prokaryotic transcription

Question 1

promoter =

Answer 1

sequence of DNA that initiates transcription of a particular gene

first base to be copied into mRNA is the last base in the promoter sequence

Question 2

terminator =

Answer 2

sequence of DNA that marks the end of agene/operon

last base to be copied into mRNA is the last base in the terminator sequence

Question 3

RNA polymerase =

Answer 3

enzyme which uses a single-stranded DNA template to synthesise a complementary strand (in the 5’ to 3’ direction)

Question 4

initiation of transcription =

Answer 4

promoter binding and promoter melting by RNA pol

• RNA pol binds to the promoter with single base pair precision
• RNA pol separates the DNA strands, providing a single-stranded template

Question 5

elongation of transcription =

Answer 5

making of mRNA by RNA pol

• template strand of DNA is read by RNA pol and complementary nucleotides are used to build mRNA (with the same sequence as the coding strand but with Uracil instead of Thymine)
• chain grows from 5’ to 3’

Question 6

termination of transcription =

Answer 6

dissociation of RNA pol + mRNA from DNA

• terminator signals that the RNA transcript is complete
• transcript is released from the RNA pol

Question 7

how RNA polymerase binds to the promoter…

Answer 7

• RNA pol reads the sequence before melting (normally finds promoter in the major groove)
• at physiological temperatures, DNA is constantly affected by BROWNIAN MOTION = erratic random movement of particle as a result of continuous bombardment from surrounding molecules = causes occasional base flipping in/out but not breaking of strands
• hydrogen bonds holding bases together can open spontaneously to allow RNA pol to bind but covalent phosphodiester bonds do not break

Question 8

can detect transcription…

Answer 8

using nucleoside triphosphates containing radioactive alpha phosphate = RNA backbone (phosphodiester bonds) becomes radioactive

Question 9

why is initiation the most regulated process…

Answer 9

transcription doesn’t require a primer - instead relies on accurate binding of RNA pol (single base pair precision)

transcription factors regulate RNA pol binding

Question 10

prokaryotic and eukaryotic transcription work by the same basic principles, but IN PROKARYOTES…

Answer 10

• transcription and translation are coupled (no nucleus)
• DNA is not packaged into histones = DNA is accessible to RNA pol
• genes are often combined into groups (operons) = genes are regulated by one promoter = one mRNA is polycistronic (encodes for several polypeptides)
• Shine-Dalgarno
• mRNA is not processed = no splicing (as prokaryotes don’t have introns), no capping of the 5’ end and no polyadenylation of the 3’ end

Question 11

Shine-Dalgarno =

Answer 11

a ribosomal binding site in prokaryotic mRNA located 8 bases upstream of thee start codon AUG

• helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning it with the start codon

Question 12

how is RNA pol discovered…

Answer 12

in vitro transcription assay

1) make cell lysate - centrifuge prokaryote to collect cells, then add lysozyme for cell lysis, centrifuge to get rid of cell debris and DNA - leaving the cell extract containing RNA pol
2) invent a transcription assay - add nucleoside triphosphates (ATP, UTP, CTP, GTP), radioactive alpha phosphate and DNA template to the lysate = makes radioactive labelled RNA
3) purify - use the transcription assays and chromatography to isolate bacterial RNA pol from lysate (collect protein fractions)
4) find which fraction contains RNAP by seeing which fraction has radioactive labelled RNA
5) check by running fraction on SDS-PAGE = RNA pol has 4 bands (4 types of polypeptides)

Question 13

prokaryotic RNA pol core:

Answer 13

made up of 5 subunits ( two α subunits, a β subunit, a β′ subunit and a ω subunit)

DNA helix enters through a primary channel between β and β′

nucleoside triphosphates enter through a secondary channel formed by β′

in the active site, Mg2+ is held in place with ionic bonds with 3 aspartate residues

Question 14

what happens in the RNAP core during initiation…

Answer 14

DNA makes a 90-degree turn inside the core by bouncing against a ‘wall’ created by the β′ subunit = this creates a transcription bubble when the DNA melts (unwinds)

Question 15

functions of the beta prime (β′) subunit -

Answer 15

does most of the work in elongation…

• nucleoside triphosphate addition through a secondary channel
• holds the Mg2+ in the active site with ionic bonds with 3 aspartate residues
• holds the template strand in the transcription bubble
• creates a wall to bend the DNA 90-degrees

Question 16

functions of the beta (β) subunit -

Answer 16

• holds the coding strand in the transcription bubble up and out of the way
• flap domain forms an exit channel for RNA = forms a ring around the RNA
• antibiotic resistance = some mutations in the beta subunit make RNA pol resistant to binding of antibodies

Question 17

importance of RNA exit channel and crab claw shape of RNAP core -

Answer 17

keep RNA clamped on DNA

Question 18

functions of the alpha 1 and 2 (α) subunits -

Answer 18

assembly of RNAP

promoter recognition

Question 19

function of the omega (ω) subunit -

Answer 19

helps β′ fold correctly

Question 20

sigma factor:

Answer 20

core RNAP need sigma factor ( a transcription factor) to bind and melt promoters to initiate transcription

required for accurate transciption initiation

core RNAP + sigma factor = RNAP holoenzyme

Question 21

sigma is comprised of:

Answer 21

four alpha-helical domains joined together by flexible linkers

these domains are re-lined up between the jaws of RNAP on the back side of RNAP (facing upstream)

Question 22

key steps of prokaryotic transcription initiation by RNAP holoenzyme:

Answer 22

1) promoter search
2) promoter binding (formation of closed complex)
3) nucleation of melting
4) DNA bending
5) promoter melting (formation of open complex)
6) synthesis of the first phosphodiester bond (initiation)
7) DNA scrunching
8) promoter escape and elongation

Question 23

step 1 of prokaryotic transcription initiation by RNAP holoenzyme: PROMOTER SEARCHING

Answer 23

RNAP aimlessly diffuses in 3D until it accidentally bumps into a random DNA segment = slow

RNAP then reads the DNA major groove in 1D with its sigma domain = speeded up by facilitated diffusion

reads without melting the DNA until it finds the promoter

Question 24

step 2 of prokaryotic transcription initiation by RNAP holoenzyme: PROMOTER BINDING

Answer 24

closed complex formation

promoter element is recognised by the helix-turn-helix of sigma domain 4 - promoter binds in the major groove of the hexameric promoter DNA sequence

DNA is not yet melted

Question 25

step 3 of prokaryotic transcription initiation by RNAP holoenzyme: NUCLEATION OF PROMOTER MELTING

Answer 25

adenine is the nucleation base

thermal fluctuations cause the adneine base to flip out of the DNA helix and get captured in the hydrophobic pocket

Question 26

step 4 of prokaryotic transcription initiation by RNAP holoenzyme: DNA BENDING

Answer 26

sigma region 2 captures a flipped base and DNA bends around this flipped base by thermal fluctuations - the DNA ends up between the two jaws of RNA pol

Question 27

step 5 of prokaryotic transcription initiation by RNAP holoenzyme: FORMATION OF ‘OPEN COMPLEX’

Answer 27

promoter unwinding/melting = template and coding strands separate = transcription bubble is formed

Question 28

step 6 of prokaryotic transcription initiation by RNAP holoenzyme: INITIATION

Answer 28

synthesis of the first phophodiester bond

Question 29

step 7 of prokaryotic transcription initiation by RNAP holoenzyme: DNA ‘SCRUNCHING’

Answer 29

ssDNA strand gets pulled inside the RNA pol cleft, like two coiled springs

5’ end of nascent RNA cannot get out because sigma domain 4 is stuck in the RNA exit channel

takes a long time for RNA pol to leave the promoter (takes many attempts)

Question 30

step 8 of prokaryotic transcription initiation by RNAP holoenzyme: PROMOTER ESCAPE AND ELONGATION

Answer 30

after many attempts, RNA pol kicks out sigma domain 4 with the growing RNA 5’ end and escapes the promoter

Question 31

why is it difficult for RNAP to escape the promoter:

Answer 31

RNAPs do not use primers. So, the RNAP holenzyme has to bind to promoters very tightly, to align its active site precisely

Tight interactions take energy and time to break

It often takes RNAP 100+ failed attempts to escape the promoter
(“abortive initiation”) and commit to elongation of RNA

The energy for escape is accumulated in the scrunched state (in which the DNA is pulled inside and wound up like a spring), and is then released in a single powerful burst, launching RNAP down the DNA

Question 32

prokaryotic transcription initiation by RNAP holoenzyme can be blocked by…

Answer 32

…antibiotic rifampicin (anti-tuberculosis antibiotic) - binds to the beta subunit of RNA pol (inside RNA exit channel) an prevents formation of transcription longer than 2 nucleotides - bacteria can acquire resistance with due to mutation