Biol 405: Lecture 19 (11/9/15)

Front 1

Describe the experiment

Back 1

RNA polymerase and LacR were allowed to incubate for 10 minutes allowing for the open promoter complex to form (initiation). Heparin was added to prevent any new initiation along with ATP, UTP, GTP and other buffer components. Five minutes later, 32P-labeled CTP was added with and without IPTG. 10 minutes passed and the results were ran on a gel and exposed to x-ray film

Front 2

Column A represents which of the following scenarios?

A. LacR- and IPTG+

B. LacR+ and IPTG-

C. Lacr+ and IPTG+

D. Lacr- and IPTG-

Back 2

D. Lacr- and IPTG-

Front 3

Column B represents which of the following scenarios?

A. LacR- and IPTG+

B. LacR+ and IPTG-

C. Lacr+ and IPTG+

D. Lacr- and IPTG-

Back 3

B. LacR+ and IPTG-

Front 4

Column B represents which of the following scenarios?

A. LacR- and IPTG+

B. LacR+ and IPTG-

C. Lacr+ and IPTG+

D. Lacr- and IPTG-

Back 4

C. LacR+ and IPTG+

Front 5

What conclusion was drawn from the experiment performed?

A. LacR and RNA polymerase are mutually exclusive

B. Transcription initiation occurs in the presence of LacR

C. Transcription elongation occurs in the presence of LacR

D. Promoter clearance occurs in the presence of LacR

Back 5

B. Transcription initiation occurs in the presence of LacR

Front 6

T/F: Herparin is a polyanion that binds ot any free RNA polymerase which prevents initiation events

Back 6

True

Front 7

Column B represents LacR+ and IPTG-. Explain why you would see no run off in this column

Back 7

LacR is bound to the DNA and since there is no IPTG then transcription cannot finish and so you would see no run-off.

Front 8

The conclusion based on the experiment was that an open promoter complex was formed even in the presence of LacR. What would happen if the Open Promoter Complex was not formed? How would this affect the column?

Back 8

If the open promoter complex was not formed than transcription initiation did not occur. Thus, Column C would look like Column B.

Front 9

Curve "A" represents which of the following

A. Heparin+

B. Repressor+

C. No DNA

D. No Additions

Back 9

D. No Additions

Front 10

Curve "B" represents which of the following:

A. Heparin+

B. Repressor+

C. No DNA

D. No Additions

Back 10

A. Heparin+

Front 11

Curve "C" represents which fo the following:

A. Heparin+

B. Repressor+

C. No DNA

D. No Additions

Back 11

B. Repressor+

Front 12

What conclusions were made based on the experiment?

Back 12

LacR behaves like heparin meaning that both blocks the reassociation of RNA polymerase onto the DNA

Front 13

T/F: Heparin and LacR blocks RNA Polymerase reassociating to the DNA through the same mechanisms

Back 13

False. While both block RNA polymerase reassociation they do so by ifferent mechanisms.

Front 14

T/F: The Lac Operon contains 3 operators

Back 14

True

Front 15

Where is Operator1 of the Lac Operon located?

A. In the Untranscribed region

B. The promoter region

C. In the Coding Region

D. None of the above

Back 15

B. The Promoter Region

Front 16

Where is Operator2 of the Lac Operon located?

A. In the Untranscribed region

B. The promoter region

C. In the Coding Region

D. None of the above

Back 16

C. In the Coding Region

Front 17

Where is Operator3 of the Lac Operon located?

A. In the Untranscribed region

B. The promoter region

C. In the Coding Region

D. None of the above

Back 17

A. In The Untranscribed Region

Front 18

What is the negative control of the Lac Operon?

A. High Lactose, High Glucose

B. High Lactose, Low Glucose

C. Low Lactose, High Glucose

D. Low Lactose, Low Glucose

Back 18

A. High Lactose, High Glucose

Front 19

What is the positive control of the Lac Operon?

A. High Lactose, High Glucose

B. High Lactose, Low Glucose

C. Low Lactose, High Glucose

D. Low Lactose, Low Glucose

Back 19

B. High Lactose, Low Glucose

Front 20

What main problem is faced when Glucose and Lactose are high in concentration?

Back 20

When glucose levels are high, there is no need to make more glucose. However, since Lactose levels are also high, LacR will not repress Lac transcription and so because of this Glucose will be produced from the hydrolization of lactose to glucose by Beta-galacosidase (product of Lac Transcription)

Front 21

CAP binds to the promoter region of lac during what situation?

A. High Lactose; High Glucose

B. High Lactose; Low Glucose

C. Low Lactose; Low Glucose

D. Low Lactose; High Glucose

Back 21

A. High Lactose; High Glucose

Front 22

T/F: Lac mRNA is abundant when CAP is bound to the promoter region

Back 22

False. When CAP is bound to the promoter region, Lac mRNA trascription is low

Front 23

When glucose levels are high, there is no need to make more glucose. However, since Lactose levels are also high, LacR will not repress Lac transcription and so because of this Glucose will be produced from the hydrolization of lactose to glucose by Beta-galacosidase (product of Lac Transcription). How is this avoided?

Back 23

CAP binds to the promoter region when lactose and glucose concentrations are high. This severly reduces lac mRNA.

Front 24

Which of the following is NOT true?

A. CAP binds to the promoter region of the lac DNA during high Lactose and high Glucose concentration resulting in very little mRNA

B. cAMP binds to CAP when concentrations of Lactose are high and concentrations of glucose are low resulting in the abundance of Lac mRNA

C. cAMP mediates the response of cells tog lucose levels by working with CAP encoed by the CRP gene

D. The LacR repressor acts very similarly to Heparin preventing reassociation of RNA polymerase to the DNA.

E. None of the above

Back 24

E. None of the above

Front 25

CAP is encoded by what gene?

Back 25

CRP (Cyclic-AMP Receptor Protein)

Front 26

Which of the following is NOT true based on the experiment provided?

A. Mutant CAP bound 10 time liess tightly to DNA than Wild Type CAP

B. Wild Type CAP produced greater concentrations of Beta-galactosidase than the mutant.

C. Beta-Galactosidase synthesis could be stimulated with the addition of Wildtype CAP.

D. None of the above

Back 26

D. None of the above

Front 27

T/F: CAP-cAMP directly promotes formation of an open promoter complex.

Back 27

False. Although it appears that CAP-cAMP promotes formation of an open promoter complex, it in fact stimulates the formation of the close promoter complex which drives towards the open promoter complex.

Front 28

T/F: Rifampicin blocks elongation by blocking the formation of phosphodiester bonds in RNA synthesis.

Back 28

False. Rifampicin blocks initiation by blocking the formation of the first phosphodiester bond in RNA synthesis. It does not affect elongation.

Front 29

T/F: Rifampicin blocks weak promoter/RNAP interaction as in an open promoter complex, but not strong interaction as in a closed promoter complex

Back 29

False. Rifampicin blocks weak promoter/RNAP interaction as in a closed promoter complex, but not strong interaction as in an open promoter complex.

Front 30

The image depicts the 1st mechanism of positive regulation by CAP-cAMP. Which of the following parts of the mechanism does CAP-cAMP directly effect?

A. Kb

B. [RPc]

C. K2

D. [RPo]

Back 30

A. Kb

Front 31

Which of the following parts of the mechanism represents the constant for the open promoter complex?

A. Kb

B. [RPc]

C. K2

D. [RPo]

Back 31

D. [RPo]

Front 32

Which of the following parts of the mechanism represents the limiting step?

A. Kb

B. [RPc]

C. K2

D. [RPo]

Back 32

A. Kb

Front 33

Describe the 2nd mechanism of positive regulation by CAP-cAMP.

Back 33

The Lac operon has 2 promoters. Promoter 1 is the major efficient promtoer while promoter 2 is inefficient. However, both are equally able to bind to the RNAP. This poses a problem, if both are equally able to bind to RNAP than there is a chance that RNAP will bind to promoter 2 which will produce a minimal amount of product for the same amount of resources as that of promoter 1. This is solved by the introduction of CAP-cAMP which blocks association of the RNAP to promoter 2 thus encouraging the association of RNAP to promoter 1. As a result, more RNAPs become available for transcription from promoter 1

Front 34

How many promoter regions does the Lac Operon contain?

Back 34

2

Front 35

How much evidence can be said to support the first mechanisms of positive regualtion by CAP-cAMP?

Back 35

6

Front 36

What six evidence was used to support the first mechanism of positive regulation of CAP-cAMP?

Back 36

1. Cosediemnt of the two RNAP complexes in the presence of the cAMP

2. Can be chemically cross-linked to each other when bound to the cis-acting element.

3. Foot-printing and genetic data

4. CAP mutations that affect interaction with RNAP decrease activations

5. Deleting the carboxyl terminal domain of the alpha-subunit (AlphaCTP) of RNAP prevents activation by CAP-cAMP

6. The X-RAy crystallography data indicate that the ARI of CAP and the alphaCTP of RNAP do touch each other

Front 37

As stated by the third line of evidence, foot-printing and genetic data support of the first mechanism of positive regulation by CAP-cAMP. How exactly does foot-printing and genetic data support the first mechanism?

Back 37

A genetic chemical study was performed where one of the cis-acting elements (binding site) was defined by footprinting experiments as the CAP binding site. When they mutated the L1 they prevented stimulation of transcription by CAP and cAMP, but it still allowed a low level of transcription. They found that the CAP binding site was made of TGTGA and since the -35 box of the lac operon was different from other strong promoters and did not contain a UP-element, it required the help of CAP-cAMP for high level transcription. Therefore, without the CAP binding site, CAP-cAMP cannot bind to the DNA resulting in only low level transcription. This is how transcription is regulated.

Front 38

As stated by the fourth line of evidence, CAP mutations that affect interaction with RNAP decreases activation, support the first mechanism of positive regulation by CAP-cAMP. How exactly does the fourth line of evidence support the first mechanism?

Back 38

The C-terminal domain of the alpha subunit of the RNAP interacts with CAP-CAMP via the Interaction Site of the Activation Region I (ARI). If the ARI is mutated, CAP and RNAP cannot interact.

Front 39

T/F: CAP stimulates the transcription of over 100 promoters

Back 39

True

Front 40

The following table helped to support which of the following lines of evidence for the 1st mechanism of positive regulation of CAP-cAMP?

A. Evidence 1

B. Evidence 2

C. Evidence 3

D. Evidence 4

E. Evidence 5

F. Evidence 6

Back 40

E. Evidence 5

Front 41

Describe the experiment performed to obtain the data found on the table

Back 41

They took two promoters, P1 and UV5, and placed them under two conditions, with cAMP-CAP and without cAMP-CAP. They used three RNAP enzymes to run the DNA. One was the wildtype (Alpha-WT), The second contained a deletion from the C-terminal end from 257-329aa (Alpha-256), and the third contained a deletion from 236-329 aa (Alpha-235). They ran the enzymes under both conditions. They found that for the wildtype, transcription increased for the P1 promoter when cAMP-CAP was present while the transcription for UV5 did not change. This indicates that UV5 is a very strong promoter, while P1 is not and thus requires the presence of cAMP-CAP to enhance transcription. When the alpha-256 and the alpha-235 mutant enzymes were used as the RNAP. they found only a slight increase in transcription and even a decrease in transcription. This indicates that the c-terminal domain is not necessary for reconstituting an active RNA polymerase (since transcripts were produced) but it was necessary for stimulation by CAP-cAMP.

Front 42

Which of the following is NOT true in regards to the table provided?

A. The wildtype RNAP yielded the highest transcription product for P1 due to the presence of the cAMP-CAP

B. P1 showed consistently low transcription (lower than UV5) due to a lack of the C-terminal domain which is essential for high transcription

C. C-terminal deletion is not necessary for reconsistuting an active RNAP but it is necessary for stimulation by CAP-cAMP.

D. By removing both cAMP-CAP and the C-terminal domain you can substantially decrease transcription

Back 42

B. P1 showed consistently low transcription (lower than UV5) due to a deletion of the C-terminal domain which is essential for high transcription

Front 43

Determine the activation fold for the second row (alpha-256)

Back 43

Taking the first two columns:

P1=53

UV5= 766

53/766=0.069

0.069*100=6.9

% of p1/UV5 without cAMP-CAP=6.9

Taking the next two columns:

P1=62

UV5=723

62/723=0.086

0.086*100=8.6

% of P1/UV5 without cAMP-CAP=8.6

Taking the two percentages

8.6/6.9=1.24

Activation = 1.2

Front 44

T/F: Nuclear run-off was used as part of the experiment

Back 44

True

Front 45

What conclusions were drawn as a result of the experiment

Back 45

The C-Terminal domain of the alpha subunit is not necessary for reconstituting an active RNA polymerase, but it is necessary for stimulation by CAP-cAMP.

Front 46

Based on previous experiments RNAP alone is able to produce high level transcription so long as the promoter is strong. If, however, the promoter is not strong what two additional units may be used to enhance transcription?

Back 46

1. Activator binding site capable of binding CAP proteins

2. UP element.

Front 47

If the C-Terminal of the alpha-subunit is deleted which of the following will likely occur?

A. CAP-cAMP dimerization will stabilize the RNA polymerase resulting in high level transcription

B. The N-terminal end of the Alpha-subunit will bind to the UP element resulting in high level transcription

C. The ARI will interact with the N-terminal end of the alpha-subunit resulting in high level transcription

D. Basal level transcription will be achieved so long as the promoter is not a strong promoter.

Back 47

D. Basal level transcription will be achieved so long as the promoter is not a strong promoter.

Front 48

Describe Fig 7.18

Back 48

DNA that are bent move much slower through gel electrophoresis than DNA that is straight. Further more the more bent the DNA, the slower it is. DNA that is bent in the middle will move slower than DNA bent at the end. Using this information, Wu and other prepared DNA fragments of the same length with CAP-binding sites at different position. If the CAP binding caused a bend in the DNA, than they would move at different rates through gel electrophoresis. When running the gels they found that Cap binding did indeed induce DNA bending and furthermore they found that the more pronounced the bend, the greater the difference in migration meaning that those at the lowest mobility were in fact bending the DNA at 90 degrees.

Front 49

T/F: Lac Operon does not contain a UP element

Back 49

True

Front 50

Which of the following will occur in regards to the Lac Operon when Glucose is present but lactose is not?

A. Repressor will bind to the operator resulting in no transcription

B. The repressor will not bind. CAP will not bind. Results in basal level transcription

C. Repressor will not bind. cAMP-CAP binds to the activator site. Results in high level transcription

Back 50

A. Repressor will bind to the operator resulting in no transcription

Front 51

Which of the following will occur in regards to the Lac Operon when Glucose is present and lactose is present?

A. Repressor will bind to the operator resulting in no transcription

B. The repressor will not bind. CAP will not bind. Results in basal level transcription

C. Repressor will not bind. cAMP-CAP binds to the activator site. Results in high level transcription

Back 51

B. The repressor will not bind. CAP will not bind. Results in basal level transcription

Front 52

Which of the following will occur in regards to the Lac Operon when Glucose is not present and lactose is present?

A. Repressor will bind to the operator resulting in no transcription

B. The repressor will not bind. CAP will not bind. Results in basal level transcription

C. Repressor will not bind. cAMP-CAP binds to the activator site. Results in high level transcription

Back 52

C. Repressor will not bind. cAMP-CAP binds to the activator site. Results in high level transcription

Front 53

Explain how glucose regulates cAMP levels in cells

Back 53

Adenylate cyclase is needed to convert ATP to cAMP. When glucose is present it indirectly blocks the formation of adenylate cyclase. This then prevents cAMP from formining.

Front 54

Which of the following is NOT true in regards to the lac operon?

A. The lac operon is controlled by two sugar sensors (one for lactose via the repressor, the other for glucose via the regulation of adenylate cyclase activity)

B. The lac operon is controlled by two transcription factors (LacR and CAP) plus the RNA polymerase

C. The Lac operon is controlled by three groups of Cis-acting elements (CAP binidng sequence, RNAP binding sequence, and operator)

D. The Lac Operon is controlled by Two small molecules (Allolactose and cAMP)

E. None of the above

Back 54

E. None of the above

Front 55

What two molecules control the lac operon?

Back 55

Allolactose and cAMP

Front 56

What three groups of cis-acting elements control the lac operon?

Back 56

CAP binding sequence, RNAP binidng elements, and operators

Front 57

What two transcription factors including the RNAP controlles the lac operon?

Back 57

LacR and CAP

Front 58

The image represent sthe lac operon regulatory network however some parts are missing. Match the following components to the correct number parts shown on the image.

A. Operator

B. CAP-cAMP Complex

C. CAP BS

D. Adenylate Cylase

E. Beta-galactosidase

Back 58

1=E. Beta-Galactosidase

2=D. Adenylate Cyclase

3=B. CAP-cAMP Complex

4=C. CAP BS

5=A. Operator