Initiation of mRNA Transcription in Eukaryotes

Question 1

RNA polymerase

Answer 1

The enzymes that perform transcription. This enzyme catalyzes the formation of the phosphodiester bonds between the nucleotides. It moves step-wise along the DNA, unwinding the DNA helix just ahead of the active site, in order to expose a new region of the template strand for complementary base-pairing. RNA polymerases do not require a primer

Question 2

Differences between transcription and DNA replication (2)

Answer 2

1. Unlike the new DNA strand, the RNA strand doesn’t remain hydrogen bonded to the DNA template strand. Behind the region where nucleotides are being added, the RNA chain is displaced and the DNA helix re-forms
2. RNA molecules are much shorter than DNA molecules, because they are copied from a limited region of DNA

Question 3

Sigma factor

Answer 3

Transcription initiation factor. It is a subunit that associates with the core RNA polymerase (in bacteria). It helps RNA polymerase to read the signals in DNA and know where to begin transcribing. Binds at -35 and -10, with transcription starting at +1

Question 4

RNA polymerase holoenzyme

Answer 4

A bacterial enzyme that works with sigma factor. They form a complex that adheres weakly to DNA, and the holoenzyme generally rapidly slides along the DNA molecule and dissociates. When the holoenzyme encounters a promoter, the complex binds more tightly

Question 5

Promoter

Answer 5

A special sequence of nucleotides that indicates the starting point for RNA synthesis. Called consensus sequences in prokaryotes. RNA polymerase binds tightly here because the sigma factor makes specific contacts with the bases on the outside of the DNA double helix

Question 6

Initiation of transcription in prokaryotes (6)

Answer 6

1. The RNA polymerase holoenzyme tightly binds at the promoter
2. Opens the double helix to expose a short stretch of nucleotides on each strand called the transcription bubble
3. The transcription bubble is stabilized by sigma factor binding to the unpaired bases on one of the exposed strands
4. The other exposed DNA strand acts as a template strand for base-pairing with ribonucleotides
5. RNA polymerase synthesizes RNA in a step-wise fashion- the polymerase moves forward one base pair for every nucleotide added
6. Transcription ends at the terminator

Question 7

Terminator

Answer 7

The DNA sequence where transcription ends (in prokaryotes). RNA polymerase halts and releases the newly made RNA molecule and the DNA template.

Question 8

How do terminating signals stop RNA polymerase in prokaryotes?

Answer 8

For most genes, a termination signal consists of a string of A-T nucleotide pairs, which is preceded by a DNA sequence that becomes a hairpin structure when it base-pairs. The hairpin helps to disengage the RNA transcript from the active site

Question 9

Consensus nucleotide sequence

Answer 9

All bacterial promoters contain related nucleotide sequences, which allows them to be recognized by the sigma factor. The consensus nucleotide sequence is derived by comparing many sequences with the same basic function and tallying up the most common nucleotides found at each position. Acts as an “average” for a large number of nucleotide sequences.

Question 10

Eukaryotic polymerases

Answer 10

RNA polymerase 1, 2, and 3.

Question 11

RNA polymerase 1 and 3

Answer 11

Transcribe the genes encoding transfer RNA (tRNA), rRNA, and various small RNAs

Question 12

RNA polymerase 2

Answer 12

Transcribes most genes, including all of those that encode proteins

Question 13

Differences in bacterial and eukaryotic RNA polymerase function (2)

Answer 13

1. Bacterial RNA polymerase only needs one transcription initiation factor (sigma). Eukaryotic require many of these factors, called general transcription factors
2. Eukaryotic transcription initiation must take place on DNA packaged into nucleosomes and higher-order forms of chromatin structure. These are absent in bacterial chromosomes

Question 14

General transcription factors

Answer 14

The many factors required for transcription initiation in eukaryotes. They help to correctly position RNA polymerase at the promoter, pull apart the DNA strands so transcription can begin, and release RNA polymerase from the promoter to start elongation

Question 15

General transcription factors for RNA polymerase 2 (5)

Answer 15

1. TFIID
2. TFIIB
3. TFIIF
4. TFIIE
5. TFIIH

Question 16

TFIID

Answer 16

Made of TBP and TAF subunits. It recognizes the TATA box and recognizes other DNA sequences near the transcription start part- regulates DNA binding by TBP

Question 17

TFIIB

Answer 17

Recognizes the BRE element in promoters, accurately positions RNA polymerase at the start site of transcription

Question 18

TFIIF

Answer 18

Stabilizes RNA polymerase’s interaction with TBP and TFIIB. Helps attract TFIIE and TFIIH

Question 19

TFIIE

Answer 19

Attracts and regulates TFIIH

Question 20

TFIIH

Answer 20

Unwinds DNA at the transcription start point, phosphorylates Ser% of the RNA polymerase CTD. Releases RNA polymerase from the promoter

Question 21

TATA box

Answer 21

A DNA sequence making up the promoter, that signals the start of transcription in eukaryotes. It gets its name because it is primarily composed of A and T nucleotides. Located 25 nucleotides upstream from the transcription start site. The general transcription factors assemble at promoters when TFIID binds here

Question 22

General transcription factors assembly process (4)

Answer 22

1. TFIID binds to the TATA box. The TFIID subunit TBP recognizes it
2. TFIID binding causes a distortion in the DNA of the TATA box
3. The distortion brings DNA sequences closer together, allowing for next steps to begin
4. Other factors assemble, along with RNA polymerase 2, forming the transcription initiation complex

Question 23

Initiation of transcription in eukaryotes (5)

Answer 23

1. TFIID binds to the TATA box, which allows for the adjacent binding of TFIIB
2. DNA distortion is produced by the binding of TFIID
3. The rest of the general transcription factors, as well as RNA polymerase, assemble at the promoter
4. TFIIH uses the energy from ATP hydrolysis to open the DNA double helix at the transcription start point, exposing the template strand. It also phosphorylates RNA polymerase 2
5. RNA polymerase is released from the general factors, elongation can begin

Question 24

Phosphorylation of RNA polymerase 2

Answer 24

TFIIH phosphorylates RNA polymerase 2 during transcription. This causes a conformational change so the polymerase is released from the general factors and the elongation phase of transcription can begin. The polymerase is phosphorylated at the C-terminal domain (CTD), which extends from the polymerase molecule

Question 25

Other promoters (3)

Answer 25

1. INR
2. BRE
3. DPE

Question 26

Sigma factor domains (4)

Answer 26

Sigma 1-4. Sigma 4 binds -35, sigma 2 binds -10 (closed complex in RNA polymerase). When sigma 2 binds it -10, it initiates DNA unwinding. Template DNA is fed through the active site for transcription (open complex)

Question 27

Elongation factors

Answer 27

Both bacterial and eukaryotic RNA polymerases require elongation factors, which are proteins that decrease the likelihood that RNA polymerase will dissociate before it reaches the end of a gene. They associate with RNA polymerase shortly after initiation and help the polymerase to move through the wide variety of different DNA sequences found in genes

Question 28

How do eukaryotic RNA polymerases deal with chromatin structure?

Answer 28

They are aided by ATP dependent chromatin remodeling complexes, which can travel with the polymerase and actively relax DNA or seek out an “rescue” a stalled polymerase. Histone chaperones can also partially disassemble nucleosomes in front of a moving RNA polymerase

Question 29

As RNA Pol transcribes, what can happen?

Answer 29

DNA supercoiling, which will stall transcription. Topoisomerases relieve/prevent DNA supercoiling

Question 30

Termination of eukaryotic transcription

Answer 30

Transcription termination occurs in a reaction coupled to RNA 3′-end processing. Both ends of eukaryotic mRNA are modified- by capping on the 5’ end and by polyadenylation of the 3’ end. These modifications allow the cell to assess whether both ends of an mRNA molecule are present and if the message is intact before the mRNA is exported from the nucleus. RNA splicing can also join different portions of the protein-coding sequence