343 Outline for B M B 400 at PSU
343 Outline for B M B 400 at PSU
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Date Created: 02/06/15
EME mu m Faun m 15 mm and Negnve Tmmcnptmnal Camml m m s M s Mm m1 mp Gm mgmm Serum cmpm IS PoslnvIz AND mmva conmm snownm m 1 owner om mu A m and gemnl camments Opams Anwpnun s a mm at mummy ngmmd gm H mm mnunl gums gemmlyemadmgenzymzs xeguhtm39ygulgsmnmdmg e g amvatmsax npnssals and leguhmxy sks such as pmmm and apenmls Negative vnsus pmhve mntznl a m ype uf cumml x mm w m lespanse at m apemn mm m Emmy pmmms pnsem n m m case afmanve canhvl m gems m m apemnaxz expxzssed unless may ale 5mm m hya npzssmpmm Thw m apemn wm m ma an camm vely m gems wm m expxzssed mm m npnsmr m mama c m m case at gamma camml m gems ale expxzssedanly whnan acme ngulamx pmum e g an amvamx x mm m m apemnwl m ma mwm m mm nguhmry pmm x mm m macuvamd hm 51 Pm vs nagahve mm Mum newmy Exlmp e av Malay prawn 5 maulMy gm nlose Emu Wequot mam Poshw wnlra 0mm Aawm DvemnOFF Ngewmnm DpmnDFF Revessar UpemuON BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac 3 Catabolic versus biosynthetic operons a Catabolic pathways catalyze the breakdown of nutrients the substrate for the pathway to generate energy or more precisely ATP the energy currency of the cell In the absence of the substrate there is no reason for the catabolic enzymes to be present and the operon encoding them is repressed In the presence of the substrate when the enzymes are needed the operon is induced or de repressed Table 412 Comparison of catabolic and biosynthetic operons Operon encodes Absence of Effect Presence of Effect catabolic enzymes substrate repressed substrate derepressed induced biosynthetic product induced product repressed enzymes For example the lac operon encodes the enzymes needed for the uptake lactose permease and initial breakdown of lactose the disaccharide 5 D galactosyl 1 gt4 D glucose into galactose and glucose catalyzed by 5 galactosidase These monosaccharides are broken down to lactate principally via glycolysis producing ATP and from lactate to C02 via the citric acid cycle producing NADH which feeds into the electron transport chain to produce more ATP oxidative phosphorylation This can provide the energy for the bacterial cell to live However the initial enzymes lactose permease and S galactosidase are only needed and only expressed in the presence of lactose and in the absence of glucose In the presence of the substrate lactose the operon in turned on and in its absence the operon is turned off Anabolic or biosynthetic pathways use energy in the form of ATP and reducing equivalents in the form of NADPH to catalyze the synthesis of cellular components the product from simpler materials e g synthesis of amino acids from small dicarboxylic acids components of the the citric acid cycle If the cell has plenty of the product already in the presence of the product the the enzymes catalyzing its synthesis are not needed and the operon encoding them is repressed In the absence of the product when the cell needs to make more the biosynthetic operon is induced E g the tip operon encodes the enzymes that catalyze the conversion of chorismic acid to tryptophan When the cellular concentration of Trp or Trp tRNA rP is high the operon is not expressed but when the levels are low the operon is expressed BMB 400 Part Four 7 I Chpt 15 Positive and Negative Transcriptional Control at lac 4 Inducible Versus repre ible operons a Inducible operons are turned on in reponse to a metabolite a small molecule undergoing metabolism that regulates the operon Eg the lac operon is induced in the presence of lactose through the action of a metabolic by product allolactose b Repressible operons are switched off in reponse to a small regulatory molecule Eg the trp operon is repressed in the presence of tryptophan Note that in this usage the terms are defined by the reponse to a small molecule Although lac is an inducible operon we will see conditions under which it is repressed or induced Via derepression Table 413 Inducible vs repressible operons Defined by response of operon to a metabolite small molecule Type of Examples operon Presence of Effect Metabolite O39peron Inducible metabolite ON lactose ac Repressible metabolite OFF Trp 17p BMB 400 PartFour r I Chpt 15 Positive and Negative Transcriptional Control at lac B Map of the E coli kw operon Figure411 Induced derepressed lac operon Prometer Structural Bperamr Iacz lacY lacA genesamp 4 h i regulatory sitesin l transcription operon AUG UAA AUG UAA AUG UAA Polycistronic l translation mRNA Bgalactosidase lactose B39galaCIOSide permeaSe transacetylase l Promoters p binding sites for RNA polymerase from which it initiates transcription There are separate promoters for the lac gene and the lacZYA genes 2 Emerator o binding site for repressor overlaps with the promoter for lacZYA 3 Repressor encoded by ad gene 4 Structural genes lacZYA 1ch encodes Bygalactosidase which cleaves the disccharide lactose into galactose and glucose 1ch encodes the lactose permease a membrane protein that faciltitates uptake of lactose 1ch encodes Sigalactoside transacetylase the function of this enzymes in catabolism of lactose is not understood at least by me BMB 400 Part Four 7 I Chpt 15 Positive and Negative Transcriptional Control at lac C Negative central The ac operon is under both negative and positive control The mechanisms for these will be considered separately 1 ln negative control the lacZYA genes are switched off by repressor when the inducer is absent signalling an absence of lactose When the repressor tetramer is bound to 0 lacZ YA is not transcribed and hence not expressed Figure 412 Repressed lac npernn Promoter ac Bperator acZ ac ac4 i i i t T 08 acrepressor Repressorbinds to the operatorin the absence of the inducer a metabolite of lactose and blocks transcription of the ac operon 2 When inducer is present signalling the presence of lactose it binds the repressor protein thereby altering its conformation decreasing its af mty for 0 the operator The dissociation of the repressoriinducer complex allows lacZ YA to be transcribed and therefore expressed Figure 413 Inductinn nf the lac npernn by derepressinn Promoter ac Bperator acZ my Am i l I t I t t I t J nducer alloiactose 2 i Aacrepressornolonger Binds operator Promoter ac BPeratoquot aCZ ac acA 5 h t l t I i i Operon is expressed BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac D Inducers 1 The natural inducer or antirepressor is allolactose an analog of lactose It is made as a metabolic by product of the reaction catalyzed by 5 galactosidase Usually this enzyme catalyzes the cleavage of lactose to galactose glucose but occasionally it will catalyze an isomerization to form allolactose in which the galacose is linked to C6 of glucose instead of C4 2 A gratuitous inducer will induce the operon but not be metabolized by the encoded enzymes hence the induction is maintained for a longer time One of the most common ones used in the laboratory is a synthetic analog of lactose called isopropylthiogalactoside IPTG In this compound the S galactosidic linkage is to a thiol which is not an efficient substrate for 5 galactosidase E Regulatory mutants Regulatog mutations affect the amount of all the enzymes encoded by an operon whereas mutations in a structural gene affects only the activity of the encoded single polypeptide 1 Repressor mutants a Wild type strains lacI are inducible b Most strains with a defective repressor lacI39 are constitutive ie they make the enzymes encoded by the lac operon even in the absence of the inducer c Strains with repressor that is not able to interact with the inducer lacIS are noninducible Since the inducer cannot bind the repressor stays on the operator and prevents expression of the operon even in the presence of inducer BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac d Deductions based on phenotypes of mutants Table 414 Phenotypes of repressor mutants Genotype 5 galactosidase IPTG IPTG transacetylase IPTG IPTG Conclusion Inducible lacZ encodes 5 galactosidase Constitutive Non1nduc1ble 5 gtI in trans 1 The vvild type operon is inducible by IPTG 2 A mutation in lacZ affects only S galactosidase not the transacetylase or other products of the operon showing that lacZ is a structural gene 3 A mutation in lacI affects both enzymes hence lad is a regulatory gene Both are expressed in the absence of the inducer hence the operon is constitutively expressed the strain shows a constitutive phenotype 4 In a merodiploid strain in which one copy of the lac operon is on the chromosome and another copy is on an F factor one can test for dominance of one allele over another The Wild tvpe lacI is dominant over lacI39in trans In a situation where the only functional lacZ gene is on the same chromosome as lacI39 the functional lacI still causes repression in the absence of inducer 5 The lacIS allele is noninducible 6 In a merodiploid the lads allele is dominant over vvild tvpe in trans The fact that the product of the lac gene is trans acting means that it is a diffusible molecule that can be encoded on one chromosome but act on another such as the F chromosome in example d above In fact the product of the lac gene is a repressor protein BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac 2 Operator mutants a Defects in the operator lead to constitutive expression of the operon hence one can isolate operator constitutive mutations abbreviated 00 The vvild type at is inducible Mutations in the operator are cis acting they only affect the expression of structural genes on the same chromosome 1 The merodiploid ItocztItotz this is an abbreviation for lacltoclacztlacI0lacZ expresses S galactosidase constitutively Thus 00 is dominant to at when 00 is in cis to lacZ 2 The merodiploid FOCZItotzt is inducible for S galactosidase expression Thus 0 is dominant to 00 when 0 is in cis to lath 3 The allele of 0 that is in cis to the active reporter gene ie on the same chromosome as 61th in this case is the one whose phenotype is seen Thus the operator is cis acting and this property is referred to as cis dominance As in most cases of cis regulatory sequences these are sites on DNA that are required for regulation In this case the operator is a binding site for the trans acting repressor protein BMB 400 Part Four 7 I Chpt 15 Positive and Negative Transcriptional Control at lac F Interactions between operator and repressor 1 Sequence of operator a The operator overlaps the start the site of transcription and the promoter b It has adyad symmetg centered at11 Such adyad symmetry is commonly found within binding sites for symmetrical proteins the repressor is a homotetramer c The sequence ofDNA that consititutes the operator was defined by the position of of mutations as well as the nucleotides protected from reaction with eg DMS upon binding of the repressor Figure 414 Interactions between operator and repressor Constitutivemutations A TGTTA C T T ACAAT G A 1 o 1 T o T 2 0 gt gt gt 5 TGTTGTGTGGAATTGTGAGC ATAACAATTTCACACA 3 ACAACACACCTTAACACTC CTATTGTTAAAGTGTGT f4 4 lt lt Dyad axis Nucleotide in c repressor ct With Promoter 2 Reprasor a Purification 1 Increase the amount of repressor in the starting material by over expression A wildrtype cell has only about 10 molecules of the repressor tetramer Isolation and purification of the protein was greatly aided by use of mutant strain with uprpromoter mutations for ac so that many more copies of the protein were present in each cell This general strategy of overrproducing the protein is widely used in purification schemes Now BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac the gene for the protein is cloned in an expression vector so that the host bacteria in this case makes a large amount of the protein often a substantial fraction of the total bacterial protein 2 Assays for repressor 1 Binding of radiolabeled IPTG gratuitous inducer to repressor 2 Binding of radiolabeled operator DNA sequence to repressor This can be monitored by the ability of the protein DNA complex to bind to nitrocellulose whereas a radiolabeled mutant operator DNA fragement 00 plus repressor will not bind Electrophoretic mobility shift assays would be used now in many cases 3 This ability of particular sequences to bind with high affinity to the desired protein is frequently exploited to rapidly isolate the protein The binding site can be synthesized as duplex oligonucleotides These are ligated together to form multimers which are then attached to a solid substrate in a column The desired DNA binding protein can then be isolated by af nity chromatography using the binding site in DNA as the affinity ligand b The isolated functional repressor is a tetramer each of the four monomers is the product of the lac gene ie it is a homotetramer c The DNAbinding domain of the lac repressor folds into a helixturn helix domain We will examine this structural domain in more in Chapter 111 It is one of the most common DNA binding domains in prokaryotes and a similar structural domain the homeodomain is found in some eukaryotic transcriptional regulators 3 Contact points between repressor and operator a b Investigation of the contact points between repressor and the operator utiblized the same techniques that we discussed previously for mapping the binding site of RNA polymerase on the promoter e g electrophoretic mobility shift assays does the DNA fragment bind DNase footprints where does the protein bind and methylation interference assays methylation of which purines will prevent binding Alternative schemes will allow one to identify sites at which methylation is either prevented or enhanced by the binding of the repressor These techniques provide a biochemical defintion of the operator 2 binding site for repressor The key contact points see Figure 414 1 are within the dyad symmetry 2 coincide in many cases with nucleotides that when mutated lead to constitutive expression Note that the latter is a genetic definition of the operator and it coincides with the biochemically defined operator BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac 3 tend to be distributed symmetrically around the dyad axis 11 4 are largely on one face of the DNA double helix The partial overlap between the operator and the promoter initially suggested a model of steric interference to explain the mechanism of repression As long a repressor was bound to the operator the polymerase could not bind to the promoter But as will be explored in the next chapter this is not the case RNA polymerase can bind to the lac promoter even when repressor is boudn to the lac operator However the polymerase cannot initiate transcription when juxtaposed to the repressor 4 Conformational shift in repressor when inducer binds a The repressor has two different domains one that binds to DNA quotheadpiecequot containing the helix tum helix domain and another that binds to the inducer and other subunits called the quotcore These are connected by a quothingequot region These structural domains can be distinguished by the phenotypes of mutations that occur in them lacI d prevents binding to DNA leads to constitutive expression lacIS prevents binding of inducer leads to a noninducible phenotype Binding of inducer to the quotcorequot causes an allosteric shift in the repressor so that the quotheadpiecequot is no longer able to form a high affinity complex with the DNA and the repressor can dissociate go to one of the many competing nonspecific sites BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac G Positive control quotcatabolite repressionquot 1 Catabolite repression a Even bacteria can be picky about what they eat Glucose is the preferred source of carbon for E coli the bacterium will consume the available glucose before utilizing alternative carbon sources such as lactose or amino acids Glucose leads to repression of expression of lac and some other catabolic operons This phenomenon is called catabolite repression 2 Two components are needed for this form of regulation a b cAMP 1 In the presence of glucose the cAMP inside the cell decreases from 10 4 M to 10 7 M A high cAMP will relieve catabolite repression 2 cAMP synthesis is catalyzed by adenylate cyclase product of the cya gene ATP gt cAMP PPi Catabolite Activator Protein 2 CAP 1 Product of the cap gene also called crp cAMP receptor protein 2 Is a dimer 3 Binds cAMP and then the cAMP CAP complex binds to DNA at speci c sites BMIB 400 Part Four e1 Chpt 15 Positive and Negative Tmnscriptional Control at lac 3 Binding site for cAMPeCAP a In the lac opemn the binding site is a region ofahcnt 20 hp located jnst npstieam from the picmotei from 52 to 42 h The penameiTGrGA is an essential element in recognition For the lac opemn the binding site is a dyad with that seqnence in both sides othe dyad c Coth points hetwen cAMPCAP and the DNA aie close to or coincident with mnatio ns that iendei the lac pmmotei no longer responsive to CAMP CAP d cAMPeCAP binds on one face of the helix Figure 415 Binding site for cAMPCAP Mutations that make A T promoter nonresponsive T A to CAP T T 7 O 6 O 5 O b gt 5 ATGTGAGTTAGCTCACACATT 3 TACACTCAAEC GTGTGTAA Dyad axis leotides in c act CAMPCAP Promoter BMB 400 Pan Four 71 Chpt 15 Positive and Negative Transcriptional Coan at lac 4 Bindin of cAMPyCAP to its site will enhance efficiech of transcrimion initiation at gmmouer a The lac pioinomis not a panicularly sLmng pioinoier The sequence at e10 TATGTT does not mamh the consensus TATAAT at two posin39ons b 1n the piesence of cAMIPeCAP the RNA polymerase will initiaue mscripiion nioie efficienuy c The lac UV5 pioinom is an uprpmmomr inumion in which the 710 region mamhes the consensus The lac operon driven by the UV5 pioinomwin achieve high level inducn39on without cAMPeCAP butthe wildrtype promoier iequiies cAMIPeCAP for high level inducu39on Figure 416 Regulawry region of lat openon including CAP binding site Activator binding site Promoter Operator H lt gt UV5 mutation up TATAAT TTTACA TATGTT a gtlt3 72 52 35 1 0 CAMPCAP Repressor RNA polymerase BMB 400 Part Four 7 I Chpt 15 Positive and Ngative Tmscdpljonal Control at In 5 Mode of ac on of CAMPCAP a Direct positive interaction with RNA polmerase The Cterrninus of the a subunit is required for RNA polymerase to be activated by cAMRCAR This will be explored in more detail in Chapter 16 b CAMPrCAP bends lhe DNA about 90 gure 417 DNA top helical structure 395 bent by the CAP dimer mr m an rs 1 Iiianimus nprnmnrrrr I Some generali s l Represms activators and polymerases interact primarily with one face of the DNA double helix 2 Regulatory proteins such as activators and repressors are frequently symmetrical and bind syrnrneuical sequences in DNA 3 RNA polymerases are not symmetrical and the promoters to which they bind also are asyrnrneuical This confers directionalig on uanscripu39on BMB 400 151 152 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac Questions for Chapter 15 Positive and Negative Transcriptional Control Amber mutations are one class of nonsense mutations They lead to premature termination of translation by alternation of an amino acid encoding codon to a UAG terminator e g CAG Gln may be changed to UAG stop amber The phenotype of such amber mutants can be suppressed by amber suppressor genes which are mutant tRNA genes that encode tRNAs that recognize UAG codons and allow insertion of an amino acid during translation Which genes or loci in the lac operon can give rise to amber suppressible mutations POB Negative regulation In the lac operon describe the probable effect on lacZ gene expression of a Mutations in the lac operator b Mutations in the lad gene c Mutations in the promoter 153 154 Consider a negatively controlled operon with two structural genes A and B for enzymes A and B an operator gene 9 and a regulatory gene B In the wild type haploid strain grown in the absence of inducer the enzyme activities of A and B are both 1 unit In the presence of an inducer the enzyme activities of A and B are both 100 units For parts a d choose the answer that best describes the enzyme activities in the designated strains Uninduced Induced EnzA EnzB EnzA EnzB a R0CAB a 1 1 100 100 b 1 100 100 1 c 50 50 100 100 b R 0AB a 1 1 100 100 b 100 100 100 100 c 100 0 100 0 c R0CABR0AB a 2 2 200 200 b 51 51 200 200 c 200 2 2 200 d R 0ABR0AB a 2 2 200 200 b 2 101 2 101 c 200 200 200 200 POB Positive regulation A new RNA polymerase activity is discovered in crude extracts of cells derived from an exotic fungus The RNA polymerase initiates transcription only from a single highly specialized promoter As the polymerase is purified its activity is observed to decline The purified enzyme is completely inactive unless crude extract is added to the reaction mixture Suggest an explanation for these observations BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac 155 Consider a hypothetical regulatory scheme in which citrulline induces the production of urea cycle enzymes Four genes citA citB citC citD affecting the activity or regulation of the enzymes were analyzed by assaying the vvild type and mutant strains for argininosuccinate lyase activity and arginase activity in the absence cit or presence cit of citrulline In the following table wild type alleles of the genes are indicated by a under the letter of the cit gene and mutant alleles are indicated by a under the letter The activities of the enzymes are given in units such that 1 the uninduced wild type activity 100 the induced activity of a wild type gene and 0 no measurable activity In the diploid analysis one copy of each operon is present in each cell Strain lyase activig arginase act number genes 7cit cit 7cit cit Haploid A B Q Q 1 1 100 1 100 2 100 100 100 100 3 0 0 1 100 4 100 100 100 100 5 1 100 0 0 Diploid A B C D A B C D 6 1 100 1 100 7 1 100 2 200 8 100 100 100 100 9 1 100 100 100 Use the data in the table to answer the following questions a What is the phenotype of the following strains with respect to lyase and arginase activity A single word will suffice for each phenotype Lyase activity Arginase activity Strain 2 Strain 3 Strain 4 Strain 5 Strain 6 b What can you conclude about the roles of citB and citD in the activity or regulation of the urea cycle in this organism Brief answers will suffice c What is the relationship recessive or dominant between wild type and mutant alleles of citA and citC Be as precise as possible in your answer d What can you conclude about the roles of citA and citC in the activity or regulation of the urea cycle in this organism Brief answers will suffice BMB 400 156 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac Consider a hypothetical operon responsible for synthesis of the porphyrin ring the heterocyclic ring that is a precursor to heme cytochromes and chlorophyll Four genes or loci porA porB porC and porD that affect the activity or regulation of the biosynthetic enzymes were studied in a series of haploid and diploid strains In the following table wild type alleles of the genes or loci are indicated by a under the letter of the par gene or locus and mutant alleles are indicated by a under the letter The activities of two enzymes involved in porphyrin biosynthesis o aminolevulinic acid synthetase and 6 aminolevulinic acid dehydrase referred to in the table as ALA synthetase and ALA dehydrase were assayed in the presence or absence of heme one product of the pathway The units of enzyme activity are 100 non repressed activity of the wild type enzyme 1 repressed activity of the vvild type enzyme in the presence of heme and 0 no measurable activity In the diploid analysis one copy of each operon is present in each cell Strain ALA synthetase ALA dehyd number par heme heme heme heme Haploid A B Q Q 1 100 1 100 1 2 100 100 100 100 3 0 0 100 1 4 100 1 0 0 5 100 100 100 100 DiPIOidi ABQDABLD 6 100 1 100 1 7 200 101 100 100 8 200 2 100 1 9 100 100 100 1 Use the data in the table to answer the following questions a Describe the phenotype of the following the strains with respect to ALA synthetase and ALA dehydrase activities A single word will suffice for each phenotype ALA synthetase ALA dehydrase Strain 2 Strain 3 Strain 4 Strain 5 Strain 6 b What is the relationship dominant or recessive between wild type and mutant alleles of the four genes and which strain demonstrates this Please answer in a sentence with the syntax in this example quotStrain 20 is repressible which shows that mutant grkl is dominant to vvild type quot porA Strain is which shows that BMB 400 Part Four I Chpt 15 Positive and Negative Transcriptional Control at lac porA is porB Strain is which shows that porB is porC Strain is which shows that porC is porD Strain is which shows that porD is c What is the role of each of the genes in activity or regulation of porphyrin biosynthesis Brief phrases will suffice d Is this operon under positive or negative control
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