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Information Transfer > IT4: How is DNA copied and maintained? > Flashcards

IT4: How is DNA copied and maintained? Flashcards

1
Q

What are Hoogsteen base pairs?

A

When the central bond of a nucleotide base pair rotates to form a syn-anti conformation, allowing G-T base pairing and formation of a triplex helix.

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2
Q

What is a G-quadruplex?

A

A secondary structure that occurs in genomes of repetitive G-C sequences that are associated with human diseases. They occur due to Hoogsteen base pairing that allows for 4-stranded DNA formation.

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3
Q

Compare type I to type II topoisomerases.

A

Type I: causes a single break.
Type II: breaks both strands of DNA

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4
Q

Compare a transversion to a transition mutation.

A

Transversion: purine to pyrimidine (or vice versa)
Transition: purine to purine; pyrimidine to pyrimidine

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5
Q

Why is uracil exclusively found in RNA?

A

The only difference between thymine and uracil is a methyl group and functionally they are identical. The reason uracil is kept exclusively in RNA is due to the possibility of deamination from cytosine to uracil. This alters the binding capacity of the base and will result in a substitution. By using thymine in the DNA, the cell can recognize any uracil in the DNA as arising from deamination of cytosine.

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6
Q

What are the two common types of tautomerizations that can occur in DNA?

A

Amino (-NH2) to imino (=NH)
Keto (-C=O) to enol (=C-OH)

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7
Q

Describe nucleotide excision repair in prokaryotes.

A
  1. UvrA and UvrB recognise the damage.
  2. UvrC cleaves either side of the damaged base.
  3. UvrD unwinds the DNA, flipping out the damaged DNA and leaving a single-stranded DNA gap.
  4. The 3’OH can be used to synthesize a new strand incision, and this is sealed using DNA ligase.
    In eukaryotes, DNA polymerases and ligases are involved, but the mechanisms are the same.
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8
Q

Describe base excision repair.

A

BER repairs DNA damage caused by the removal or a single base.
1. Glycosylase enzymes recognise the damaged base and create an AP site.
2. AP endonuclease cleaves the phosphodiester backbone at the AP site, creating a single strand break.
3. Removal of the sugar phosphate group.
4. DNA polymerase fills in the gap.
5. Ligation.

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9
Q

What are ADP-ribosyltransferases?

A

A group of enzymes that catalyze the transfer of ADP-ribose from NAD+ to specific amino acids.
Poly-ADP-ribosyltransferases (PARPs) can transfer many ADP-ribose units, facilitating the recruitment of other repair factors to sites of DNA damage.

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10
Q

How does homologous recombination work in bacteria?

A

In bacteria, the RecBCD complex recognizes the break and begins to unwind the DNA and resection the ends to leave a 3’OH overhang for DNA synthesis.
This ssDNA region is coated by RecA, promoting strand invasion. A double-Holliday junction is formed which requires a nuclease to resolve.

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11
Q

How does homologous recombination work in human somatic cells?

A

In humans, the Mre11-Rad50-Xrs2 complex recognizes the break and begins to unwind the DNA and resection the ends to leave a 3’OH overhang for DNA synthesis.
This ssDNA region is coated by Rad51 (loaded by BRCA2), promoting strand invasion. A double-Holliday junction is formed which requires a nuclease to resolve.

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12
Q

Describe the process of non-homologous end-joining.

A
  1. Recognition and binding of the broken ends: the NHEJ machinery (Ku70/80) recognizes the broken ends of the DNA strands and binds to the DNA ends.
  2. End processing: the broken ends of the DNA strands are processed by exonucleases and other enzymes to remove any damaged or unpaired bases and to generate a small amount of single-stranded overhangs.
  3. Ligation: the processed DNA ends are then ligated together by a ligase enzyme, restoring the integrity of the DNA molecule.
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13
Q

What is the MRN complex?

A

Mre11 - endonuclease
Rad50 - ATPase activity
Nbs1 - scaffold to facilitate Mre11 and Rad50 interactions

The MRN complex plays a crucial role in DSB repair.

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14
Q

Compare yeast and mammalian NHEJ.

A

Yeast:
- MRX complex
- No DNA-PKcs

Mammalian:
- MRN complex
- Uses DNA-PKcs

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15
Q

What can cause replication fork barriers?

A
  • Non-histone proteins
  • Collisions with transcription complexes and associated DNA-RNA hybrids
  • DNA damage
  • Sequences that can fold into none B-form structures
  • Chromosome structure and organisation
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16
Q

Describe DNA fibre analysis.

A

DNA molecules are labeled with thymidine analogues and then spread on a slide. The labeled DNA fibers can be visualized and the distances between the labeled segments measured to give an indication of rate of DNA replication.

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17
Q

How can replication fork barriers be studied?

A
  • DNA fibre analysis
  • ^^ using DNA damage agents and inter-crosslinking agents
  • 2D gel electrophoresis
  • Reporter systems
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18
Q

What strategies are used to cope with replication fork barriers?

A
  • preventing arrest
  • separate transcription from replication (temporally or spatially)
  • skipping DNA damage
  • replication fork regression
  • rescue of arrested forks
  • break-induced replication
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19
Q

What is the ATR activation pathway?

A

A cellular signaling pathway that responds to DNA damage and replication stress. ATR protein kinase recognizes ssDNA generated at stalled replication forks, leading to recruitment of DNA damage response proteins.

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20
Q

What is PrimPol?

A

Primase-polymerase is a DNA polymerase that synthesizes short RNA and DNA primers for polymerases. It’s involved in the repair of DNA damage as it can bypass lesions, allowing replication to continue past the site of damage.

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21
Q

Describe break-induced repair.

A

DNA repair mechanism that repairs double-strand breaks in DNA by using an intact homologous chromosome as a template. This occurs when one end of a DNA ds-break invades the template, forming a D-loop.
BIR can be error-prone if the template is non-homologous.

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22
Q

How is break-induced repair associated with APOBEC proteins?

A

APOBEC proteins can induce DNA damage and breas through deamination of cytosines. The resulting uracil causes DNA damage. BIR can be used as a repair mechanism to fix these breaks, but can itself be error prone in the presence of APOBEC-induced DNA damage. This is because the invading strand may anneal with a non-homologous template.
Hence studies show that APOBEC-induced mutations can contribute to the development of cancer.

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23
Q

What is the DNA damage response and what are the 3 outcomes of it?

A

The DDR is a complex network of cellular pathways that detect, signal, and repair DNA damage.
The 3 major outcomes are:
1. DNA repair
2. Cell cycle arrest (allows time for DNA repair)
3. Apoptosis (if DNA damage cannot be repaired)

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24
Q

What is the SOS response?

A

The SOS response is a bacterial response to the accumulation of ssDNA that arise when DNA damage blocks replication fork progress. RecA recognizes the damage and inactivates the repressor of SOS response genes, LexA. It is an error-prone system that contributes significantly to DNA changes observed in a wide range of species.

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25
Q

What are the 5 checkpoints in the cell cycle?

A

G1 restriction point
G1/S
Replication and S-phase checkpoints
G2
Spindle (Anaphase)

26
Q

What do checkpoints ensure?

A
  • DNA integrity is maintained through replication
  • DNA is replicated only once per cell cycle
  • DNA replication is complete before mitosis
  • Each stage of mitosis is complete before moving onto the next stage
  • DNA damage is repaired before the genome is duplicated in S-phase or segregated in mitosis
27
Q

How do prokaryotes regulate DNA replication?

A
  1. Blocking DnaA binding OriC by SeqA.
  2. Negative regulation of DnaA transcription.
  3. Hydrolyzing ATP-bound DnaA by Hda, preventing DnaA from binding.
  4. Diluting DnaA concentration by datA sequestration
28
Q

Explain the cell fusion experiment by Rao and Johnson.

A

EXPERIMENT: A G1 cell (2N) and an S cell (2-4N) were fused together and this triggered the G1 nucleus to enter S phase…there must be something produced by the S phase cell that induces this. This was called S-phase promoting factor (SPF).
The same experiment was repeated using a G2 cell instead of G1. The G2 nucleus doesn’t replicate until it has gone through mitosis. There was also a delay into mitotic entry as the G2 cell waited for the S cell to enter G2 phase. This delay was a checkpoint.

29
Q

What is the proposed model for licensing factors?

A

Some licensing factor is not able to pass through the nuclear membrane, but is able to associate with chromatin in mitosis when the nuclear membrane has been degraded. It stays there to activate initiation of DNA replication. After DNA replication, it’s quickly disrupted or inactivated. These features guarantee that the genome only replicates once during S phase.

30
Q

What does the initiation of replication involve?

A
  • ARS contains the origin of replication in the yeast genome.
  • Origin recognition complex (ORC) is bound to the origin throughout the cell cycle.
  • Cdc6 levels increase in late mitosis/early G1, binding ORC at the origin.
  • Geminin is ubiquitinated by the APC at the metaphase-anaphase transition, releasing Cdt1.
  • MCM2-7 helicase loading on the origin via Cdc6/Cdt1.
  • G1/S kinases phosphorylate Cdc6, leading to its degradation.
  • Cdc45/GINS and firing factors are loaded, forming the PIC.
  • Cdc7 and CDK activity phosphorylates the firing factors and the MCM to allow replication to start.
  • Early origins can be fired by cyclin B5/6 with Cdk1; late origins need cyclin B5-Cdk1.
  • Rad53 inhibits Cdc7 and CDK to prevent origin firing in the presence of DNA damage.
  • Cdt1 is degraded in S phase and negatively regulated in G2 by geminin.
31
Q

What is the timing of origin firing determined by?

A
  1. Chromosomal context i.e., heterochromatin.
  2. Affinity of the origin for firing factors.
32
Q

Compare prokaryotic and eukaryotic replication origins.

A

Prokaryotes:
- Modular
- AT rich
- Usually one per circular chromosome.
e.g., OriC

Eukaryotes:
- Modular
- Has additional regulation
- AT rich
- Multiple origins per linear chromosome
E.g., ARS in yeast

33
Q

Give the steps in DNA replication.

A
  1. Origin recognition
  2. Recruitment of helicase loaders.
  3. Helicase loading.
  4. Melting of DNA
  5. ssDNA stabilization
  6. Recruitment of primase
  7. Priming (iRNA then iDNA)
  8. Loading processivity factors
  9. Hand-off to elongation polymerases
  10. Fork progression and coordination
  11. Okazaki fragment processing
  12. Replisome disassembly
  13. Termination
  14. Resolution
  15. Telomere replication
34
Q

Compare helicase loading in prokaryotes and eukaryotes.

A

Prokaryotes:
- DnaC loads DnaB onto the lagging strand.

Eukaryotes:
- Cdc6-Cdt1 loads MCM2-7 onto the leading strand.

Both use ATP hydrolysis.

35
Q

What is the primase counting mechanism?

A

A model for how the eukaryotic primase complex determines the number of RNA primers to synthesize during DNA replication. It proposes that the primase complex counts the number of DNA polymerases present at the replication fork to determine the number of RNA primers needed to initiate DNA synthesis.
This ensures the primer is long enough for polymerase to elongate.

36
Q

What is primase hand-off?

A

Eukaryotic primase is tightly bound to DNA pol-alpha and places the 3’OH of the RNA primer into the active site of the DNA polymerase. This blocks the primase from elongating the current primer and frees up the primase to start again.
DNA pol-alpha then synthesizes the initiator DNA for elongation.

37
Q

How is the sliding clamp loaded onto the DNA in eukaryotes and prokaryotes?

A

Eukaryotes:
- ATP-RFC recruits PCNA.
- RFC uses ATP hydrolysis to open PCNA and load it onto the DNA.

Prokaryotes:
- ATP-gamma complex recruits B clamp.
- Y complex uses ATP hydrolysis to open B clamp and load it onto the DNA.

38
Q

What is the role of the sliding clamp?

A

The sliding clamp, also known as the processivity factor, is a protein complex that helps DNA polymerase to maintain contact with the DNA template during DNA replication. The sliding clamp is loaded onto the DNA by clamp loader complexes, which are ATP-dependent molecular machines.

39
Q

How does the sliding clamp differ between eukaryotes and prokaryotes?

A

Eukaryotes:
- Heterotrimeric
- Requires additional proteins for loading

Prokaryotes:
- Homodimer
- Only uses Y complex

40
Q

What are B-type DNA polymerases?

A

B-type DNA polymerases are often involved in the replication of the lagging and leading strand during eukaryotic DNA replication, and they possess 3’ to 5’ exonuclease activity that allows for proofreading and correction of errors in DNA synthesis.

41
Q

How do accurate DNA polymerase active sites compare to inaccurate polymerases?

A

Accurate DNA polymerases have solvent-inaccessible active sites that ensures tight fits around the correct base pairs. Inaccurate polymerases have solvent-accessible active sites.

42
Q

What are the steps in polymerisation and proofreading?

A
  1. Templated addition of dNTP
  2. Phosphodiester bond formation and PPi release
  3. Incorrect dNTP incorporated -> helical distortion
  4. Flip into proofreading active site of DNA pol to excise the dNTP
  5. Correct dNTP added
43
Q

How is DNA replication coordinated in both eukaryotes and prokaryotes?

A

Eukaryotes:
- CMG complex (Cdc45-MCM-GINS)
- Cdc45 recruits additional replication factors
- GINS stabilizes CMG to enhance activity

Prokaryotes:
- Tau complex (DnaB-DnaC-tau)
- Tau stabilizes the DnaB helicase and coordinates multiple helicases.

44
Q

Describe Okazaki fragment processing in eukaryotes.

A
  1. Pol displaces the short flap
  2. Fen1 cuts the flap
  3. Pol fills gap
  4. DNA ligase joins

If a long flap is made, Dna2 makes it small flaps for FEN1 to cleave.

45
Q

What is the role of FEN1 and Dna2 in eukaryotic OFP?

A

FEN1: removes the short flap RNA primers displaced by DNA polymerase.
Dna2: endonuclease used to cleave long flaps of iDNA. It creates short flaps than can be processed by FEN1.

46
Q

Compare the toolbelt model to the sequential switching model of PCNA action.

A

PCNA is a trimer with 3 hydrophobic pockets and each can bind a protein partner. The toolbelt model shows DNA pol-delta synthesizing and when it reaches an OF, the PCNA spins around to put Fen1 in a position to deal with it. The same then occurs with the ligation. The other model is the sequential switching model that involves each enzyme out-competing the other to bind to PCNA, one at a time. Competitive interactions have been shown experimentally.

47
Q

How is replication terminated in prokaryotes?

A

Replication termination occurs at specific sites on the chromosome called ter sites. These ter sites bind the Tus protein which acts as a counter-helicase, preventing the replication helicase from moving further down the DNA strand. This creates a replication fork barrier that leads to the dissociation of the replication machinery from the DNA, terminating replication.

48
Q

How is replication terminated in eukaryotes?

A

Replication termination is not a well-defined process in eukaryotes and is thought to occur when replication forks from neighboring origins meet and fuse, completing the replication of the entire chromosome. This process is called replication fork convergence, and it requires the coordination of several proteins and checkpoint pathways to ensure the faithful and complete replication of the genome.

49
Q

Describe site-specific termination in eukaryotes.

A

Transcription and replication polymerases collide without barriers in highly transcribed regions (e.g., rDNA loci) of the genome, so these contain multiple termination sites that bind proteins to prevent collision.
It also occurs at sites where genetic information must be retained e.g., the MAT locus in yeast.

50
Q

How is the eukaryotic replisome distinguished from dormant replisomes?

A

Active replisomes will contain Cdc45 and GINS i.e., be in the CMG complex formation. This can be targeted for removal by p27-dependent ubiquitination.

51
Q

Describe the end replication problem.

A

This problem arises because the lagging strand, which is synthesized in short Okazaki fragments, requires a primer to initiate DNA synthesis, and there is no 3’ end beyond the final RNA primer at the end of the lagging strand to extend the DNA. This results in the loss of a small portion of DNA from the end of the chromosome with each round of replication.
In eukaryotes, this problem is addressed by the presence of telomeres which protect the chromosomes and lead to cellular senescence as they shorten.

52
Q

How is catenated DNA formed?

A

When supercoiling passes from being in front of replication machinery to being behind the machinery, it can cause intertwining of the replicated DNA, forcing it to become catenated.

53
Q

Compare type I and type II topoisomerases.

A

Type I: cuts once
Type II: cuts twice

54
Q

What are anaphase bridges?

A

Anaphase bridges occur when the sister chromatids fail to separate completely and remain connected by a bridge of chromatin material. These bridges can cause loss or gain of genetic material.

55
Q

What are the 3 quantitative issues faced by DNA replication?

A
  1. Over-replication (leads to gene amplification)
  2. Under-replication (leads to gene loss…TSGs…)
  3. Linking replication to cell division
56
Q

What are indels?

A

Indels, short for “insertions and deletions,” are small mutations in DNA that involve the insertion or deletion of one or more nucleotides.

They’re often caused by replication slippage in nucleotide repeat regions.

57
Q

Describe the process of DNA mismatch repair.

A
  1. MutS scans the DNA to find the mismatched base pair.
  2. MutS recruits MutL.
  3. Excision of mismatched base pair.
  4. Filling in of gap.
  5. Proofreading by DNA polymerase.
58
Q

Why are immune checkpoint therapies good at targeting tumors with MMR mutations?

A

Immune checkpoint therapies work by blocking these immune checkpoints, which can enhance the activity of immune cells and help them recognize and attack tumor cells. In the case of tumors with MMR mutations, the accumulation of mutations and production of neoantigens can make these tumors more susceptible to attack by the immune system. By blocking immune checkpoints, immune checkpoint therapies can help activate immune cells and promote an immune response against these tumors.

59
Q

How do E. coli and mammals discriminate between parental and new strand in MMR?

A

E. coli:
Methylation status - the parental strand is methylated but the daughter strand isn’t.

Mammals:
MutL nicking is directed by the PCNA and how it’s loaded onto the DNA.

60
Q

How does mammalian MMR differ from E. coli MMR?

A
  1. MutS and MutL function as heterodimers that have different mispair binding specificities.
  2. Not MutH homologues - nicking is by MutL homologues.
  3. Strand discrimination is via PCNA loading, not methylation status.
61
Q

Give two examples of trinucleotide repeat disorders.

How do trinucleotide microsatellites differ from non-expandable microsatellite sequences?

A

Fragile X syndrome:
FMR1 gene expansion causes methylation of DNA and transcriptional silencing.

Huntington’s:
Formation of poly(Q) tracts causes aggregation.

Trinucleotide microsatellites undergo multiple expansions at once, rather than just an increase or decrease in length by one repeated motif.

62
Q

How can MMR promote trinucleotide repeat expansions?

A

Potentially by inducing small expansions that undergo iterative cycles until large expansions arise.

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