L6 - Damage Repair and Recombination Flashcards

(80 cards)

1
Q

What four ways can damage occur to DNA

A

Deamination
Depurination
Pyrimidine Dimer Formation
DNA breaks

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2
Q

What three types of chemical reactions would cause DNA damage

A

Oxidation, hydrolysis and methylation

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3
Q

What way would oxidation damage the DNA

A

At C=C double bonds (gain of oxygen or loss of hydrogen)

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4
Q

How would a hydrolysis reaction damage the DNA

A

Splitting a bond using water - can liberate from amino (NH2 groups)

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5
Q

Describe how methylation would damage the DNA

A

Nitrogen (N) can be methylated to NH2 sponatenously on any of the four nucleotides

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6
Q

What bases are purines

A

A or G

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7
Q

What bases are pyrimidines

A

C or T

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8
Q

Describe how a depurination occurs

A

Nitrogenous base removed from the deoxyribose sugar - using water

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9
Q

Describe how deamination occurs

A

Change from cytosine to uracil (will pair with A)

Uses water and releases ammonia

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10
Q

Describe how depurination can cause mutation

A

Nucleotide is missing therefore there is a deletion in one of the strands

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11
Q

Describe how deamination can cause mutation

A

C–>U so now pairs with A, point mutation of a base pair in a strand

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12
Q

Describe how a deamination may be repaired

A

By base excision repair

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13
Q

What is the enzyme involved in base excision repair which recognises the damage

A

Uracil DNA glycosylase

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14
Q

In base excision repair what is the sugar removed with

A

Apurinic/apyrimidinic endonucleases

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15
Q

In base excision repair what is the phosphate removed using

A

Phosphodiesterase

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16
Q

In base excision repair the removal of the sugar and phosphate causes –>

A

Formation of primer:template junction –> act of DNA polymerase
Nick is the sealed using DNA ligase

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17
Q

What occurs during pyrimidine dimer formation

A

When two adjacent pyrimidines are able to form a covalent linkage

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18
Q

What does pyrimidine dimer formation lead to

A

Arrest of DNA replication
Misreading by polymerases
Unable to complementary base pair with purines

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19
Q

Describe the process of nucleotide excision repair

A

Excision nucleases cleave the single strnaded DNA with the defect
DNA helicase removes the damaged segment
Formation of primer template junction leads to activation of DNA polymerase then the nick is sealed by DNA ligase

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20
Q

What disease is as a result of nucleotide excision repair deffects

A

Xeroderma Pigmentosum

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21
Q

What are the 7 xeroderma pigmentosum genes

A

XPA XPC XPD XPF XPG

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22
Q

What do the xeroderma pigementosum genes all encode

A

Proteins which are involved with nucelotide excision repair

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23
Q

What are the E.coli homologues of the xeroderma pigmentosum genes

A

UvrA UvrB UvrC UvrD

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24
Q

What is nucelotide excision repair targetted to

Why is this the case

A

DNA sequences which are being actively transcribed

By the physical coupling of RNA polymerase to the repair mechanisms

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25
Why are ds breaks dnagerous
Mean that large fragments of chromosomes can be lost
26
What are the two methods of repairing a double strand break
Non-homologous end joining | Homologous recombination
27
Why is non homologous end joining not be the ideal way to repair DNA
No specificity - so likely to cause mutation | Can cause deletions
28
Describe the process of non homologous end joining
Accidental double stranded breaks Loss of nucleotides due to degredation End joining gives a deletion mutation
29
Describe the process of homologous recombination
Double stranded break Resection at the 5' end so can be used as a template and anneal with the undamaged chromosome Binding protein RecA promotes strand invasion of the undamaged template by one strand from a damaged DNA molecule that acts as a primer Formation of heteroduplex facilitates templated synthesis of the one stand Newly synthesised strand from its template and reaneals to partner strand - 2nd strand synthesis and two nicks sealed by ligation
30
What are BRCA2, ATM and Fanconi Anaemia all examples of
Genes involved in homologous recombination
31
What are BRCA2 mutations associated with
Breast, ovarian and prostate cancer
32
What are ATM mutations associated with
Ataxia telangiectasia - leukameia and lymphoma
33
What occurs once homologous recombination is inneffective
The cell becomes critically dependent on other mechanisms of repair
34
If a cell that has a mutation in the genes involved in homologous recombination is then treated with an inhibitor for base excision repair What is the clinical relevance of this
Too much DNA damage and the cell will die - synthetically lethal Can be used to treat cancer
35
What drugs are specific inhibitors of poly ADP ribose
Olaparib and lumpaza
36
What does homologous recombination machinery mediate
Genetic recombination in meiotic cells
37
What is the function of SPO11 in homologous recombination
Makes the initial clevage
38
What is the function of Mre11 in homologous recombination
Resection at the 5' end
39
What does RecA mediate
Strand invasion
40
What does homologous recombination result in
Formation of the double holliday junction
41
What is the resolution of the double holliday junction if only internal strands are broken and rejoined
Then no recombination
42
What is the resolution of the double holliday junction if both internal and external strands are cut
Recombinattion
43
Recall the purine bases
Adenine, guanine
44
Recall the pyrimidine bases
Thymine, cytosine, uracil
45
What are the four main types of DNA damage
Deamination, depuration, pyrimidine dimers, DNA breaks
46
What are the three main methods of DNA repair
Base excision repair, nucleotide excision repair, homologous recombination
47
What chemical modifications can lead to DNA damage
Hydrolysis, oxidation or random uncontrolled methylation
48
What is the most frequency types of DNA damage
Hydrolytic depurination and deamination of bases
49
What happens when cytosine is deaminated and what are the downstream effects of this
Converted to uracil. This will still pair with guanine but during replication the guanine will be replaced with an adenine, leading to a nucleotide substitution from CG-->TA
50
Explain how deamination of cytosine is achieved
The NH2 group attached to the purine ring of cytosine at position 4 is replaced with a carbonyl group
51
Explain the process of depurination
The carbon to nitrogen bond between the carbon position one in the deoxyribose sugar and the nitrogen in the purine ring is hydrolysed. This releases the base and results in the loss of a base pair, nucleotide deletion, in one of the daughter DNA molecules produced during replication
52
Which DNA damage(s) is base excision repair used for
Deamination and depurination
53
Explain the process of DNA excision repair
The damage base is first removed by DNA glycosylase which cleaves off the inappropriate base(s). The deoxyribose sugar is then removed by an enzyme known as apurinic/apyrimidinic endonuclease. Lastly the phosphate is removed by the phosphodiesterase enzyme. This leads to a ssDNA break in the polynucleotide chain. This gap is then recognised by DNA polymerase which then extends over this gap which acts as a primer template junction. Finally, DNA ligase uses ATP hydrolysis to provide the energy required to seal the nick
54
What are pyrimidine dimers and what causes them
Pyrimidine dimers are caused by the covalent linkage of benzene rings in two adjacent pyrimidine bases. This is usually caused by UV light
55
Which base is primarily affected by UV light and more prone to form dimers
Thymine
56
What is the results of the formation of pyrimidine dimers
This arrests DNA replication or can cause mis-reading of the DNA sequence by DNA polymerase
57
Nucleotide excision repair is only used to repair pyrimidine dimers, T or F
F – it is used to repair a variety of different types of DNA damage including pyrimidine dimers
58
Pyrimidine dimers can only occur between identical adjacent pyrimidine bases, T or F
F – it can be the same pyrimidine or different ones (i.e. T-C, T-T, C-C)
59
Explain the process of nucleotide excision repair
An excision endonuclease enzyme cleaves the single strand of the DNA containing the defect by cleaving either side of the dimer etc. This creates a polynucleotide fragment from the DNA molecule that contains the defect which is then removed by DNA helicase. DNA polymerase then extends the primer template junction created to replace the excised sequence. This is followed by resealing of the nick mediated by DNA ligase.
60
Xeroderma pigmentosum is a disease caused by defective nucleotide excision repair machinery, what is the effect of this on patients
XP renders patients extremely sensitive to sunlight-induced skin cancer
61
How many different genes have been identified as being responsible for xeroderma pigmentosum, give some examples
7 different genes including XPA, XPC, XPD, XPF and XPG
62
What are the products of genes encoded by the genes mutated in xeroderma pigmentosum
These genes encode proteins that participate in the nucleotide excision repair pathway
63
What is significant about the function of the XP genes
Their function is tightly coupled and targeted to regions of the genome that are highly transcribed. This acts as a surveillance system to signify regions of the genome that are transcribed. RNA polymerase required for transcription is physically coupled to the DNA repair machinery
64
Uvr genes are homologues of the XP genes found in yeast and are also transcription coupled, T or F
T
65
dsDNA breaks are often caused by non-ionising radiation, T or F
F – they are caused by ionising radiation
66
Why are dsDNA breaks especially dangerous
Large fragments of chromosomes can be lost
67
What two methods are there of repairing dsDNA breaks
Non-homologous end joining, homologous recombination
68
What can make DNA molecules susceptible to dsDNA breaks
If the DNA has already suffered a ssDNA break due to base excision repair or nucleotide excision repair. This may weaken the DNA structure
69
What is meant when we refer to non-homologous end joining being quick and dirty
Non-homologous end joining is a far from optimal repair mechanism. It is carried out quickly but in itself, can induce mutations
70
Explain the process of non-homologous end joining
The ends of the double stranded break are rendered flush with loss of bases via degradation from the ends of the strands. These flush ends are then ligated together. However, this leads to a loss of DNA sequence due to the degradation that occurs prior to ligation.
71
Homologous recombination also isn’t a perfect repair mechanism, T or F
F – homologous recombination provides a perfect repair of dsDNA breaks and is an accurate and preferred method of repair
72
What does homologous recombination rely on
Using the intact DNA sequence information in the undamaged homologous chromosome
73
Explain how homologous recombination is mediated
Firstly, the 5’ ends are resected by exonuclease to create single strands that can be used to prime DNA synthesis when annealed to a template strand from the complimentary chromosome. The DNA-binding protein RecA promotes strand invasion of the undamaged template molecule by one strand from the damaged DNA molecule that acts as a primer. This forces complimentary binding with the same sequence in the undamaged sister chromatid. This forms a heteroduplex structure between the dsDNA helix of the sister chromatid and the single stranded sequence from the damage DNA molecule. This facilitates templated DNA synthesis of one strand by DNA polymerase. DNA polymerase synthesises across the damaged region by reading information out of the undamaged sister chromatid. The newly synthesised DNA then dissociates from its template and re-anneals to its original partner strand allowing second strand synthesis and formation of a pair of staggered singles stranded nicks. These nicks are then ligated by DNA ligase
74
Give an example of a gene involved in homologous recombination that when mutated can cause cancer
BCRA2 – causing breast, ovarian and prostate cancer
75
What happens in patients with mutations in genes needed for homologous recombination
Homologous recombination is defective and thus these cells become critically dependant on other pathways for correct DNA repair i.e. base excision and nucleotide excision repair
76
How can this defective recombination pathway seen in some cancers be harnessed as a cancer treatment
Combined inactivation of homologous recombination by mutations and the inhibition of base or nucleotide excision repair by anticancer drugs make cells that suffer DNA damage unable to make adequate repairs to their DNA and die. This is known as synthetic lethality.
77
Homologous recombination is required to mediate genetic recombination and diversity created in meiosis, T or F
T
78
Explain how a Double Holliday Junction is created during genetic recombination in meiosis
Two chromatids exist with different alleles for two genes (AB and ab). Spo11 endonuclease makes an initial cleavage in chromatid to create a dsDNA break. Mre11 then carries out a 5’ resection to create a targeted dsDNA break with a 3’ overhang in the DNA molecule. RecA then mediates strand invasion by this strand on the other chromosome creating a heterocomplex/triple helix structure. DNA polymerase then synthesises across the gap between the end of the 3’ overhang from the invading strand and the 3’ end of the other chromosome. The 3’ end is then relegated with the original strand to create a double holliday junction whereby the invading strand from one helix is forced into interactions with the complimentary strand in the helix of the other chromatid and vice versa.
79
How many ways can the Double Holliday Junction be resolved
2
80
Explain how different resolutions of the Double Holliday Junction account for whether homologous recombination occurs or not
If the internal strands are broken and re-joined than the same DNA strands are broken and re-joined at each junction and recombination is not achieved as each chromatid will contain the same sequence of alleles it did prior to cleavage. If external and internal strands are broken and re-joined then different DNA strands are broken and re-joined at each junction. Hence recombination is achieved. This recombination of allelic forms results in chromatids with different combinations of alleles (Ab and aB)