DNA replication Flashcards

1
Q

Give an overview of DNA replication?

A

DNA replication is semi-conservative
DNA polymerases use ssDNA as templates to synthesise a complementary strand from correctly paired nucleotides
The OH- on the growing chain nucleophilically attacks the triphosphate and as a result releases pyrophosphate (P2O7 4-)

The chain is extended in the 5’ to 3’ direction

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2
Q

Overview - where does DNA replication occur?

A

It occurs at replication forks
This is a branch in a replication eye/bubble, that forms when two parental strands separate to allow synthesis of a complementary daughter strand

This replication is bidirectional - the replication forks move apart

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3
Q

Overview - what type of replication model is used in DNA replication?

A

Semidiscontinuous replication - the parental strands are replicated in different ways

Leading strand - continuously synthesised in it’s 5’ to 3’ direction
Lagging strand - discontinuously synthesised in the 5’ to 3’ direction forming Okazaki fragements (later joined by DNA ligase)

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4
Q

Overview - what is required for initiation of DNA polymerase?

A

RNA primers - DNA synthesis extends from them

Leading strand needs only one primer
Lagging strand needs multiple primers

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5
Q

Describe prokaryotic DNA?

A

Double stranded circular DNA
Single replicon - replicated as a single unit
Replicated from a single origin
Bi-directional - goes both ways around the circle to reach the terminus

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6
Q

Describe the different DNA polymerases in prokaryotes?

A

Pol I, Pol II and Pol III
Structure looks like a hand - when the ‘fingers’ and ‘thumb’ move together it triggers the catalysis

Pol I and Pol II - repair
Pol III - replicase

All add nucleotides in the 5’ -> 3’ direction
All can remove nucleotides in the 3’ -> 5’ direction - using exonuclease
Pol I can remove nucleotides in the 5’ -> 3’ direction - using exonuclease

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7
Q

What does Pol I do in prokaryotic replication?

A

Pol I has two hydrolytic sites: 3′ - 5′ exonuclease and a 5′ - 3′ exonuclease

3′ - 5′ exonuclease activity allows Pol I to edit its mistakes
Proof reading = high fidelity (accuracy)

5′ - 3′ exonuclease activity allows Pol I to remove RNA primers on the lagging strand and replace with DNA (nick translation)

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8
Q

How does polymerase maintain correctly paired nucleotides?

A

DNA polymerase senses complementary base pairs via sequence-independent interactions
This occurs in an active site pocket, with a shape complementary to Watson-Crick base pairs
The shallower minor groove allows protein side chains to form H bonds with N atoms of purine bases and O atoms of the pyrimidine bases
If bp were mismatched these interactions would be distorted/disrupted

DNA pol only forms catalytically closed conformation when the correct dNTP binds

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9
Q

What does the DNA polymerase catalytic mechanism involve?

A

Two metal ions - normally Mg2+ ligands, bound to two Asp side chains in the palm domain

Metal ion A - bridges the α-phosphate of dNTP and the primer’s 3′-OH group
Role: activates primers 3′-OH group for a nucleophilic attack

Metal ion B - is bound to all 3 phosphates of dNTP
Role: orient its bound triphosphate group and electrostatically shield its negative charges

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10
Q

Describe initiation P1 of prokaryotic replication?

A

DnaA.ATP (initiator protein) binds to 9-mer tandem repeats (AT rich) at the OriC (origin of replication)
This causes strain and denaturation within the 13-mer repeat (forms a melted region)
DnaC (DNA helicase loader) loads 2 hexameric complexes of dnaB (DNA helicase), one to each end of the ‘melted’ region
dnaB helicase ‘unwinds’ the DNA, translocating along the lagging strand, breaking the H bonds between the nitrogenous bases - this uses hydrolysis of NTP for energy (called overhand mechanism)

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11
Q

Describe initiation P2 of prokaryotic replication?

A

Single-stranded DNA binding protein (SSB) stabilises the ssDNA, preventing reannealing and forming secondary structures (stem loops)
SSB interacts with ssDNA via electrostatic attraction and interacts with the bases via stacking interaction
Primosome moves in the 5’ to 3’ direction, momentarily reversing migration to synthesise short RNA primers (11 nt) - this is needed to initiate each Okazaki fragment
DNA polymerase III holoenzyme can now attach to the primer

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12
Q

What keeps DNA polymerase III holoenzyme in place?

A

Sliding clamp - it promotes pol III processivity

The β subunit of pol III forms a ring around the DNA that functions as a sliding clamp, preventing pol III from diffusing away
It also greatly increases the rate of nucleotide polymerization
The γ complex = clamp loader

The dimeric ring is a pseudosymmetrical six-pointed star
35 Å diameter hole - doesn’t enter the major/minor grooves = minimises association with DNA
Interior = positively charged
Exterior = negatively charged

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13
Q

Describe elongation of prokaryotic replication?

A

The strands are synthesised in the replisome (2 pol II enzymes)
Leading strand = continuous replication (5’-3’)
Lagging strand - discontinuous replication (5’-3’)

A primer is continually formed on the lagging strand before each Okazaki fragment - occurs by loop formation
The δ subunit in the clamp loader acts as a ‘wrench’ moving the sliding clamp to the next primer

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14
Q

Prokaryotic replication elongation - how are the RNA primers removed?

A

RNA primers are replaced by pol I catalysed single-strand nicks, with the nicks sealed by DNA ligase
OR
RNaseH degrades the RNA then pol I removed the primers 3’-ribonucleotide

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15
Q

How does DNA ligase work?

A
  1. The adenylyl group of NAD+ is transferred to the amino group of an enzyme Lys residue to form an unusual phosphoamide
  2. The adenylyl group of this activated enzyme is transferred to the 5′-phosphoryl terminus of the nick to form an adenylylated DNA
    AMP is linked to the 5′-nucleotide via a pyrophosphate (not usual phosphodiester bond)
  3. DNA ligase catalyzes the formation of a phosphodiester bond by attack of the 3′-OH on the 5′-phosphoryl group, thereby sealing the nick and releasing AMP
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16
Q

Describe termination P1 of prokaryotic replication?

A

A Tus protein binds at 1 of 10 nearly identical, nonpalindromic, 23 bp terminator sites (Ter sites)
The bound Tus protein prevents strand displacement by DNA helicase = arresting replication fork advancement
Tus protein = 309 residue monomer

Ter sites act as one-way valves, so replication forks can enter the terminus region but not leave it
Therefore the bidirectional forks will meet in the terminus even if one arrives ahead of the other
This is highly conserved

17
Q

Describe termination P2 of prokaryotic replication?

A

When the two replication forks collide at the termination site, the strands become covalently linked = two closed double stranded chromosomes
The catenated circular dsDNAs need to be separated to form 2 daughter cells
Decatenation of the circular DNAs uses type II topoisomerase named topoisomerase IV

It binds both DNA helices and breaks open the DNA and forms a transient covalent attachment, interrupting the helix and forming a protein gate
The gate opens and shuts to let the second DNA helix pass

18
Q

What feature of prokaryotic DNA is required for replication?

A

Only methylated strands of DNA undergo replication
The newly synthesised daughter strand isn’t methylated - called hemi-methylated DNA
There is a lag period before it becomes methylated as limited DAM (DNA adenine methylase)
SeqA binds hemi-methylated regions - preventing replication until it is methylated

Once origins become fully methylated, they are competent for initiation

19
Q

Describe eukaryotic replication?

A

DNA replication is bidriectional and semi conservative
Takes place during the S phase of the cell cycle
Much bigger genomes and replication fork is x10 slower than prokaryotes, so we use multiple origins of replication
Cells must fully divide before replication can begin again
Takes around 8 hours in a human eukaryote
Nucleosomes reassemble behind the replication forks

20
Q

What are the eukaryotic DNA polymerases involved in replication? functions?

A

All in the B-family polymerases:
a - extends the primer in 5’ to 3’ direction, no exonuclease activity = no proofreading and moderately processive

ẟ - leading and lagging strand replicase, is in a complex with PCNA - proliferating cell nuclear antigen, it has exonuclease activity and high processivity

ε - leading strand replicase and has exonuclease activity, no PCNA and high processivity

21
Q

What is a special DNA polymerase in eukaryotes?

A

DNA polymerase γ

It replicates the mitochondrial and chloroplast genome

22
Q

Describe initiation P1 of eukaryotic DNA replication?

A

Takes place during: G1 of the cell cycle
ORC (origin replication complex), 6 subunits, binds to ORI (origin of replication
ORC recuits CDC6 + CDT1 proteins, CDC6 drives the recruitment of MCM-2-7 (helicase) = currently inactive
This is the pre-replicative complex (pre-RC)

23
Q

Describe initiation P2 of eukaryotic DNA replication?

A

Takes place during S phase of the cell cycle
Transition to S phase is mediated by CDK + DDK (kinases), they phosphorylate CDC6 and MCM-2-7
CDC6 = degraded, CDT1 is inhibited by Geminin
Phosphorylated MCM recruits CDC45 + GINs = CMG complex = active MCM helicase
MCM unwinds/melts the DNA breaking H bonds forming ssDNA
CMG recruits RPA, pol a and primase
RPA binds to both strands of ssDNA to prevent annealing
Primase adds a primer to the leading strand and multiple to the lagging strand
Polymerase a works from the primers adding a few deoxynucleotides to initiate DNA synthesis

24
Q

Describe elongation P1 in eukaryotic DNA replication?

A

RFC, a clamp loader, loads PCNA onto the template strand near the primer, displacing pol a after adding 15 nt, allowing pol ẟ to bind
Pol ẟ replicates the leading strand continuously and lagging strand discontinuously forming Okazaki fragments

Topoisomerase II, will relieve any stress on DNA strands

25
Q

Describe elongation P2 in eukaryotic DNA replication?

A

Pol ẟ displaces the primer first and will then start synthesising DNA on the lagging strand
The RNA primers are removed by RNase H1 which cleaves the phosphodiester bonds between ribonucleotides and deoxynucleotides and then FEN1 (flap endonuclease 1) removing the final adjacent nucleotide
FEN1, recuited by PCNA, effectively proofreads pol a’s work
Pol a synthesises the gaps at primer sites
DNA ligase then joins the Okazaki fragments

26
Q

Describe termination of eukaryotic DNA replication?

A

Due to multiple origins of replication there are many MCM complexes working at once
When MCM complexes meet/converge a SCF E3 ubiquitin ligase complex binds - catalysing the polyubiquitination of MCM-7 lys residues
This is recognised by p97 or VCP, a protein remodeller/segregase enzyme
It segregates/causes disassembly of CMG complex, MCM is now inactive and the cofactors are also removed
= termination of DNA replication

27
Q

What is the issue with the lagging strand in eukaryotic replication?

A

When the end primer is removed only a 5’ end is exposed and pol a can’t synthesise deoxynucleotides in the gap
This could lead to the shortening of the chromosomes in each replication

28
Q

How are teleomeres formed?

A

Telomeres = ends of chromosomes
Consists of 1000+ tandem repeats of short G-rich sequences
It is synthesised and maintained by telomerase, a ribonuleoprotein (protein and RNA)
Telomerase adds nucleotides only to the parent strand
Once the parent strand is longer, primase and pol a come in to fill the gap
Primase is removed and the ‘overhang’ can be removed by FEN1

It’s subunit TERT is homologous to reverse transcriptase

29
Q

What do telomeres form?

A

G-Quartets - cyclic tetramers formed from guanine rich polynucleotides
G-rich overhangs of telomeres fold back on themselves to form hairpins
Two of these associate in an antiparallel way forming stable complexes of stacked G-quartets

They serve as binding sites for capping proteins - help regulate telomere

30
Q

What are some key differences between prokaryotic and eukaryotic replication?

A

E is linear and P is circular
E has multiple origins of replication, P has one
E has much smaller Okazaki fragments 200bp and P has longer ones 2000 bp
E replication is confined to S phase and can’t reinitiation until cell divides
P has continuous replication, which can initiate before first round is completed

31
Q

Why is DNA replicated at different rates?

A

Transcriptionally active genes replicate early in S phase
Repressed genes and heterochromatin replicate late in S phase
Inactive X chromosome replicates very late

Chromatin accessibility strongly influences the use of origins