Goals of today’s lecture 1) Describe the basics of prokaryotic gene regulation -operons, negative and positive regulation 2) Illustrate the use of genetics in understanding cellular processes 3) Cover some aspects of DNA-binding proteins Gene Expression • Regulating the amount of active gene product (protein) by: – Control of gene expression at the level of transcription – Control of at the level of translation – Post-translational control • Constitutive expression = gene product made continuously • Regulated expression = gene product made on demand; expression can be induced or repressed Gene Expression & Information Flow Replication Transcription Translation Post-translation RNA Protein Protein* DNA • Replication = produce exact copy of DNA for mitosis (cell division) or reproduction (pass to the next generation) • Transcription = transcribe DNA code into RNA (uses same ‘language’ of nucleic acids) • Translation = translate nucleic acid code into a sequence of amino acids (the primary structure of polypeptides) • Post-translational modification = chemical modification to activate a protein so it can function in the cell Regulation of Gene Expression in Prokaryotes Catabolic Metabolism • E. coli use many sugars for metabolism • Glucose is the preferred source of carbon for E. coli –Why? • If glucose is not available, bacteria can break down lactose to generate glucose Figure 17-2b-setup Figure 17-2c-results • E. coli produces high levels of b-galactosidase, the enzyme that cleaves lactose to glucose + galactose, only when lactose is present in the environment. Thus, lactose (actually it is a metabolite of lactose) acts as an inducer—a molecule that stimulates the expression of a specific gene. • Jacques Monod found that b-galactosidase is not expressed in E. coli cells grown in medium containing glucose or glucose + lactose but only in medium containing lactose and no glucose. Genetic screening unmutagenized cells mutagenized cells mutagen 105 cells Mutant that can’t utilize lactose as a carbon source each cell has a different mutation How to find the needle in the genetic haystack? Which mutations effect lactose utilization? Replica plating allows the identification of genes that are essential to utilize lactose ~5000 colonies/plate • Three classes of E. coli mutants defective in lactose metabolism were isolated (Table 17.1): lacZ, lacY, and lacI. lacI is a constitutive mutant—one that has lost the ability to regulate expression of a particular gene because it produces a product at all times, not just when an inducer is present. Two proteins are critical for E. coli to use lactose and one is critical for regulation of their expression Galactoside permease E. coli b-Galactosidase Glucose Galactose Lactose Plasma membrane lacl product Section of E. coli chromosome lacl b-Galactosidase Galactoside permease lacZ product lacY product lacZ Far away on the chromosome lacY Model for Operons in Prokaryotes • Portion of DNA including a set of genes involved in a specific metabolic pathway • Single regulatory region (operator + promoter) • Generates single polycistronic RNA • Repressor binds the operator and blocks RNA polymerase • Repressor is the product of a regulatory gene AUG 5’ UGA AUG lacZ UAA lacY 3’ Figure 17-6a Negative control: Regulatory protein shuts down transcription TRANSCRIPTION No negative control… RNA polymerase Gene sequence Regulatory protein With negative control… No transcription Figure 17-6b Positive control: Regulatory protein triggers transcription No positive control… No transcription RNA polymerase Regulatory protein With positive control… TRANSCRIPTION Gene sequence • Jacob and Monod proposed that the lacI gene produces a repressor (the LacI+ protein) that exerts negative control over the lacZ and lacY genes. The repressor was thought to bind directly to DNA near or on the promoter for the lacZ and lacY genes (Figure 17.7). Repressor present, lactose absent: Repressor binds to DNA. No transcription occurs. The repressor blocks transcription Repressor synthesized DNA Normal lacl gene Repressor present, lactose present: Lactose binds to repressor, causing it to release from DNA. Transcription occurs (lactose acts as inducer). Repressor synthesized Normal lacl gene lacZ lacl+ RNA polymerase bound to promoter (blue DNA) TRANSCRIPTION BEGINS b-Galactosidase Permease mRNA lacZ lacl+ RNA polymerase bound to promoter (blue DNA) No repressor present, lactose present or absent: Transcription occurs. No functional repressor synthesized Mutant lacl gene lacY lacY Lactose-repressor complex TRANSCRIPTION BEGINS mRNA b-Galactosidase Permease Lacl – lacZ RNA polymerase bound to promoter (blue DNA) lacY The Trp operon is also under negative control, but with a twist When tryptophan is present, transcription is blocked. Repressor Tryptophan No transcription Operator RNA polymerase bound to promoter When tryptophan is absent, transcription occurs. TRANSCRIPTION 5 genes coding for enzymes involved in tryptophan synthesis RNA polymerase bound to promoter Figure 17-10 lac operon trp operon Catabolism (breakdown of lactose) Anabolism (synthesis of tryptophan) Repressor Lactose Repressor Tryptophan Tryptophan binds to repressor Lactose binds to repressor Lactoserepressor complex releases from operator Operator Tryptophanrepressor complex binds to operator Operator Transcription of lac operon TRANSCRIPTION No more transcription of trp operon Why doesn’t beta-galactosidase get induced in media containing Glucose? Catabolite Repression CAP regulates lac operon positively and requires cAMP for DNA binding When cAMP is present: cAMP binds to CAP and the cAMP-CAP complex binds to DNA at the CAP site. RNA polymerase binds the promoter efficiently. Transcription occurs frequently. When cAMP is absent: CAP does not bind to DNA. RNA polymerase binds the promoter inefficiently. Transcription occurs rarely. cAMP CAP FREQUENT TRANSCRIPTION CAP site Operator lacZ lacY lacA RNA polymerase bound tightly to promoter (blue DNA) CAP INFREQUENT TRANSCRIPTION CAP site Operator lacZ RNA polymerase bound loosely to promoter (blue DNA) lacY lacA Cyclic AMP (cAMP) is synthesized when glucose levels are low Glucose inhibits the activity of the enzyme adenylyl cyclase, which catalyzes production of cAMP from ATP. Glucose inhibits this enzyme ATP Adenylyl cyclase cAMP Two phosphate groups The amount of cAMP and the rate of transcription of the lac operon are inversely related to the concentration of glucose. CAP HIGH glucose concentration INACTIVE adenylyl cyclase LOW cAMP CAP does not bind to DNA Infrequent transcription of lac operon (Cell continues to use glucose as energy source.) cAMP CAP LOW glucose concentration ACTIVE adenylyl cyclase HIGH cAMP CAP-cAMP complex binds to DNA Frequent transcription of lac operon (Cell uses lactose if lactose is present.) Dual Regulation of lac operon • Negative control by lac repressor >> needs the inducer (lactose) to inactivate the lac repressor • Positive control by CAP (activated by high [cAMP] resulting from low [glucose]) >> determines rate of transcription if the operator is NOT blocked by the repressor Figure 17-15 lac operon Repressor Promoter INFREQUENT TRANSCRIPTION Glucose HIGH Lactose LOW CAP site Operator lacZ lacY lacA RNA polymerase bound loosely to promoter INFREQUENT TRANSCRIPTION Glucose HIGH Lactose HIGH CAP site Operator RNA polymerase bound loosely to promoter lacY lacA Inducer-repressor complex FREQUENT TRANSCRIPTION Glucose LOW Lactose HIGH lacZ CAP site Operator RNA polymerase bound tightly to promoter lacZ lacY lacA Figure 17-11-1 DNA FOOTPRINTING Radioactive atom DNA Repressor protein No repressor 1. Generate fragments 2. Divide fragments 3. Cut fragments with nuclease from the DNA region of interest, such as the lac operon of E. coli. Attach a label to end of fragments. into two samples. Add repressor protein to one sample. The repressor will bind to the operator. to produce fragments of different lengths. Repressor protects operator DNA from nuclease cleavage. Figure 17-11-2 DNA FOOTPRINTING “Footprint” No cuts occurred in the DNA region protected by the repressor. This region must be the operator. Largest fragments (cut far from label) 4. Load fragments into two lanes in a gel. Sort by size via electrophoresis. (The fragments with a label will be visible.) Smallest fragments (cut close to label) A DNA sequencing reaction can be used to determine the sequence of the “footprint.” You isolate an E. coli mutant that has high expression of the lacZ gene even in the absence of lactose (media has no glucose). You think you have a mutation in the lacI repressor, but this turns out not to be the case. Which of the following do you think best explains your mutant? A) B) C) D) Mutation in the lacZ gene that increases its activity. Mutation in the O1 binding site that reduces lacI binding. Mutation in the O1 binding site that increases lacI binding. Mutation inactivating the lacY gene. Summary of Prokaryotic Gene Regulation • Prokaryotic genes that code for enzymes in a specific metabolic pathway are clustered in groups and regulated together = operon • Lac operon discovered by Jacob & Monod • Key advantage: single ‘on-off’ switch to coordinate gene expression • Switch = operator that controls access of RNA polymerase to the promoter • Repressor protein binds to the operator • Repressor proteins are subject to regulation (positive or negative) by the metabolic substrates and products of the pathway.