BIOLOGY 305 1st Edition Lecture 21 Outline of Last Lecture I. MutationsA. ProofreadingB. Post-Replication RepairC. Spontaneous MutationsD. Induced MutationsII. Seven Forms of Repair: OverviewA. Mismatch Repair, Direct ReversalB. Base Excision Repair, Nucleotide Excision RepairC. Translesion Synthesis, NHEJ, RecombinationIII. Types of Mutations Affecting Coding and their EffectsIV. DiseasesV. Vocabulary and Sample QuestionsOutline of Current Lecture I. Introduction to Regulatio via The Lac Operon II. Two Tools to Modify and Analyze OperonsIII. Analysis of MutantsIV. SummariesA. Lacl Repressor MutationsB. Phenotype Based on InducerC. How to solve lac operon questions given GenotypeD. Summary of the Lac Operon and its GenesV. Catabolite RepressionVI. DNAse I FootprintingVII. Vocabulary and Sample QuestionsCurrent LectureRecall: Key Traits of DNA as genetic material1) Diversity of structure2) Ability to mutate3) Replication4) Regulated expressionI. Introduction to Regulation via The Lac OperonRNA Polymerase binding during transcription (initiation) is the most highly regulated stepLac operon was discovered by Francois Jacob, Adre Lwoff, Jacques Monod in 1965What is an operon? Example of operon in polycistronic DNA:lac operon – a very well-understood gene for regulations:Glucose is the preferred food source of E. Coli. But in the presence of lactose and absence of glucose, E. Coli will induce expression of enzymes necessary for lactose metabolismLac Z: a gene that allows metabolism (cleavage) of lactose using b-galactosidaseLac Y: encodes for a permease in the cell membrane that allows lactose to diffuse insideLac A: encodes for transacetylase (function in lactose metabolism is unknown)Positive vs. Negative Regulation:lactose is the inducer in lactose metabolismRegulation is regulated: the lacl gene encodes for the Lac repressorAllosteric changes in the repressorRepressor without inducer  repressor binds DNA site  Transcription OFFRepressor + inducer  no longer binds to operator  Transcription ONRepressor is thus allosteric: irreversible interaction with another molecule causes conformational change in three-dimensional shape and change in activityII. Two Tools to Modify and Analyze Operons – 2 Goals:1) Need to modify what lactose operon does2) Need to be able to visualize this change for analysisTool 1) IPTG - a lactose analog that cannot be hydrolyzed:Note: You must be careful to look at the correct time point or it may look like there’s no activityInducer = lactose, but the activity for the operon eliminates lactoseTool 2) X-gal – a substrate for beta-galactosidase, ideal for visualizing activity:When X-Gal is digested by beta-Gal., galactose and a blue dye formsThus, blue color = evidence of B-Gal. activityExample: Mutants in lactose metabolism in E. ColiYou are studying lactose metabolism in E. Coli. You isolate of number of mutants and plated them on X-gal containing plates with out without IPTG present… what should you expect?Wild-type: inducible: so without IPTG, colonies are white. With IPTG, colonies are blue because B-Galactose is activeLacZ-, Mutant 4: uninducible: not subject to change in environment, because even with IPTG present, colonies are not metabolizing X-Gal and are whiteMutant 2 and 3: constituitive: not subject to being turned on or off and is expressed continuously regardless of environment (hence X-Gal is meabolized in both plates)III. Analysis of MutantsUnediploid – haploid for all genes except the lactose operon, has a short plasmid that contains acopyQuestion to ask: Do the mutations act in cis or trans?cis – a region of DNA or RNA that regulates gene expression on the same DNA moleculeThe lac operon is cis-acting: the region is bound by the lac repressor which prevents transcription of the adjacent genes on the same DNA moleculetrans – are diffusible factors, usually proteins, affects gene expression of the whole cell because they can modify distant genes than the original gene transcribed to create them, does not affect gene expression in the mitochondriaAnalysis of Mutant 3: 1st combination = mutant in lacZ plasmid = constitutivethere is continuous expression in both plates of -/+ IPTG, hence both = blue2nd combination = mutant in both = induciblewhatever is produced by F’ is able to induce expression in other3rd = mutation is inducible, 3+ is able to repress lac Z geneWhat is mutated? The repressor, as the repressor is trans-acting and there is loss-of-functionThe operator and promoter act in cisAnalysis of mutant 2:1st: constitutive = expresses genes without presence of an inducer2nd: constitutive = continuous expression3rd: Inducible, 2+ is able to repress lac Z geneSuper-repressor – able to repress lactose operons without association with IPTGWhat is mutated?The operator (which is cis-acting) and the operator cannot be bound by the repressorAnalysis of Mutant 4What is mutated? Trans-acting and is also a super-repressorIV. SummaryA. Lacl Repressor MutationsLacl + = lactose binding and DNA bindingLacl - = loss of function mutation, can’t bind DNALacl s = always bind DNA, can’t bind lactose, gain of functionLac O = operator, binding site for the repressorLac P = promoter, binding site for RNA polymeraseLac I = repressor protein, binds to operator to block RNA polymerase bindingB. Phenotype Based on Inducer Note: genotypes are written in order: repressor (i), promoter (p), operator (o), LacZ (z), LacY (y) | = repressor, O = operator, Logic:1) repressor- + acts in trans (repressor non-functional, operator functional)2) repressor- s acts in trans (repressor always functional, operator functional = no expression)3) operator- + acts in cis (repressor functional, operator always functional = always expresses)4) operator- c acts in cis (repressor functional, operator always functional)5) repressor- s cannot bind operator-c (repressor always functional, operator always functional)C. How to Solve a Lac Operon Question given genotype:1) Look at the promoter first - do you have a good copy of the promoter (p+)? If not, RNA polymerase cannot get transcription started and that chromosome is a bust.2) Have p+? Next look at how the repressor and the operator interact:Here are a few scenarios:i+p+o+ : the repressor binds to the operator in the absence of lactose and inhibits transcription of the z and y genes. BUT, when lactose is around, it kicks the repressor off of the operator, and the z