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DNA replication Flashcards

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Biology 101 flashcards
1
Q

Where does DNA replication begin

A

Origins of replication

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2
Q

What binds to DnaA boxes at origins to initiate replication

A

Initiator proteins

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3
Q

How many origins do bacteria have?

A

1

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4
Q

Have many origins do eukaryotes have and why?

A

Many, because larger chromosomes have to be replicated

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5
Q

Describe the origin in E.coli

A

OriC, 245bp, contains many DnaA boxes with adjacent rich AT sequence

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6
Q

Explain in depth initiation in bacteria

A
  1. Initiator proteins bind at DnaA boxes
  2. ATP is attached DnaA forms a spiral, multisubunit complex
  3. DNA wraps itself around the DnaA complex, causing DNA to bend and rich AT regions to unwind
  4. DnaC helicase loader assembles with DnaB helicase
  5. DnaC helicase loader binds to DnaA on the origin
  6. DnaC helicase loader places DnaB helicase around ssDNA at the origin
  7. DnaC helicase loader disassociates from DnaB helicase
  8. Any unwound DNA bound by SSB
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7
Q

What is known about eukaryotic replications origins

A

Origins bound by ORC (Origin replication complex)

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8
Q

What factors affect binding to ORC

A

Genetic changes
Chromatin configuration
Histone modification

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9
Q

Explain Eukaryotic DNA initiation

A
  1. ORC binds to the DNA at the origin
  2. Cdc6 and Cdt1 are both recruited as DNA helicase loaders
  3. 2 ring-shaped MCM2-7 hexamers are arranged head-to-head with orientation around dsDNA
  4. ORC disassociates from the complex after the 2 hexamers are arranged.
  5. Forms the MCM2-7 helicase prereplicative complex which is inactive in G1 and encircles both strands
  6. The MCM2-7 complex is phosphorylated by DDK in S-phase to recruit Cdc45 and Sld3
  7. Sld2 and Sld3 phosphorylated by CDK in order to form the GINS complex
  8. This then forms the CMG complex which is a replicative helicase which opens the bound dsDNA and unwinds it to ssDNA encircling the origin
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10
Q

What comes after DNA unwinding?

A

Priming

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11
Q

What is the enzyme used for priming in bacteria

A

DNA primase is DnaG
Single subunit
Encodes for 10-30nt RNA
Then passes on to DNA pol III

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12
Q

What is the enzyme used for priming in eukaryotes

A

Polymerase a-primase
4 subunits
2 subunit primase encode for RNA 10 nucleotides long
2 a-subunits then encodes for short piece of DNA and add this DNA to the RNA (30-100bp)

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13
Q

Can DNA polymerase be used de novo and why?

A

No, because it requires a 3’OH for synthesis which is brought about by a primer

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14
Q

Can RNA polymerase be used de novo?

A

Yes

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15
Q

What does association of a sliding clamp do?

A

Increase processivity

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16
Q

What is the rate per second of base pairs read and added by DNA polymerase III

A

10bp/sec without clamp
1000sbp/sec with clamp

17
Q

What is the sliding clamp in eukaryotes and it’s structure?

A

Proliferating Cell Nuclear Antigen (PCNA) - Contains peptide motifs of 8 amino acids associated with each subunit

18
Q

What is the sliding clamp in bacteria and it’s structure?

A

Beta-protein
- Contains a small peptide motif which interacts with core enzyme on DNA pol III
- Core enzyme contains a and e subunits which contain polymerase and exonuclease activity respectively, and are stabilised by 0

19
Q

Structure of bacterial sliding clamp

A

2 beta subunits, which 3 similar domains

20
Q

Structure of eukaryotic sliding clamp

A

3 different, but related polypeptides, with 2 similar domains

21
Q

Ring shape diameter by sliding clamp

A

3.5nm

22
Q

Clamp loader in bacteria

A

5 subunit complex called gamma complex
Contains 3 gamma subunits, 1 delta subunit, and one delta’ subunit

23
Q

Clamp loader in eukaryotes

A

Replication Factor C (RFC)
Consists of 5 different, but related polypeptide subunits

24
Q

What are some of the clamp loader subunits

A

AAA+ ATPases

25
Q

Explain in detail function of the sliding clamp

A
  1. The clamp loader has low affinity for the sliding clamp when not bound to ATP
  2. Once bound to ATP, the clamp loader will bind to the sliding clamp, forming a spiral shape and opening the clamp
  3. The clamp loader/sliding clamp complex has high affinity towards the primer template junction
  4. Binding of complex to the junction promotes stimulation of ATPases which close the sliding clamp and release the clamp loader
  5. Sliding clamp still stays in association with the DNA
  6. Sliding clamp recruits DNA polymerase to start elongation
26
Q

How does clamp loading facilitates polymerase switching?

A
  1. Polymerase a-primase complex synthesizes RNA and initiator DNA. Complex dissociates.
  2. Replaced by one of the processive eukaryotic DNA pols delta or pol epiphron
  3. Handover from Pol a-primase to DNA Pol delta or DNA pol epiphron is known as polymerase switching
  4. Replicative polymerase (delta or epiphron) is recruited by the sliding clamp
  5. DNA pol epiphron replicates the leading strand and DNA pol delta the lagging strand
  6. Mechanism ensures that replicative DNA pols are loaded onto DNA at the right time and in the right place to begin elongation.
27
Q

What way do leading and lagging strand move

A

Leading: Towards the fork movement direction
Lagging: Away from fork movement direction

28
Q

How does movement of leading and lagging strand coordinated at replication fork?

A

The lagging strand template loops itself around

29
Q

Where are leading and lagging strand coordinated in bacteria?

A

The replisome which contains DnaB helicase, DnaG primase, and DNA polymerase III holoenzyme

Lagging and leading strands are tethered together by binding to flexible Tau protein subunits of clamp loader complex

Replisome ensures the DNA pol III core enzyme remains associated with fork even though it’s released at the end of Okazaki fragments

30
Q

What coordinates leading and lagging strands in eukaryotes?

A
  1. Ctf4
  2. Leading strand DNA polymerase epiphron and lagging strand DNA polymerase delta are both present at replication fork - Not linked by Tau proteins
  3. Multisubunit Ctf4 acts as a hub to couple CMG helicase, DNA polymerase epiphron and DNA polymerase a-primase at the fork
31
Q

Okazaki fragments in bacteria

A
  1. Okazaki fragments join after synthesis of DNA
  2. DNA polymerase III synthesises DNA up until the next RNA primer is reached, when it disassociates
  3. The sliding clamp is still attached to the DNA
  4. DNA polymerase I is recruitecd to remove the RNA primer (5’ to 3’ exonuclease activity) and replace it with DNA
  5. DNA polymerase I nicks the DNA
  6. DNA ligase seals the nick
32
Q

Okazaki fragments in eukaryotes

A
  1. DNA polymerase delta continues to synthesise DNA even after meeting the RNA primer
  2. This displaces the existing RNA primer and the DNA surrounding it, forming a flap
  3. Fen1 (Flap endonuclease 1) then cleaves the flap DNA
  4. DNA ligase I then seals the nick
33
Q

Termination in bacteria

A
  1. termination of replication is caused by Ter sites at the zone of termination
  2. Each ter site is bound to by a tus molecule
  3. The tus molecule bound to ter site stalls replication in one direction
  4. TerC is most commonly used (first encountered by clockwise fork)
  5. If an anticlockwise fork encounters a clockwise fork stalled by the tus molecule on ter site, then replication finishes as the replisomes will cross past each other
  6. DNA polymerase I and DNA ligase I will both complete replication as termination is completed
34
Q

Topoisomerases in bacterial termination

A
  1. Type 1A topoisomerases (which cleave 1 strand and guide the other through the gap) can separate incompletely replicated molecules
  2. Type II topoisomerases (cleave both DNA strands) are required to separate completely replicated molecules
35
Q

Why must increased torsional stress be removed during termination

A
  • Increased supercoiling of fork makes strand separation more difficult as forks approach each other
  • Topoisomerases break DNA to remove excess supercoiling and resolve overwinding
36
Q

Termination in Eukaryotes

A
  1. Termination occurs at multiple sites
  2. When replication forks converge, CMG move past each other on leading strand
  3. CMG transitions to encircling dsDNA
  4. DNA polymerase delta, Fen1, and DNA ligase 1 recruited to complete maturation
  5. dsDNA remains entwined, type II topoisomerase separates dsDNA
37
Q

Problem of replicating a completely linear chromosome in eukaryotes

A
  1. RNA primer removal on lagging strand leaves a gap that can’t be filled
  2. Dissolution of replication fork on leading strand leads to loss of terminal okazaki fragments if lagging strand is not also complete
  3. Leads to erosion of chromosome, ends over several rounds of replication
38
Q

How do telomeres protect ends of linear chromosomes

A
  • Eukaryotic chromosomal ends have simple sequence repeats
  • 50bp to > 20,000bp length
    G-rich strand extends 5’ to 3’ towards chromosome end, and terminates in a ssDNA tail ~75-300nt

Genetics (18 decks)

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