Transcription

Question 1

List the order in which transcription factors bind to form the eukaryotic PIC.

Answer 1

TFIID, TFIIA, TFIIB, TFIIF + RNA Polymerase II, TFIIE, TFIIH, mediator

Question 2

TFIID

Answer 2

• 14 subunits
• 3 lobes
• TBP and TAFs
• TAF1 and TAF7 bind to DPE and MTE to form D complex

Question 3

TBP

Answer 3

• TATA box binding protein
• binds to minor groove and forces open DNA
• highly conserved
• causes DNA to bend 80-90 degrees

Question 4

TFIIA

Answer 4

• 2 subunits
• replaces TAF1 on TBP binding surface
• TFIIA + TAF1 translocate to TATA box, triggering rearrangement of TFIID
• DA complex

Question 5

TFIIB

Answer 5

• 1 subunit
• binds to BREu and BREd to stabilize TFIIA+TBP+DNA complex
• DAB complex

Question 6

BRE

Answer 6

• TFIIB recognition elements

- has flexible domain called B ribbon that hooks RNA polymerase into place on the DNA

Question 7

TFIIE

Answer 7

• 2 subunits
• forms “clamp” around DNA by binding to RNA polymerase stalk, wrapping around DNA and binding to TFIIF on the other side

Question 8

TFIIH (+XPB, XPD subunits)

Answer 8

• 10 subunits (most complex GTF) divided into TFIIH core and CAK (cyclin-dependent kinase activating complex)
• catalyzes transition of PIC from closed complex to open complex (enzymatic activity)
• XPB: translocates DNA
• XPD: unwinds DNA

Question 9

TFIIF

Answer 9

• 2 subunits

- recruits RNA polymerase II to PIC

Question 10

mediator

Answer 10

• 24 subunits
• 3 lobes (head, middle, tail)
• relays input from enhancers/activators to PIC

Question 11

RNA polymerase II

Answer 11

• 12 subunits
• transcribes mRNA
• subunit 1 has CTD tail involved in transcription initiation

Question 12

RNA polymerase holoenzyme

Answer 12

• prokaryotes
• core and sigma factor
• core catalyzes 3’-5’ phosphodiester bonds (polymerization) and consists of 2 alpha, beta, beta prime and omega subunits
• sigma factors recognize promoter

Question 13

What is each of the 5 subunits of the core enzyme of the RNA polymerase holoenzyme responsible for?

Answer 13

• 2 alpha: recognize UP element
• beta: polymerase activity
• beta prime: binds to DNA
• omega: stability

Question 14

List the steps of prokaryotic transcription initiation.

Answer 14

• sigma factor attaches to core enzyme
• sigma factor binds to -10 and -35 binding sites
• 2 alpha subunits bind to UP element
• sigma factor loads DNA into beta and beta prime subunits
• complementary bases added to +1 through beta prime entrance channel
• first couple transcripts (~10 nucleotides) aborted due to blocking of exit channel by sigma factor
• elongation begins after sigma factor pushed off

Question 15

CTD (phosphorylation)

Answer 15

• C-terminal domain of RNA polymerase II subunit 1
• phosphorylated by TFIIH with the help of mediator, occurs at 5th serine in 7 residue sequence YSPTSPS
• involved in hnRNA processing

Question 16

sigma factor opens the DNA helix from ___ position to ___ position

Answer 16

-10 to +10

Question 17

RNA polymerase reads DNA _’ to _’, and synthesizes RNA _’ to _’.

Answer 17

• reads DNA 3’ to 5’

- synthesizes RNA 5’ to 3’

Question 18

Describe intrinsic termination.

Answer 18

• prokaryotic
• mRNA folds into hairpin due to self-complementary G/C rich region
• hairpin pulls adjacent U rich region
• since U rich region is weakly bound to DNA, it is easily pulled off

Question 19

Describe rho-dependent termination.

Answer 19

• rho protein attaches all 6 subunits on mRNA (it is a homohexamer)
• rho protein adopts lock washer conformation to feed mRNA through its central channel
• rho protein translocates 5’ to 3’ along mRNA and pulls off weak U bonds when it reaches RNA polymerase

Question 20

How is eukaryotic transcription terminated?

Answer 20

• does not have specific termination signal

- AAUAAA sequence in 3’ UTR of hnRNA signals 3’ trimming and polyadenylation (poly-A tail)

Question 21

rut (rho utilization site)

Answer 21

• ~70 bases long
• C rich
• where the rho protein initially loads onto mRNA