Module 4 Sections 1-3

Question 1

list all tRNAs, rRNAs, snRNAs, snoRNAs

Answer 1

snRNAs = small nuclear RNAs, play a role in gene regulation - form part of the spliceosome that is important for mRNA processing

mRNA = only type of coding RNA – determines the amino acid sequence of a protein

rRNA = ribosomal, present during translation

tRNA = transfer, present during translation

snoRNA = small nuclear RNA, involved in the processing of rRNAs

miRNA – micro RNAs, limit translation by binding to 3’-end of target mRNAs

IncRNA = long non-coding RNAs, can be important regulatory RNAs

Question 2

transcription vs DNA replication characteristics

Answer 2

uses template strand: both

has initiation, elongation and termination: both

proceeds in the 5’-3’ direction: both

requires primers for initiation: DNA replication

has specific start and stop sites: both

is selective: transcription

rNTP’s are used as building blocks: transcription

both strands are used simultaneously as a template: DNA replication

Question 3

the template strand

Answer 3

RNA polymerase uses 3’-5’ template DNA strand to make RNA - RNA will be complementary to the template strand but MATCH the sequence of the coding strand, replacing T with U

Question 4

transcriptional start site

Answer 4

Where RNA polymerase starts reading

Distinct from translational start site

Which encodes the first amino acid of the protein

Question 5

translational start site

Answer 5

Generally AUG (met)

Located many nucleotides upstream the initiation codon

Region between TSS and start codon = the 5’ untranslated region (5’UTR)

Question 6

bacterial RNA polymerase holoenzyme subunits

Answer 6

5 polypeptide subunit core with 3 subunits (B, B’, w) and 2 alpha subunits

Question 7

sigma factor

Answer 7

acts as a transcriptional initiating factor and adds DNA binding selectivity

Question 8

main mechanic of RNA synthesis

Answer 8

3 aspartic acid residues in the RNA pol active site, which capture and coordinate two Mg2+ ions

Question 9

Mg2+ ions role in RNA synthesis

Answer 9

1 of them interacts with the phosphate groups of the NTP and the other acts to bring the 3’OH of the last added nucleotide in close enough proximity to the incoming rNTP of a nucleophilic attack reaction to take place on the alpha phosphate, releasing PPi

Question 9

does RNA polyermase have a built-in proofreading center

Answer 9

no

Question 10

RNA polymerase alternative proof-reading mechanisms

Answer 10

1. kinetic proofreading
2. nucleolytic proofreading

Question 11

kinetic proofreading

Answer 11

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

When recognized by the RNA pol, it can stall until pyrophosphorolysis reverses the reaction at this base pair, allowing for the correct rNTP to be paired

Question 12

nucleolytic proofreading

Answer 12

RNA polymerase must backtrack to the mismatched base to fix the error

Must reverse its direction by a few nucleotides in the template and uses its intrinsic endonuclease activity to hydrolyze the phosphodiester backbone of the transcript upstream of the error, removing the incorrect base

Question 13

kinetic vs nucleolytic proofreading

Answer 13

kinetic = the polymerase stalls, allowing pyrophosphorolysis to remove the incorrect base

nucleolytic = the polymerase backtracks and hydrolyzes the phosphodiester bond upstream of the incorrect base

Question 14

phases of transcription

Answer 14

1. initiation
   Occurs as RNA polymerase binds to specific DNA sequences called promoters

These sequences contain specific elements and are located upstream of the TSS

2. elongation
   the process of adding nucleotides to the growing RNA strand
3. termination
   the release oft he product RNA when the polymerase reaches the end of a gene or other transcription unit

Question 15

5 main steps in the process of transcription

Answer 15

1. The polymerase binds the promoter, assisted by sigma factors in prokaryotes or transcription factors in eukaryotes
2. Open complex is formed, in which bound DNA is partially unwound near a region of 10bp upstream of (ahead of) the transcription start site
3. Transcription is initiated within the complex, leading to a conformational change that converts the complex to the form required for elongation
4. Promoter clearance, involving movement of the transcription complex down the DNA template and away from the promoter, leads to the formation of a tightly bound elongation complex
5. Once elongation begins, RNA polymerase 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

Question 16

sigma factors in transcriptional initiation

Answer 16

a transient subunit of the bacterial RNA polymerase that directs the enzyme to the promoter - different sigma factors are specific for different promoters

Question 17

most common sigma factor

Answer 17

O^70, 70kDa as molecular weight

Question 18

consensus sequence

Answer 18

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

Question 19

-35 and -10 regions

Answer 19

important interaction sites for O^70
-10 region: 5’-TATAAT-3’
-35 region: 5’-TTGACA-3’

Question 20

upstream promoter

Answer 20

AT-rich recognition element between -40 and -60 in the promoters of highly expressed genes bound by one of the subunits of RNA polymerase

Question 21

importance of consensus sequences, UP sequences, and spacer

Answer 21

The efficiency with which RNA polymerase binds to a promoter and initiates transcription is determined by the –10, -35 and UP sequences (and the spacer), and the distance of the UP element from the transcription start site

Question 22

spacer length importance

Answer 22

any changes to the spacer length (larger or smaller) will reduce transcription

2 turns of a double helix allows the orientation of consensus sites to be on the same side of the helix

Question 23

increase in spacer length

Answer 23

longer than 2 turns of a double helix = consensus sites are no longer on the same side of the helix = transcription is reduced

Question 24

steps in bacterial transcription

Answer 24

1. initiate bacterial transcription - Sigma factor must bind to the promoter region, contacting the –35 and –10 sites - The enzymes is in the closed conformation - N-terminus of the sigma factor is blocking the DNA entry channel 2. open conformation - Pincers of the Pol enzyme closes around DNA, and the N-terminus moving from the active site cleft - Independent of ATP for all sigma factors except O&54, which is ATP-dependent 3. abortive initiation - RNA polymerase does not require a primer - Must bind and hold 2 nucleotides in place for long enough to catalyze the phosphodiester bond - After this bond and for the first 8-10 bonds, there is a high chance that the polymerase will release the transcript without extending it further - Energetically favourable for the RNA polymerase to release the 8-10 base pair RNA transcript from the initiation complex - Holoenzyme will begin NA synthesis again on the same template 4. elongation - RNA polymerase can hold onto the transcript long enough to extend it beyond 10 nucleotides - At this point, the RNA becomes stable and clears the promoter to enter the elongation phase - RNA polymerase will continue along the DNA template until it reaches a termination signal - Elongation allows for the promoter sequence to clear from the polymerase - Weakens the hold of the sigma factor, which falls off the polymerase 5. termination - Enzyme releases the DNA template only when it encounters a termination sequences - Polymerase is processive, moving smoothly along the template, synthesizing the complementary RNA strand and dissociating only when the transcript is complete

Question 25

open vs closed conformation

Answer 25

closed: bound DNA is intact open: bound DNA is partially unwound upstream of the transcription start site

Question 26

RNA polyermases in eukaryotic cells

Answer 26

1. RNA Pol I - Ribosomal RNA (rRNA) precursors - Responsible for transcribing rRNAs - 80% of all transcription 2. RNA Pol II - Messenger RNA (protein coding genes) - Responsible for the synthesis of mRNA 3. RNA Pol III - Smaller functional RNAs (tRNAs, snRNAs)

Question 27

how many subunits does RNA Pol have for eukaryotes

Answer 27

12

Question 28

other uses for transcription inhibitors

Answer 28

some species rely on transcription inhibitors for natural biodefense Ex: mushroom Amanita phalloides (the death cap) produces a-amanitin, a cyclic peptide that disrupts eukaryotic mRNA synthesis by blocking Pol II

Question 29

general structure of eukaryotic promoter

Answer 29

Core promoter that consists of a region that is recognized by the general transcription factors which are able to recruit RNA Polymerase Has both upstream and downstream regulatory sequences Bind gene-specific transcription factors that can act as activators (or repressors) of transcription at that locus

Question 30

differentiation

Answer 30

the process where a cell changes from one cell type to another

Question 31

transcription factors involved in the differentiation of cells

Answer 31

1. developing organisms starting out in a pluripotent state and move to a a higher state like a blood cell, brain cell, pancreatic cell, etc - they turn on specific subsets of genes to drive them towards their terminal fate (committed to a particular function, can no longer divide) - done by genes that regulate the transcription of other genes by binding to promoters and interacting with the transcriptional machinery 2. using skin fibroplast to change a cell's gene expression - involves induced pluripotency by pushing cell back to its embryonic stem cell state 3. cellular reprogramming - using genetic engineering to force the fribroblast t oexpress the TFs or providing the fibroblasts with the actual TF protein

Question 32

pluripotent state

Answer 32

stem cells that can differentiate into cells derived from any of the 3 germ layers

Question 33

the TATA binding protein

Answer 33

binds to TATA box near position -30 of the promoter region

Question 34

TATA box sequence

Answer 34

5'-TATAAAA

Question 35

purpose of TATA box/protein

Answer 35

plays a major role in transcription initiation

Question 36

how are TATA-binding proteins recruited

Answer 36

through proteins called TBP-associated factors (TAFs) that recognize other promoter sequences

Question 37

structure of the TATA binding protien

Answer 37

Saddle-shaped protein that is able to create a sharp bend in the DNA upon binding TBP sits on the DNA double helix like a saddle, with an extended beta sheet and loop stirrups in contact with the minor groove of the TATA box sequence

Question 38

is the structure of the TATA binding protein conventional

Answer 38

No, it is unconventional - most DNA-binding proteins recognize DNA by inserting alpha helixes into the major groove

Question 39

base pairs that are favoured in recognition sequence of TBP

Answer 39

A=T base pairs are favoured in the recognition sequence More easily distorted to allow opening of the minor groove

Question 40

Pol II promoters have variety of

Answer 40

transcription factors 1. TFIID (composed of TBP and TAFs) 2. FIIB 3. TFII proteins

Question 41

purpose of EMSA and DNA footprinting

Answer 41

to determine the presence of a transcription factor

Question 42

EMSA

Answer 42

electrophoretic mobility shift assay

Question 43

how EMSA works

Answer 43

1. Fragments of DNA of a known sequence are incubated with the protein of interest and then analyzed on a nondenaturing polyacrylamide gel 2. Visualized by staining with a dye or covalently attacking a radioactive phosphate group at one end 3. Free DNA fragments migrate more quickly through the gel than DNA bound by protein - A shift in migration of DNA from fast to slow in the presence of protein indicates a direct binding interaction between the protein and DNA

Question 44

EMSA results

Answer 44

an increase in protein concentration = DNA bound protein grows longer, as well as there is less unbound DNA

Question 45

what gel is used for EMSA

Answer 45

nondenaturing, since the addition of SDS (gel used for SDS-PAGE) would denature the protein, and disrupt formation of the DNA-protein interaction

Question 46

DNA footprinting goal

Answer 46

to map the exact nucleotide bases in contact with the bound protein once a direct DNA-binding interaction has been established

Question 47

DNA footprinting steps

Answer 47

1. DNA of interested is amplified and radiolabeled at one end. One of the two PCR primers must be radiolabeled on the 5' end to provide a point of reference for visualization 2. Cleavage occurs at sites that are not physically protected by bound proteins. Each piece of DNA is cleaved once, generating a set of fragments that represent all possible cleavage products (except DNA that is protected by the bound protein) 3. Resulting fragments are separated by gel electrophoresis and detected by exposing the gel to film which detects radioactive emissions from the labeled DNA fragments. Any gaps (DNA protected by bound proteins) produces a "footprint" that indicates the boundaries of the protein-binding site relative to the radiolabeled end of the DNA sequence of interest

Question 48

DNA footprinting results

Answer 48

footpring is identified by analyzing the sites of nuclease cleavage in the DNA before and after adding the protein

Question 49

processing of pre-mRNA in prokaryotes

Answer 49

Can be directly transcribed, sometimes while still being synthesized Directly translated by ribosomes No compartmentalization (dividing into sections) of these processes

Question 50

processing of pre-mRNA in eukaryotes

Answer 50

Transcription and pre-mRNA processing takes place in the nucleus Mature mRNA has to exit the nucleus to the cytoplasm to be translated

Question 51

structure of mature mRNA

Answer 51

5' cap 3' poly(A) tail coding region - composed of the exons spliced together with no introns included

Question 52

purpose of 5'capping

Answer 52

prevents degradation from the 5' end

Question 53

which polymerase helps with 5'capping and how

Answer 53

RNA Pol II synthesizes single-stranded ribonucleic acids that are vulnerable to degradation at either end by exoribonucleases

Question 54

what is the 5' cap made of

Answer 54

residue of 7-methyl-guanosine, 5'-5'-triphosphate linkage, and additional methyl groups

Question 55

5'-5'-Triphosphate Linkage

Answer 55

7-Methyl-Guanosine is linked to the 5'-terminal residue of the mRNA through an unusual 5',5'-triphosphate linkage Generated by an enzyme called guanylyltransferase (capping enzyme) which forms the 5' cap through condensation of a molecule of GTP

Question 56

additional groups added to 5' capping

Answer 56

methyl groups, added at the 2' hydroxyl of the ribose sugars at the first and second nucleotides adjacent to the cap

Question 57

steps of 3'Polyadenylation

Answer 57

1. RNA polymerase will transcribe the poly(A) site at the terminal end of the transcript - The mRNA site where cleavage occurs is marked by 2 sequence elements - Highly conserved sequence 5'-AAUAAA located 10 to 30 nucleotides on the 5' side (upstream) of the cleavage site - A less-well defined sequence rich in G and U residues located 20-40 nucleotides downstream from the cleavage site 2. Polyadenylation factors bind the poly(A) signal upstream of the cleavage site - Subsequent cleavage generates the free 3'-hydroxyl group that defines the end of the mRNA 3. A residues are immediately added to the frere 3'-OH of the reaction - Bound by poly(A) binding proteins (PABPs) which physically protect the 3' mRNA from exoribonucleases

Question 58

major difference between mRNA processing in bacteria & eukaryotes

Answer 58

splicing of pre-mRNA

Question 59

do prokaryotes have introns

Answer 59

no

Question 60

alternative splicing

Answer 60

exons in the primary transcript from a single gene are spliced together in several combinations to produce different mRNAs = different polypeptides

Question 61

what does alternative splicing give

Answer 61

different polypeptides = genetic diversity

Question 62

oligo dT primers

Answer 62

using a poly(T) sequence as a primer in the reverse transcription reaction - 3' poly(A) tail is not present on rRNA

Question 63

characteristics of the genetic code

Answer 63

1. Degenerate - Several codons have the same meaning - 64 unique ways to combine 4 different nucleotides in a triple codon sequence, but only 20 common amino acids - Multiple codons encode for the same amino acid - 61 coding amino acids, 3 specifies the termination of translation amino acids 2. Robust - Degeneracy provides DNA with the ability to absorb single-base mutations with minimal consequences to the protein sequences it encodes 3. Universal - All organisms use the same genetic code with a few minor modifications - Once the code evolved, it resisted change

Question 64

rules of the genetic code

Answer 64

1. It is non-overlapping 2. It does not contain pauses 3. The genetic code is read in triplets 4. It is read in a linear fashion 5. The code is resistant to mutations

Question 65

non-overlapping vs overlapping codons

Answer 65

non-overlapping: a single nucleotide substitution would change only 1 codon = 1 amino acid change overlapping: a single nucleotide substitution would change 3 codons = protein would have 3 consecutive amino acid changes

Question 66

the genetic code does not contain pauses experiment & result

Answer 66

Brenner and Crick used mutagens known as acridines to induce mutations in the B gene of the T4 bacteriophage - Acridines produce mutations through the insertion or deletion of a single base pair into DNA result: Most mutations completely inactivated the B gene - Suggested that not 1 but many amino acids were changed by these mutations

Question 67

frameshit mutation

Answer 67

insertion or deletion of 1 or more paired nucleotides, changing the reading frame of codons during protein synthesis

Question 68

translation is linear rule experiment

Answer 68

looked at the synthesis of hemoglobin, using extracts from rabbit reticulocytes (immature red blood cells)

Question 69

transition mutation

Answer 69

a purine is replaced by another purine or a pyrimidine is replaced by another pyrimidine

Question 70

transition mutations positions

Answer 70

1. third position - rarely cause a change 2. second position - determines whether it encodes a polar (purine) or hydrophobic (pyrimidine) amino acid - tend to conserve the chemical nature 3. first position - results in an amino acid change, but chemically similar to the original

Question 71

silent mutation

Answer 71

Single-base substitutions Non deleterious and non-harmful Do NOT result in an amino acid replacement

Question 72

missense mutation

Answer 72

Single-base substitution Replaces one amino acid with another Can be deleterious or non-deleterious

Question 73

nonsense mutation

Answer 73

Far more harmful than silent mutations Result in a termination codon Abort protein synthesis Results in an incomplete protein that is rarely functional

Question 74

non-stop

Answer 74

loss of a stop codon