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Goals of today s lecture

Gene Expression. Regulating the amount of active gene product (protein) by:Control of gene expression at the level of transcriptionControl of at the level of translationPost-translational controlConstitutive expression = gene product made continuouslyRegulated expression = gene product made on

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Goals of today s lecture

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    1. Proks and euks regulate gene expression differently Why not have all genes in a cell expressed at the same time? Because there are many different genes (liver cell genes, heart cell genes, etc.) (if e.coli had alllll of its genes expressed at the same time, it would take too much energy to have all the genes just sitting there waiting to be used. It needs diffferent genes at different times. Sooooo genes have to be regulated.) difference between proks and euks on how they do this.Proks and euks regulate gene expression differently Why not have all genes in a cell expressed at the same time? Because there are many different genes (liver cell genes, heart cell genes, etc.) (if e.coli had alllll of its genes expressed at the same time, it would take too much energy to have all the genes just sitting there waiting to be used. It needs diffferent genes at different times. Sooooo genes have to be regulated.) difference between proks and euks on how they do this.

    2. 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 Induced- turned on. Repressed- turned off. Constitutive expression this refers to certain genes that are ALWAYS being expressedInduced- turned on. Repressed- turned off. Constitutive expression this refers to certain genes that are ALWAYS being expressed

    3. 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 Post-translation- where protein can be activated or repressedPost-translation- where protein can be activated or repressed

    4. Regulation of Gene Expression in Prokaryotes

    5. 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 Glucose is preferred source cuz we have GLYCOLYSIS that is designed already to break down glucose. Energetically less expensive for cell to use glucose as energy source. So when it is given lactose, beta-galactosidase can break this down into galactose and glucose for further breakdown. Galactose can be turned into glucose with the use of 1 ATPGlucose is preferred source cuz we have GLYCOLYSIS that is designed already to break down glucose. Energetically less expensive for cell to use glucose as energy source. So when it is given lactose, beta-galactosidase can break this down into galactose and glucose for further breakdown. Galactose can be turned into glucose with the use of 1 ATP

    6. Figure 17-2b-setup

    7. Figure 17-2c-results Treatment 1e.coli produces VERY LITTLE B-GALACTOSIDASE (book says NONE. Dr. cad says VERY LITTLE) The presence of lactose can induce the production of high production of b-galactosidase. And glucose blocks the production of b-galact.Treatment 1e.coli produces VERY LITTLE B-GALACTOSIDASE (book says NONE. Dr. cad says VERY LITTLE) The presence of lactose can induce the production of high production of b-galactosidase. And glucose blocks the production of b-galact.

    8. 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. Top bulletthe result of the experiment.Top bulletthe result of the experiment.

    9. The bunch of circles on left are 100,000 cells. Then you add a mutagen so that (on right) all the cells are mutagenized differently! And you are hoping that at least ONE cell in that group is a mutant that cant utilize lactose as a carbon source. Cuz thats what you wanna find! So how do you find that ONE CEL? Next pageThe bunch of circles on left are 100,000 cells. Then you add a mutagen so that (on right) all the cells are mutagenized differently! And you are hoping that at least ONE cell in that group is a mutant that cant utilize lactose as a carbon source. Cuz thats what you wanna find! So how do you find that ONE CEL? Next page

    10. Replica plating allows the identification of genes that are essential to utilize lactose So you want to figure out which colonies can grow on lactose. The ones that grow on lactose are able to metabolize lactose and you want to find which ones those are. When you use the block, you can tell which colonies those are because they are going to be in the same spot, because you just press the block down and lift it up and press it down in the same spot on a lactose replica plate. The lactose replica plate did NOT have any colonies on it prior to pressing down the block. So you want to figure out which colonies can grow on lactose. The ones that grow on lactose are able to metabolize lactose and you want to find which ones those are. When you use the block, you can tell which colonies those are because they are going to be in the same spot, because you just press the block down and lift it up and press it down in the same spot on a lactose replica plate. The lactose replica plate did NOT have any colonies on it prior to pressing down the block.

    11. In lacI beta-galac. Is on all the time. In lacI beta-galac. Is on all the time.

    12. Lac Z codes for B-galact. which breaks down lactose Lac Y codes for galactoside permease which allows the entrance of lactose into the cell ^^^these two are found VERY CLOSE to each other on the chromosome. Lac I codes for a Lac repressor. ^^far away from others. The operator is found in-between LAC I and lac Z and so lac I is always on and lac Z can be blocked.Lac Z codes for B-galact. which breaks down lactose Lac Y codes for galactoside permease which allows the entrance of lactose into the cell ^^^these two are found VERY CLOSE to each other on the chromosome. Lac I codes for a Lac repressor. ^^far away from others. The operator is found in-between LAC I and lac Z and so lac I is always on and lac Z can be blocked.

    13. 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 These two are found on ONE SINGLE molecule of RNA. Although there are two start & stop codonsThese two are found on ONE SINGLE molecule of RNA. Although there are two start & stop codons

    14. Figure 17-6a Negative control- polymerase can not transcribe cuz regulatory protein binds in front of it and blocks itNegative control- polymerase can not transcribe cuz regulatory protein binds in front of it and blocks it

    15. Figure 17-6b Positive control opposite. Promotors can undergo positive or negative controlPositive control opposite. Promotors can undergo positive or negative control

    16. 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).

    17. Lac I is always on. The repressor is always on. In the absence of lactose, the repressor binds to the promotor and it helps to keep lacZ and lacY off! Cuz they are not needed. When there is lactose, it binds to the repressor, so that the repressor cannot be binded to the DNA so lacZ and lacY can be transcribed. If a mutant lacI gene, it doesnt bind to dna and transcription occurs anywaybad.Lac I is always on. The repressor is always on. In the absence of lactose, the repressor binds to the promotor and it helps to keep lacZ and lacY off! Cuz they are not needed. When there is lactose, it binds to the repressor, so that the repressor cannot be binded to the DNA so lacZ and lacY can be transcribed. If a mutant lacI gene, it doesnt bind to dna and transcription occurs anywaybad.

    19. Figure 17-10 So difference between the two because--- When the operon is present, it binds to the repressor in both. But in lactose, it causes the release from the operator. And in trp, it causes the complex to bind to the operator. Because you already have trp. So you dont want more.So difference between the two because--- When the operon is present, it binds to the repressor in both. But in lactose, it causes the release from the operator. And in trp, it causes the complex to bind to the operator. Because you already have trp. So you dont want more.

    20. Catabolite Repression Glucose can block the production of b-galac. Glucose can block the production of b-galac.

    21. CAP regulates lac operon positively and requires cAMP for DNA binding High levels of glucose---NO CAMP OR CAP! If lactose, the repressor releases and binds to lactose. CAP allows RNA pol. To bind really good to the DNA and then, if glucose is presentno cap. And not good binding to DNA and not very often transcription. cAMP and CAP make transcription more efficient and this can only happen if NO glucose. Makes sense cuz you really need the transcription of lacZ and lacY because then you have Adenyl cyclase is the molecule that lets glucose in and if no glucose, it changes the structure of ATP and dephosphorylates ATP into cyclic-AMP.High levels of glucose---NO CAMP OR CAP! If lactose, the repressor releases and binds to lactose. CAP allows RNA pol. To bind really good to the DNA and then, if glucose is presentno cap. And not good binding to DNA and not very often transcription. cAMP and CAP make transcription more efficient and this can only happen if NO glucose. Makes sense cuz you really need the transcription of lacZ and lacY because then you have Adenyl cyclase is the molecule that lets glucose in and if no glucose, it changes the structure of ATP and dephosphorylates ATP into cyclic-AMP.

    22. Remember what the cell wants! In the presence of high glucose, you have low cAMP. Which is what you want. If glucose level is low, you get high cAMP. And cyclic AMP is needed for CAP. And CAP pushes the synthesis of the lac operon.Remember what the cell wants! In the presence of high glucose, you have low cAMP. Which is what you want. If glucose level is low, you get high cAMP. And cyclic AMP is needed for CAP. And CAP pushes the synthesis of the lac operon.

    23. 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

    24. Figure 17-15

    25. Figure 17-11-1 A piece of dna corresponding with lac operon and you label it with radioactivity. Next step: high circle = protein added low cirlce= free dna. Next step- the nuclease digestion involves only ONE cut per molecule and the cut were at random. If you take those pieces and run it on the gel you see the next page. And because there are all different sizes of the dna, its called dna laddering. A piece of dna corresponding with lac operon and you label it with radioactivity. Next step: high circle = protein added low cirlce= free dna. Next step- the nuclease digestion involves only ONE cut per molecule and the cut were at random. If you take those pieces and run it on the gel you see the next page. And because there are all different sizes of the dna, its called dna laddering.

    26. Figure 17-11-2 The lower one is the one without the represor protein because there are not places where the nuclease is not able to cut the dna. The higher one has a section without any cuts cuz the repressor protein protected the dna from being cut there. Called a fotoprint They found the exact sequence for the operator thru this experiment. The lower one is the one without the represor protein because there are not places where the nuclease is not able to cut the dna. The higher one has a section without any cuts cuz the repressor protein protected the dna from being cut there. Called a fotoprint They found the exact sequence for the operator thru this experiment.

    27. Know that this sequence isa mirror-image symmetry.Know that this sequence isa mirror-image symmetry.

    28. A motif is part of a protein. A- the part that binds to dna of a protein. (shown in c) A- there is a long a-helix, followed by a sharp turn, and then a shorter a-helix (helix-turn-helix) This helix-turn-helix is part of the repressor. The repressor has four equal parts and each part has the helix-turn-helix.A motif is part of a protein. A- the part that binds to dna of a protein. (shown in c) A- there is a long a-helix, followed by a sharp turn, and then a shorter a-helix (helix-turn-helix) This helix-turn-helix is part of the repressor. The repressor has four equal parts and each part has the helix-turn-helix.

    30. When lac repressor is bound, polymerase cannot possibly transcribe. Ok so when the repressor binds, it makes the DNA form a loop, so the rna pol. Cannot transcribe cuz its physically blocked.When lac repressor is bound, polymerase cannot possibly transcribe. Ok so when the repressor binds, it makes the DNA form a loop, so the rna pol. Cannot transcribe cuz its physically blocked.

    31. Lactose = inducer. Lactose = inducer.

    32. I think this is a Good example of an exam-type question that I might ask on.. On the exam.Dr. Cadigan. Answer- B Because the operator codes for the repressor, so if the operator is effed up, the repressor cannot bind.I think this is a Good example of an exam-type question that I might ask on.. On the exam.Dr. Cadigan. Answer- B Because the operator codes for the repressor, so if the operator is effed up, the repressor cannot bind.

    33. 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.

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