DNA damage and repair Flashcards

1
Q

What effects do abasic site have on DNA strand?

A

i) Loss of coding info (lack base)
ii) no site for polymerase (aldehyde chain is not a substrate for DNA pol)
iii) cause DNA backbone cleavage (once converted to aldehyde form)

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2
Q

Types of DNA damage

A

i) spontaneous (naturally)
–> ROS, free radicals (interaction between DNA and chemical agents during mechanism)

ii) Environmental (e.g. UV, X-rays)
–> alkylating agents

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3
Q

Major form of nucleotides with abasic sites

A

hemiacetal

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4
Q

What happens if nucleotide with abasic site is converted to minor form?

A

from hemiacetal to aldehyde
–> susceptible to beta elimination

–> form 5’ phosphate and 3’ aldehyde chain (not substrate for DNA pol)

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5
Q

How does depurination or depyrimindation occur?

A

With help of water, base is removed form nucleotide through hydrolysis
–> form abasic sites (highly mutagenic)

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6
Q

Difference between DNA damage and mutation

A

Damage: alteration in DNA structure
Mutation: the change in DNA can be replicated and inherited

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7
Q

How is ROS formed?

A

From oxidative phosphorylation –> e- leaking
–> superoxide (O2-)
—> H2O2 through SOD
——> hydroxyl radicals (reduction)

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8
Q

What pairs with 8-oxo-G?

A

anti –> still pair C (steric clash is found)
syn –> pair with A (new H bond donor at N7)

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9
Q

Why would 8-oxo-G possibily cause DNA mutation?

A

can change from 8-oxo-G:C to 8-oxo-G:A
–> later become heritable

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10
Q

Types of spontaneous DNA damage

A

i)Depurination/depyrimidination
ii)Deamination
iii)Oxidative damage (ROS)

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11
Q

What is one common oxidative leision in DNA?

A

8-oxo-G, carbonyl grp added C8’ of guanine

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12
Q

What happens during deamination of C?

A

C to U as amino grp removed on C
–> H atom on NH2 replaced with O (H bond acceptor)

–>C:G –> U:G (damage)

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13
Q

Where does UV damage usually occur within DNA?

A

2 pyrimindine bases close tgt

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14
Q

100,000 per cell per day

A

UV damage everyday per cell

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15
Q

Two products formed from UV damage on DNA

A

Cyclobutane pyrimidine dimer (CPD)
–> can be between two thymine as well

6,4 photoproduct
–> bond formed between C4’ and C6’ of two pyrimidine bases

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16
Q

structural feature for cyclobutane pyrimidine

A

Formation of 4 carbon ring
–> pyrimidine bases brought close tgt

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17
Q

What does O6MeG base pairs with?

A

C and T (50% for each)

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18
Q

What happens after O6 methylated in guanine?

A

Before: O6 as acceptor, H in N7 as donor

After: CH3 has no effect, N7 becomes acceptor

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19
Q

Structural feature of 6,4 photoproduct

A

bond formed between C4’ of one and C6’ of another

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20
Q

500 per cell per day

A

Frequency of lesions per cell from depyrimidination everyday

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21
Q

What types of CPD can cause mutagenesis?

A

CT and CC
–> as AA would also be inserted on the opposite side
–> change their sequence

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22
Q

10,000 per cell per day

A

Depurination

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23
Q

Example of alkylation damage

A

O6 in guanine methylated

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24
Q

Types of environmental DNA damage

A

i) UV damage (CPD, 6,4 photoproduct)
ii) Alkylation damage (O6MeG)

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25
Q

Frequency of lesions per cell from alkylation everyday

A

5000 per cell per day

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26
Q

Why does thymine dimer not cause mutagenesis?

A

TLS pol for UV prefer to add AA to opposite side
–> no sequence change

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27
Q

What type of enzyme is MGMT?

A

suicide enzyme
–> can only be used once as methyl group attached will inactivate the enzyme

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28
Q

Residues in MGMT important for function

A

arginine bind to minor groove
–> base flips out into active site of MGMT

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29
Q

Function of DNA glycosylase in base excision repair (BER)

A

recognize damaged base
cleave N-glycosidic bond

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30
Q

Frequency of lesions per cell from depyrimidination everyday

A

500 per cell per day

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31
Q

5000 per cell per day

A

Frequency of lesions per cell from alkylation everyday

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32
Q

Frequency of lesions per cell from deamination everyday

A

100 per cell per day

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33
Q

Frequency of lesions per cell from depurination everyday

A

10,000 per cell per day

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34
Q

Frequency of lesions per cell from UV everyday

A

100,000 per cell per day

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35
Q

What system can repair O6MeG?

A

Direct repair by MGMT

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36
Q

characteristics of DNA glycosylase for BER

A

each can only recongize one type of damaged base

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37
Q

What is SP-BER?

A

Short patch-BER
–> only one nt replaced

38
Q

Function of 5’dRP Lyase

A

cleave 5’ deoxyribose phosphate

39
Q

Function of 5’ AP endonuclease

A

5’ AP site is cleaved

40
Q

Function of TFIIH in NER

A

recruit 2 DNA helicases (XPB, XPD) to unwind DNA

41
Q

Enzymes needed for base excision repair (BER)

A

DNA glycosylase
5’ AP endonuclease
5’ dRP lyase
DNA polymerase + DNA ligase

42
Q

Function of XPF

A

cleave 5’ of lesion

43
Q

What removes the damaged base during base excision repair (BER)?

A

DNA glycosylase

44
Q

Major targets of glycosylases in base excision repair

A

i) Uracil
ii) methylated damage
iii) 8-oxo-G

45
Q

Major steps in base excision repair (BER)

A

i) recognize damaged base by DNA glycosylase
ii) cleave damaged base by cleaving N-glycosidic bond (DNA glycosylase)
iii) Damaged base released
iv) AP endonuclease cleave 5’ of AP site
v) 5’ dRP lyase remove 5’deoxyribose phosphate
vi) DNA polymerase and ligase to produce correct base and fix the nicks

46
Q

What happens after helicase start to unwind DNA in NER?

A

one hit the lesion
–> other helicase keep unwinding until around 20 bp is unwound

47
Q

Predominant BER pathway in mammals?

A

SP-BER (Short patch)
–> only one nt replaced

48
Q

How does nucleotide excision repair recognize damage in DNA?

A

distortion of DNA helix or
H bond disruption

49
Q

what does nucleotide excision repair usually repair?

A

chemical damages such as UV damage, bulky addicts (complex alkylation damage)

50
Q

Function of XPC in humans

A

recognize distortion in DNA
kinks DNA –> signal to recruit TFIIH

51
Q

Function of XPA, XPG

A

cleave 3’ of lesion

52
Q

Function of RPA in nucleotide excision repair

A

bound to ssDNA
engage both XPG and XPA

53
Q

What repairs the strand after being removed in nucleotide excision repair?

A

Repaired by DNA replication machinery
(RFC, PCNA, Pol delta)

54
Q

Souce of double stranded breaks

A

i) ionizing radiation
ii) ROS
iii) type 2 topoisomerase
(formed covalent complex which has DSB if not fixed)
iv) meiosis (which requires DSB)

55
Q

In what process is the double holiday junction found?

A

Homologoys recombination (HR)

56
Q

Why is type 2 topoisomerase a source for DSB?

A

it has a covalent complex which has DSB if not fixed

57
Q

Why is meiosis a source for DSB?

A

it needs DSB during the process

58
Q

What protein decides that NHEJ would be used to repair damage?

A

KU, which prevents resection from happening if first bind to DNA

59
Q

Function of RAD51 paralogs in HR

A

stabilizes RAD51 filament

60
Q

Function of RAD51 in HR

A

Catalyze pairing between 3’OH ssDNA and repair template
— strand invasion

61
Q

Characteristics of non-homologoys end-joining (NHEJ)

A

lack resection
rejoin the ends tgt with minimal processing

62
Q

What protein decides that HR would be used to repair damage?

A

If MRN first bind to DNA before KU
–> causes 5’ end resection

63
Q

Function of Exo1 in HR

A

help with extensive resection
–> continue until repair template is found

64
Q

Choice of repair template for HR

A

i) sister chromatid
ii) homologous choromosome (might convert one allele to another)
iii) region of homology in another chromosome

65
Q

Proteins involved in end resection of HR

A

MRN, CtIP
RPA
Exo1

66
Q

Steps for end resection in HR

A

i) MRN form dimer, with one CtIP on each end
–> the two catalyze short ranged resection
ii) after 5’ end cleaved, RPA starts binding to ssDNA
–> CtIP leaves DNA
iii)Exo1 joins for extensive resection
–> continue until repair template is found
iv) Exo1 leaves DNA strand once repair template is found

67
Q

Function of RAD51 in HR

A

catalyze pairing between 3’OH ssDNA and repair template
–>strand invasion

68
Q

Function of RAD54 in HR

A

remove RAD51
–> allow DNA polymerase to access the 3’ end

69
Q

Describe the experiment to test for role of BRCA2 in HR

A

i) created an artificial DSB + homologous template that is radiolabelled
ii) some add BRCA2 while others don’t
iii) if there is repair, would detect two bands
–> if lack repair, only one band for dsDNA

70
Q

Results of the experiment for BRCA2

A

Lane with RAD51 + RPA
–> only one band
–> RPA inhibits RAD51
Lane with RAD51, RPA and BRCA2
–> able to restore strand, two bands
–> BRCA2 stimulates RAD51

71
Q

Describe steps in strand invasion of HR

A

i)RPA still remain bound after Exo1 leaves
ii) RPA then replaced with RAD51, that binds to BRCA2
–> catalyze strand invasion, allow pairing between 3’OH ssdNA and repair template
iii)RAD51 paralogs added to stabilize RAD51 filament
iv) RAD54 later joins in to remove RAD 51
–> allow DNA Pol to access 3’ end

72
Q

Describe major steps in Homologous recombination (HR)

A

i) 5’ end resection after MRN binds
ii)D loop formed between damaged and repair template
iii) strand invasion (BRCA2, RAD51)
iv) 2nd end capture –> double holiday junction (dHJ)
v) nuclease cleave the overlapping regions

73
Q

Describe steps for single-strand annealing (SSA)

A

i) flaps formed after 5’ end resection
ii) flaps cleaved
iii)Gaps filled and ligated

74
Q

Describe steps for NHEJ

A

i) KU need to bind DNA first
synpasis, forming rlly small regions of homology
ii) end processing
–> remove overhangs to present terminals
iii) ligation (DNA ligase)

75
Q

Possible consequences to using NHEJ for repair

A

may result in deletions or insertions depending on how u base pair the ends tgt

76
Q

What eukaryotes is genome engineering first done in?

A

budding yeast
(yeast prefer HR over NHEJ)

77
Q

Better genome editing method developed from yeast

A

i) introduce DSB onto chromosomes
ii) 3’ overhangs form on both templates and chromosome
iii) undergo single stranded annealing (SSA)

78
Q

Characteristics of improved genome editing in yeast

A
  • no need RAD51 –> no strand invasion
  • double holiday junction not found
  • need to introduce DSB in target gene in order for SSA to work
79
Q

Components in CRISPR-Cas 9

A

i) Cas 9 (introduce site-specific DSB)
ii) crRNA (DNA site specificity)
iii) tracrRNA (bind to crRNA, structural purposes)
–> II and III now usually combined into one, called guide RNA (gRNA)

80
Q

What does guide RNA consist of?

A

crRNA –> DNA site specificity
tracrRNA –> structural

81
Q

Function of Cas 9

A

induce site specific DSB

82
Q

How does CRISPR induce DSB at specific site?

A

Cas9 + sgRNA
–> scan dsDNA to find PAM
–> then see if sequence adjacent is complementary to guide sequence

83
Q

Function of CTD in Cas9

A

PAM recognition region
–> disordered without sgRNA binding

84
Q

Function of HNH domain in Cas 9

A

inhibit RuvC when not bound to sgRNA

once activated –> cleave the target strand

85
Q

Function of RuvC domain in Cas9

A

cut the nontarget strand (with PAM) when activated (bind to sgRNA)

86
Q

Major steps in CRIPSR

A

i) sgRNA binds to Cas 9
(conformational changes)
ii) start the search for PAM, then see if sequence adjacent is complementary to guide sequence (once PAM found)
iii) start to unwind DNA once confirm the complementarity
iv) HNH cleave on target strand, while RuvC cleave on nontarget strand

87
Q

Conformational change in Cas9 after sgRNA binds

A

no sgRNA
- PAM recognition site in CTD disordered
- HNH not in catalytic conformation (inhibit RuvC)

with sgRNA
- orders PAM recognition region
- forms a channel to target DNA

88
Q

Two pathways to repair DSB from CRISPR

A

Non-homologous end joining
Homologous repair

89
Q

How to cause preferentially HR to repair DSB from CRISPR?

A

supply with donor DNA as repair template

90
Q

Examples of using CRISPR in therapy

A

engineer T cells with CCR5 deletion with CRISPR
–> HIV resistant

introduce HR template containing CCR5 deletion

91
Q

what edits did scientist do to make T cells HIV resistant?

A

CCR5 deletions