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BHCM218 > Module 04 - Section 01 > Flashcards

Module 04 - Section 01 Flashcards

Transcription

1
Q

Where can you find snRNA?

A

it is part of the spliceosome

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2
Q

What are snoRNAs involved in?

A

modifying the ribosome

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3
Q

How much % of the genome codes for proteins?

A

1.5%

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4
Q

How much % of the genome is transcribed?

A

85%

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5
Q

What are characteristics shared by both processes of transcription and DNA replication? (4)

A

(1) Uses a template strand as the blueprint for forming a new polynucleotide
(2) Has initiation, elongation & termination steps, which can be described in terms of the mechanisms and use of distinct factors
(3) Proceeds in the 5’-3’ direction
(4) Has specific “start” and “stop” sites

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6
Q

What are characteristics that are different between the processes of Transcription and DNA replication? (4)

A

(1) DNA replication requires primers for initiation while transcription does not
(2) In DNA replication, both strands are used simultaneously as a template, not in transcription
(3) Transcription is “selective”, meaning not all of the DNA is transcribed at all times but only specific regions. While DNA replication replicates the entire genome once started
(4) rNTPs are used as building blocks in transcription while dNTPs are used in DNA replication

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7
Q

What is the convention of direction of the template strand in transcription?

A

RNA polymerase uses the 3’-5’ DNA strand as the template for making RNA - Which means RNA will be complementary to the template strand, and will matche the sequence of the coding strand replacing T with U

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8
Q

Are transcriptional start site (TSS) and Translation start site at the same location?

A

No, typically AUG(codes for methionine) is the translational start site, while the TSS can be located many nucleotides upstream to the initiation codon

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9
Q

What is the region between the TSS and start codon called?

A

5’ untranslated region (or 5’UTR)

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10
Q

What is the enzyme that catalyzes Transcription

A

RNA polymerase

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11
Q

How many RNA polymerase do eukaryotes have and what are they called?

A

3, RNA Pol I, RNA Pol II, RNA Pol III

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12
Q

How many RNA polymerase do Prokaryotes have?

A

1

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13
Q

How many subunits does the core of bacterial RNA polymerase have? what are they?

A

5

Beta, beta-prime, omega (the one that looks like w) and two alpha subunits

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14
Q

What does the bacterial RNA polymerase core must bind to form RNA Pol I holoenzyme

A

sigma factor

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15
Q

What is the role of the sigma factor in the RNA Pol I holoenzyme?

A

Acts as a transcriptional initiating factor and adds DNA binding selectivity

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16
Q

What is the general 3D structure of the RNA polymerase?

A

Clamp that wraps around DNA (according to the course material) – I think it could look like a right hand too

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17
Q

Where does the DNA enter the RNA polymerase?

A

Between the Beta (top one) and Beta prime (bottom one) subunits - in between the 2 parts of the clamp (or fingers and thumb of right hand)

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18
Q

Where does the RNA exit RNA polymerase?

A

Hole at the top of the clamp (or like between ring and little finger)

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19
Q

Where does the rNTP enter the RNA Pol I ?

A

from the bottom of the clamp (or like between the pinky and thumb at the end of the palm)

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20
Q

How are the DNA strands opened?

A

There is a “pin structure” under the top clamp (or like a wart on the inside of finger) which open up the DNA strands and form a 17 base pair DNA bubble

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21
Q

How big is the DNA bubble in transcription?

A

17 bp

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22
Q

Describe how RNA and DNA move through RNA Pol I (7 steps)

A

(1) DNA enters between the clamp (between beta and beta prime)
(2) rNTPs enter from the bottom (between beta prime and alpha)
(3) Pin structure opens up DNA double helix
(4) rNTP are added to template strand
(5) Formation of a transient RNA-DNA helix with the template strand
(6) RNA exits out of channel (top between beta and alpha)
(7) DNA strand re-anneal together

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23
Q

Describe the mechanics of RNA synthesis (reaction) (3 characteristics)

A

(1) 3 aspartic acid residues in the RNA Pol active site which capture and coordinate 2 Mg2+ ions
(2) One Mg2+ interacts with phosphate groups of the rNTP, the other Mg2+ brings the 3’OH of the last added nucleotide in close enough proximity to the incoming rNTP for a nucleophilic attach reaction on the alpha phosphate - release of PPi
(3) Process is assisted by the hydrogen bonding between the incoming rNTP and the template DNA, bc it allows precise alignment of the active site, and also the inner structure of the enzyme forces precise alignment

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24
Q

Is RNA Pol built for speed or accuracy?

A

Speed, synthesizes 50-90 nucleotide per second

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25
Q

Why is RNA Pol’s error rate higher than DNA Pol?

A

RNA Pol does not have a built-in proofreading center - error is 1 in 10,000 to 1 in 100,000 vs 1 in 1,000,000 for DNA pol

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26
Q

What are the 2 proofreading mechanisms for RNA Pol ?

A

(1) Kinetic Proofreading

(2) Nucleolytic Proofreading

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27
Q

Describe Kinetic Proofreading

A

If incorrect nucleotide is added = proper H-bonding isn’t formed, which causes fraying at the DNA-RNA duplex

  • If recognized by RNA Pol, it can stall until pyrophosphorolysis reverses the reaction at this base pair
  • based on reversible enzyme kinetics of the polymerization (learned in Module 02)
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28
Q

Describe nucleolytic proofreading

A

RNA Pol transcribed past mismatched base and must “backtrack” to fix the error
RNA Pol must reverse direction by a few nucleotides in template, and use its intrinsic endonuclease activity to hydrolyze the phosphodiester backbone of the transcript upstream of the error, removing the incorrect base

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29
Q

Describe the difference between Kinetic and nucleolytic proof reading

A

Kinetic: polymerase stalls if it incorporates a mismatched base, which allows the removal of the incorrect base by pyrophosphorolysis
Nucleolytic: Uses inherent endonuclease activity of polymerase, backtracks a few nucleotides and hydrolyzes phosphodiester bon upstream of mismatch base

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30
Q

What are the 3 Phases of Trancription?

A

(1) Initiation
(2) Elongation
(3) Termination

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31
Q

describe the initiation phase of transcription

A

Occurs as RNA Pol binds to specific DNA sequences called promoters (contains specific elements and are located upstream of the TSS

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32
Q

describe the elongation phase of transcription

A

process of adding nucleotides to the growing RNA strand

33
Q

describe the termination phase of transcription

A

Release of the produc RNA when polymerase reaches the end of a gene or other transcription unit

34
Q

What are the 5 steps of the process of transcription?

A

(1) Polymerase binds the promoter assisted by transcription factors (TFs) - prokaryotes have sigma factors - Closed complex is formed in which DNA is intact
(2) Formation of transcription bubble, by partially unwinding DNA about 10 bp upstream of start of transcription site
(3) Trancription initiated within complex leading to conformational change that convert the complex to the form required for elongation
(4) Promoter clearance incolving movement of the transcription complex down the DNA template and away from promoter = formation of tightly bound elongation complex
(5) RNA becomes a highly efficient enzyme completing synthesis of the transcript before dissociating from the DNA template, and then recycling for a new round of transcription

35
Q

What is the role of sigma factors in the initiation of bacterial transcription?

A

Offers context specific transcription, it is inefficient to express all genes all the time and also, some gene products are only helpful under specific circumtances (RNA Pol is the screwdriver sigma factors are the bits)

36
Q

Which sigma factor is the most common in E.coli?

A

sigma 70

37
Q

What is a consensus sequence?

A

A sequence of nucleotides or amino acids that has a similar structure and function in different organism

38
Q

What is the -35 and -10 regions

A

two short sequences that have similarities around positions -10 and -35

39
Q

What is an upstream Promoter (UP)?

A

AT-rich recognition element which occurs between positions -40 and -60 in the promoters of certain highly expressed genes - Bounds to one of the alpha subunits of RNA polymerase

40
Q

What determines the efficiency at which RNA polymerase binds to a promoter and initiates transcription?

A

-10, -35 and UP sequences, the spacing between -10/-35 and UP, and the distance of UP element from transcription start site

41
Q

How long is the spacer between the -10 and -35?

A

17-20 nucleotides (roughly 2 turns of a double helix, which allows the orientation of these consensus sites to be on the same side of the helix, which is helpful for recognition by sigma factor in 3D)

42
Q

Why can spacer length affect amount of transcription?

A

Because if they are on the same side of the helix, it helps recognition by sigma factor in 3D - 17-20 bp is approx 2 turns so they will end up on the same side.. a little longer and a little shorter and they might not be in the same plane

43
Q

Describe the steps of the process of initiation (4 steps)

A

(1) Sigma factors binds to the promoter region contacting the -35 and -10 sites. RNA Pol is in the closed conformation. N-terminus is blocking DNA channel entry
(2) Complex converts to the open conformation by (a) Pincers of Pol close around DNA and (b) N-terminus moving from active site cleft
(3) No Primer required, RNA Pol binds 2 nucleotides in place long enough to catalyse the phosphodiester bond
(4) Enters Elongation phase past 10 bonds, which continues until termination signal is reached

44
Q

Which sigma factor can induce the conformation change of RNA Pol to “open” without ATP?

A

All sigma factors EXCEPT sigma-54 which is ATP-dependent

45
Q

When & why is RNA Pol likely to release the transcript without extending it further?

A

The first 8-10 phosphodiester bonds of transcription, because it is energetically favourable

46
Q

What is abortive initiation?

A

When the transcription stop within the first 8-10 phosphodiester bonds

47
Q

When does the sigma factor release the promoter?

A

When polymerase has cleared the promoter, as it weakens the hold of the sigma factor which often falls off

48
Q

When is RNA pol said to be processive?

A

in the elongation phase

49
Q

What is the difference between the open complex and the closed complex?

A

Closed: bound DNA is intact
Open: Bound DNA is partially unwound upstream of the transcription start site

50
Q

What are 2 reasons for transcription complexity in eukaryotes?

A

Organization of chromosomal DNA is far more intricate than bacterial
Eukaryote are more complex organisms in general (not all genes are “on” in all cells at the same time eg skin vs neuron)

51
Q

What is the difference between RNA Pol I, II & III?

A

they are responsible for transcribing a different subset of RNA

52
Q

What is the most important product of RNA Pol I?

A

rRNA precursors (80% of all transcription)

53
Q

What is the most important product of RNA Pol II?

A

mRNA (protein coding gene)

54
Q

What are the main products of RNA Pol III?

A

Smaller functional RNAs such as tRNAs and snRNAs

55
Q

What are the RNA Pol I subunits in their prokaryote equivalent?

A

alpha: RPC5/RPC9
beta prime: RPA
Beta: RPA
omega: RPB6

56
Q

What are the RNA Pol II subunits in their prokaryote equivalent?

A

alpha: RPB3/RPB11
beta prime: RPB
Beta: RPB
omega: RPB6

57
Q

What are the RNA Pol III subunits in their prokaryote equivalent?

A

alpha: RPC5/RPC9
beta prime: RPC1
Beta: RPC2
omega: RPB6

58
Q

alpha-amanitin is a specific inhibitor in a mushroom, what is it used for?

A

Natural biodefense

59
Q

What is the significance of two proteins having conserved structure?

A

Evolutionary conservation, as is seen between eukaryotic and bacterial polymerases, indicates the structure’s importance

60
Q

Beside RNA Pol, what are other conserved structures?

A

human genomes encodes an estimated 20,000-25,000 genes

61
Q

Approximately how many genes are encoded by the human genome?

A

About 24,000 genes

62
Q

How are individual cell types defined in terms of the genome?

A

Individual cells are defined by the genes they express (about 15,000)

63
Q

What is the equivalent of sigma factors in eukaryotic cells?

A

Transcription factors (TFs)

64
Q

What is the typical eukaryotic promoter layout? (2 characteristics)

A

(1) Core promoter consists of a region that is recognized by the “general transcription factors” which are able to recruit RNA Pol
(2) Both an upstreamd and downstream regulatory sequences, which bind gene-specific transcription factors that can acts as activators (or repressors) of transcription at that locus

65
Q

How are transcription factors involved in the differentiation of cells?

A

Different transcription factors define individual cell types by driving a specific pattern of gene expression during differentiation.

66
Q

How was transcription factors and differentiation used in medicine?

A

Scientists have developed ways to reprogram cell’s gene expression profile which converts one differentiated cell type into another

67
Q

What are the two ways “cellular reprogramming” can be done?

A

(1) Using genetic engineering to force the fibroblast to express the TFs
(2) Providing the fibroblasts with the actual TF protein

68
Q

Which terminally differentiated cells are generally used for cellular reprogramming? why do you think this is?

A

Skin fibroblast are used most often since they are easy to obtain and grow in the lab

69
Q

Do eukaryote have distance constraints as it pertains to the distance between promoters and regulatory sequences?

A

No it can be 1000s of bp up or downstream

70
Q

What is the TATA binding protein? (4 things)

A

(1) Protein that binds to a 5’-TATAAA sequence (called TATA box) near position -30
(2) used by all 3 RNA Pol
(3) Major role in transcription initiation
(4) Only present in 25% of the genes

71
Q

Describe the structure of the TATA binding protein

A

(1) saddle-shaped protein
(2) Able to create a sharp being in the DN upon binding
(3) sit on the DNA double helix much like a saddle with an extended beta sheet and loop stirrups with the minor groove of the TATA box sequence

72
Q

How does TBP recognize DNA?

A

Bends the DNA by positioning two pairs of Phe residue side chains between base pairs at each end of the recognition sequence
Bending opens and widens the minor groove, enabling hydrogen bonding between protein side chain and the minior-groove edges of the DNA bases

73
Q

How do most DNA-binding protein recognize DNA?

A

By inserting alpha helices into major groove

74
Q

How do you now if a transcription factors binds a specific regulatory region of DNA? (2 techniques)

A

(1) electrophoretic mobility shift assay (EMSA)
- if a particular protein can bind to a particular stretch of DNA
(2) DNA footprinting
- Where a transcription factor binds on a particular stretch of DNA

75
Q

Describe Electrophoretic Mobility Shift Assay (EMSA)

A

(1) fragments of DNA of a known sequence are incubated with the protein of interest and then analyzed on a nondenaturing polyacrylamide gel
(2) DNA used in experiment is visualized either by staining with a dye or by covalently attaching a radioactive phosphate group at one end
(3) free DNA fragments migrate more quickly through the gel than DNA bound by protein (shift from fast to slow in the migration indicates a direct binding)

76
Q

Why is SDS unable to be used in EMSA?

A

A nondenaturing gel has to be used in EMSA, since the addition of SDS would the protein and disrupt formation of the DNA-protein interaction. Remember the influence of protein structure on its binding properties

77
Q

Describe DNA footprinting

A

(1) DNA of interest must be amplified and radiolabelled at one end. PCR primers must be radiolabeled on the 5’end to provide a point of reference for the visualization step. Unbound and DNA-protein complexes are incubated with nucreases
(2) Cleavage occurs only at sites that are not physically protected by the presence of bound protein. Conditions are carefully controlled so that each piece of DNA is cleave just once on average generating a set of fragments that represent all possible cleavage products
(3) resulting fragments are separated by gel electrophoresis and detected by exposing the gel to film which detects radioactive emissions from the labelled DNA fragments. Any gaps in the cleavage sites where the protein associates with the DNA produces a “footprint” that indicated the boundaries of the protein-binding site relative to the radiolabeled end of the DNA sequence of interest

78
Q

How are the results of DNA footprinting analyzed?

A

Footprint identified by analyzing the site of nuclease cleavage in the DNA before and after adding the protein, its darker where it has bound and has not been cleaved

BHCM218 (24 decks)

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