Prokaryotic Transcription

Question 1

Describe biochemistry of prokaryotic transcription

Answer 1

Incoming phosphates are named alpha, beta, and gamma (beta and gamma leave after attachment

RNA builds 5’-3’, DNA template is 3’-5’

You can replace gamma or alpha phosphate with a labeled phosphate

Question 2

Promoter Consensus Sequence

Answer 2

-35 and -10. they are named in reference to their location in relation to the start site. -35 is higher in C-G, whereas -10 is high in A-T which makes it easy for RNA polymerase to separate DNA strand

Question 3

Footprinting

Answer 3

Identifies DNA-binding sites for protein by their protection against nicking

Question 4

RNA Polymerase composition

Answer 4

Holoenzyme: 2-alpha, 2-beta, 1-sigma
Core: 2-alpha, 2-beta

Question 5

Early transcription events

Answer 5

1. form the closed promoter complex, core plus sigma makes holoenzyme
2. form the open promoter complex
3. incorporate 1st few nucleotides
4. promoter clearance, sigma leaves
5. elongation

Question 6

Properties of Sigma

Answer 6

1. Confers promoter specificity
2. Stimulates tight DNA binding
3. Stimulates initiation
4. Can be reused

Question 7

sigma-factor tread-milling

Answer 7

Once the 1st 10 bases are transcribed, sigma releases and attaches to new core to create new holoenzyme. Leads to multiple RNA polymerases transcribing at once. Produces christmas tree effect

Question 8

Torsional stress

Answer 8

Gyrase un-winds DNA strand at leading edge before going into polymerase.
Topoisomerase re-winds DNA strand following transcription.

Question 9

Transcription termination

Answer 9

Rho independent

Rho dependent

Question 10

Rho independent transcription termination

Answer 10

A sequence of inverted repeats (C-G rich) leading up to a multi-T sequence causes formation of a hairpin which leads to de-stabilization of the enzyme interaction

Question 11

Rho dependent transcription termination

Answer 11

1. Decreases net rate of RNA synthesis
2. Reduces size of RNA products
3. Releases RNA products from DNA
   Rho binds to transcript at rho loading site, pursues polymerase, hairpin forms, polymerase pauses, rho catches up, rho releases transcripts and causes termination

Question 12

Common gene regulation themes

Answer 12

1. Selective usage of promoters is mediated by the binding of protein to specific DNA sequences( - proteins turn genes off, + proteins turn genes on)
2. Promoter binding influences promoter utilization by:
   a. Blocking the binding or movement of other proteins
   b. Promoting the binding of other proteins
   c. Interacting with basal transcription apparatus
3. Interaction of regulatory proteins with DNA must be reversible and accessible to regulation
4. Coordinated regulation of gene expression can be accomplished by propagation of common binding sites for regulatory proteins

Question 13

General features of regulated gene expression unique to prokaryotes

Answer 13

1. Transcription and translation are coupled
2. Genes can be arranged in operons
3. The half life of mRNA is normally on the order of minutes

Question 14

General features of regulated gene expression unique to eukaryotes

Answer 14

1. Gene accessibility - chromatin
2. Transcript Transport/Localization
3. Multicellularity

Question 15

Lactose utilization by E. coli

Answer 15

Lactose is brought in through a permease, beta-galactosidase converts to galactose and glucose (fuel for cell)

Question 16

Lac operon organization

Answer 16

6000bp total

1. Promoter
2. lacI - Repressor, 1040bp
3. Operator, 82bp
4. lacZ - beta-galactosidase, 3510bp
5. lacY - Permease, 780bp
6. lacA - Transacetylace, 825bp

Question 17

Lac operon description

Answer 17

1. lacI makes mRNA that makes repressor protein
2. Repressor binds to operator and prevents transcription of z, y, and a
3. When inducer (lactose) is added it binds to repressor protein, preventing it from binding to operator
4. Absence of active repressor allows RNA polymerase to access promoter and transcribe mRNA for z, y, and a proteins

Question 18

PaJaMo experiment

Answer 18

Addition of lacI and lacZ wild type genes to lacI and lacZ deficient mutants stimulates synthesis of beta-galactosidase for a short time, but lacI eventually represses synthesis. When inducer is added to same model, beta-galactosidase continues to increase.

Question 19

Problems with Lac Operon Regulation

Answer 19

1. Formation of allolactose (inducer) requires β-galactosidase activity
2. Entry of lactose into the cell requires the lac permase
3. The binding of regulatory proteins to DNA is not infinitely strong
4. Coordinate regulation with glucose availability
5. Transcription of the Lac operon is inhibited when glucose is present
6. Catabolite Activator Protein is a positive acting regulatory protein

Question 20

CAP

Answer 20

Catabolite Activator Protein is a positive acting regulatory protein

Question 21

Describe Lac Operon Regulation

Answer 21

1. Glucose present (cAMP low); no lactose
   a. lacI produces repressor which binds to operator
   b. CAP does not bind
   c. Gene off, no mRNA
2. Glucose present (cAMP low); lactose present
   a. lacI produces repressor, lactose (inducer) binds to repressor, repressor does not bind to operator
   b. CAP does not bind
   c. Minimal mRNA
3. No glucose present (cAMP high); lactose present
   a. lacI produces repressor, lactose (inducer) binds to repressor, repressor does not bind to operator
   b. CAP binds to promoter, recruits RNA polymerase
   c. Abundant mRNA