult to understand 1 Suppressor mutations fall into two main categories a Those occuring within the same gene as the original mutation but at a diiTerent site in this gene intragenic suppression b Those occuring in another gene intergem39c suppression i Genes that cause suppression of mutations in other genes are called suppressor genes N ii Both of the types of suppression considered here work by causing the production of good copies of the protein made inactive by the original harmful mutation iii Intergenic suppression is generally for missense mutation 1 Its effect can sometimes be reversed through an additional missense mutation in the same gene iv Intragenic suppression is usually for frameshi mutations b Intergenic Suppression involves Mutant tRNAs i Suppressor genes do not act by changing the nucleotide sequence of a mutant gene instead they change the way the mRNA template is read ii In E coll suppressor genes are known for each of the three stop codons 1 They act by reading a stop codon as it if were a signal for a speci c amino acid there are three wellcharacterized genes that suppress the UAG codon a One suppressor gene inserts Serine one inserts a Glutamine and a third inserts a Tyrosine at the nonsense point i In each of the three UAG suppressor mutants the anticodon of a tRNA species speci c for one of these amino acids has been altered l The tyrosine suppressor arises by a mutation with a tRNATyr gene that changes the anticodon from GUA 339 AUGS39 to CUA 3 AUC5 thereby enabling it to recognize UAG codons 2 The serine and glutamine suppressor tRNAs also arise by single base changes in their anticodons iii The discovery that cells with nonsense suppressors contain mutationally altered tRNAs raised the question of how their codons corresponding to these tRNAs could continue to be read normally 1 Tyrosine UAG suppressor the answer comes from the discovery that three separate genes code for tRNATyr a One codes for the major tRNATyr species whereas the other two are duplicate genes encoding for a species present in smaller amounts i One or the other of the two duplication genes is always the site of the suppressor mutation 2 No dilemma exists for UGA suppression which is mediated by a mutant for of tRNAT p the suppressing tRNAT p retains its capacity to read UGG tryptophan codons while also recognizing UGA stop codons a This is possible because the anticodon was changed from CCA in the wild type to UCA in the mutant tRNAT p and wobble rules allow recognition ofA or G in the 339 position ofthe codon by U in the 539 position of the anticodon c Nonsense Suppressors Also Read Normal Termination Signals i The ct of nonsense suppression can be viewed as a competition between the suppressor tRNA and the release factor 1 When a stop codon comes into the ribosomal A site either readthrough or polypeptide chain termination will occur depending on which arrives rst d Provi i P Suppression of UAG codons is efficient in the presence of the suppressor tRNA more than half of the chainterminating signals are read as specific amino acid codons i Bacteria E 001139 can handle this misreading of the UAG stop codon because UAG is used infrequently as a chain terminating codon at the end of open reading frames ng the Validity of the Genetic Code A classic and instructive experiment in 1966 helped to validate the genetic code well before DNA sequencing was possible 1 The experiment was based on the construction of genetic recombination of a mutant gene of phage T4 that harbored a usually suppressing pair of insertion and deletion mutations 2 The gene in question encoded a cellwalldegrading enzyme called lysozyme chosen because it is small easy to purify and its complete amino acid sequence was known When the amino acid sequences of the mutant NHzThr Lys Val His His Leu met Ala Ala LysCOOH and wild type NHzThr Lys Ser Pro Ser Leu Asn Ala Ala LysCOOH were compared they were found to differ by a stretch of five amino acids in hold 1 This suggested that the insertion and deletion mutations had scrambled a short stretch of codons in the message of the mutant If the genetic code is elucidated in biochemical experiments is valid then it should be possible to identify a set of codons for the wildtype sequence Ser Pro Ser Leu Asn that when properly aligned and bracketed with an insertion at one end and a deletion at the other would specify the mutant amino acid sequence a The solution exists which requires a deletion of a nucleotide at the 539 end of the coding sequence and the insertion of a nucletodie at the 3 end39 5 Lys Ser Pro Leu Ala I COOH AAA AGU CCA H GUU GC 339 5 The Codei AAA GUC CAU H UUA GC 339 NH2 s Nearly Universal Lys Val His Leu Ala VCOOH a To understand the conservative nature of the code consider what might happen if a mutation changed the genetic code b Sequences of the regions known to specify proteins have revealed the following differ i iii ences between the standard and mitochondrial genetic codes UGA is not a stop signal but codes for tryptophan l Anticodon of mitochondrial tRNATrp recognizes both UGG and UGA Internal Methionine is encoded by both AUG and AUA In mammalian mitochondria AGA and AGG are not Arginine codons but specify chain termination 1 Therefore there are four stop codons iV a UAA b UAG c AGA d AGG In uit y mitochondria AGA and AGG are also not Argenine codons but spec ify s erine First Position 5 End U C A G Third Position 3 End UUU UUC UUA UUG F GAA F GAA L UAA L UAA UCU UCC39 UCA UCG S UGA S UGA S UGA S UGA UAU Y GUA UAC Y GUA UAA STOP UAG STOP UGU C GCA UGC c GCA UGG W UCA CUU CUC CUA CUG L UAG L UAG L UAG L UAG CCU CCC CCA CCG P UGG P UGG P UGG P UGG CAU CAC CAA CAG H GUG H GUG Q UUG Q UUG CGU R UCG CGC R UCG CGA R UCG CGG R UCG AUU AUC I GAU 1 GAU CAU AUG 1 M CAU ACU ACC ACA ACG T UGU T UGU T UGU T UGU AAU AAC AAA AAG N GUU N GUU K UUU K UUU AGU S GCU AGC S GCU AGA STOP AGG STOP GUU GUC GUA GUG V UAC V UAC V UAC V UAC GCU GCC GCA GCG A UGC A UGC A UGC A UGC GAU GAC GAA GAG D GUC D GUC E UUC E UUC GGU G UCC GGC G UCC GGA G UCC GGG G UCC CIDgtOC1 CDDgtOC1 CIDgtOC1 CIDgtOC1 Note Letters in parenthesis are the anticodon codes 5 gt3


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