18.1

MOLECULAR BIOLOGY Dr. Cynthia Haseltine

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

CLASS ACTIVITY

#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

Diauxic growth of E. coli

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

The -galactosidase reaction

CLASS ACTIVITY

#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

#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