DNA damage and repair Flashcards

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1
Q

What is the definition of DNA damage?

A

change to regular chemical structure of the DNA double helix

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2
Q

Give four examples types of DNA damage?

A
  • break in phosphodiester backbone
  • loss of base from deoxyribose sugar
  • alteration to the structure of the base
  • mis-matched base pairs
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3
Q

How can DNA damage lead to a mutation?

A

if damaged DNA undergoes further rounds of replication, this can result in a fixed mutation

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4
Q

What is the definition of a mutation?

A

permanent heritable change in the sequence of an organism’s genome

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5
Q

Give the two types of mutations and give examples of each

A

Point mutation = alteration, insertion, deletion of one or more bases at a time

Chromosome mutation = translocation, deletion, insertions

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6
Q

What is a transition point mutation?

A

purine replaced by a purine
or
pyrimidine replaced by a pyrimidine

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7
Q

What is a transversion point mutation?

A

Purine replaced by a pyrimidine, or vice versa

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8
Q

What is a missense point mutation?

A

codon now codes for a different amino acid

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9
Q

What is a nonsense point mutation?

A

codon now codes for a stop codon

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10
Q

What are the stop codons?

A

TAA, TAG, TGA

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11
Q

What are the three types of effects of base-pair substitution mutations?

A

Neutral = substituted amino acid has similar chemical properties to the original amino acid

Silent = base substitution in third position so codon encodes for the same amino acid (degeneracy of DNA)

Frameshift = caused by insertion or deletion results in change in all downstream codons

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12
Q

What is the difference between a forward mutation, a reverse/reversion mutation, and a suppressor mutation?

A

forward mutation = wild type gene changes to a mutant gene

Reversion mutation = mutant gene changes back to wild type gene

Suppressor mutation = changes sequence at a different location from the original mutation in a way that compensates for the original mutation and restores function

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13
Q

What are the two types of reversion mutations that can occur?

A
  • True reversions = restores sequence to code for the wildtype amino acid
  • partial reversions = changes the amino acid sequence but restores the function of the protein
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14
Q

What are the two types of suppressor mutations that can occur?

A

Intragenic = occurs within the same gene

Intergenic = occurs within a different gene

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15
Q

What does spontaneous mean in terms of mutation?

A

Occurs without exposure to exogenous agents

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16
Q

What does Lamarckism tell us about mutations?

A
  • tells us that they are adaptive

i. e. they develop as a response to a particular environment to allow adaption

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17
Q

Are mutations random or adaptive? why?

A

Random
- mutations occur spontaneously throughout growth but may confer a selective advantage in a particular environment, allowing the effect of the mutation to be observed

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18
Q

What experiment was carried out by Luria and Delbruk? what did this show about mutation?

A

1) Grow E.coli in a batch culture
2) plate out some of the batch culture on nutrient medium and isolate single colonies and re-suspend them in liquid media
3) plate the individual colonies as a lawn and add T4 bacteriophage
4) count the number of colonies = colony growth = resistance

Showed:
- large variation in number of resistant colonies between individual plates = number of resistant colonies depended on when, during growth in batch culture, the mutation occurred.

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19
Q

How do random spontaneous mutations arise?

A

Occurs due to replication of pre-mutagenic DNA

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20
Q

What are the two main causes of endogenous premutagenic damage to DNA? give examples of each

A

DNA replication errors:

1) nucleotide or template tautomerism causing mismatches
2) replication slippage (causes incorporation or deletion of extra bases)

Endogenous DNA damage (within the cell):

1) base deamination
2) base loss (depurination, depyrimidination)
3) base modification due to exposure to metabolic products (alkylation, oxidative damage)

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21
Q

What four mechanisms are used to ensure the high fidelity of DNA synthesis during replication?

A

1) base pairing
2) DNA polymerase - base selection, 3’-5’ exonuclease activity
3) Accessory proteins (single stranded binding protein)
4) post-replication mismatch repair

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22
Q

True or false: mismatches are pre-mutagenic lesions?

A

True (a further round of replication would created a fixed mutation)

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23
Q

Tautomerised adenine resembles which other nitrogenous base? what is the result of this?

A

Resembles guanine

result: cytosine added opposite the adenine tautomer and if replicated again results in a T-A to C-G mutation

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24
Q

How do DNA looping-out errors spontaneously generate addition or deletion mutants?

A

when the new strand loops out , results in an additional base in the new strand

When the parental strand loops out, this results in the loss of a base from the new strand

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25
Q

Where do DNA looping-out errors typically occur?

A

repetitive dna sequences

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26
Q

What is depurination? what are results?

A

endogenous DNA damage: hydrolysis of bond between sugar and base results in removal of the purine base, leaves an apurinic (SP) site premutagenic lesion

  • results in random base substitution or base-skipping during replication leading to substitution or deletions - further round of replication causes mutation
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27
Q

Which base loss endogenous damage is more common: depurination or depyrimidination?

A

Depurination

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28
Q

What is deamination? what are results using cytosine as an example?

A

Endogenous DNA damage: removal of the amino group from a base.

Cytosine deamination results in uracil, which can be repaired but if not, when replicated, the uracil will base pair with an adenine, changing the C-G to a T-A fixed mutation.

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29
Q

What is the results of deamination of methylated cytosine (5-MeC)?

A

Produces thymine

results in G-C to A-T mutation

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30
Q

What is alkylation? what are results using guanine as an example?

A

endogenous DNA damage: addition of methyl or ethyl group to sugar or base by endogenous alkyl donors

results: guanine able to be methylated at the O6 position to produce O6-methylguanine which can base pair with thymine. If this premutagenic lesion is replicated, this produces a G-C to A-T mutation.

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31
Q

What is oxidative DNA damage? what are results using guanine as an example?

A

ROS attack bases

ROS attack of guanine can produce 8oxoG premutagenic lesion, which can base pair with either C or A depending on the isomeric form. If base pairs with A, this can result in a GC to AT mutation

Results of ROS damaged bases can either have no effect, be mutagenic, or leads to strand breakage/block replication

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32
Q

Why is there a high amount of oxidative DNA damage in mitochondrial DNA?

A

ROS leak from the ETC

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33
Q

What are the two main causes of exogenous mutagens? give examples of each

A

Chemical mutagens:

  • base analogues
  • base modifying agents (alkylating agents)
  • Intercalating agents

Physical mutagens:

  • ionising radiation (X-rays)
  • UV radiation
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34
Q

What is an example of a base analogue? what are the results of this mutagen?

A

5-bromouracil (5BU)

results: behaves like thymine normally, but when tautomerised, it behaves like cytosine and base pairs with guanine.
changes AT to CG transition mutation

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35
Q

What are the effects of the base-modifying agent nitrous acid?

A

exaggerate endogenous deamination processes by creating acidic conditions that promote deamination

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36
Q

What are the results of G, C, and A deamination DNA damage?

A

Deaminated G produces Xanthine which base pairs with cytosine so has no effect

Deaminated C produces U, which base pairs with adenine so produces a CG-AT transition mutation

Deamination A produces Hypoxanthine, which base pairs with C so produces a AT-CG transition mutation

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37
Q

Give three examples of base-modifying agents

A
  1. nitrous acid
  2. hydroxylamine
  3. methylmethane sufonate (MMS)
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38
Q

What are the effects of the base-modifying agent hydroxylamine? give an example

A

hydroxylates and changes the bases

  • E.g. hydroxylation of cytosine results in a CG-AT transition mutation
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39
Q

What are the effects of the base-modifying agent MMS? Give an example

A

methylation of bases, which blocks replication or introduces mutations upon further replication

  • E.g. methylation of guanine results in O6-methylguanine that base pairs with thymine to produce a GC-AT transition mutation
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40
Q

Which base is particularly susceptible to DNA damage?

A

guanine

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41
Q

What is an intercalating agent? give an example of an intercalating agent? what are the results of this mutagen?

A

Have a flat, planar structures that insert into the minor groove of the helix, resulting in partial unwinding and insertion or deletion of additional bases upon replication.

E.g. ethidium bromide

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42
Q

What is benzopyrene and how does it become a carcinogen?

A

Benzopyrene is a procarcinogen in the environment but metabolised in the organism into a carcinogen benzopyrene dihydrodiol epoxide. It is this BPDE that binds to the DNA and induces pre-mutagenic lesions.

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43
Q

Which UV radiation is most able to cause affect DNA, why?

A

UVC because DNA absorbs UV radiation with a peak absorbance at 254 nm (within the UVC absorbance range)

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44
Q

What is the wavelength range for UVC, UVA and UVB?

A
UVC = 200-280 nm 
UVB = 280-320 nm
UVA = 320-400 nm
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45
Q

What type of DNA damage is caused by UV radiation?

A

bulky DNA damage:

  • 6-4 photoproducts
  • cyclobutane dimers
46
Q

How do cyclobutane dimers form?

A

UV radiation causes formation of intra-strand cross-linked pyrimidine dimers - adjacent pyrimidines (T and C) are linked via a 5-5 bond and 6-6 bond

47
Q

How do 6-4-photoproducts form?

A

UV radiation causes formation of a 6-4 bond between adjacent pyrimidines which distorts the helix and partially unwinds it

48
Q

How are linked pyrimidines (bulky DNA damage) replicated?

A
  • linked pyrimidines cannot be accommodated into the active site of the replicative DNA polymerases, causing replication fork stalling.
  • trans-lesion synthesis (TLS) DNA polymerases with a more relaxed active site can accommodate the bulky DNA but are more error prone/low fidelity so introduces mismatches than can lead to mutation
49
Q

What two ways can ionising radiation (X-rays and gamma rays) cause DNA damage?

A

1) directly = direct interaction of radiation energy with DNA
2) indirectly = ionisation of cellular water produces reactive oxygen species that attack the DNA and cause oxidative damage

50
Q

What are the major effects of ionising radiation on DNA?

A

Base damage

Single-strand or double-strand breaks (DSB particularly lethal)

51
Q

Describe the Ames test: purpose, method

A

Purpose: test whether a particular chemical is mutagenic

Method:

  • Salmonella strain that is a His auxotroph plated on MM lacking His
  • add suspected mutagen and incubate
  • look for colony growth = any growth indicates a reversion mutation has occurred within the gene encoding for His biosynthesis.
52
Q

What may need to be added to the Ames test for chemicals that are converted into carcinogens within the organism?

A

liver enzymes - allows the conversion to occur

53
Q

What are the two main methods of DNA damage repair? Give examples of each

A

Excision of damaged DNA:

  1. mismatch repair (MMR)
  2. Base excision repair (BER)
  3. Nucleotide excision repair (NER)

Direct reversal of DNA damage:

  1. Repair of O6-alkylguanine
  2. Enzymatic protoreactivation
54
Q

Describe the process of mismatch repair (MMR) system that operates in bacteria?

A

MutSLH mismatch repair system

  • MutH is bound at a hemi-methylated GATC site (where parental stand is methylated but new strand is not)
    1. MutS interacts with the beta-sliding clamp and binds to mismatch when encounters one
    2. MutS then recruits MutL and the second lobe threads the DNa through and binds to the next MutH at a hemi-methylated site.
    3. This activates MutH and causing it to nick the new strand either 5’ or 3’ of the mismatch
  1. Nicked strand in unwound by UvrD helicase and digested back to the mismatch by a exonuclease:
    - if nick is 5’ of mismatch, the 5’ to 3’ exonuclease is used (RecJ nuclease or exonuclease VII)
    - if nick is 3’ of the mismatch, the 3’ to 5’ exonuclease is used (exonuclease I)
  2. the DNA is re-synthesised by DNA polymerase III
55
Q

How can mismatch repair (MMR) correct replicative insertions or deletions?

A
  • repetitive regions of DNA can form hairpin loops on template strand that can be skipped during replication of the new strand (deletion)
  • repetitive regions of DNA can form hairpin loops on the new strand that can result in base insertion

MMR can detect and repair hairpins:
- newly synthesised DNA is degraded, hairpin unfolds, new strand can be re-synthesised.

56
Q

What can result from defects in MMR?

A

increased rate of spontaneous mutation (mutator phenotype)

cancer

57
Q

Who discovered the MMR system?

A

Paul Modrich

58
Q

Describe the process of base excision repair (BER) system?

A

1) DNA glycosylase cleaves between damaged base and the sugar to leave an abasic site
2) Abasic sites are repaired by apurinic/apyrimidinic (AP) endonucleases- cleave the backbone wither side of the abasic sugar
3) the gap is filled by the DNA polymerase and ligated

59
Q

Which enzymes are involved in BER? give three examples

A

DNA glycosylases (specific for particular type of DNA damage)

E.g.

  • 8-oxoguanine DNA glycosylase
  • Uracil-DNA glycosylase
  • Failsafe glycosylase (T:G mismatch glycosylase)
60
Q

Who discovered the BER system?

A

Tomas Lindahl

61
Q

Describe the process of nucleotide excision repair (NER) system in E. coli?

A

1) UvrAB complex recognises and binds to the bulky DNA lesion
2) UvrA is released and UvrB has helicase activity that unwinds the damages region
3) UvrC is recruited and cleaves either side of the damage,
4) UvrD removes the excised oligonucleotide
5) DNA polymerase III binds to the 3’ OH and resynthesises the using the undamaged strand as a template
6) ligation of the gap.

62
Q

Which DNA repair mechanism has specificity for helix distortion (bulky DNA damage) and is not specific for a type of DNA damage?

A

NER

63
Q

Which group of genes are involved in the NER in E. coli?

A

Uvr genes: A, B, C, D.

64
Q

Who discovered the NER system?

A

Aziz Sancar and Tomas Lindahl

65
Q

The NER proteins are _________________ between bacteria and eukaryotes, but the mechanism _________________

A

not conserved

is conserved

66
Q

The NER proteins are _________________ between bacteria and eukaryotes, but the mechanism _________________

A

not conserved

is conserved

67
Q

What two pathways does NER operate in?

A

Global genomic repair (GGR) - recognises and repairs damage anywhere in genome

Transcription-coupled repair (TCR) - genes that are actively transcribed (signalled by stalling of replication fork)

68
Q

What can result due to defects in NER? Give an example

A

Mutator phenotype

Example:
- Xeroderma Pigmentosum (XP) is an inherited NER defect = sun sensitive, high predisposition to skin cancer as unable to repair UV DNA damage

69
Q

How is alkylation damage repaired by direct reversal of DNA damage mechanisms? (in E.coli)

A

E.coli has alkyltransferase (Ada / O6-methylguanine-DNA methyltransferase)

  • the alkyl group from the alkylated guanine or backbone is transferred to Ada (at specific Cys residues) and inactivates it
  • methyl-Ada acts as a transcriptional activator and stimulates production of more Ada and the AlkA glycosylase
  • AlkA glycosylase removes the methylated guanine bases during the first step of BER
70
Q

How are pyrimidine dimers repaired by direct reversal of DNA damage mechanisms?

A

Enzymatic photoreactivation by photolyase enzyme

  • photolyases specific for CPDs or 6-4 photoproducts contain two chromophores (an antennae pigment that absorbs photon and a catalytic cofactor (FADH-)
  • electron transferred from the FADH- to the UV-induced lesion causing dimer splitting, then electron transferred back to FADH.
71
Q

What are the only organisms that don’t have enzymatic photoreactivation as a means to directly repair UV-induced lesions?

A

Placental mammals

72
Q

What happens if altered/absent bases are not repaired before replication?

A
  • normal replicative polymerases cannot replicate damaged DNA so replication forks stall
  • DNA damage response induced
  • TLS polymerases replicate some of the damaged DNA
73
Q

Which polymerases allow for DNA damage tolerance? How?

A

TLS DNA polymerases (Y family)

  • more open, flexible active site (but low fidelity)
  • lack proof-reading 3’-5’ exonuclease activity
  • higher error rate (error prone is better than no replication)
  • allows continued replication at unrepaired lesion
  • risk of insertion of incorrect base leading to mutation
74
Q

How do A and B family polymerases differ from Y- family (TLS polymerases?

A

Surface of the polymerases:

  • A and B tightly contact base in template and incoming nucleotide
  • Y contact less tightly to allow accommodation of damaged bases to continue synthesis (low fidelity)

Little finger domain:
- Y- family polymerases have an additional little finger domain that contacts DNA close to the lesion site (is specific for particular type of DNA damage)

75
Q

How are TLS polymerases recruited to the replication fork?

A

interaction with the sliding clamp

76
Q

What are the signals in prokaryotes and eukaryotes that induce the DNA damage response?

A

1) Stalled replication fork leads to LARGE REGIONS OF ssDNA as helicase continues to unwind but replicative polymerases cannot synthesise
2) single-strand nicks causes replication fork collapse which can lead to DOUBLE-STRANDED BREAKS.

77
Q

What are the three results of the DNA damage response?

A

1) increase DNA repair proteins
2) arrest cell cycle
3) trigger apoptosis in multicellular organisms

78
Q

Describe the DNA damage response in bacteria

A
  • stalled replication fork exposes ssDNA which is bound by RecA forming a filament and becomes activated
  • activated RecA cleaves LexA repressor
  • cleaved LexA cannot bind to the operator of the SOS genes anymore so SOS genes are transcribed and translated
79
Q

Give examples of the genes for DNA repair proteins that are under SOS regulation in bacteria

A
  • UvrA, B, D = NER
  • RecA, RuvA, RuvB = recombination repair of strand breaks
  • TLS polymerases (polIV and polV)
  • DinI = DNA mimic involved in supressing DNA damage response following repair
80
Q

How is the DNA damage response suppressed when DNA is repaired?

A
  • following repair, ssDNA decreases, reducing RecA filament assembly and LexA cleavage
  • DinI binds and sequesters RecA
  • LexA now able to dimerise and bind to the operator s of SOS genes and represses their transcription.
81
Q

Describe the DNA damage response to stalled replication forks / large amounts of ssDNA in eukaryotes

A
  • exposed ssDNA is bound by DNA damage sensor RPA on lagging strand (normally removed as lagging strand is replicated, but stalling of replication fork exposes large amount of ssDNA so RPA accumulates)
  • RPA recruits transducer regulator kinase ATR via ATRIP, specific sliding clamp loader 9-1-1, and Rad6/Rad18 complex
  • 9-1-1 recruits TOPBP1, and these activate ATR to phosphorylate downstream targets
  • Rad6/Rad18 monoubiquinates PCNA to reduce affinity for replicative polymerases for PCNA so they dissociate. TLS polymerases have higher affinity for ubiquitinated PCNA so are recruited to the replication fork to resume error prone synthesis.
82
Q

Give some downstream phosphorylation targets of ATR

A

1) cell cycle control - phosphorylates checkpoint kinase 1 (CHK1) to arrest cell cycle
2) Replication fork stabilisation - phosphorylates RPA, polymerases, 9-1-1 to slow progression of replication fork and further activate ATR
3) Replication origin control - phosphorylates RPA, MCM complex, PreRC to delay initiation of replication at origins

83
Q

Describe the DNA damage response to DSBs in eukaryotes (MRN pathway)

A
  • DSBs are bound by the DNA damage sensor MRN complex (interacts at break- may hold ends together)
  • MRN recruits transducer regulator kinase ATM (Ataxia Telangiectasia Mutated)
  • interaction of ATM with MRN/DSB complex leads to autophosphorylation and activation of the ATM dimer
  • activated ATM phosphorylates downstream targets
84
Q

Give some downstream phosphorylation targets of ATM

A

Phosphorylation of H2AX histones resulting in recruitment of a large protein complex including MDC1 (which is bound by MRN and ATM itself) which further activates ATM and amplifies the signal

  • this results in phosphorylation of downstream targets:
    1) cell cycle control - phosphorylates CHK1 and CHK2 to arrest pause cell cycle
    2) phosphorylation of p53 to induce apoptosis is break unrepaired
    3) phosphorylation of proteins involved in Non-homologous end joining in G1 or homology directed repair in S or G2.
85
Q

Describe the DNA damage response to DSBs in eukaryotes (KU pathway - in vertebrates)

A
  • DSB bound by DNA damage sensor KU (a heterodimer of KU70 and KU80) to ends of DSBs
  • in vertebrates, KU recruits transducer regulatory kinase DNA-PKcs to form DNA-PK
  • DNA-PK recruits proteins to join broken ends by NHEJ in G1 (leads to some loss of information at site of DSB)
86
Q

How are ATR, ATM and DNA-PKcs structurally related?

A
  • all large proteins with similar domain structures
87
Q

Which complexes control the global DNA damage response?

A

ATM-MRN-DNA

ATR-ATRIP-RPA-DNA

88
Q

Which complex (s) directly promote repair of DNA DSB breaks by NHEJ in G1?

A

KU-DNA-PKCS

89
Q

NHEJ occur predominantly in _____________ cells

A

non-dividing

90
Q

What is NHEJ and what are the pathways of NHEJ in G1?

A

NHEJ joins broken ends together, no template required, activated by KU-DNA-PKCS in G1, information lost

Pathways:

1) Nuclease digestion
- removes a few nucleotides before ligation

2) Resection to expose single stranded DNA on each strand which base pair at a complementary sequence and are ligated together and overhang processed.

91
Q

What are the results of information loss by NHEJ?

A

mutations (specifically frameshift mutations)

92
Q

What is homology directed repair and what are the two types of templates that can be used?

A

HR is the mechanism for the repair of DSB in late S phase/G2 phase using a homologous DNA sequence as a template

Two types of templates:

  • Sister chromatic in late S-phase/G2
  • Homologous chromosome in G1 (template may not be identical so results in loss of heterozygosity)
93
Q

Describe the mechanism of HR of DSBs that is conserved across all organisms

A

1) end resection = generation of ssDNA at site of DSB by helicases and exonucleases (presynapsis)

2) 3’ end of one single strand invades the intact homologous duplex (ds) to form a heteroduplex and
a D-loop in which there is pairing between the invading single strand end and a strand on the homologous duplex (synapsis)

3) repair damaged duplex by DNA synthesis using strands from intact duplex as template (postsynapsis)
- repair via synthesis-dependent stand annealing (SDSA)

4) separation of the two duplexes (postsynapsis)

94
Q

Following formation of the D-loop, how is the DSB repaired by SDSA?

A

1) DNA synthesis is initiated from the 3’ OH of the invading strand using the undamaged strand of the homologous duplex as a template
2) as synthesis proceeds, newly synthesised strand is released from the template strand and the D-loop moves along the template strand until the new DNA can base pair with the single-stranded 3’ overhang of the broken strand
3) the damaged DNA now has a region of single stranded DNA where the double-strand gap used to be
4) the gap is filled on the opposite strand using the newly synthesised strand as a template

95
Q

What two things characterise HR via SDSA?

A
  1. new DNA spans the break, and both strands are the result of conservative DNA synthesis (both strands at the break are new)
  2. No joining of DNA from the undamaged duplex to the repaired duplex
96
Q

How is the generation of ssDNA tails at DSBs (the presynaptic step) mediated in E.coli?

A

Mediated by RecBCD
- RecBCD has both helicase (unwind) and nuclease (digest) activity

1) RecBCD attaches at DSB and moves along the DNA unwinding and digesting both strands (has 5’-3’ and 3’-5’ nuclease acitvity)
2) RecBCD encounters a Chi sequence where the nuclease activity degrades just the 5’ strand, leaving a 3’ ssDNA tail
3) RecBCD facilitates RecA binding to the ssDNA tail, forming a presynaptic spiral nucleoprotein filament (which can trigger the SOS response)
4) The RecA nucleoprotein filament promotes pairing with homologous DNA by sampling stand to search for identical DNA.
5) invasion by the 3’ tail that displaces a duplex strand into the D-loop

97
Q

What is a Chi sequence?

A

An over-represented 8 bp sequence that occurs approximately every 5 kbp

98
Q

What is bacterial RecA?

A

A strand exchange recombinase enzyme

99
Q

What is eukaryotic Rad51?

A

A strand exchange recombinase enzyme

- equivalent to the bacterial RecA during HR

100
Q

Why are mutants in recBCD, Rad51 (eukaryotes), and RecA (bacteria) very sensitive to DNA damage?

A

unable to effectively repair damage

101
Q

What is the accuracy of HR?

A

90-100%

102
Q

Describe the effects of homology directed repair using a homologous chromosome as a template (as in G1)?

A
  • usually have a slightly different sequence so when used to repair a DSB, the new DNA now has the sequence of the homologue, resulting in GENE CONVERSION

–> E.g. if an individual is hereozygous for gene B (so is Bb), one chromosome has the B allele, and the other has the b allele. If a DSB occurs within gene B for one chromosome and the other chromosome is used to repair it, this will change the genotype from heterozygous to homozygous
= LOSS OF HETEROZYGOSITY

103
Q

What are the basic similarities and differences between homologous recombination and homology-directed repair?

A

Similarities: uses the same cellular machinery

Differences:

Outcomes

  • HR results in gene conversions
  • homologous recombination can result in reciprocal exchange of large DNA segments between homologous double stranded chromosomes

Processing of DSB ends

  • in HR, the second end of the DSB captures the newly synthesised strand and the two duplexes do not become linked
  • in homologous recombination, the second end of the DSB captures the displaced strand of the D-loop, which links the two duplexes together.
104
Q

How is meiotic recombination initiated in eukaryotes?

A

Spo11 introduces DSB to recruit Rad51 initiate meiotic recombination

105
Q

What are the key roles of homologous recombination in eukaryotes? (2 things)

A

1) pairs homologous chromosomes to ensure homologous chromosomes are separated in meiosis I
2) allows reciprocal exchange of random segments of maternal and paternal (non-sister) chromatids to produce unique gametes = diversity

106
Q

Describe the process of homologous recombination in bacteria

A
  • following gene transfer (conjugation, transduction, transformation) the donor DNA can be integrated
    1) introduction of a DSB and resection by RecBCD to expose 3’ ssDNA tails, which become bound by RecA
    2) RecA nucleoprotein filament promotes the invasion of the 3’ ssRNA into the homologous chromosome forming a D-loop and the 3’ OH of the primes new DNA synthesis
    3) The displaced D-loop is captured by the other free 3’ OH of the DSB and DNA synthesis occurs form the 3’ end
    4) continued DNA synthesis from both the invading strand and the captured strand results in the formation of two Holliday junctions
107
Q

What is the definition of a Holliday junction?

A

four-armed cross-stranded structures in which the participating DNA molecules are physically linked by a region of heteroduplex DNA and exchanged strands

108
Q

How does a Holliday junction move along the recombining DNA?

A

Branch migration

109
Q

How does branch migration of Holliday junctions occur in bacteria?

A

Driven by the RuvAB complex

  • all 4 arms of the Holliday junction are bound by a tetramer of RuvA, opening the junction into a square planar configuration
  • one RuvB helicase hexamer binds to each of the two opposite arms of the junction
  • RuvB moves the Holliday junction along the DNA, breaking and re-forming base pairs in one direction
110
Q

How are Holliday junctions resolved?

A

RuvAB complex recruits RuvC Holliday junction resolvase enzyme to cleave the junction:

  • horizontal cleavage of both junctions = non-recombinant products
  • horizontal cleavage of one junction and vertical cleavage of the other junction = recombinant product