BIOL2010 “Flow Of Genetic Information” Flashcards

Question

W = +ive

Answer 1

Left DNA is overwound

Answer 2

Right DNA is underwound

Answer 3

Sigma = (Lk - Lk^o)/Lk^o = delta (Lk) /Lk^o Normalised way of expressing how supercoiled a piece of DNA is, removing the effect of chain length In relaxed DNA, sigma = 0 Sign indicates type of supercoiling as before

Answer 4

Purified cellular DNA Sigma = -0.06 (euk & pro) Conversion of -ive writhe to less twist aids unwinding for transcription and replication Euk DNA -ively supercoiled around histones when forming a nucleosome

Answer 5

Helicase-based unwinding—>overwinding elsewhere Overwinding will resist replication fork movement Lk can’t change to relieve the stress without breaking the phosphodiester bond —> +ive supercoils form Also applies to very long linear DNA Needs to be removed for replication to continue

Answer 6

Cause a change in Lk Type I: cleave backbone of one strand, allowing duplex rotation and loss of -ive supercoils Type II: cleave backbone of both strands, using ATP, and introduce a -ive supercoil Subtypes: IA, IB, IIA, IIB DNA backbone reseated after manipulation

Answer 7

Phosphodiester bond is transferred to Tyr residue on enzyme, breaking one DNA strand Unbroken strand passed through the gap Phosphodiester bond transferred back to DNA, reforming the backbone on the other side Lk + 1

Answer 8

Horizontal section cut, 5’-P linked to Tyr on A subunit Vertical section passed through Introducing a -ive supercoil cancels out a +ive one —> vital for replication fork progression Lk - 2

Answer 9

Bidirectional synthesis When circular DNA is replicated, the two daughter rings are interlocked and form a catanene Topoisomerase IV: cleave, pass through, and re-seal (decatenation) - to separate rings and complete the new DNA

Answer 10

Two interlocked daughter rings of circular DNA formed by the bidirectional replication of circular chromosomes

Answer 11

Best solution: 2 polymerases, one for each direction —> extra genes needed —> loss of proofreading/editing 3’—>5’ (mutation rate increased by 100x) All DNA polymerases synthesise DNA by adding to the 3’ end of an existing ds region (primer) - works but requires complete strand separation before synthesis can start on upper strand - wasted time (separation time + synthesis time) —> loop back the lagging strand to correct direction Synthesis of both strands can occur simultaneously but semi-discontinuously

Answer 12

End result of semi-discontinuous replication is two, non-identical dsDNA molecules Leading strand has 1 primer, lagging strand has multiple primers Chromosome is replicated, but lagging strand copy is unfinished, with ‘nicks’ in the new backbone, because pol III can’t join fragments

Answer 13

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 synthesised —> size jump

Answer 14

Pulse/chase experiments - After 2s the ^3 H-TTP is removed and chased with normal TTP —> radioactivity all moves down the tube DNA ligase mutants - DNA ligase seals nicks in the backbone and can join Okazaki fragments to each other. ATP consumed - Short fragments still made, but they remain as fragments —> lagging strand can’t be finished

Answer 15

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

Answer 16

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

Answer 17

Incoming dNTP has to form correct base pair to properly fit into polymerase active site —> Induced fit —> Binding + shape discrimination Incorrect dNTP are excluded by steric collisions, even if there are hydrogen bonds possible Pol III adds wrong dNTP at a rate of 1 per 100,000 bp —> drops to 1 per 10,000,000 with proofreading activity of polymerase

Answer 18

Pol III has 3’—>5’ exonuclease activity —> can remove the last nt added if it was incorrect Mismatched nt at 3’ end won’t be in the right place to add the dNTP —> polymerase stalls The end of the strand becomes substrate for the exonuclease active site, which cleaves the terminal Phosphodiester bond, releasing a dNMP

Answer 19

Small N-terminal fragment (5’—>3’ exonuclease) Large C-terminal fragment (Klenow fragment) containing polymerase and 3’—>5’ exonuclease

Answer 20

3’—>5’: shortened strand re-engages polymerase active site —> synthesis resumes 5’—>3’: Nick translation—> can remove and replace nt ahead of it

Answer 21

rNTPs can’t be directly incorporated into growing DNA strands because the extra OH in ribose causes a steric clash

Answer 22

The fragments that need to be joined together in the leading strand of DNA

Answer 23

Pol III will incorporate U instead of T every 1 in 300 times (1 in 1200 nt synthesised) Since U shouldn’t be present in DNA it must be removed —> nicks in chain —> fragments of ~1200 nt If enzyme 4 is mutated, more U & shorter fragments

Answer 24

It suggests damage that has lead to mutation (C can be deaminated to form U)

Answer 25

Uracil-N-glycosylase (ung) - removes U —> baseless nt recognised and phosphodiester backbone cleaved by AP (apyrimidinic) endonuclease (xthA) —> nicked DNA with incorrect nt Pol I binds to nick, then removed and replaces baseless nt

Answer 26

No U removal No fragments/nicks

Answer 27

DNA ligase

Answer 28

Origin of replication Circular chromosomes and plasmids have a single ori Region of repetitive dsDNA that is rich in AT Different in different organisms/plasmids

Answer 29

Multiple DnaA products bind to 9bp repeats Use ATP to separate (melt) the duplex at 13bp repeats A DnaC protein binds to ssDNA and lids a DnaB helicase onto one strand, facing towards the 3’ end DnaC detaches and helicase moves to fork After ~65 nt have been unwound by the helicases, DnaG primase enzymes bind to them, forming a primasome Primase synthesises a ~10 nt RNA primer and detaches from helicase Single stranded binding protein (SSB) binds to exposed ssDNA preventing re-annealing First primer and SSB trigger arrival of the pol III holoenzyme at the 3’ end of the primer The clamp loser loads a beta clamp onto the DNA Pol III core binds to the beta clamp 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 30

An RNA polymerase (but not the one involved in transcription) Self-priming, adding RNA 5'-->3' on a 3'-->5' DNA strand (GTA site preferred) RNA primers are ~10 nt in length Has no editing functions (no proof reading) Activity is increased in the presence of helicase and itself (co-operativity)

Answer 31

Encoded by ssb gene Forms a tetramer Not sequence specific Leaves bases exposed when ound Binds co-operatively to ssDNA --> Proteins at the ends are bound less well and are easier to dislpace by polymerase

Answer 32

Binds beta clamp proteins Transfers the beta clamp onto DNA at primer 3' end ATP hydrolysis detaches loader

Answer 33

Encoded by dnaN gene Forms a ring dimer Not sequence specific Binds Pol III core & imparts processivity --> Goes from 10s of bp to >50,000

Answer 34

Pol III holoenzyme Primasome Occupies ~50nm area around replication fork

Answer 35

1) As helicase unwinds the duplex, primase re-binds and synthesises a new primer 2) A beta clamp is added to the primer by the clamp loader 3) As more ssDNA is created, the lagging strand starts to loop back, reversing the primer polarity 4) A Pol III core binds once enough ssDNA has emerged for the beta clamp to reach it 5) First Okazaki fragment starts. DNA is pulled by helicase and Pol III. The lagging strand loops out, picking up SSB 6) Okazaki fragment lengthens. Lagging strand loop gets longer 7) First Okazaki fragment finished. Primase re-binds helicase, then adds a second primer 8) Pol III core and beta clamp detach from DNA, releasing completed fragment

Answer 36

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 37

Ter and Tus - To prevent the forks overshooting, there are 23 bp sewuences called Ter, which bind the Tus protein Tus can be displaced by the fork only in the correct direction, otherwise the helicase stalls Topoisomerase IV - As the forks get within 200 bp of each other, there is no longer room for DNA gyrase to bind --> +ive supercoiling, which is relieved by topoisomerase IV decatenating the molecules

Answer 38

OriC contains GATC sequences, which are substrates for dam (DNA adenosine methylase) dam methylates N6 of adenine After replication, only one strand will be methylated Hemimethylated GATC sequence bind SeqA protein --> Prevents DnaA binding to OriC - The GATC sites are methylated by dam very slowly (~13 mins), so newly synthesised dsDNA remains hemimethylated and new initiation is prevented --> Binds DNA to the membrane - May assist with separation and moivement of each chromosome into its new cell

Answer 39

Hemimethylated plasmids are not replicated Methyated undergo one round of replication Unmethylated are replicated normally

Answer 40

Replication of E.coli chromosome takes ~40 mins Can reproduce every ~20 mins by starting a new round of replication before the last has finished

Answer 41

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

Answer 42

Occurs in nucleus not cytoplasm More genetic material to replicate (can be 4 orders of magnitude greater) More than 1 chromosome (46 in humans) Linear chromosomes (except mitochondria) Additional packaging (e.g. nucleosomes)

Answer 43

Multiple origins per chromosome (10s-1000s) DNA polymerases are much slower: 50 nt/s vs >1000 nt/s, 40 mins vs 8 hours Polymerases synthesising the leading and lagging strands aren't physically linked together DNA polymerases don't have 5'-->3' exonuclease Okazaki fragments are much shorter: ~165 nt vs ~2000 nt

Answer 44

Tightly linked to the cell cycle Initiation must only happen once per cycle Two step process ensures this Origins of replication arise from sections of DNA called *Autonomously Replicating Sequences (ARS)* Separated by ~30kb, so range from 10 on small chromosomes, to 1000s on the largest AT-rich consensus sequence (A region) in yeast is: 5'- T/A TTTAYRTTT T/A - 3' (Y = any pyrimidine, R = any purine)

Answer 45

An Orgigin Recognition Complex of proteins (ORC) binds to the A region - Less clear how it works in higher eukaryotes Accessory proteins (licensing factors) accumulate in the nucleus during G1. Cdc-1 & Cdt-1 bind to ORC 2 helicases are loaded by Cdt1. Cdc6/Cdt1 leave

Answer 46

Pre-replication complex must be activated Pre-replication complex (+ additional proteins + DNA polymerases) --> active initiation complex/replisome progression complex

Answer 47

Not every pre-replication complex is used in each round of replication Replication forks from one origin will pass through another origin (passive replication) ~ Polymerases ~ 5 main euk. DNA polymerases: α, β, γ, δ, ε Polymaerase α has its own primase activity, as well as polymerase, so can make its own primers - Has no proofreading 3'-5' exonuclease - Not v processive since doesn't associate with the euk. sliding clamp protein called PCNA (proliferating cell nuclear antigen) Replication factor C (RPC) loads PCNA onto the primer/DNA ready for a different Pol to bind Pol β: involved in repair Pol γ: replicates mtDNA Pol δ: (lagging strand) associates with PCNA and has proofreading Pol ε: (leading strand) associates with PCNA and has proofreading --> Pol δ & Pol ε can displace RNA primers, but can't release them as dNMPs ~ Editing ~ RNA "flap" produced by Pol δ or ε can be mostly digested by RNAse H1, with 1 RNA nt left (like E.coli version) Flap 1 endonuclease (FEN1) binds PCNA & can remove any incorrect nt, by cutting off a section of ssDNA/RNA DNA ligase fills in nick

Answer 48

End replication problem: - Affects the lagging strand - Last RNA primer may not be at the extreme 3' end of the DNA --> missed section - Even if it is at the end, it will be removed because it's RNA --> ssDNA section n-10 strand -->shortened strand locked in in next replication --> further end loss (gene shrinkage) after next replication

Answer 49

End replication solution: -DNA at 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 Once the telomeric DNA is gone (~40 replications) --> gene loss --> cells stop dividing, a point called the Hayflick limit

Answer 50

Telomerase is a ribonucleoprotein Contains an RNA strand (450nt) which has sequence: CUAACCUUAAC This acts as a template for the synthesis of new telomeric repeats, growing the telomere: - Telomerase binds and extends ssDNA --> Pol α (primase) binds and adds a primer --> Pol δ binds primer and extends it --> DNA ligase fills in nick

Answer 51

2-10 copies of circular mtDNA per mitochondrion Unidirectional replication (1 fork) Pol γ synthesises leading strand Lagging strand is RNA Okazaki fragments

Answer 52

Small circular genome Some use a 'rolling circle' mechanism to continuously synthesise new DNA Sigma (σ) replication

Answer 53

Many animal and plant viruses have genomes composed of RNA (ignores the central dogma) 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 --> highly error prone

Answer 54

RNA genome Virally encoded reverse transcriptase creates a NDA strand using RNA as template and tRNA Lys as a primer RNA half is degraded by RNase H Second NDA strand synthesised using first as template --> incorporated into host genome

BIOL2010 “Flow Of Genetic Information” Flashcards

(78 cards)