8.1 - Transcription

Question 1

How is transcription carried out in eubacteria?

Answer 1

By RNA polymerases and sigma factors that guide it to specific promoters

Question 2

What is a core RNA polymerase?

Answer 2

An RNA polymerase enzyme without the sigma factor

Question 3

What does the core RNA polymerase enzyme consist of?

Answer 3

2 alpha (2α), 1 beta (β), 1 beta prime (β’) and 1 omega (ω) subunit

Question 4

What does the C-terminus of the alpha subunit interact with and why?

Answer 4

Regulators because it is flexible and can therefore reach out.

Question 5

What is a feature of the core enzyme?

Answer 5

It itself does not recognize any specific DNA sequence and thus has the ability to transcribe RNA from non-specific initiation sequences.

Question 6

How can we get the core enzyme to intiate transcription at specific promoters?

Answer 6

With the addition of a sigma factor

Question 7

What regions does the sigma factor bind to on promoters?

Answer 7

They recognize and bind to specific sequences on the promoter, these are the -35 and -10 regions.

Question 8

What characteristic allows sigma factors to recognize and bind DNA?

Answer 8

The fact that they are elongated

Question 9

What happens to the sigma factor once transcription starts?

Answer 9

It is ejected from the holoenzyme

Question 10

What makes a promoter strong?

Answer 10

Strong promoters have an A/T rich regions that will be recognized by the alpha subunit

Question 11

What can RNA polymerases also do in the case of strong promoters?

Answer 11

The C-terminus of the alpha subunit is also involved in sequence recognition in the case of very strong promoters.

Question 12

What is an example of a strong promoter?

Answer 12

The promoter of rRNA

Question 13

Name some features of RNA polymerase

Answer 13

(1) Proteins which bind TATA box

(2) Kinases to phosphorylate subunits

Question 14

What are the beta prime subunits for in bacterial RNA polymerase?

Answer 14

They are catalytic subunits and are where the RNA is synthesized

Question 15

How is transcription initiated?

Answer 15

RNA polymerase does random walk till it comes to strong sequence it recognizes (promoter). There is binding affinity and when docked on promoter it needs to unwind DNA and whole polymerase needs to go through conformational change. This requires activation energy. Reaction goes forward due to phosphates that are hydrolyzed. This happens at certain rate because you have to overcome certain activation energy

Question 16

What is the intrinsic promoter strength?

Answer 16

A reaction between RNA polymerase, its sigma factor and DNA. It does not take into consideration other regulating factors.

Question 17

What can the intrinsic promoter strength be broken down into?

Answer 17

(1) Strength of binding

(2) Strength of unwinding DNA

Question 18

What is strength of binding?

Answer 18

How strongly does RNA pol bind DNA. This is expressed by a binding constant (Kb) which is an equilibrium constant meaning that the reaction can go back and forth (binding and unbinding

Question 19

What is strength of unwinding DNA?

Answer 19

How efficiently can RNA pol I unwind DNA at the promoter to initiate transcription. This is expressed by a rate constant called isomerization constant (Kf). Kf is not an equilibrium constant because the reaction doesn’t go back and forth

Question 20

What does a strong promoter have in terms of binding and unwinding?

Answer 20

Weak binding constant (weak Kb) but a strong isomerization constant (strong Kf).

Question 21

What do the strongest promoters have?

Answer 21

Both strong Kb and strong Kf.

Question 22

What influences the isomeration constant (Kf)?

Answer 22

The sequence of the TATA box.

Question 23

What is the formula for promoter strength?

Answer 23

promoter strength = binding x opening rate constant

Question 24

What does RNA polymerase not have when it enters the elongation step?

Answer 24

Sigma factor because it is ejected from the holoenzyme once transcription starts

Question 25

What are the 2 elongation factors that bind to RNA polymerase?

Answer 25

(1) NusA

(2) NusG

Question 26

What does NusG do?

Answer 26

NusG forms a ring around the DNA that helps the RNA polymerase to continue transcribing without falling off; so like a processivity-promoting factor (analogous to clamp in DNA replication).

Question 27

What does NusA do?

Answer 27

NusA will help eliminate secondary structures of RNA during transcription.

Question 28

Where does RNA polymerase stop transcribing?

Answer 28

At sequences called transcription terminators

Question 29

What are the two termination mechanisms in E.coli and most bacteria?

Answer 29

(1) Rho dependent termination

(2) Rho independent termination

Question 30

What does the Rho protein bind to?

Answer 30

A specific RNA sequence called transcription termination pause site

Question 31

What are transcription termination pause sites?

Answer 31

A G-C rich region on the RNA that is being synthesized

Question 32

What happens when Rho recognizes and binds to the termination pause site?

Answer 32

It will use energy from ATP hydrolysis to translocate along the RNA until it reaches the DNA-RNA helical region (3’ end of the RNA being synthesized). - On its way to the DNA-RNA helical region it will unwind the RNA from the DNA template.

Question 33

What family of proteins does Rho belong to?

Answer 33

Rho is a RNA-DNA helicase (uses ATP), it is part of the AAA+ family.

Question 34

What causes termination in rho dependent termination?

Answer 34

The contact between RNA polymerase and Rho

Question 35

How is a stem loop structure formed in rho independent termination?

Answer 35

The mRNA contains a G-C rich sequence that can base pair with itself to form a stem-loop structure (due to inverted repeats region) of 7-20bp in length.

Question 36

What is an important properties of G-C bonds?

Answer 36

C-G have significant base-stacking interactions and can form 3 hydrogen bonds between each other, resulting in a stable RNA duplex.

Question 37

What are base stacking interactions?

Answer 37

Interactions between adjacent base – pairs, very important in stability.

Question 38

What follows the stem loop structure in rho independent termination?

Answer 38

There is a sequence of uracil residues. Bonds between uracil and adenine are very weak. NusA protein, which is bound to RNA polymerase, will bind to the G-C stem loop tightly enough causing the RNA polymerase to stall when transcribing the Uracil rich region following the stem loop.

Question 39

What do the weak adenine-uracil bonds do?

Answer 39

They will lower the energy of destabilization required for the RNA-DNA duplex unwinding. As a result, RNA will unwind from the template and dissociate from the RNA polymerase