Biochemistry Exam Iv - Transcription

Front 1

________ is the first step in gene expression

Back 1

Transcription

Front 2

Synthesizes Prokaryotic RNA

Cannot recognize promoter sites

Relies on sigma element

Back 2

Core polymerase

Becomes a holoenzyme with addition of sigma subunit (required for promoter site recognition)

Front 3

Eukaryotic RNA polymerase:

18S rRNA
28S rRNA
5.8S rRNA

Back 3

RNA polymerase I

Front 4

Eukaryotic RNA polymerase:

mRNA

Back 4

RNA polymerase II

Front 5

Eukaryotic RNA polymerase:

tRNA
5S rRNA

Back 5

RNA polymerase III

Front 6

Eukaryotic RNA polymerase:

rRNA
tRNA
mRNA

Back 6

Mitochondrial RNA polymerase

Front 7

RNA polymerase (does/does not) need a primer to initiate RNA synthesis?

Back 7

Does not

Front 8

Only one DNA strand is used to produce RNA

Can determine the DNA coding strand by looking at the mRNA strand

Back 8

Within a specific region of DNA, only one of the two strands of the ds-DNA molecule serves as a template

The template strand can vary throughout the DNA molecule

Front 9

+ strand
Sense strand
Nontemplate strand

Back 9

The coding strand of DNA

Front 10

- strand
Nonsense strand
Template strand

Back 10

The noncoding strand of DNA

Front 11

RNA adds nucleotides to the 3' OH on the sugar moiety.

What direction is mRNA synthesized

Back 11

5' to 3'

Just like DNA synthesis

Front 12

1. Template binding
2. Initiation
3. Elongation
4. Termination

Back 12

Model for transcription in prokaryotes.

Front 13

Positive numbers are in the direction of __________

Back 13

transcription

termed 'downstream'

Front 14

Negative numbers are in the opposite direction of transcription.

What is this termed?

Back 14

Upstream

Front 15

Holoenzyme recognizes promoter elements _______ from the site of initiation

Back 15

Upstream

These promoter elements are called cis-acting elements found on DNA

Front 16

Does the sigma subunit scan one strand or both strands of DNA?

Back 16

Both

Front 17

Centered at ~ -10 base pairs.

Sequence:
5'-TATAAT...3': coding
3'-ATATTA...5' template

Back 17

Pribnow box

Front 18

Centered at ~ -35 base pairs

Sequence:
5'..TTGACA...3': coding
3'..AACTGT...5': template

Back 18

-35 base pair sequence.

The sigma subunit must recognize the Pribnow box and the -35 base-pair sequence.

Front 19

The bases that occur with greatest frequency at each position from a group of sequences that are thought to have a common function

Back 19

A consensus sequence

Front 20

1. When the sigma subunit bfinds a promoter site, it binds tightly to the site

2. Core polymerase opens DNA strand ______stream from bound subunit

Back 20

downstream

Front 21

1. Hairpin formation in RNA
2. A string of uridylate residues in the RNA base paired to a string of adenylate residues in the DNA

Back 21

1. Factor independent termination of transcription (prokaryotes)

The lack of GC bases in the region weakens the DNA-RNA interaction

Hairpin formatin requires palindrome

Front 22

Does not bind to RNA polymerase or DNA template: only binds to RNA molecules being produced.

Binds to rut site on RNA.

Has helicase activity (ATP)

Back 22

Factor dependent termination of transcription (prokaryotes).

Rho protein.

Disrputs interaction between RNA and DNA. It releases RNA molecules and polymerases during termination

Front 23

Eukaryotes:

Centered at +1

Consensus sequence poorly defined

Back 23

Initiator box

Front 24

Between -20 and -30 base pairs.

Sequence:
5'-TATAAAA-3':coding
3'-ATATTTT-5':template

Back 24

TATA box

Front 25

Promoter elements that are located on the DNA

Back 25

Cis-acting elements:
Initiator box
TATA box
CAAT box
GC box

Front 26

Between -40 and -100 base pairs:

Sequence:
5'-GGNCAATCT-3'
3'-CCN'GTTACA-5'

Protein factor (CAAT-binding transcription factor (CTF))

Back 26

CAAT box

CTF enhances initiation of transcription at that promoter site

Front 27

Between -40 and -100 base pairs:

Consensus sequence
5'-GGGCGG-3'
3'-CCCGCC-5'

Protein factor: Sp1

Back 27

GC box

Sp1 enhances initiation of transcription at that promoter site

Front 28

Trans-acting elements

Back 28

Increase the efficiency of transcription initiation by RNA polymerase II
- CTF
- Sp1

Front 29

Can RNA pol II bind to promoter site by itself?

Back 29

No. requires transcription factors A-H

All are trans-acting elements that bind to different regions between promoter sites and INR

Front 30

TFII__ directly interacts with TATA box

Back 30

TFIID

Front 31

1. The proteins CSA and CSB bind to a RNA polymerase stalled at a thymine dimer (or a large adduct attached to a nitrogenous base.

2. TFIIH binds and RNA polymerase, RNA, CSA & CSB are released.

3. In a series of steps, XPA, RPA, XPG & XPF-ERCC1 bind. XPG and XPF-ERCC1 will use their endonuclease activities to excise a DNA fragment containing the lesion (in this case a thymine dimer).

4. The DNA fragment containing the thymine dimer dissociates along with XPA, TFIIH, XPG, XPF-ERCC1.

Next, RFC loads PCNA onto the DNA and either Pol delta or Pol epsilon binds and fills in the missing DNA. Finally, DNA ligase 1 in the presence of ATP ligates the DNA fragments together.

Back 31

Transcription coupled repair in eukaryotes

TCR allows certain damage to DNA in the template strand to be repaired prior to transcription.

This helps to ensure the integrity of the mRNA.

Front 32

TCR is a subpathway of ______

Back 32

Nucleotide excision repair (NER)

Front 33

In the case of (prokaryotic/eukaryotic) RNA polymerase, additional proteins stimulate its
proofreading (i.e. the exonuclease) activity.

Back 33

prokaryotic

Front 34

In the case of (prokaryotic/eukaryotic) RNA polymerases I & II, additional proteins also stimulate their proofreading (i.e. exonuclease).

RNA polymerase III appears to have a high intrinsic proofreading (i.e. exonuclease) activity that does not require accessory proteins.

Back 34

eukaryotic

Front 35

1. Basepairing (mispairing 1/10^4 times)

2. Proofreading

Back 35

Fidelity of transcription.

Replication has higher fidelity due to:
1. Base pairing
2. Proofreading
3. Postreplicative repair

Front 36

1. Rifampicin
2. alpha-Amantin
3. Actinomycin D

Back 36

Inhibitors of transcription

Front 37

Binds to the β subunit of RNA polymerase (prokaryotes)

Key component of tuberculosis treatment

Back 37

Rifampicin (prokaryotes only)

Front 38

Dramatic effect on RNA
polymerase II
- Cannot make mRNA
- Cannot make proteins

RNA pol I and mt RNA pol are insensitive

Back 38

Alpha-Amanitin (eukaryotes only)

Front 39

Intercalates into DNA and inhibits transcription.

All intercalators block transcription

Back 39

Actinomycin D (both)