Transcription

Question 1

How does the chemical structure of RNA differ from DNA?

Answer 1

1) Ribose sugar instead of deoxyribose
2) Uracil instead of thymine

Question 2

In what direction does transcription proceed?

Answer 2

It goes from left to right in a 5’ –> 3’ direction.

Question 3

Why is polymerization of the rNTP to the growing RNA strand favoured in transcription?

Answer 3

The high energy bond between the alpha and beta phosphates of the incoming rNTP (ribonucleotide triphosphate) is replaced by a lower-energy phosphodiester bond.

Question 4

RNA polymerase begins transcription at gene nucleotide designated ____.

The polymerase travels “downstream” where the bases are ______.

“Upstream” bases are _______.

Answer 4

+1

positive

negative

Question 5

Where do important features like promoter sequences with transcription factors that recruit RNA polymerase lie on the transcription site?

Answer 5

Upstream!

Question 6

What are the three stages of transcription? Briefly describe them.

Answer 6

1) Initiation: polymerase binds to the promoter sequence, locally denatures the DNA, and catalyzes the first phosphodiester linkage

2) Elongation: Polymerase advances 3’->5’ down template strand (RNA is being made 5’ -> 3’), melting duplex DNA and polymerizing the RNA by adding rNTPs.

3) Termination: Polymerase recognizes a STOP site, releases the completed RNA and dissociates from DNA.

Question 7

Match the following steps of initiation:

1. Step 1
2. Step 2
3. Step 3

A. Polymerase catalyzes phosphodiester linkages of two initial rNTPs.
B. Polymerase binds to PROMOTER SEQUENCE in duplex DNA. “Closed Complex.”
C. Polymerase melts duplex DNA near transcription start site, forming a transcription bubble. “Open Complex.”

Answer 7

Step 1: Polymerase binds to PROMOTER SEQUENCE in duplex DNA. “Closed Complex.”

Step 2: Polymerase melts duplex DNA near transcription start site, forming a transcription bubble. “Open Complex.”

Step 3: Polymerase catalyzes phosphodiester linkages of two initial rNTPs.

Question 8

What do eukaryotic RNA polymerase require to find promoter regions and initiate transcription?

Answer 8

General transcription factors (associated proteins)

Question 9

(T/F) Elongation complex is quite clumsy, the polymerase tends to fall quite often.

Answer 9

False! Elongation complex is very stable; polymerase doesn’t fall until a stop site is reached.

*small genes transcribed in few mins, big genes can take hours

Question 10

Fill in the blank:
The completed RNA molecule (after the three stages of transcription) is called the ______ ______.

Answer 10

Primary transcript

Question 11

What are sigma factors found in bacterial RNA polymerase?

Answer 11

Sigma factors are INITIATION FACTORS of transcription that recognize promoter regions and recruit RNA polymerase to the site. Once transcription is initiated, sigma factor is released.

Question 12

Match the following sigma factors to their definition:

1) Sigma factor 70
2) Sigma factor 54

A) recognizes promoters of genes involved in nitrogen metabolism
B) is the best known, recognizes TTGACA…TATAAT & is for growth related genes

Answer 12

Sigma factor 70: is the best known, recognizes TTGACA…TATAAT & is for growth related genes

Sigma factor 54: recognizes promoters of genes involved in nitrogen metabolism

Question 13

How is organization of genes different in prokaryotes compared to eukaryotes?

Answer 13

In prokaryotes, genes with a common function are often arranged linearly in OPERONS and transcribed together on a SINGLE mRNA. Very few non-coding gaps of DNA (introns) in prokaryotic genomes.

In eukaryotes, genes analogs are scattered on several chromosomes. Co-regulation can not be achieved by physical linkage. They can’t be transcribed together on a single mRNA. Lots of introns!

Question 14

(T/F) In both prokaryotes and eukaryotes, transcripts must go through several processing steps to become mRNA.

Answer 14

False!
In prokaryotes, mRNAs are directly transcribed from DNA.

Question 15

(T/F) For eukaryotes, regulatory sites can be tens of kilobases upstream or downstream of the promoter.

Answer 15

True!

*Transcription factors bind to regulatory sites and affect the expression of a gene!

Question 16

The three eukaryotic RNA polymerases are ______ complex than the bacterial RNA polymerase.

Answer 16

MORE

Question 17

Eukaryotic RNA polymerases contain ____ large subunits and _______ smaller subunits, some which are common to all and others specific.

Answer 17

two

10-14

Question 18

What is the bacterial RNA polymerase composed of?

Answer 18

Has two large subunits (β’ and β) and three small subunits (two α and one ω).

Question 19

Which eukaryotic RNA polymerase contains C-terminal domain (CTD)? In which subunit?

Answer 19

RNA polymerase II

In its largest subunit (RPB1)

Question 20

Fill in the blanks:
RNA polymerases that initiate transcription have __________ CTD, while RNA polymerases that are actively transcribing have __________ CTD.

Answer 20

Unphosphorylated; phosphorylated

Question 21

Briefly describe and compare the structures of eukaryotic RNA polymerases.

Answer 21

There are three; RNA pol I, II and III.

All of them contain two large subunits, b and b’ like of bacterial large subunits.

RPB1 and RPB2 in polymerase II!

And multiple same or differing small subunits.

RNA pol II is the only one with a C-terminal domain in its RPB1.

RNA pol I and pol III contain the same α-like subunits, while pol II has different α-like subunits.

They have all share the same ω-like subunit and four other common subunits.

Question 22

Which RNA polymerase mostly transcribes mRNA?

Answer 22

RNA polymerase II

Question 23

What happens to the clamp domain of RPB1 during before and during elongation?

Answer 23

Clamp domain is open when downstream DNA is inserted into RPB1.

Clamp domain SHUTS by swinging over the cleft between the jaws of the polymerase when enzyme is in elongation mode, anchoring polymerase onto the downstream dsDNA.

*talking about eukaryotic RNA pol II

Question 24

How is transcribing RNA pol II different than free RNA pol II?

Answer 24

Position of a clamp domain; clamp is shut of the transcribing RNA pol II to trap the template DNA strand and transcript.

Question 25

What ion participates in the catalysis of phosphodiester bond formation (for transcription)?

Answer 25

Mg2+

Question 26

(T/F) RPB2 wall domain forces the template DNA entering the jaws of the polymerase to BEND before it EXITS the polymerase.

Answer 26

True!

Question 27

What complex needs to be formed to initiate transcription? What does it contain?

Answer 27

PRE-INITIATION COMPLEX; contains many general transcription factors and RNA polymerase.

TFII = transcription factors associated with RNA pol II

Question 28

(T/F) Transcription factors can only be studied in vitro.

Answer 28

False! Transcription factors can be studied in vitro (in test tubes with defined components) and in vivo using genetic techniques.

Question 29

Match the following RNA polymerases to what they transcribe:

1) RNA pol I
2) RNA pol II
3) RNA pol III

A) mRNA (encodes protein), snRNAs (splicing), siRNA + miRNA (translational control)
B) tRNA (protein synthesis), 5S rRNA (ribosome component), U6 splicing snRNA , and other small stable RNAs with unknown functions.
C) 45S precursor (pre-rRNA) of 18S, 5.8S and 28S rRNA (ribosome components, protein synthesis).

Answer 29

RNA pol I: 45S precursor (pre-rRNA) of 18S, 5.8S and 28S rRNA (ribosome components, protein synthesis).

RNA pol II: mRNA (encodes protein), snRNAs (splicing), siRNA + miRNA (translational control)

RNA pol III: tRNA (protein synthesis), 5S rRNA (ribosome component), U6 splicing snRNA , and other small stable RNAs with unknown functions.

Question 30

What are the core promoter elements in eukaryotes?

Answer 30

1) BRE: TFIIB recognition element - always found immediately near TATA box
2) TATA box
3) Initiator
4) CpG islands
5) Downstream promoter element

Question 31

Which one of the statements regarding RNA polymerase II promoter sequences is false?

1) Promoter sequences specify where polymerase initiate transcription
2) Single base change in TATA box increases gene transcription

Answer 31

2! Single base change in TATA box DECREASES gene transcription.

Question 32

Match the steps of eukaryotic transcription initiation:

1) Step 1
2) Step 2
3) Step 3
4) Step 4
5) Step 5
6) Step 6

A) TFIIB binds to DNA and TBP, melting the DNA at the start site, forming a TRANSCRIPTION BUBBLE.

B) TFIIH, with its kinase subunit, phosphorylated CTD, promoting elongation. Transcription factors released, except for TBP.

C) TFIIH binds, completing the formation of the pre-initiation complex. TFIIH is a helicase that unwinds DNA and allows polymerase to form the open complex.

D) A pre-formed complex of RNA-pol II and TFIIF (tetramer) bind. (RNA polymerase does not bind precisely at the TATA box)

E) Binding of TBP (TATA-box binding protein) to the DNA. TBP is part of TFIID, which is composed of TBP and 13 other TBP-associated factors (TAFs). TFIIA forms a complex with TFIID and TATA-box in vivo.

F) TFIIE (tetramer) binds.

Answer 32

Step 1: Binding of TBP (TATA-box binding protein) to the DNA. TBP is part of TFIID, which is composed of TBP and 13 other TBP-associated factors (TAFs). TFIIA forms a complex with TFIID and TATA-box in vivo.

Step 2: TFIIB binds to DNA and TBP, melting the DNA at the start site, forming a TRANSCRIPTION BUBBLE.

Step 3: A pre-formed complex of RNA-pol II and TFIIF (tetramer) bind. (RNA polymerase does not bind precisely at the TATA box)

Step 4: TFIIE (tetramer) binds.

Step 5: TFIIH binds, completing the formation of the pre-initiation complex. TFIIH is a helicase that unwinds DNA and allows polymerase to form the open complex.

Step 6: TFIIH, with its kinase subunit, phosphorylates CTD, promoting elongation. Transcription factors released, except for TBP.

Question 33

Match the transcription factors that form the pre-initiation complex to their definitions:

1) TFIID
2) TFIIA
3) TFIIB
4) TFIIF
5) TFIIE
6) TFIIH

A) Forms a complex with TFIID and TATA-box, stabilizing the TBP and TFIIB to the promoter. In the 1st step of eukaryotic transcription initiation.

B) Made of 9 subunits, with two key roles; helicase and kinase. In the 5th step of eukaryotic transcription initiation, the helicase unwinds the DNA and allows Pol II to form the OPEN complex. In the 6th step of eukaryotic transcription initiation, its kinase phosphorylated CTD of polymerase II, promoting elongation.

C) a complex of TBP and 13 other subunits called TBP-associated factors (TAFs). In the 1st step of eukaryotic transcription initiation.

D) A tetramer that binds tightly to pol II in a pre-formed complex, which binds to TFIIB and prevents binding of Pol II to nonspecific DNA sequences. In the 3rd step of eukaryotic transcription initiation.

E) Binds to TBP and DNA, melting the DNA at the transcription start site, forming a replication bubble. In the 2nd step of eukaryotic transcription initiation.

F) A tetramer that binds in the 4th step of eukaryotic transcription initiation, that recruits TFIIH and has helicase and ATPase activity.

Answer 33

TFIID: a complex of TBP and 13 other subunits called TBP-associated factors (TAFs). In the 1st step of eukaryotic transcription initiation.

TFIIA: Forms a complex with TFIID and TATA-box, stabilizing the TBP and TFIIB to the promoter. In the 1st step of eukaryotic transcription initiation.

TFIIB: Binds to TBP and DNA, melting the DNA at the transcription start site, forming a replication bubble. In the 2nd step of eukaryotic transcription initiation.

TFIIF: A tetramer that binds tightly to pol II in a pre-formed complex, which binds to TFIIB and prevents binding of Pol II to nonspecific DNA sequences. In the 3rd step of eukaryotic transcription initiation.

TFIIE: A tetramer that binds in the 4th step of eukaryotic transcription initiation, that recruits TFIIH and has helicase and ATPase activity.

TFIIH: Made of 9 subunits, with two key roles; helicase and kinase. In the 5th step of eukaryotic transcription initiation, the helicase unwinds the DNA and allows Pol II to form the OPEN complex. In the 6th step of eukaryotic transcription initiation, its kinase phosphorylated CTD of polymerase II, promoting elongation.

Question 34

(T/F) The conserved C-terminal domain of the TATA-box binding protein (TBP) binds to the major groove of specific DNA sequences rich in A and T, untwisting and sharply bending the double helix.

Answer 34

False! The conserved C-terminal domain of the TBP binds to the MINOR groove of specific DNA sequences rich in A and T, untwisting and sharply bending the double helix.

Question 35

(T/F) Transcription of most eukaryotic genes require participation of TBP.

Answer 35

True!

Question 36

(T/F) Because pol I and pol III transcription initiation processes are similar to pol II, they all use the same general transcription factors.

Answer 36

False! They require different general transcription factors.