IT4 - DNA Repair

Question 1

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

Answer 1

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)

Question 2

What is the role of CtIP in HR, and how is it regulated?

Answer 2

CtIP co-activates MRN for resection; phosphorylated by Cdks in S/G2 to promote HR.

Question 3

How does homologous recombination work in human somatic cells?

Answer 3

In humans, the Mre11-Rad50-NBS1 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.

Question 4

What are B-type DNA polymerases?

Answer 4

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.

Question 5

What kind of mutation rate information do MA lines provide?

Answer 5

Estimates of point mutation and indel rates per base pair per generation.

Question 6

What is the SOS response in bacteria, and why is it error-prone?

Answer 6

RecA-LexA pathway induces repair genes; error-prone due to TLS polymerases (e.g., Pol V).

Question 7

How does mammalian MMR differ from E. coli MMR?

Answer 7

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.

Question 8

What are the steps in polymerisation and proofreading?

Answer 8

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

Question 9

Give two examples of trinucleotide repeat disorders.

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

Answer 9

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.

Question 10

What is the ATR activation pathway?

Answer 10

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 such as CHK1

Question 11

Describe the process of DNA mismatch repair.

Answer 11

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.

Question 12

Which DNA repair defects are linked to neurodegeneration?

Answer 12

SSBR gene mutations (TDP1, APTX, PNKP, XRCC1) cause ataxia, neuropathy, seizures.

Question 13

What are Hoogsteen base pairs?

Answer 13

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.

Question 14

What are ADP-ribosyltransferases?

Answer 14

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.

Question 15

What are indels?

Answer 15

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.

Question 16

What challenges do replication fork barriers (RFBs) cause, and what strategies does the cell use to overcome them?

Answer 16

Challenges: RFBs can cause replication fork arrest or stalling, risking genome instability.
Cellular strategies to cope include:

Preventing fork arrest

Separating transcription and replication in time or space

Skipping over DNA damage using lesion bypass

Replication fork regression (reversing the fork)

Rescue of arrested forks via repair pathways

Break-Induced Replication (BIR) to restart collapsed forks

Question 17

Why is minimizing natural selection important in MA lines?

Answer 17

It prevents the removal of deleterious mutations, allowing them to accumulate and giving a true measure of mutation rates.

Question 18

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

Answer 18

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

Question 19

Why is uracil exclusively found in RNA?

Answer 19

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.

Question 20

What are the two main challenges to DNA replication fidelity?

Answer 20

1. DNA polymerase errors.
2. Replication fork obstacles (e.g., lesions, protein blocks).

Question 21

How is catenated DNA formed?

Answer 21

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.

Question 22

How does 5-methylcytosine deamination contribute to mutations?

Answer 22

Converts 5mC→T, creating T:Gmismatches; frequent at CpG sites, increasing C→T transitions.

Question 23

How does 53BP1/Rif1/Shieldin regulate NHEJ vs. HR?

Answer 23

Blocks resection by shielding DNA ends, favoring NHEJ; loss restores HR.

Question 24

How can MMR promote trinucleotide repeat expansions?

Answer 24

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

Question 25

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

Answer 25

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

Question 26

What histone marks recruit 53BP1 to DSBs?

Answer 26

H4K20me (constitutive) + H2AK15Ub (damage-induced by RNF168/Ubc13).

Question 27

Compare error-prone vs. error-free lesion bypass mechanisms.

Answer 27

Error-prone: TLS (Pol η/ζ). Error-free: Template switching (fork reversal, sister chromatid use).

Question 28

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

Answer 28

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.

Question 29

What are the clinical features of Xeroderma Pigmentosum (XP)?

Answer 29

UV sensitivity, skin pigmentation defects, high skin cancer risk (NER defects, XPA-XPG).

Question 30

How does homologous recombination work in bacteria?

Answer 30

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.

Question 31

What is the SOS response?

Answer 31

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.

Question 32

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

Answer 32

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.

Question 33

What is the MRN complex?

Answer 33

Mre11 - endonuclease Rad50 - ATPase activity Nbs1 - scaffold to facilitate Mre11 and Rad50 interactions The MRN complex plays a crucial role in DSB repair.

Question 34

What are anaphase bridges?

Answer 34

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.

Question 35

What is a G-quadruplex?

Answer 35

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.

Question 36

What is the main purpose of mutation accumulation (MA) lines?

Answer 36

To accurately estimate mutation rates by minimizing the effects of natural selection.

Question 37

Compare type I to type II topoisomerases.

Answer 37

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

Question 38

Compare type I and type II topoisomerases.

Answer 38

Type I: cuts once Type II: cuts twice

Question 39

Name three genome instability syndromes and their defective genes.

Answer 39

AT (ATM), NBS (NBS1), ATLD (MRE11).

Question 40

Describe base excision repair.

Answer 40

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.

Question 41

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

Answer 41

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.

Question 42

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

Answer 42

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.

Question 43

Describe nucleotide excision repair in prokaryotes.

Answer 43

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.

Question 44

What types of DNA end damage are commonly caused by ionizing radiation (IR) or ROS?

Answer 44

3'-phosphate (3'-P) 5'-hydroxyl (5'-OH) Damaged bases Gaps in DNA strands

Question 45

Explain the cell fusion experiment by Rao and Johnson.

Answer 45

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.

Question 46

How can replication fork barriers be studied?

Answer 46

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

Question 47

Compare a transversion to a transition mutation.

Answer 47

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

Question 48

Describe break-induced repair.

Answer 48

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.

Question 49

How do Chk1 and Chk2 differ in DDR signaling?

Answer 49

Chk1: ATR-dependent (replication stress). Chk2: ATM-dependent (DSBs). Both arrest cell cycle.

Question 50

How is break-induced repair associated with APOBEC proteins?

Answer 50

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.

Question 51

What is the role of PARP1 in SSB repair?

Answer 51

Binds SSBs, recruits XRCC1 via PARylation, and coordinates BER/SSBR.

Question 52

Describe DNA fibre analysis.

Answer 52

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.

Question 53

Describe the process of non-homologous end-joining.

Answer 53

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.

Question 54

How does the sliding clamp differ between eukaryotes and prokaryotes?

Answer 54

Eukaryotes: - Heterotrimeric - Requires additional proteins for loading Prokaryotes: - Homodimer - Only uses Y complex

Question 55

Compare yeast and mammalian NHEJ.

Answer 55

Yeast: - MRX complex - No DNA-PKcs Mammalian: - MRN complex - Uses DNA-PKcs

Question 56

Describe the end replication problem.

Answer 56

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.

Question 57

How does Break-Induced Replication (BIR) contribute to cancer evolution?

Answer 57

Repairs one-ended DSBs mutagenically; APOBEC-induced damage exacerbates mutations during BIR.

Question 58

How is the eukaryotic replisome distinguished from dormant replisomes?

Answer 58

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.