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

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 DNA RNAProtein Replication Transcription Translation Protein* Post-translation Gene Expression & Information Flow

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

Jacques Monod found that  -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. E. coli produces high levels of  -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.

Genetic screening unmutagenized cells mutagenized cells Mutant that can’t utilize lactose as a carbon source 10 5 cells mutagen 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.

E. coli  -Galactosidase Galactoside permease Lactose Plasma membrane Glucose Galactose Two proteins are critical for E. coli to use lactose and one is critical for regulation of their expression Section of E. coli chromosome lacl product lacl  -Galactosidase lacZ product lacY product lacZlacY Galactoside permease Far away on the chromosome

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 RepressorRepressor binds the operator and blocks RNA polymerase RepressorRepressor is the product of a regulatory gene lacZlacY 5’3’ AUG UGAUAA

Figure 17-6a Negative control: Regulatory protein shuts down transcription No negative control… With negative control… RNA polymerase Regulatory protein TRANSCRIPTION Gene sequence No transcription

Figure 17-6b Positive control: Regulatory protein triggers transcription No transcription RNA polymerase No positive control… 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 present, lactose present: No repressor present, lactose present or absent: Transcription occurs. Repressor synthesized DNA lacl + RNA polymerase bound to promoter (blue DNA) lacZ lacY TRANSCRIPTION BEGINS  -Galactosidase Permease mRNA lacZ lacY RNA polymerase bound to promoter (blue DNA) Lactose-repressor complex Repressor synthesized No functional repressor synthesized mRNA TRANSCRIPTION BEGINS  -Galactosidase Permease lacZ lacY RNA polymerase bound to promoter (blue DNA) Lacl – Repressor binds to DNA. No transcription occurs. Lactose binds to repressor, causing it to release from DNA. Transcription occurs (lactose acts as inducer). Normal lacl gene Normal lacl gene lacl + Mutant lacl gene The repressor blocks transcription

When tryptophan is present, transcription is blocked. Repressor Tryptophan No transcription Operator RNA polymerase bound to promoter When tryptophan is absent, transcription occurs. RNA polymerase bound to promoter TRANSCRIPTION 5 genes coding for enzymes involved in tryptophan synthesis The Trp operon is also under negative control, but with a twist

Figure lac operontrp operon Catabolism Anabolism (breakdown of lactose) (synthesis of tryptophan) Repressor Lactose Tryptophan Lactose binds to repressor Tryptophan binds to repressor Lactose- repressor complex releases from operator Operator Tryptophan- repressor complex binds to operator No more transcription of trp operon Transcription of lac operon TRANSCRIPTION

Catabolite Repression Why doesn’t beta-galactosidase get induced in media containing Glucose?

When cAMP is present: When cAMP is absent: RNA polymerase bound tightly to promoter (blue DNA) RNA polymerase bound loosely to promoter (blue DNA) FREQUENT TRANSCRIPTION INFREQUENT TRANSCRIPTION cAMP CAP CAP site CAP CAP site Operator lacZ lacY lacZlacY lacA 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. CAP does not bind to DNA. RNA polymerase binds the promoter inefficiently. Transcription occurs rarely. CAP regulates lac operon positively and requires cAMP for DNA binding

Glucose inhibits the activity of the enzyme adenylyl cyclase, which catalyzes production of cAMP from ATP. The amount of cAMP and the rate of transcription of the lac operon are inversely related to the concentration of glucose. ATP Adenylyl cyclase Glucose inhibits this enzyme cAMP Two phosphate groups Infrequent transcription of lac operon (Cell continues to use glucose as energy source.) CAP does not bind to DNA CAP LOW cAMP INACTIVE adenylyl cyclase HIGH glucose concentration LOW glucose concentration ACTIVE adenylyl cyclase HIGH cAMP CAP CAP-cAMP complex binds to DNA Frequent transcription of lac operon (Cell uses lactose if lactose is present.) Cyclic AMP (cAMP) is synthesized when glucose levels are low

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 lac operon Promoter Repressor INFREQUENT TRANSCRIPTION CAP site CAP site CAP site FREQUENT TRANSCRIPTION Operator RNA polymerase bound loosely to promoter RNA polymerase bound loosely to promoter RNA polymerase bound tightly to promoter Glucose HIGH Glucose LOW Lactose LOW Lactose HIGH lacZ lacY lacA Inducer-repressor complex

Figure DNA FOOTPRINTING Radioactive atom DNA Repressor protein No repressor 1. Generate fragments from the DNA region of interest, such as the lac operon of E. coli. Attach a label to end of fragments. 2. Divide fragments into two samples. Add repressor protein to one sample. The repressor will bind to the operator. 3. Cut fragments with nuclease to produce fragments of different lengths. Repressor protects operator DNA from nuclease cleavage.

Figure 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) Smallest fragments (cut close to label) 4. Load fragments into two lanes in a gel. Sort by size via electrophoresis. (The fragments with a label will be visible.) 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)Mutation in the lacZ gene that increases its activity. B)Mutation in the O1 binding site that reduces lacI binding. C)Mutation in the O1 binding site that increases lacI binding. D)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.