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Genetics > GENE 8: protecting the genome > Flashcards

GENE 8: protecting the genome Flashcards

1
Q

What are the two basic sources of DNA damage?

A

Reactive chemicals and radiation

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2
Q

Give an example of reactive chemicals and how they arise

A

Free radicals -> basic metabolism (oxidise DNA)

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3
Q

What are free radicals?

A

molecules with an unpaired electron

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4
Q

What are the most common type of free radicals? and how are they formed?

A

Reactive oxygen species (ROS)

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5
Q

Name some ROS

A 
oxygen
superoxide anion
peroxide
hydrogen peroxide
hydroxyl radical
hydroxyl ion
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6
Q

What are the forms of endogenous DNA damage?

A

ROS (oxidation)
spontaneous hydrolysis
alkylation by endogenous alkylating agents

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7
Q

What are the types of UV rays and which are the worst

A

UVC

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8
Q

What other types of radiation are there?

A

X rays and radioactive elements

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9
Q

What is ionising radiation?

A

Radiation that causes DNA damage

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10
Q

What two ways can IR damage DNA

A
  • directly
  • indirectly from generation ROS from h20
    IR work as atomic and subatomic particles with sufficient energy dislodge electrons from the atoms they collide with
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11
Q

What type of DNA damage are ionising and UVB radiation?

A

Exogenous physical

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12
Q

Name exogenous chemicals that can cause DNA damage

A
  • Environmental pollutants (tobacco, pesticides, car fumes)
  • Natural toxins (fungal aflatoxins [SAM1])
  • Dietary chemicals (cooking/curing: nitrosamies)
  • Anti-cancer drugs (cisplatin)
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13
Q

Give and explain an example of endogenous physical damage

A

Mechanical damage can arise from errors in chromosome replication or segregation causing chromosomes to be torn apart by the mitotic spindle apparatus (unscheduled replication of centromeric DNA)

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14
Q

Describe mechanical damage at the centromere due to bad replication

A

Failure of controlled replication origin firing may cause re-replication in the centromeric region, resulting in one sister with two chromatids - depending on how the spindles attach, the outcome may be chromosome mis-segregation or physical breakage of chromosomal DNA.

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15
Q

Give examples of endogenous biological genome damage

A
  • DNA replication errors
  • transposons (can move from site to site within a genome causing insertional or excisional mutagenesis)
  • Chromosome mis-segregation
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16
Q

Give an example of exogenous biological DNA damage

A

Retroviruses inserting their genome into the genome of their host cell

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17
Q

How do DNA replication errors arise?

A
  1. Nucleotide mis-incorporation by DNA polymerase
  2. Replication slippage during microsatellite DNA replication (resulting in addition or removal of a base pair) leading to insertion deletion loops of unpaired bases
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18
Q

Describe chromosome segregation failure

A

In normal segregation, each pair is pulled to opposite poles of the cell by spindles (microtubules) attached to centromeric proteins (kinetochores). Each daughter cell will therefore receive one copy of the top chromosome. failed segregation leads arises from one sister that fails to attach to the spindle. As a result, when both sisters will move to a single pole, one daughter cell will receive two copies and the other will receive none.

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19
Q

ROS are formed endogenously by ______

A

oxidative phosphorylation

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20
Q

Genome damage is caused if ______ is triggered before mitotic spindles are properly attached to ______

A

anaphase

kinetochores

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21
Q

Name three mechanisms our cells use to prevent endogenous genomic damage

A
  1. Neutralising reactive oxygen species (ROS)
  2. Avoiding DNA replication errors
  3. Avoid chromosome segregation errors
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22
Q

Explain the main cellular enzymatic antioxidant defense mechanism of neutralising ROS

A
  1. Superoxide dismutase (SOD) catalyses the decomposition of superoxide radicals into H2O2 and O2
  2. Catalase (CAT) or Glutathione peroxidase (GPx) converts H2O2 into water
  3. In the GPx reaction: GSH (reactant) is oxidised to form a dimer GSSG (product) which is the reduced back to GSH by glutathione reductase (GH)
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23
Q

How are DNA replication errors avoided in the body?

A

The DNA polymerases of the replisome (DNA pol α, 𝛿 and ε) have inbuilt proof-reading activity. During chain elongation they can sense nucleotide mis-incorporation and use their 3’-to-5’ exonuclease activity to remove the offending nucleotide before resuming synthesis.

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24
Q

By what mechanism are chromosome segregation errors prevented?

A

Spindle Assembly Checkpoint (SAC) prevents aberrant chromosome segregation at anaphase and any resulting aneuploidy or chromosome breaks

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25
Q

Explain the mechanism of how chromosome segregation errors are avoided

A
  1. The SAC uses specialised proteins in the kinetchore (a centromeric protein complex) to sense the spindle tension. When this is too low because of incorrect attachment of spindles, a signal is sent that keeps APC/C in an inactive state.
  2. APC/C is responsible for triggering anaphase by promoting the degradation of cohesin, the protein that holds sister chromatids together. In this way anaphase can only proceed when all chromosomes are correctly attached to spindles.
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26
Q

Explain the role of Spindle Assembly Checkpoint (SAC)

A

In early mitosis, misaligned chromosomes (lacking proper spindle attachments) activate the SAC to prevent premature onset of chromosome segregation. Only when all chromosomes achieve chromosome bi-orientation (correct spindle attachments) is the SAC inactivated so that chromosome segregation is triggered.

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27
Q

Name the main types of DNA damage

A
  • Base modifications
  • Single stranded break
  • Bulky adducts
  • Mismatches and IDLS
  • Inter-strand X-links
  • Double Strand Breaks
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28
Q

How do base modifications arise?

A

Oxidation (ROS/IR)
Hydrolysis
Alkylation

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29
Q

Consequences of base modifications?

A

Point mutations

Replication stalling

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30
Q

Repair mechanisms of base modifications?

A

Base excising repair

Direct repair

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31
Q

Single Strand Break causes

A

ROS/IR

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32
Q

Consequences of SSB

A

Converted to double strand breaks by DNA replication.

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33
Q

Repair pathway of SSB

A

Break Excision Repair

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34
Q

How are bulky adducts formed?

A

UV light induces the formation of thymine dimers in DNA which distort the DNA duplex

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35
Q

How intra-strand crosslinks formed?

A

Bifunction alkylating agents form cross-links between bases on the same on the same strand (or between strands or between DNA strand and a protein)

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36
Q

What are the causes of bulky adducts and intra-strand crosslinks?

A

UVB or alkylating

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37
Q

What are the consequences of bulky adducts and intra-strand crosslinks?

A

Replication stalling

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38
Q

What is the repair pathway bulky adducts and intra-strand crosslinks?

A

Nucleotide Excision Repair

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39
Q

What are the causes of Mismatches and IDLs?

A

Replication errors

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40
Q

What is frequency of Mismatches and IDLs?

A

one per 10^8 nucleotides

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41
Q

What is the consequence of mismatches and IDLs?

A

Replication stalling

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42
Q

What is the repair pathway for mismatch and IDLs?

A

Mismatch repair (MMR)

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43
Q

How are inter-strand crosslinks formed?

A

Bi-functional alkylating agents

44
Q

What are the consequences of inter-strand crosslinkages?

A

Replication stalling and cell death

45
Q

What is the repair pathway for inter-stand crosslinkages

A

Homologous recombination (HR)

46
Q

What are the causes of DSB?

A

ROS/IR
Mechanical Breaks
Replication of SSB

47
Q

What are the consequences are DSB?

A

If unrepaired may lead to small or large mutations including chromosomal deletions, inversions, translocations and chromosome loss.

48
Q

What is the repair pathway for DSB?

A 
Non-homologous end-joining (NHEJ)
Homologous Recombination (HR)
49
Q

Detail the process of oxidation of guanine to 8-oxoguanine

A

The base guanine can be oxidised by ROS to form 8-oxoguanine (8-OG), but this can be repaired by base excision repair

50
Q

What happens if 8-OG is not repaired by excision repair?

A

It pairs with A instead C during DNA replication

51
Q

What is depurination?

A

hydrolysis of guanine to form an abasic site

52
Q

Detail depurination

A

The removal of a purine (G/A) base from DNA by hydrolysis leaving an abasic site (no base present)

53
Q

What is the consequence of depurination?

A

It blocks replication and transcription and needs repairing by base excision repair

54
Q

How can cytosine be converted to uracil?

A

Hydrolytic deamination

55
Q

What is deamination? Why does it prevent DNA replication?

A

the removal of an amino group, if this happens to a cytosine, it becomes a uracil, it is thought uracil in DNA blocks replication and must be repaired by base excision repair

56
Q

How can 5-methyl-cytosine be converted to thymine?

A

hydrolytic deamination

57
Q

What happens if 5-methyl-cytosine is converted to thymine?

A

Becomes a fixed mutation in the DNA sequence

58
Q

Name four different types of base modifications

A

1) Oxidation of guanine to 8-OG
2) Depurination: hydrolysis of guanine -> abasic site
3) Hydrolytic deamination of cytosine to uracil
4) Hydrolytic deamination of 5-methyl-cytosine to thymine

59
Q

CPDs are formed by ____ and can ____ DNA replication

A

UV light

block

60
Q

How is UV damage repaired?

A

Nucleotide excision repair (around 24 bases removed and replaced)

61
Q

What are the different types of DNA repair?

A

Direct repair} damaged nucleotide
Excision repair} damaged nucleotide/base – removed section
Mismatch repair} replication error
Non-homologous end joining} double strand break
Homologous end joining} not covered here

62
Q

Describe the process of base excision repair

A

a. The damaged base is detected and excised from a nucleotide by a specific DNA glycosylase, which first flips the damaged base out of the helix and then cuts the b-glycosidic bond between the base and sugar, leaving a baseless site, called an AP site.
b. An AP endonuclease cuts the phosphodiester bond on the 5’ side of the AP site.
c. And then the other 3’ side is cut, either by the endonuclease itself or by a phosphodiesterase, so the base-less sugar is removed
d. The resulting gap is filled in by a DNA polymerase and sealed by a DNA ligase

63
Q

how are double stranded breaks fixed?

A

A repair process called non-homologous or homologous recombination

64
Q

What is the repair process for non-homologous recombination?

A

a. A pair of proteins called Ku bind to broken DNA ends
b. The individual Ku proteins also have an affinity for one another, which brings them and the two broken ends of the DNA molecule into proximity
c. The DNA fragments are joined back together by a DNA ligase

65
Q

What can base excision repair (BER) fix?

A

DNA with damaged or missing bases, or with a single strand breaks. BER has to deal with many different types of altered base including uracil, 8-oxo-G and complete loss of base.

66
Q

Why is BER very versatile?

A

It can choose from multiple glycosylase enzymes, each recognising a different kind of damaged base.

67
Q

What do the last two steps in BER contribute to? the repair of IR-induced SSBs that have been recognised by a protein call polyADPribose polymerase (PARP).

A

As we will see later, PARP turns out be an important therapeutic target.

68
Q

How does nucleotide excision repair work?

A

1) XPC recognises damaged DNA to initiate the global repair pathway. If the damage is within a transcribed region, it is recognised by RNApol II to start the transcription coupled pathway.
2) DNA is unwound around the damaged region by helicases (XPB and XPD).
3) Endonucleases (XPF and XPG) make an incision several nucleotides upstream and downstream of the damage region.
4) 25-30 nucleotides are excised around the damage region. DNA is re-synthesised and sealed by DNA polymerase and DNA ligase.

69
Q

What is xeroderma pigmentosum from?

A

This is an autosomal recessive disease caused by mutations in one of the genes encoding these proteins. XP patients suffer from sensitivity to UV light resulting in corneal ulcerations and dry skin prone to blistering and cancer.

70
Q

How is damage for nucleotide excision recognised?

A

It depends on whether it is in a region transcribed by RNApolII (Transcription Coupled NER) or elsewhere in the genome (Global NER).

71
Q

Name two diseases associated with defective NER

A

Cockayne’s syndrome (CS)

72
Q

What is CS characterised by?

A

By short stature and neurological dysfunctions and mild skin sensitivity. This is an autosomal recessive disorder caused by mutations in the protein CSA or CSB, which you can see in the diagram above. They appear to be required for post-translational modifications of RNApol II.

73
Q

What does mismatch repair (MMR) fix?

A

It repairs mismatched bases and insertion-deletion loops (IDLs) resulting, respectively, from nucleotide mis-incorporation or replication slippage.

74
Q

How is mismatch repair distinguished from excision repair?

A

the strand requiring re-synthesis has no recognisable base damage. It is simply an incorrectly placed base. The MMR system must distinguish the newly synthesised strand with the error from the parental strand and repair only the former. In eukaryotes, it appears that the newly replicated strand is recognised on the basis of temporary nicks generated during replication.

75
Q

Explain the mechanism of MMR in eukaryotes

A
  • Base mismatches or small IDLs (represented by a red triangle) are recognised by repair enzymes MutSα and MutLα.
  • MutSα and MutLα diffuse away from the mismatch to reach nicks in the newly synthesised strand bound by either PCNA or RFC.
  • An exonuclease (EXO1) is recruited to degrade the newly-synthesis strand towards and through the mismatching region.
  • Finally, PCNA recruits a DNA polymerase to re-synthesise the gapped region, leaving nicks to be sealed by DNA ligase.
76
Q

What is Hereditary Nonpolyposis Colorectal Cancer (HNPCC)?

A

An inherited in an autosomal dominant fashion as a result of mutations in any of several genes encoding MMR proteins, such as MutSα and MutLα.

77
Q

When does HR occur?

A

is an important part of meiosis. It also occurs in mitosis where is acts as a DSB repair mechanism.

78
Q

What is the purpose of meiotic HR? and what does it occur between?

A

Between chromosome homologues and generates genetic diversity by forming crossover (CO) products.

79
Q

What is the purpose of mitotic HR?

A

It occurs between sister chromatids, using the intact sister as a template to repair DSBs in its sister.

80
Q

What are the key steps of HR?

A

a) Following a double strand break, exonuclease proteins bind to both ends of exposed DNA. The DNA is cleaved to give single stranded overhangs.
b) Single stranded overhangs are protected by RPA.
c) BRCA1 and BRCA2 then help RAD51 bind the DNA, replacing RPA.
d) The DNA end, bound by RAD51, invades homologous DNA on the sister chromatid. DNA polymerase can then extend and repair the lost information. When a strand invades in this way it forms a Holliday Junction (HJ).
e) Further proteins (not shown) remove the HJs to separate the repair DNA strands form the template. One important protein involved at this stage is called BLM. Finally, DNA polymerase and ligase activities fill and reseal any remaining gaps.

81
Q

How is the importance of the HR pathway shown?

A

By the loss of viability in cultured mammalian cells when key HR proteins, including RAD51 and BRCA2, are depleted. Furthermore, the genes for some of HR proteins, including BRCA1 and BRCA2 are mutated in inherited or acquired cancers.

82
Q

How is HR more accurate than NHEJ?

When is HR not so accurate and is a possible cause of genome instability?

A

By using a sister chromatid template to repair the DSB. Aberrant HR involving inappropriate templates (e.g. repeat sequences) is a known source of genome instability. In one mechanism to minimise this, HR is inactive during G1 when sister chromatids are unavailable.

83
Q

When can NHEJ occur?

A

Can occur at any stage of the cell cycle and is inherently error-prone.

84
Q

What are the steps of NHEJ mechanism?

A

a) NHEJ begins with the binding of Ku70/Ku80 heterodimers to each of the broken ends of DNA. If the ends are already blunt and undamaged they can be ligated immediately by DNA ligase.
b) If the ends require processing, then Ku recruits the catalytic subunit of a DNA protein kinase (DNA-PKcs) to form active DNA-PK complex.
c) DNA-PK recruits and activates Artemis exonuclease to trim DNA ends, and DNA polymerase to add nucleotides in a template-independent manner. These proteins may generate blunt ends.
d) DNA ligase is recruited.
e) Processed DNA ends are ligated by DNA ligase. The net result is either accurate repair of the DSB or repair with insertion/deletion (indel) mutations at the site of the original DSB.

85
Q

What happens if NHEJ is done inappropriately?

A

It can be genome destabilising, many translocations associated with cancer appear to have arisen by aberrant NHEJ of DNA ends derived from different chromosomes.

86
Q

What important role does NHEJ play?

A

in adaptive immunity by mediating the rearrangement of the immunoglobulin and T cell rector genes. For this reason, inherited mutations in the genes encoding certain NHEJ proteins can cause various inherited immunodeficiency syndromes.

87
Q

What pathway must ditiguish the parental and newly replicated strands?

A

MMR

88
Q

What pathways resynthesise one DNA strand using the other as a template?

A

BER
NER
MMR

89
Q

What pathway invloves more than one DNA double helix

A

HR

90
Q

What pathways require multiple proteins that each recognise a different type of DNA

A

BER

91
Q

What types of proteins are invloved in DNA damage response (DDR) pathways?

A

sensors, mediators, transducers and effectors

92
Q

Name two sensor proteins that detect

a) DSB
b) SSB

A

a) MRN

b) RPA

93
Q

What type of proteins are H2AX, BRCA1? What do they do?

A

Mediator proteins
They help recruit and activate transducer kinases ATM (Ataxia Telangiectasia Mutated) and ATR (ATM-Related). A positive feedback loop between mediators and transducers leads to the maintenance and amplification of this signal.

94
Q

What do producer kinases phosphorylate?

What do these proteins go on to phosphorylate?

A

Effector kinase Chk1 and Chk2 that phosphorylate further effector proteins that control transcription factors (e.g. E2F1, p53) apoptosis (e.g. p53) and cell cycle arrest (e.g. Cdc25).

95
Q

What is the choice of the outcomes of apoptosis and cell cycle arrest determinant on?

A

On the severity of the DNA damage and the strength and duration of the induced kinase response and associated phosphorylations.

96
Q

How may p53 act?

A

Tumour suppressor and transcription factor

  • in transcription-dependent or -independent ways to promote reversible cell cycle arrest or apoptosis. Cdc25 proteins are Chk1/Chk2 activated are phosphatases that down-regulate Cdk activity to arrest cell cycle progression.
97
Q

Why do you think it might be advantageous to arrest the cell cycle in response to DNA damage?

A

Cell cycle arrest allows DNA damage to be repaired before the DNA is replicated (G1/S checkpoint) or transmitted to daughter cells (G2/M checkpoint). It also avoids complications such as stalling of DNA replisome at damaged DNA.

98
Q

What is the advantage of a cell undergoing apoptosis in response to DNA damage?

A

Apoptosis eliminates the high risk of tumorigenesis from cells where damage was too severe to be properly repaired.

99
Q

How can DNA damage help with cancer treatment?

A

Ionising radiation and chemotherapeutic drugs: inducing DNA damage, particularly double strand breaks. Because cancer cells grow and divide more frequently that normal cells, they are more susceptible to such agents.

100
Q

What are the side effects of inducing DNA damage from cancer treatment?

A

treatments have many side effects, such as killing the normally fast-growing cells in the bone marrow, digestive tract and hair follicles, leading to infections, sickness and hair loss. More seriously, DNA damage in normal cells increases the risk of developing treatment-related cancers. Chemotherapy for cancer is also hampered by cancers that develop resistance to the drug.

101
Q

What type of cancers can you treat with synthetic lethality?

A

Cancers with a known defect in DNA repair, such as a mutation in a BRCA gene.

102
Q

What new cancer treatment is being developed to target sythetic lethality?

A

To treat such cancers without harming normal cells, drugs have been developed that inhibit PARP.

103
Q

Can you predict the effect of a PARP inhibitor on DNA repair?

A

It will impair the ability of BER to repair single strand breaks.

104
Q

What happens to unrepaired SSBs?

A

They are converted into DSBs during DNA replication.

105
Q

So what will be the effect of PARP inhibition on normal cells and on BRCA1-deficient cancer cells?

A

Normal cells will be able to repair the DSB by HR, but the cancer cells will not because BRCA1 is required for DSBR by HR. Normal cells will therefore survive while the cancer cells will prone to undergo apoptosis. Cell death from synthetic lethality as induced by inhibition of Poly(Adenosine Diphosphate [ADP]–Ribose) Polymerase 1 (PARP1).

106
Q

What is the name of the PARP inhibitor used to treat BRCA1 or BRCA2 deficient ovrian cancers?

A

Olaparib

Genetics (10 decks)

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