Semester 1 Flow of Genetic Information Flashcards

Question

Supercoil removal via Topoisomerase Type I mechanism

Answer 1

Phosphodiester bond is transferred to Tyr residue on enzyme, breaking one DNA strand The unbroken strand is passed through the gap The phosphodiester bond is transferred back to DNA, reforming the backbone on the other side

Answer 2

Supercoiled DNA runs faster than linear DNA in an agarose gel

Answer 3

DNA replication is bidirectional, so with circular DNA the DNA Polymerases can collide with each other Also, as the DNA is seperated and expanded, two duplexes will be created that when finished, will be entwined with one another When circular DNA molecules are replicated, the two daughter rings are interlocked and form a CATANENE

Answer 4

Topoisomerase IV can be used Cleaves 1 duplex, the second duplex passes through and then the cleavage is sealed Process called decatenation

Answer 5

DNA strands are antiparallel: one 5'→3' and the other 3'→5' DNA polymerase can only add nucleotides in the 5'→3' direction The leading strand is continuously synthesized in the direction of the replication fork The lagging strand is synthesized discontinuously in short segments called Okazaki fragments To facilitate this, the lagging strand loops out, forming a structure that allows the DNA polymerase to synthesize the strand in the 5'→3' direction, but effectively in the opposite direction of replication fork movement

Answer 6

End result of semi-discontinuous replication is two, non-identical dsDNA molecules Chromosome is replicated, but lagging strand copy is unfinished, with ‘nicks’ in the new backbone, because Pol III can’t join fragments

Answer 7

Okazaki investigated the synthesis of DNA over time using radioactivity 1) E.coli culture infected with T4 bacteriophage 2) Add 3H-TTP (tritiated thymidine) so that the new DNA strands created will be radioactive 4) Mapped out sizes of radioactive ssDNA over time using density centrifugation 5) Small new strands of radioactive DNA were always present, as pictured 6) Even after full size strand had been created, new smaller strands were still present, proving size of radioactive ssDNA doesn’t simply increase smoothly over time. Short pieces are always present

Answer 8

Short fragments are repeatedly being made on the lagging strand as new sections of DNA are unwound After a few seconds they are joined to each other and to non-radioactive pieces that had already been synthesized Results in a size jump

Answer 9

Chase/Pulse experiments carried out After 2 seconds the 3H-TTP is removed and “chased” with normal TTP Radioactivity all moves down the tube as pictured The observations supported the hypothesis that the lagging strand is synthesised discontinuously in short fragments, which are later connected to form a continuous strand.

Answer 10

DNA Ligase consumes ATP to join Okazaki fragments of DNA together When DNA Ligase mutant is used instead of DNA Ligase, the fragments can not be joined together This results in density centrifugation only showing short, small okazaki fragments and no fully formed DNA fragments, as pictured

Answer 11

Not completely DNA Primer is RNA We know this because tiny fragments are left over after using DNAse on Okazaki fragments

Answer 12

1) Incorrect nucleotide added 2) RNA nucleotide instead of DNA nucleotide 3) Nicks in backbone (fragments)

Answer 13

The incoming dNTP has to form the correct base pair to fit properly into the polymerase active site DNA Polymerase has an induced fit mechanism Has binding and shape discrimination Incorrect dNTP are excluded by steric collisions, even if there are hydrogen bonds possible

Answer 14

Pol III adds the wrong dNTP at a rate of 1 per 100,000 bp (error rate of 10-5) Would be ~ 46 point mutations per E.coli replication Error rate drops to 10-7 with proofreading activity of polymerases

Answer 15

Proofreading in DNA polymerase is the enzyme's ability to detect and correct mistakes during DNA replication When DNA polymerase inserts an incorrect nucleotide opposite the template strand, it can recognise this error and correct it

Answer 16

Pol III has 3’-5’ exonuclease activity - Can remove the last nucleotide added, if it was incorrect A mismatched nt at the 3’ end won’t be in the right place to add the next dNTP so: - The polymerase stalls The end of the strand become substrate for the exonuclease active site, which cleaves the terminal phosphodiester bond, releasing a dNMP

Answer 17

An enzyme with multiple activities: - 5’-3’ polymerase (synthesis) - 3’-5’ exonuclease (proofreading) - 5’-3’ exonuclease (nick translation) When pol I is treated with protease it produces 2 main fragments: Small N-Terminal Fragment - Contains 5'-3' exonuclease Large C-Terminal Fragment (Klenow Fragment) - Contains polymerase - Contains 3'-5' exonuclease

Answer 18

Nick translation Binds to nick in synthesised dna fragment Will then remove nucleotides ahead of it and replace with correct nucleotide Doesn't remove nick, just moves it further along

Answer 19

Yes, there is RNA at the 5' end, and a nick at their 3' end Pol I binds to the nicks in in the fragments, but Pol III does not RNA primers are removed by Pol I 5'-3' exonuclease Pol I detaches after 100bp and a new nick is left behind Pol I starts joining together the fragments while Pol III is still expanding the strand

Answer 20

It is a protein that removes RNA in DNA strands It cant cut between RNA and DNA nucleotides, so leaves a gap between the end 2 Pol I comes along and fills the gap

Answer 21

Leading strand also consists of fragments that need to be joined together Strand is grown continuously and nicks are added in afterwards

Answer 22

The synthetic pathway for making dTTP, includes dUTP Enzyme 7 can sometimes add a dUTP instead of a dTTP During synthesis, Pol III will incorporate U instead of T, 1 out every 300 times (once every 1200 nt synthesized) by accident Since U shouldn’t be present in DNA it must be removed: - Nicks in chain - Fragments of ~1200nt If enzyme 4 is mutated: - More U & shorter fragments

Answer 23

When U is accidentally added in place of a G there is no problem When U is formed by deamination of C, mutations are caused Presence of U in DNA is therefore offensive as it suggest damage and therefore it must be removed

Answer 24

Base is removed by: - Uracil-N-Glycosylase (gene = ung) Baseless nucleotide is recognised and phosphodiester backbone cleaved by: - AP(apyrimidinic) endonuclease (xthA) Results in nicked DNA with incorrect nucleotide

Answer 25

DNA Ligase seals nicks in DNA Can't do RNA - DNA nicks (wont accidentally join in the RNA primer)

Answer 26

Circular chromosomes and plasmids have a single origin of replication (ori) The origin is a region of repetitive dsDNA that is rich in A-T Different in different organisms/plasmids e.g. oriC in E.coli chromosome oriC is about 245bp long, split into 2 main sections: - 3, 13-bp repeats - 4, 9-bp repeats

Answer 27

It is a protein that multiple of bind to 9bp repeats in the ori, and cause it to coil which in turn induces unwinding at the 13-bp repeats It uses ATP to separate (melt) the duplex at the 13bp repeats

Answer 28

DnaC is a protein that binds to the ssDNA created by DnaA and loads a DnaB helicase onto one strand, facing the 3' end The DnaC then detaches and the helicase moves to fork

Answer 29

After 65 nucleotides have been unwound by the DnaB helicases, DnaG primase enzymes bind to the helicases, forming a 'primosome' Primosomes are the functional units that create Primase

Answer 30

Creates a primer for DNA polymerase to bind to Primase is an RNA polymerase (but not the one involved in transcription) It is self priming, adding RNA 5'-3' on a 3'-5' DNA strand (GTA site preferred) RNA primers are -10 nucleotides in length It has no editing functions (no proof reading) Activity is increased in the presence of helicase and itself (co-operativity) Primase synthesises a 10 nucleotide RNA primer and detaches from helicase

Answer 31

Single stranded binding protein (SSB) binds to exposed ssDNA preventing re-annealing - Encoded by ssb gene - Forms a tetramer - Not sequence specific - Leaves bases exposed when bound - Binds co-operatively to ssDNA - Therefore, proteins at the ends are bound less well and are easier to displace by polymerases

Answer 32

First primer and SSB trigger arrival of the Pol III holoenzyme at the 3’ end of the primer

Answer 33

The clamp loader on Pol III loads a beta-clamp onto the DNA Pol III core then binds to the beta-clamp

Answer 34

Binds β clamp proteins Transfers the β clamp onto DNA at primer 3’ end ATP Hydrolysis is required to detach the clamp loader from the beta-clamp

Answer 35

Encoded by dnaN gene Forms a ring dimer Not sequence specific Binds to the Pol III core and imparts processivity Changes ability of Pol III from doing 10s of basepairs at a time to >50,000 Pol III on its own isnt any better than Pol I at staying on DNA, beta-clamp improves this

Answer 36

Pol III travels to replication fork, synthesising the leading strand as it goes and displacing SSB As it catches the helicase, a replisome forms

Answer 37

It is a region around the replication fork that contains these proteins: - Pol III Holoenzyme - Primosome It occupies around 50nm area around the replication fork

Answer 38

As helicase unwinds the duplex, primase re-binds and synthesises a new primer A β clamp is added to the primer by the clamp loader A Pol III core binds once enough ssDNA has emerged for the β clamp to reach it Primer 1 bound, with correct polarity

Answer 39

First Okazaki fragment starts DNA is pulled by helicase and Pol III The lagging strand loops out, picking up SSB The first Okazaki fragment is finished

Answer 40

Primase re-binds helicase, then adds a second primer Pol III core and β clamp detach from DNA, releasing completed fragment Primer 2 gets a β clamp Looping process repeats for primer 2

Answer 41

Pol I binds the end of the first Okazaki fragment and replaces the RNA with DNA DNA ligase seals the nick Process repeats for each fragment

Answer 42

Ter and Tus To prevent the forks overshooting there are 23 bp sequences called Ter The Ter sequences bind the Tus protein Tus can be displaced by the fork only in the correct direction, otherwise the helicase stalls

Answer 43

As the forks get within 200bp of each other, there is no longer room for DNA gyrase to bind This results in positive supercoiling, which is relieved by topoisomerase IV decatenating the molecules The topoisomerase IV binds to one of the strands of dsDNA, cuts it, passes the other dsDNA strand through it, and then rejoins the original strand

Answer 44

OriC contains GATC sequences (within the 13bp repeats), which are substrates (binding sites) for dam (DNA adenosine methylase) The 'A' within the GATC sequences in the 13bp repeats will have a methyl group attached to it dam methylates N6 of adenosine After replication, only one strand will be methylated (template strand) (referred to as hemimethylated, one strand methylated, one strand not methylated) Hemimethylated GATC sequence bind SeqA protein

Answer 45

Prevents DnaA binding to OriC The GATC sites are methylated by dam very slowly (~13 mins) So newly synthesized dsDNA remains hemimethylated and new initiation is prevented SeqA also binds DNA to the membrane, localising it to the membrane (assists with separating two chromosomes during cell division)

Answer 46

Both use helicases to unwind the duplex and create replication forks Both use SSBs to hold the ssDNA apart Both use RNA primers for the polymerase/clamp Both use various DNA polymerases for synthesis of the new DNA on leading and lagging strands

Answer 47

Eukaryotic occurs in the nucleus not the cytoplasm Eukaryotes have much more genetic material to replicate (can be 4 orders of magnitude greater) More than 1 chromosome in eukaryotes Linear chromosomes (except mitochondria) in eukaryotes Eukaryotes have more packaging (nucleosomes, histones) Eukaryotic replication has multiple origin points (10s-1000s), replication starts from multiple points DNA polymerases are much slower in eukaryotes (50nt/s vs 1000nt/s) Polymerases synthesising the leading and lagging strands aren't physically linked together in eukaryotes Eukaryotes do NOT have any polymerases with a 5'-3' exonuclease Okazaki fragments are much shorter in eukaryotes (165nt vs 2000nt)

Answer 48

Tightly linked to the cell cycle Initiation must only happen once per cycle A two step process ensures this ^

Answer 49

G1 Phase (Gap 1): - This is the first phase of the cell cycle - During G1, the cell grows in size, carries out its normal functions, and prepares for DNA replication. - At the end of G1, the cell decides whether to continue the cell cycle and divide or enter a non-dividing state called G0. S Phase (Synthesis): - In the S phase, DNA synthesis occurs - The cell replicates its genetic material by duplicating the chromosomes G2 Phase (Gap 2): - After DNA replication in the S phase, the cell enters the G2 phase. - During this phase, the cell continues to grow, synthesizes proteins, and prepares for mitosis (or meiosis if it's a germ cell). M Phase (Mitosis or Meiosis): - The M phase is when the actual cell division occurs - It includes mitosis or meiosis

Answer 50

Origins of replication arise from sections of DNA called Autonomously Replicating Sequences (ARS) Autonomously Replicating Sequences (ARS) are separated by ~30kb, so range from 10 on small chromosomes, to 1000s on the largest ARS contain an AT-rich consensus sequence In yeast (A region): 5'- (T/A)TTTAYRTTT(T/A) -3’ Where: - Y = any pyramidine - R = any purine

Answer 51

In yeast, an Origin Recognition Complex of proteins (ORC) binds to the A region of the ARS It is unclear how this works in higher eukaryotes

Answer 52

Accessory proteins (licensing factors) accumulate in the nucleus during G1 Cdc6 and Cdt1 (examples of licensing factors) bind to ORC

Answer 53

Two helicases are loaded by the licensing factors The licensing factors then leave Pre-replication complex is formed

Answer 54

During the S phase, additional proteins + DNA polymerases are added, and the pre-replication complex is activated This results in a active initiation complex/replisome progression complex

Answer 55

Not every pre-replication complex is used in each round of replication Replication forks from one origin will pass through another origin (passive replication)

Answer 56

5 main eukaryotic DNA polymerases: α, β, γ, δ, ε Polymerase α is involved in making primers Polymerase β is involved in repair Polymerase γ replicates mtDNA Polymerase δ (lagging strand) associates with PCNA and has proofreading Polymerase ε (leading strand) associates with PCNA and has proofreading Pol δ & Pol ε can displace RNA primers, but can’t break phosphodiester bonds, and cant release them as nucleotides

Answer 57

Polymerase α has its own primase activity, as well as polymerase, so can make its own primers (main role is making primers) It has no proofreading 3’-5’ exonuclease It is not very processive since it does not associate with the eukaryotic sliding clamp protein called PCNA (proliferating cell nuclear antigen)

Answer 58

It is the eukaryotic sliding clamp protein PCNA Replication factor C (RPC) loads the PCNA onto the primer/DNA ready for a different Pol to bind

Answer 59

Pol δ & Pol ε displace the primers and push them to the side generating an RNA 'flap' The RNA "flap" produced by Pol δ or ε can be mostly digested by RNAse H1, with 1 RNA nucleotide left at the end Flap 1 endonuclease (FEN1) binds PCNA and can remove any incorrect nt, by cutting off a section of ssDNA/RNA DNA ligase fills in nick generated by FEN1

Answer 60

Affects the lagging strand Last RNA primer may not be at the extreme 3’ end of the DNA which results in a missed section

Answer 61

They are the end replication solution The DNA at the 3’ ends of chromosomes doesn’t encode genes Instead it is multiple repeats (100s-1000s) of a simple sequence: TTAGGG The last 20-200nt (species dependent) at the 3’ end are ssDNA made of these repeats These repeated ends are called telomeres The more the chromosomes are replicated, the shorter the telomeres become

Answer 62

The Hayflick limit is the maximum number of times normal human somatic cells can divide As the chromosome is replicated, the telomere shortens, until there is no telomeric DNA Once telomeric DNA is gone, gene loss occurs and the cell stops dividing

Answer 63

They can synthesise new telomeric DNA Telomerase is a ribonucleoprotein It contains an RNA strand (450 nt) which has the sequence CUAACCUAAC This acts as a template for the synthesis of new telomeric repeats, growing the telomere

Answer 64

There are 2-10 copies of circular mitochondrial DNA per mitochondrion Mitochondrial DNA undergoes unidirectional replication (1 replication fork) Pol γ synthesises the leading strand Lagging strand is unknown but thought to be RNA Okazaki fragments

Answer 65

They have a small circular genome Uses a 'rolling circle' mechanism to continuously synthesise new DNA Called Sigma (σ) replication - Start off with circular DNA molecule - Nick is deliberately created in DNA - Nick acts as binding site for polymerase - Pol displaces existing strand and replaces with newly synthesised one

Answer 66

Many animal and plant viruses have genomes composed of RNA Have an RNA-dependent RNA polymerase called RNA replicase encoded by the viral genome The plus strand RNA is copied directly to make the minus strand, which is used as a template for making more plus strand Can be self-priming – no primer required No proof-reading meaning it is highly error prone

Answer 67

Contain RNA genome Virally encoded reverse transcriptase creates a DNA strand using RNA as template and tRNALys as a prime RNA half is degraded by RNase H Second DNA strand synthesized using first as template and is then incorporated into host genome

Answer 68

Something wrong with DNA that may lead to: - Mutations - Unable to replicate

Answer 69

Abnormal base pairs Chemical adducts Base pair mismatches Double-stranded breaks Single-stranded breaks Abasic site Thymine dimers DNA cross-links Nucleotide insertions Nucleotide deletions

Answer 70

Replication errors Tautomerisation Deamination Depurination

Answer 71

Intercalating agents Base analogues Deaminating agents Alkylating agents Oxidising agents Radiation, U/V

Answer 72

Normal replication introduces the wrong base once every 10^7 bp, even with proofreading There is a good chance this error will be repaired Repair systems reduce error rate to 10^-10 Some repetitive regions cause slippage of the growing strand insertion of more repeats which can't be repaired Slippage is particularly prevalent with trinucleotide repeats

Answer 73

Tautomerization is a reversible chemical process in which a DNA base temporarily changes its structure, leading to the incorrect pairing of nucleotides during DNA replication In the example, the shifted enol form of T binds G instead of A The mutations are consolidated on next replication

Answer 74

It is the loss of a base amino group C --> Uracil (pairs A) A --> Inosine/Hypoxanthine (pairs C) G --> Xanthine (pairs C less strongly) C, A and G deaminations can be repaired - Around 100 C are converted to U per day - Around 1 a day per cell for A and G

Answer 75

It is a regular Cytosine base that has had a methyl group attached to it When deaminated, 5-Methyl-C turns into a T This T generated, can not be distinguished from a normal T, and therefore cannot be repaired

Answer 76

Occurs mainly in A or G It is the cleavage of base-sugar bond Forms an abasic site (apurinic) Around 10,000 purine glycosidic bonds hydrolyse/cell/day Around 600 pyrimidine glycosidic bonds hydrolyse/cell/day 4x more likely in ssDNA Mutation is consolidated on next replication

Answer 77

Intercalating agents Base analogues Deaminating agents Alkylating agents Oxidising agents

Answer 78

They are chemicals that insert themselves between bases, and distort the DNA They look similar to bases, so are able to insert themselves e.g. ethidium bromide Intercalating agents cause frameshift mutations - Insertions - Deletions

Answer 79

These are chemicals that are very similar in structure to bases For example, bromouracil (T analogue) (pictured) Because the structure is so similar, they are incorporated into the DNA These base analogues are more prone to tautomeric shifts

Answer 80

Alkylating agents add carbon groups (alkyl) to nucleotides e.g: - Nitrosamines - Methyl bromide Most common is G --> O6-Methyl-G, which is G with an additional methyl group O6-Methyl-G pairs T Alkylation can also speed up depurination

Answer 81

They are chemicals that remove amino groups from the bases e.g: - Nitrous acid - Nitrosamines - Nitrite - Nitrate Far quicker than spontaneous deamination

Answer 82

Cause of most mutations Any chemical that makes bonds or breaks bonds e.g: - Free radicals - Superoxide ions (O2-) - H2O2 Many possible nucleotide alterations e.g. (OH- free radical) + G ---> 8-oxo G 8-oxo G pairs A

Answer 83

It can cause UV-induced formation of dimers between adjacent thymines Stops correct base-pairing in polymerases

Answer 84

Breaks bonds & creates free radicals cause: - Single or double-strand breaks - Bases chemically altered, linked or detached This is the leading source of mutation Number of mutations is in direction proportion to the radiation dose

Answer 85

Damage can be directly repaired Damage that can’t be directly repaired: → Remove and replace the affected DNA

Answer 86

Sequence repair - Direct reversal/repair - Base Excision Repair (BER) - Nucleotide Excision Repair (NER) - Mismatch Repair (MMR) Molecule repair - Homologous recombination - Non-homologous end repairH

Answer 87

Errors in sequence must first be detected by scanning the DNA Checking if it’s: → Not a DNA nt → The wrong DNA nt for the base pair Specialised proteins either directly reverse the error, or edit the DNA to reset that section of sequence

Answer 88

Where the damage has converted A,C,G,T to something else, it may be possible to convert back to the original nt e.g. Demethylation following alkylation e.g. Removal of crosslinks following UV light damageH

Answer 89

Occurs via specific reversal proteins e.g. O6-methylguanine-DNA methyltransferase (MGMT) Transfers methyl/ethyl group from G to a Cys residue on itself: → G restored

Answer 90

Via DNA photolyase Absorbs blue light and breaks T-T internucleotide bonds, using FADH → 2 Ts restored Mammals have to use another system for repairing thymine dimers,as they don’t have this enzyme

Answer 91

It is a removal of individual bases (local correction) Group of >6 DNA glycosylases which recognise abnormal bases and cleave them from the deoxyribose, creating an abasic site Glycosylases flip out bases for closer checking If the base is found to be incorrect, the sugar-base bond is cleaved, but UDGase remains attached to DNA

Answer 92

T has a methyl group and the U doesn't That methyl group clashes with the tyrisine residue on the active site There is a separate TDGase for GT pairs If Tyr is mutated to Ala, both T and U will be removed by the enzyme

Answer 93

U in DNA --> Base removed by Uracil-N-glycosylase (ung) --> Baseless nt recognised and phosphodiester backbone cleaved by AP(apyrimidinic) endonuclease (xthA) --> Results in nicked DNA --> Pol I nick translation restores T + DNA ligase seals nick (pol beta in eukaryotes)

Answer 94

Removal of oligonucleotide fragments from one strand (bulk correction) Triggered by changes in the physical structure of the duplex as a result of damage Corrects any of the types of sequence damage e.g. thymine dimer removal Achieved by a protein complex called UvrABC exinuclease in E.coli (many more proteins in eukaryotes)

Answer 95

Detection and removal of incorrect base pairs Principle: - A mismatched pair will distort the helix - This can be detected by specialist proteins - Incorrect base removed It must somehow know which one of the two is the wrong one: strand-directed mismatch repair

Answer 96

Hemi-methylation provides the information on which strand is parent (correct) and daughter 1. MutH binds to unmethylated GATC at OriC, identifying the daughter strand. 2. MutS binds to a distorted site on the duplex 3. MutL binds to MutS 4. MutL/MutS complex travels back to the origin and activates MutH 5. MutH cleaves daughter strand (nicked) 6. Specialized helicase and exonucleases remove nt until past the distortion 7. Pol III fills in missing nt. DNA ligase seals nick

Answer 97

Eukaryotes have several homologues of MutL and MutS e.g. MSH1-6 No homologues of MutH, but they don’t use hemimethylation replication tags either

Answer 98

Sometimes a risk of mutation is better than the alternative, so there are polymerases that can add nt where processive, proof-reading polymerases cannot e.g. Pol IV&V in E.coli and at least 8 Pols in eukaryotes → Blockage is bypassed, but the nt added may not be correct

Answer 99

1) Homologous recombination - Following replication, while the sister chromatids are still joined one can be used as a template to repair the other. 2) Non-Homologous end-joining (NHEJ) - A protein complex binds the naked ends of duplex fragments and recruits DNA ligase IV, which can ligate both strands – but it does it blindly, to any two pieces of DNA, with loss of some nt

Answer 100

Individuals show dry, parchment-like skin (xeroderma) and many freckles (pigmentosum) Increased sensitivity to UV light 1000-fold increased risk of skin cancer --> Due to inherited defects in one of eight distinct genes responsible for components of the NER complex

Answer 101

Individuals exhibit a predisposition to colon cancer (2-3% of all colon cancer cases) Due to defects in the human equivalents of the MutS/L MMR system (MSH2 and MLH1) Leads to the accumulation of mutations throughout the genome

Answer 102

DNA contains the code for proteins DNA is in the nucleus Proteins are made in the cytoplasm Requires mRNA

Answer 103

Replication + transcription + translation all happens in the cytoplasm in prokaryotes

Answer 104

- Better regulation of the protein production process - Fidelity of DNA replication is very important. Having an intermediate step prevents interference and potential additional mistakes - Proteins can still be made when DNA is being replicated.

Answer 105

RNA is a linear polymer like DNA Residues linked by phosphodiester bonds Contains ribose not deoxyribose RNA is single stranded Contains: A (adenine) G (guanine) C (cytosine) but not T (thymine); uses U (uracil) instead U base-pairs with A (just like A:T base pair)

Answer 106

Although RNA is single stranded, it can fold into specific structures This involves base-paring and covalent bonds This allows some RNA molecules to have structural and catalytic functions

Answer 107

rRNA: Ribosomal RNA, form the basic structure of the ribosome and catalyse protein synthesis tRNA: Transfer RNAs, central to protein synthesis as adaptors between mRNA and amino acids Non-coding RNA, e.g. microRNAs, long non-coding RNAs mRNA: Messenger RNA, codes for proteins (3-5%)

Answer 108

1. Transcription control Regulation of transcription is the most common form of control of protein production 2. RNA processing control 3. RNA transport and localization control 4. Translation control 5. mRNA degradation control 6. Protein activity control (Long non-coding RNA) - (>200 nt, not transcribed. Involved in regulation through binding RNA, proteins and DNA thus influencing important interactions)

Answer 109

Separates the 2 strands of DNA Uses one of the DNA strands as a template for RNA synthesis Does not require a primer It is very accurate 1/10,000 bases Moves along the gene in a 5’ to 3’ direction Synthesises a complementary RNA copy of the DNA template strand

Answer 110

Many RNA copies are made at the same time Multiple RNA polymerases per gene Multiple transcripts per gene - As soon as the first RNA polymerase starts to move down the gene, the next polymerase can bind and initiate RNA transcription

Answer 111

RNA Polymerase binds, has helicase activity to begin with, breaking hydrogen bonds and unwinding DNA Does this as it moves in 5' to 3' direction, forming a transcription bubble, which opens up so that the template strand can be copied Free nucleotides join and are added to complement the template strand and generate a copy of the coding strand Length of the bubble is 12-14bp Reaction rate approx 40 bases per sec

Answer 112

1. Initiation - Template recognition - Polymerase locates promoter - Polymerase unwinds the DNA - Transcription begins 2. Elongation - Polymerase places RNA in exit hole - Once RNA is 10 or more base pairs long - Polymerase conformation change - tightens grip 3. Termination - Termination sequence - Polymerase separates - RNA released

Answer 113

Differences in initiation Differences in promotor Different RNA Polymerases

Answer 114

Constitutive Genes: – Housekeeping genes Regulated genes: e.g. Changes in food sources Switches on genes that encode enzymes which are needed to metabolise that sugar e.g. Changes in environmental stresses (pH and temp) Switches on genes that encode proteins which help the bacterium survive

Answer 115

Single unit that controls multiple genes Genes encoding for proteins in the same pathway are located adjacent to one other and controlled as a single unit that is transcribed into a polycistronic RNA - No introns

Answer 116

Deals with changes in lactose Switches on genes that encode enzymes which are needed to metabolise that sugar LacZ --> LacY --> LacA Contains three genes: - B-Galactosidase - Permease - Transacetylase

Answer 117

Control of transcription most commonly occurs at transcription initiation Promoters are the sites of transcription initiation in the DNA Promoters are recognised by RNA polymerase by having a consensus (common pattern of) DNA sequence The first is a hexamer (6bp) at -35 Second is a TATAAT sequence at -10 (Pribnow box) - Asymmetric (only in 1 DNA strand), hence RNA polymerase knows which way to go!

Answer 118

Holoenzyme (complete enzyme) consists of 5 types of subunit (6 units) - 2 alpha subunits (40 kD) – enzyme assembly - beta and beta' = form catalytic centre - Omega subunit – enzyme assembly and stability - Sigma factor (32- 90 kD) – binds promoter Core enzyme = 2 alpha, beta, beta', omega The core enzyme has a general affinity for DNA- this is known as loose binding positively charged (Mg2+ and Zn2+ bound ions which has affinity for the negatively charged DNA The σ (sigma) unit ensures RNA polymerase only binds at promoter sequences It increase 1000 X binding strength There is enough σ unit for 1/3 of all of the RNA polymerases

Answer 119

e.g., Heat Shock σ32 is induced by high temperatures σ32 is induced by the accumulation of unfolded proteins σ32 recognises a different -35 and -10 sequence

Answer 120

Lag Phase - No increase in number of living cells Log phase - Exponential increase in number of living bacterial cells Stationary phase - Plateau in number of living cells; rate of cell division and death roughly equal Death or decline phase - Exponential decrease in number of living bacterial cells sigma70 in log phase sigma38 replaces it near stationary phase

Answer 121

A regulon is a group of genes that are regulated as a unit This is different from an operon. Commonly studied regulons in bacteria are those involved in response to stress such as heat shock. The heat shock response in E. coli is regulated by the sigma factor σ32 (RpoH), whose regulon has been characterized as containing at least 89 open reading frames. Hence, a regulon involves a more diverse set of genes that are regulated at the same time.

Answer 122

Initiation: RNA polymerase binds to the promoter - Consensus DNA sequence . - Regulates transcription. 2. Elongation: RNA polymerase makes an RNA copy complementary to the template strand. - Bacterial polymerase: only 1, with 5 units (2x a, b, b’, w ,σ). - Operons, no introns, no post-transcriptional modification of RNA. 3. Termination: RNA polymerase stops when it recognises termination sequences (prokaryotes).

Answer 123

RNA polymerase binds to the promoter Unwinds the DNA Transcription begins Regulates transcription

Answer 124

RNA polymerase makes an RNA copy complementary to the template strand Bacterial RNA polymerase: only 1, with 5 units (sigma)

Answer 125

RNA polymerase stops at a termination sequence RNA polymerase separates RNA released

Answer 126

Transcription is controlled by regulatory proteins (transcription factors) a) Negative regulation - transcriptional repressors - Repressor binds to sites known as Operator sites - Stops RNA polymerase binding b) Positive regulation - Transcriptional activators - Activator binds to specific site - Helps RNA polymerase bind

Answer 127

Regulated by both negative and positive regulators Negative = lac repressor (lacI) Positive = Catabolite activating protein (CAP) LacI, the lac repressor, is said to be constitutively expressed hence it is always on hence lacI is always present The lacI gene encodes the lac repressor The lac repressor is the protein that binds down stream and its job is to control transcription of the lactose operon

Answer 128

Lac repressor is produced Lac repressor binds to the operator Lac operon transcription blocked

Answer 129

Allolactose (derived from lactose metabolism) is produced Allolactose binds to the lac repressor This induces a conformational change in LacI Repressor cannot bind to operator

Answer 130

Lac repressor binds to the operator site which overlaps the start site of transcription The sequence where lac repressor binds and the first A is where transcription starts and this is labelled plus 1

Answer 131

Lac repressor binds to the operator site which overlaps the start site of transcription The lac repressor must bind 2 out of 3 sites - Repressor can bind to O1 and O2 or O1 and O3 - But not to O2 and O3

Answer 132

When glucose is available it is used in preference to other sugars (e.g., lactose) Low Glucose --> cAMP High --> CAP Active High Glucose --> cAMP Low --> CAP Inactive When cAMP is high it binds to CAP and causes conformational change allowing it to bind upstream of the -35 sequence and stabilizes RNA polymerase

Answer 133

Termination sequences in DNA tell RNA polymerase when to stop transcription Also known as intrinsic terminators These are run of A-Ts in the template strand in around 50% genes

Answer 134

Stem loops formed by inverted repeats E.g. CCCCXXXGGGG Inverse sequences result in folding of strand You find these in the 3 prime control region of a gene The stem-loop structure causes RNA polymerase to pause When the Polymerase pauses the RNA:DNA hybrid unravels from the weakly bonded A: U terminal region The sequence of the hairpin and length of the U rich region determine the efficiency of termination (2 -90%)

Answer 135

Rho is a prokaryotic transcription protein ~275kD hexamer Each subunit has a RNA binding domain & an ATP hydrolysis domain: moves along the RNA Also requires an inverted repeat Stalls the RNA polymerase Rho is a helicase Unwinds DNA:RNA hybrids

Answer 136

Expressing all genes at once is a waste of resources. Biology prefers not to waste energy as organisms are always in competition. Those organisms that can save energy while still carrying out their essential functions have an advantage over other competing organisms. Overproduction of some proteins are likely to be toxic. A simple example is ion channels or transporters. Knowing how systems are regulated allows us the ability to manipulate the system. Protein over-expression systems are used extensively in the Biotech/Pharma industries. Development and growth requires certain groups of proteins to be present at certain times. This can define how cells ultimately become tissue, how tissues become organs and how organisms are formed. Responding to external stimuli can require changes in the proteome. For example, we all produce more melanin when exposed to UV light, apart from the congenital absence of melanin in an animal (albinism) which is known to reduce the survivability of animals.

Answer 137

Some genes are constitutively expressed - on all the time e.g., lacl. Other genes are regulated in a positive and negative way e.g., see details of the lac operon.

Answer 138

Prokaryotes: - Lagging strand Eukaryotes: - Leading strand

Answer 139

Eukaryotes seperate transcription (nucleus) and translation (cytoplasm) which allows for more control and complexity

Answer 140

Eukaryotes have 3 types of RNA polymerases RNA Pol 1 - Structurally important RNA Pol 2 - All genes that encode proteins - snoRNA (small nucleolar - non-coding) - snRNA (small nuclear - pre-mRNA splicing) RNA Pol 3 - Structurally important

Answer 141

Core promoter elements: - TATA Box (-25) - Inr (+1) - DPE (+25) Regulatory elements: Proximal - CAAT Box - GC Box Distal - Enhancer Elements

Answer 142

Only 30% of genes have TATA Box Other 70% lack it and are known as TATA-less These genes have an Inr (initiator) and a DPE (downstream promoter element) usually located downstream (generally +28 and +32) and contain sequence AGAC Recognised by TF2I

Answer 143

TATA Box and INR (initiator) make up the 'Core Promoter Region' INR: - Simplest functional promoter - Can initiate basal transcription in absence of TATA box - Conserved Y is pyrimidine - Consensus sequence YYANWYY in humans, where, Y = C/T, W = A/ T, N=A/C/G/T, and +1 is underlined. - The INR element facilitates binding of TFIID TATA Box: - Conserved sequence TATAAAA - Approx 25 bp upstream of initiator - Found in ~30% of promoters - Core promoter similar concept in Pribnow box (prokaryotes) and TATA box (eukaryotes).

Answer 144

Eukaryotic RNA Pol II does not bind directly to the TATA box region of the DNA, it requires proteins called "General Transcription Factors (GTF)" to position it Each transcription factor is a complex of polypetides TFIID begins whole process Need to assemble an RNApol II pre-initiation complex This positions the RNA pol II over transcription start sites.

Answer 145

General transcription factors for RNA Pol II (TFII) bind to the TATA box The first transcription factor added to the TATA box is TFIID The TFIID complex consists of TBP (TATA binding protein) and TAF (TATA associated factors) and acts as a scaffold for remainder of the preinitiation complex Next, TFIIA binds followed by TFIIB TFIIB is able to then recruit RNA Pol II to the TATA Box TFIIF accompanies the RNAPol II and stabilises it TFIIE and TFIIH bind and make up the initiation complex Pre- initiation complex (PIC) is now assembled

Answer 146

The function of TFIIH is a helicase, it melts the DNA and separates the strands A short sequence of pre-mRNA is added to the gap TFIIH and TFIIE also have kinase activity Phosphorylation of the C terminal domain of RNApol II by TFIIH then occurs After phosphorylation, TFs are released from the complex The phosphorylation also causes a conformational change in RNApol II, which tightens the grip it has, also allows binding of new factors (elongation factors) that increase efficiency

Answer 147

Cell fate is determined by: - Gene expression - Cell-cell/cell-matrix interactions - External factors e.g. hormones WHICH CAUSES: - Unique transcriptional factor (TFs) environment for each cell type

Answer 148

Transcription can be enhanced by the binding of transcription factors to sites upstream of the PIC Upstream sequence elements (USE) Examples: GC Box: - GGGCGG - Binds SP1 TF CAAT Box: - GG(T/C)CAATCT - Binds CAAT box TF These must be in same orientation as the RNA pol II initiation site Upstream position is important.

Answer 149

Regulatory sequences that act at a distance Cis acting (up to 1Mb away) with reversible orientation Bound by activator proteins These interact with the mediator complex Encourage binding of RNApol II Enhancers work because DNA structure is complex and twisted around itself, so regions distant in sequence may be physically close to each other

Answer 150

- They can activate transcription when placed thousands of bp away from the TATA box - They act in either orientation - Can act when placed upstream or downstream of the TATA box, or when placed within an intron

Answer 151

Determine whether transcription occurs Determine cell specificity Confer response to specific timed stimuli Structure is central, determined by the exact sequence of amino acids Mechanisms: - How they initiate transcription can be difficult to determine due to TFs being expressed at low levels in cells

Answer 152

TFs are proteins made up of amino acids hence their 3D structure is important to their function Modular structure: - One region binds DNA - All the amino acids for DNA binding in 1 region - Another region binds to other components - Also contains activation or inhibitory domains

Answer 153

- Zinc fingers - Helix turn helix - Basic binding domains Binding mostly based on a-helices fitting into DNA grooves

Answer 154

Contains a loop of 23 aa Usually have multiple zinc fingers per TF The linker between the fingers is 7-8 aa a-helix contacts the major groove of DNA Often multiple zinc fingers involved in binding the specific DNA sequence. Zn2+ ion does not directly interact with the DNA but is essential for the folding of the finger. Zinc fingers bind both to the major and minor grooves.

Answer 155

Two helices held at a fixed angle Recognition helix binds major groove of DNA Bind DNA as dimers, so the 2 recognition helices are separated by one turn of the DNA helix

Answer 156

Transcription factors with basic binding domains cannot bind to DNA alone Transcription factors with basic binding domains must dimerise