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Transcript of 18.1
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MOLECULAR BIOLOGY Dr. Cynthia Haseltine
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June 5, 2014
What to do when you arrive Take a quarter sheet of paper
Todays class
Transcriptional control in prokaryotes
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TRANSCRIPTIONAL CONTROL IN PROKARYOTES
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CLASS ACTIVITY
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#1. A. Why is it important to control gene expression in
prokaryotes?
B. What are the advantages of controlling gene expression at transcription?
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Gene regulation Allows prokaryotic cells to:
To conserve energy
To respond to changes in the environment
Most prokaryotic control is at the level of transcription initiation Mediated by regulatory proteins
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Regulatory proteins Bind to DNA near sequences they regulate
Can have positive or negative effects
Activators - recruit
Repressors - block
Figure 18.1
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Regulatory proteins May have allosteric effects on RNAP Example: to allow
isomerization
May have allosteric effects on DNA Example: to position
promoter sites
Figure 18.2
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Distant interactions DNA looping allows proteins to interact
DNA-bending proteins facilitate looping
Figures 18.3 & 18.4
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OPERONS
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Diauxic growth of E. coli
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Genes of the lac Operon Genes are grouped:
lacZ = -galactosidase lacY = galactoside permease lacA = galactoside transacetylase
All 3 genes are transcribed together producing 1 mRNA, a
polycistronic message that starts from a single promoter Note: Each cistron, or gene, has its own ribosome binding site and can be transcribed by separate ribosomes that bind independently of each other
Figure 18.5
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The -galactosidase reaction
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CLASS ACTIVITY
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#2. When glucose is present, the lac genes are not fully expressed, even in the presence of inducer. This is called catabolite repression. (a) Why does it make biological sense to have the lactose operon under negative control by Lac repressor? Why does it make biological sense to have the lactose operon controlled by catabolite repression? (b) It is commonly stated that lactose induces the lac operon. However, allolactose, which is a product of basal galactosidase activity on lactose, is the actual inducer molecule. Devise an experiment to prove this.
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#2 (continued) (d) CAP is necessary to turn on several sugar operons (including the arabinose, lactose, maltose, and galactose operons). Cells with mutations in CAP cannot efficiently metabolize any of these sugars. On plates that contain a sugar and tetrazolium (an indicator dye), colonies are white if that sugar is metabolized and red if it is not. This kind of plate is often used to screen for cells that cannot metabolize a particular sugar. How could you use these plates to isolate CAP mutants? (e) You find that you obtain two classes of mutants with this screen. The first class of mutants is CAP mutants. What do you think the second class could be?
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Lac operon control Figure 18.6
Involves regulatory proteins:
Lac repressor
CAP
(activator)
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Control sequences
Lac repressor binds operator Blocks RNAP binding to promoter
CAP binds CAP site
Facilitates RNAP binding to promoter
Figure 18.8
How could you identify binding sites experimentally?
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CAP binding to DNA Lac promoter is weak
-35 is not optimal No UP-element
CAP binds CAP site and CTD of RNAP
Helps RNAP bind to promoter
Figures 18.9 & 18.10
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Helix-turn-helix Common DNA-binding motif
One -helix associates with major groove
One -helix associates with backbone
Found in CAP and lac repressor
Figure 18.11
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Lac repressor binding to DNA
Can bind DNA as a tetramer
To two of three possible operator sites
Requires DNA looping
Figures 8.7 & 18.12
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Allosteric regulation of repressor
Lac repressor regulated by allolactose
Acts as a lactose sensor
Prevents lac repressor from binding operator
Recall: What is allosteric
regulation?
Figure 18.13
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Allosteric regulation of activator
CAP regulated by cAMP
Acts as a glucose sensor (cAMP is high when glucose is low)
Allows CAP to bind to CAP site
Figure 18.14
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CLASS ACTIVITY
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#3. An operon in E. coli is controlled by a repressor that binds at two operator sites (O1 and O2). In the presence of the appropriate inducer, a transcription rate of 100 is observed, but in the absence of inducer, the transcription rate falls to 5. If either of the two sites is mutated so that the repressor cannot bind, then the transcription rate is observed to be 100. Additionally, if base pairs are inserted between the two sites, the level of transcription is found to vary with the size of the insert. Briefly explain this data.
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Molecular biology June 5, 2014Transcriptional control in prokaryotesClass activitySlide Number 5Gene regulationRegulatory proteinsRegulatory proteinsDistant interactionsoperonsDiauxic growth of E. coliGenes of the lac OperonThe b-galactosidase reactionClass activitySlide Number 15Slide Number 16Lac operon controlControl sequencesCAP binding to DNAHelix-turn-helixLac repressor binding to DNAAllosteric regulation of repressorAllosteric regulation of activatorClass activitySlide Number 25