Ruf 2 - Main Points: Lecture 1 Elements of a promoter: -35...

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Unformatted text preview: Main Points: Lecture 1 Elements of a promoter: -35 and -10 regions, site of transcription initiation, importance of DNA sequence and spacing between elements Prok. RNAP: subunits, sigma factor as specificity factor, “core” vs. “holoenzyme” Termination of transcription: Rho-dependent, -independent How Do We Know What defines a Promoter? • 1. DNA sequence conservation • 2. Genetics • 3. Biochemistry DNA sequence conservation of upstream regions of bacterial genes HYPOTHESIS: If a promoter is more like the consensus sequence, then it will be recognized more often and is therefore a more efficient promoter 1. Sequence conservation: rRNA promoter is strong and more similar to consensus “weak” promoters are less similar to consensus 1. Genetics: mutations that change the sequence change promoter activity 2. Biochemistry: direct protein-DNA binding experiments, measure strength of interaction and correlate to promoter activity HYPOTHESIS: since sigma factors control promoter specificity, if there are different sigma factors there can be groups of coordinately regulated genes, each with its own “consensus” recognition site for targeting core polymerase The σ factor is the first level of specificity - whole classes of genes are turned on or off depending on which s factor is being expressed at a given time a. σ70 (RpoD) is the housekeeping/normal factor for most genes b. other σ factors: σs - stationary phase, starvation, stress; σH - heat shock and other stresses; sporulation, flagellar synthesis, etc. Why Regulate Gene Expression? Some proteins are required in large amounts (translation elongation factors) Some proteins are only present at a few copies/cell Requirements for gene products changes over time a. Environmental: different food sources b. Developmental/Differentiation c. Cell specialization 7 Processes that determine Protein concentration in a cell 1. The σ factor is the first level of specificity - whole classes of genes are turned on or off depending on which σ factor is being expressed at a given time a. σ (RpoD) is the housekeeping/normal factor for most genes Specificity of Transcription Initiation b. other σ factors: σ s - stationary phase, starvation, stress; σ - Constitutive Expression: unvarying expression. Housekeeping Genes: genes whose proteins are required all the time, are expressed at a constant level in virtually all cells. Although made constitutively, concentration of these proteins varies widely due to differences in polymerase/promoter interactions Regulated Gene Expression: concentrations change in response to molecular signals: induction (activation) or repression Types of Proteins that regulate transcription initiation 1. Specificity Factors: alter the specificity of RNAP for a given promoter (Sigma factors in bacteria) 1. Repressors: Prevent access of RNAP to promoter. Bind to specific sites on DNA (operators) near promoters 1. Activators: Enhance interactions between RNAP and promoters. Bind to specific sites usually adjacent to promoters that are weakly (or not at all) by RNAP An inhibitor is present and prevents transcription inhibitor= repressor Antagonist of repressor= inducer, needed to allow initiation of transcription An effector molecule activates promoter= activator Some systems are regulated by both: 2 regulators that respond to different conditions Bacterial genes involved in metabolizing or synthesizing metabolites which may or may not be present in the environment involves the linked expression of genes from an OPERON The Operon A gene cluster and promoter, plus additional sequences that function together in regulation A single transcription unit containing more than 1 gene Advantages: Disadvantages: Lac Operon is under both negative and positive regulation In the absence of lactose: In the presence of lactose: Therefore: How: How: But:: How: Lac Operon encodes 4 proteins: 1 regulatory and 3 enzymes Promoter for I Regulatory protein Promoter Operator 3 enzymes O P IGe PLZee Y n A n I L e n G n Ge e Ge e A T G 5U ’A G 3 ’ + 1 + 1 5A ’U G UA GT AG A U G UA GT AG A U G 3 ’ U G A mA R N -35 “weak” -10 Ip t i re on (e e o rp s r r s) Zr t i pe on Yr t i pe on ( - ac s a ) ( e e e βltis ga od e p m ra s ta p t rn o) sr Ar t i pe on ( ac s e ga od lti ta a t l s rn c y e sea) Function of Lac Operon encoded enzymes Y: lac permease: transport of Lac into cells Z: disaccharide cleavage Encoded by Lac Y gene Encoded by Lac Z E. Coli Growing on Lactose as Carbon & Energy Source Activity is induced 1000-fold when lactose is present and glucose is absent from the medium. L cg n a oe βGl o L c -a ra ut k pa e o go t r rw h ( o e“ e s r ” s m ma ue ) Ad a d Lc 10 - 00 0X 0X 1 “n u to ” i d ci n Te i m L cOeo a pr n O P L Z ee Y ee A ee L Gn Gn Gn AG T 5 AG ’U U A AG GT AG U U A AG G T AG U 3 ’ UA G + 1 P I Gn I ee mN RA Lac Operon: the players in repression/induction Promoter for I Regulatory protein O P IGe PLZee Y n A n I L e n G n Ge e Ge e A T G 5U ’A G 3 ’ + 1 + 1 5A ’U G UA GT AG A U G UA GT AG A U G U G A A lactose derivative within the cell, allolactose, 3 ’ mA R N serves as inducer (product of a B-gal dependent side-reaction) Ip t i re on (e e o rp s r r s) Zr t i pe on Yr t i pe on ( - ac s a ) ( e e e βltis ga od e p m ra s ta p t rn o) sr Ar t i pe on ( ac s e ga od lti ta a t l s rn c y e sea) No Lactose: Lactose Present: Analysis of Lac mutants b t n rm l L cu ta eIn u tio u o a a p k dc n L cg n a oe β ao Lc -G l r a u ta e pk og w r ro th (s m “ e s re ) o e ma u ” A dL c da 1 0 -1 0 X 0X 00 “ dc n in u tio ” Te im L cO e n a p ro PO L L ZG n ee YG ne AG ne e e Use a colorless substrate which turns blue (or yellow) + 1 P 1. Can cells grow? I I G ne e 2. Is lac Z functional? AG T U AAG G U A AG G UA G T T upon cleavage. 5 AG ’U AG U AG U 3. Is LacY functional? Use radioactive lactoseA(14C), is it taken3’into cells? mN R Mutant: P I I G ne e L cO e n a p ro PO L L ZG n ee YG ne AG ne e e AG T 5 AG ’U U AAG G T AG U U A AG G T AG U 3 ’ UA G + 1 Zg n “ u tio ” n β a o g w e e m ta n o -G l r ro th bt o a a p k dc n Observation: u n rm l L cu ta eIn u tio mN RA Lac uptake β ao Lc -G l r a u ta e pk og w r ro th (s m “ e s re ) o e ma u ” A dL c da 1 0 -1 0 X 0X 00 “ dc n in u tio ” L cg n a oe B-gal and growth Te im L cO e n a p ro PO L L ZG n ee YG ne AG ne e e AG T 5 AG ’ U U AAG G T U A AG G T AG U AG U + 1 Zg n “ u tio ” n β a o g w e e m ta n o -G l r ro th b t n rm l L cu ta eIn u tio u o a a p k dc n UA G P I I G ne e mN RA 3 ’ LacZ-: 14C enters cells (must still be making Y), no B-Gal Yg n “ u tio ”n L cu ta e o g w e e m ta n o a p k r ro th b t n rm l β a In u tio u o a -G l d c n AG T 5 AG ’U U AAG G T U A AG G T UA G AG U AG U + 1 Mutant: P I IGn ee L cO e n a p ro PO L L ZG n ee YG n AG n ee ee mN RA 3 ’ Observation: B-gal L cg n a oe β ao Lc -G l r a u ta e pk og w r ro th (s m “ e s re ) o e ma u ” A dL c da 1 0 -1 0 X 0X 00 “ dc n in u tio ” Lac uptake, growth Te im Yg n “ u tio ”n L a u e k o g w e e m ta n o a cO tan r ro th Lc p ro e p b t n rm l β Oa In u tio AGene u PoI Gene P L ZGend YGene aL l ecn -G I A T U AAG U A AG G T T LacY-: no 14C enters cells, B-gal presentG(mustGstill be making Z) 5 ’ AG U AG U AG U UA G L cg n a oe 3 ’ + 1 mN RA Gene Complementation in E. coli P I I Gene P OL Z Gene L ATG Y Gene A Gene UGA ATG UGA UGA ATG P I I Gene P OL Z Gene L ATG Y Gene A Gene UGA ATG UGA UGA ATG Neither one can grow on Lactose (They are Lac -) But... If you make a “diploid” Gene Complementation in E. coli P I I Gene P OL Z Gene L ATG Y Gene A Gene UGA ATG UGA UGA ATG P I I Gene P OL Z Gene L ATG Y Gene A Gene UGA ATG UGA UGA ATG One makes β -gal, the other makes Lac Permease and the two diffusible protein products make the Diploid cell Lac + Gene non-Complementation in E. coli P I IS Gene P OL Z Gene L ATG Y Gene A Gene U GA ATG UGA UGA ATG P I I- Gene P OL Z Gene L ATG Y Gene A Gene U GA ATG UGA UGA ATG Neither one makes β -gal, they both make Lac Permease and the Diploid cell is still Lac - ...
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