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AMEG Chaz > DNA Repair > Flashcards

DNA Repair Flashcards

1
Q

What is the difference between mono- and multigenic diseases?

A

Multigenic- mutations in multiple genes, the combination causes the clinical phenotype
Monogenic- mutations in a single gene that are more severe, and tend to be recessive so they escape selection pressure

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2
Q

What is replication stress?

A

Any process that obstructs DNA replication or interfere with the replication forks
Unresolved replication stress may lead to genetic instability

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3
Q

Give specific examples of monogenic diseases in the DNA damage response

A

Fanconi anaemia- variations on the FANC gene- bone marrow failure, skeletal defects, genome instability, myeloid leukaemia
Seckel syndrome- ATR, ATRIP- growth retardation, dwarfism, microcephaly, mental retardation, genome instability
Ataxia Telangiectasia- ATM- genome instability, T/B cell leukaemia/lymphoma
Ataxia Telangiectasia-like disease- Mre11- neurodegeneration, ataxia, genome instability
Bloom syndrome- BLM- premature ageing, growth retardation, genome instability, leukaemia/lymphoma

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4
Q

Describe Seckel1

A

Seckel syndrome is a group of monogenic inherited disorders
-severe microcephaly, growth retardation, mental retardation, dysmorphic features, skeletal abnormalities
Rare autosomal recessive disorder
ATR kinase or ATRIP
Are functionally related proteins that form part of the DDR
Found at replication forks, are crucial in maintaining genome stability
Defects in phosphorylation if Chk1 and H2AX
Increased numbers of centrosomes
Rapidly proliferating CNS progenitor cells cannot deal with stress without ATR/ATRIP

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5
Q

Describe Bloom syndrome

A

Growth retardation, photosensitivity, Telangiectasia, hypo- and hyper pigmented skin, predisposition to malignancies, chromosomal instability
BLM is a helicase involved in resolving intermediates of DNA repair and removing proteins from DNA
➡️ reflects on replication, genome instability, increased genetic exchNge between chromosomes

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6
Q

How would you identify where a disease gene is?

A

Use genetic markers to analyse inherited regions- linkage
Linkage is the non-random association of traits passed from parents to offspring including restriction enzyme markers and SNPs
Requires at least two generations with affected and unaffected individuals
Genotype using marked and analyse regions conserved between affected individuals and carriers
Evaluate all genes in that region to identify the causative mutation

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7
Q

What is the principle behind using linkage when identifying disease genes?

A

If two sequences are close together they are likely to be inherited together- less likely to split up during recombination
Use restriction sites (trace inheritance of a particular pattern of restriction fragments) SNPs (SNP arrays- anneal to probes- measure fluorescent signal) and microsatellites

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8
Q

What are the cyclin and Cyclin dependent kinases involved at the different stages of cell cycle?

A 
G1- Cyclin D1-3
Cdk2,4,5,6
G1-S- Cyclin E 
Cdk2
S- Cyclin A
Cdk2
G2 and M- Cyclin A and B
Cdk2
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9
Q

What are cell cycle checkpoints?

A

A point in eukaryotic cell cycle where progress through the cycle can be halted until conditions are suitable for the cell to proceed to the next stage
They prevent the propagation of deleterious genetic errors

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10
Q

Describe the G1-S checkpoint

A

Prevents entry into S phase and the replication of damaged DNA
Targets G1 phase Cdk activity
Controlled by ATM
DSB➡️MRN➡️ ATM-P➡️
➡️p53-P➡️p21- cdk2 inhibitor
➡️Chk2-P➡️cdc25A-P➡️ degraded so it does not remove the phosphates from Cdk2➡️ cdk2-P inactive

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11
Q

Describe the regulation of p53

A

Normally MDM2, bound to USP7 and inactive p53, targets and ubiquitinates p53 for destruction
If there is a DSB ATM-P inactivates MDM2 and blocks the interaction with USP7, it is ubiquitinated and degraded
ATM also activates Chk2 which activates p53 (WIP1 also removes phosphorylations on p53)
p53 promotes the transcription of p21 but build up of p53 in the cell activates MDM2 in a negative feedback loop

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12
Q

Name two important serine residues on p53 for phosphorylation

A

Serine 15 and 20
Regulated by ATR, ATM, DNA-PK, chk1 and chk2
Decrease interaction with MDM2 increase transcriptional activity

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13
Q

Describe the intra-S phase checkpoint

A

Prevents replication of damaged DNA
Both elongation and late origin firing are inhibited
Target replication associated cdk activity (cdk2/7)
Controlled by ATM/ATR dependent DDR pathway
DSB➡️MRN ➡️ MDC1➡️ ATM-P➡️ chk2-P➡️
Single strand DNA of the replication fork➡️ 9-1-1/TOPBP1➡ ATR/ATRIP-P➡️ chk1-P➡️
➡️cdc25A-P is degraded➡️ cdk2-P prevents exit out of S phase

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14
Q

Name some proteins, sites of phosphorylation and associated diseases

A 
ATM- S367, T1885, S1981, S2996- AT
RAD50- S635- NBS-like disorder
BRCA1- S1387- Fanconi anaemia complementarity group S
ATR/ATRIP- T1989 on ATR- Seckel syndrome
Chk1- S317, S345
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15
Q

Describe the G2/M phase checkpoint

A

Prevents entry into mitosis with DNA damage
Targets G2/M phase cdk activity eg. Cyclin B and Cdk2
Controlled by ATM/ATR dependent DDR
Resected DSB (single DNA strand)➡️
9-1-1/TOPBP1 ➡ATR/ATRIP-P ➡️
MRN➡️ MDC1➡️ ATM-P➡️
➡️chk1 and 2-P➡️ cdc25B/C-P➡️ cdk1-P (inactive)
(➡14-3-3 exported from the nucleus)

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16
Q

What can replication cause?

A

Replication forks that are not moving properly can be converted into double strand breaks
Most likely to happen during S phase and also mitosis

17
Q

How are replication forks converted into DSBs?

A

Stalled forks means that the polymerase on the lagging strand can fall off
Then endonuclease processes it into a one ended DSB by cutting the single strand DNA at the ‘mouth’ of the fork

18
Q

When are regressed replication forks found in normal cell division?

A

They are holiday junctions that make up chiasmata during recombination in meiosis
Endonuclease normally cuts Holliday junctions and a regressed fork is seen as the same structure and so it is cut

19
Q

Describe the direct/alternative replication fork collapse

A

Single strand break on the template that is not repaired
Enzyme encounters break and falls off template
SSB transformed into a DSB

20
Q

How are collapsed replication forks dealt with?

A

DSB➡️ ATM pathway
ssDNA➡️ ATR pathway
Degradation of cdc25A and phosphorylated cdk1, 2 lead to cell cycle arrest
Either homologous recombination or apoptosis

21
Q

Describe homologous recombination of a one ended DSB

A

A 3’ over hang is required EXO1, Mre11 and BRCA1
Rad51 and BRCA2 finds the homologous sequence in the sister chromatid strand
Forms a D-loop
HJ resolution by HJ resolvases (GEN1, MUS81/EME1, SLX1/SLX4)
Re-establishes the replication fork

22
Q

How would you target homologous recombination for cancer therapy?

A

Familial cancers with mutation in enzymes such as BRCA1,2 and FANCN,J have a HR defect
Tumours are sensitive to PARP1 inhibitors (SSB repair enzyme) normal somatic cells still have one wild type copy
Cancer cells die of unrelated spontaneous SSBs

23
Q

What is the significance of oncogenes in replication stress?

A

Oncogenes slow down replication forks- spontaneous replication stress demonstrated by cancer cells- eventually the cell will go into mitosis with replication unfinished- anaphase bridge between chromatids
Leading to genomic instability, fragmentation and other damage

24
Q

How would you target endogenous replication stress for cancer therapy?

A

ATR inhibitors- ATR is a principle mediator of the G2/M cell cycle checkpoint
It prevents premature entry into mitosis with incomplete replication or DNA damage
When inhibited the cell will eventually go into mitosis unfinished and this will lead to mitotic catastrophe and apoptoisis

25
Q

Describe non-homologous end joining

A

Does not require homologous template and can occur at any phase of cell cycle
Mostly accurate but small deletions can occur at the site of the DSB
Essential for cell viability
Required for immune system development
Ku70/Ku80 is recruited to the break and recruits PAXX which forms a molecular bridge between the ends
The bridge complex recruits DNA-PK and Artemis processes the ends so they can be ligated
DNA ligase4 and XRCC4 form a ligation complex and stably recruits DNA polymerase with Ku

26
Q

Describe alternative DNA end joining

A

If Ku cannot be recruited then Mre11 will recognise the complex which looks for microhomology, recruits CtIP and exo1 processes ends to the microhomology
Flap removal (by FEN1) can mean that a significant amount of DNA is lost
Extremely error prone and has been linked to the formation of chromosomal translocations

27
Q

Describe homologous recombination

A

Requires homologous template so can only occur during S phase
Very accurate
Mre11 recruits CtlP which promotes DNA resection by Exo1 and Dna2
53BP1 complex limits resection
RPA is a ssDNA binding protein that recruits RAD51 that recruits BRCA1, BRCA2, PALB2
RAD51 searches for the homologous sequence and DNA pol uses that as a template
The Holliday junction can be resolved by the BLM complex so there is no cross over or by a structure-specific nuclease complex that cuts the junctions and can give cross over or no crossover

28
Q

Name the protein that repairs DNA using NHEJ

A

Ku70/Ku80 heterodimer

29
Q

How does the MRN complex repair DSB?

A

Homologous recombination

30
Q

Describe how Ku binds to a DSB

A

Does not require a homologous template
Can occur at any phase in the cell cycle
Accurate but small deletions can occur
Essential for cell viability
Required for immune system maturation
Ku is shapers like a clamp and slides onto the broken ends of the DNA which are then tethered together by PAXX molecules that interact with both Kus and each other
Ku binds and activates DNA-PK which transphosphorylate each other and phosphorylates H2AX

31
Q

How does the MRN complex bind to a DSB?

A

Involved in 15-20% of DSB repairs in S and G2 phase
Extremely accurate
A MNR complex binds to each side and binds together through Mre11 and Rad50 dimers to hold the DNA together
ATM molecules at brought to the site and in close enough proximity to each other to transphosphorylate each other and phosphorylate H2AX which binds a phosphorylated MDC1 which binds NBS1
PARP1 binds to NDS1 and Mre11 complex

32
Q

Which factors affect the method of DNA DSB repair and how?

A

High cdk activity has a positive effect on MRN dependent repair -phosphorylates CtIP which binds to NBS1 at the BRCT1 domain - suppresses Ku activity along with ATM
Euchromatin is more permissive to Ku repair
Heterochromatin can only be repaired by the ATM/MRN complex
Complexity of the DSB: Greater complexity – MRN, Lower complexity – Ku
4 reasons

33
Q

How to assay for NHEJ

A

Using a nuclease called I-Scel one, to induce a specific DSB break. This nuclease has a long specific recognition sequence that is not found normally in the human genome. This sequence must therefore be inserted. It is a GFP gene, a promoter and a Puromycin selection gene in the middle. Essentially these are flanked by the recognition sequence. If you transfect the nuclease into the cells it will cut the DNA to form a DSB either side of the Puro gene. This will bring the GFP gene right next to the promoter if it is repaired by NHEJ.

34
Q

HR assay?

A

GFP gene with an I-Scel cut site so that the gene is not active. The plasmid also contains a bit of this gene, which is identical to the region of the GFP gene (iGFP). If the gene is cut to produce a DSB at the I-Scel, the iGFP acts as a homologous sister chromatid template, as it is identical to the region where the break has occurred. If there is HR between the non-functioning GFP gene and the fragment, then the GFP gene can be repaired back to wild type and the cells would show up green. Again FACS can be used, with the RFP plasmid, the GFP plasmid and the nuclease plasmid, to calculate the RFP:GFP ratio

AMEG Chaz (7 decks)

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