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Genetics > Gary Sharples > Flashcards

Gary Sharples Flashcards

1
Q

What is the difference between :

Dna A and dna A

A

The one with the capital is a protein the one in lower case is a gene.

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2
Q

What was the Meselson-Stahl experiment? What did it conclude?

A
  • Bacterial cells grown for several generations on a medium containing heavy N15 nitrogen medium (DNA therefore contained heavy N15 DNA).
  • the cells were then transferred to a lighter N14 medium.
  • at various time the DNA was extracted and dissolved in a solution of caesium chloride and spun in centrafuge.
  • conch gradient is established.
  • first generation contained mix of light and heavy.
  • second generation contained both LL and LH .
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3
Q

How was the discovery of bidirectional replication carried out?

A

Using Radioisotopes, the density was the same either side of the origin of replication. (If unidirectional it would only occur one side).

  • grow e.coli in media without radioactively lab led thymidine.
  • grow in 3H thymidine of low specific activity.
    Grow 3H thymidine of high specific activity.
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4
Q

Explain the process of bidirectional replication

A

When DNA is stored in double stranded circle covalent lay closed. Replication begins at the origin of replication.
Semi conservative replication proceeds in 2 directions.
- Per circle there are 2 replication forks, each one with leading and lagging strand.

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5
Q

What is the difference between leading and lagging strand synthesis?

A

Leading strand is synthesised continually in the 5’ to 3’ direction.
Lagging strand is synthesised discontinuously in short pieces known as Okazaki fragments. Still in the 5’ to 3’ direction.

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6
Q

What do all DNA polymerases add nucleotides to the 3’ end of a growing strand?

A
  • a template: strand to copy
  • a primer: another strand annealed to the template tot supply a 3’ OH group
  • dNTPs: deoxynucleoside triphospahte.
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7
Q

How is DNA polymerase capable of correcting mistakes?

A

It has a 3’ to 5’ exon lease activity, this means it only makes mistake 1/5*10^7 bp

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8
Q

Explain the process of DNA polymerase exotic lease activity.

A
  1. Polymerase miss airs dC with dT.
  2. Polymerase repositions the miss paired 3’ terminus into the 3’ to 5’ exonuclease site.
  3. Exonuclease hydrolysed the miss paired dC
  4. The 3’ terminus repositions back to the polymerase site
  5. Polymerase in-cooperates the correct nucleotide
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9
Q

How is DNA replication primed?

A

By RNA, the RNA primer (oligonucleotides) are synthesised by DNA primase.
- DNA polymerase then extends the RNA primers.

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10
Q

In bacteria what synthesises the RNA primers?

How long are the oligonucleotides that are synthesised?

A

DNA primase DnaG.

10-12 base pairs long.

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11
Q

How is the primer removed?

A

The primer is removed by DNA polymerases 5’ to 3’ exonuclease activity (this is different from the 3’ to 5’ used to correct bases). It syntheses a new DNA strand as the RNA is removed.

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12
Q

How are nicks in the DNA backbone sealed?

A

DNA ligase seals the nicks in the backbone between Okazaki fragments by catalysing the formation of a phosphorites tear bond. E.coli DNA ligase utilises nicotine midge (NAD) as a cofactors in this reaction.

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13
Q

How are parent strands separated at the replication fork?

A

DnaB helically. A 5’ to 3’ helicase that unwinds the parental DNA as it moves.

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14
Q

How does DnaB helicases structure enable it to carry out its function?

A

It consists of 6 identical subunits that forms a ring.
Of the 6 identical ATP binding sites 2 opposing ATP bind ATP tightly. 2 are more likely to bind ADP and phosphate and 2 are empty.
- these may inter convert as ATP is hydrolysed- this creates a ripple effect that continually runs around the ring.
- because of these conformaitonal changes the loops in the centre of the ring for DNA binding cause ossolation up and down.
- it’s thought the ossolation gloop may pull DNA though the central hole thus unwinding the double stranded helix.

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15
Q

Why is DnaG primase bound to DnaB helicases?

A

It ensures that DnaG is in the correct position for RNA primer synthesis. The activities of both protein are stimulated by this interaction .

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16
Q

What polymerase is used as the replicative polymerase in E.coli?

A

DNA polymerase III

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17
Q

What is OriC?

A

A 245 bop sequence necessary for replication.

  • 5 repeats of a 9bp sequence make up the R1-5 for key initiator protein DnaA
  • 3 additional repeats are bound by DnaA only when complex with ATP.
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18
Q

What is DUE at OriC?

A

The DNA unwinding element is called DUE. It is AT rich and the site where DNA is opened for replication.

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19
Q

How is the initiation of DNA replication regulated?

A

Hda stimulates ATP hydrolysis by DnaA to aid dissemble after initiation.
Dam methylates adenine at GATC in e.Coli

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20
Q

How is DnaB loaded onto the fork?

A

DnaB is loaded onto the replication fork by DnaC, it opens the DnaB ring. ATP hydrolysis releases DnaC leaving DnaB bound to the ring.

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21
Q

What is SSB?

A

An accessory protein.

  • It stops unwound parent strands from reanealing.
  • protects DNA strand from degradation or damage.
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22
Q

What does DNA gyrase (topoisomerase II) do?

A

Alleviates supercooling by cutting 2 strands of a DNA molecule and passing another though the break before reasealsing the cut strands.

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23
Q

What do the 3 DNA polymerases do in E.coli?

A

DNA polymerase 1: removes RNA primer and replaces it with DNA
DNA polymerase 2: filling gaps following repair of DNA damage
DNA polymerase 3: chromosome replication

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24
Q

What is the purpose of the B sliding clamp on DNA polymerase III?

A

Homodimer which encircles DNA.

Imparts process it’s by ensuring that the core polymerase does not fall off during DNA synthesis.

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25
Q

How is the B clamp loaded on to the lagging strand at each Okazaki fragment?

A 
The y (gamma) complex of DNA pol III.
- binding of ATP allows opening of the clamp whilst ATP hydrolysis allows clamp release and ring closure.
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26
Q

In very brief terms explain how DNA replication occurs in E.Coli.
8 steps

A
  1. DnaA opens the duplex at OriC.
  2. DnaC loads DnaB helicases.
  3. DnaG synthesises RNA primer on the lagging strand.
  4. Alpha subunit synthesise DNA on leading and lagging strand.
  5. Tau subunit ensures dimerisation of core polymerase subunits.
  6. Y complex loads B clamp at each Okazaki fragment
  7. B clamp encircles DNA and binds complex to fork.
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27
Q

How long does it take E.coli to replicate?

A

Should take around 40 mins but it can divide at much faster rates of around 20-25 minutes. 900 nt per second.

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28
Q

Why does DNA homologously recombine?

A 
Conserves genetic identity
- aligns chromosomes at meiosis 
- repairs DNA breaks
- restores stalled replication forks
Generates genetic diversity 
- rearranges genes within a genome 
- exchanges homologous partners 
- incorporates foreign DNA
- Contributes to acquisition of new traits
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29
Q

Define; Crossover/ splice/ recombinant

A

Recombination where flanking markers are exchanged

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30
Q

Define; non-crossover/ patch/ non-recombinant

A

Recombination where flanking markers are not exchanged

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31
Q

Define; gene conversion

A

A mismatched DNA sequence from one heteroduplex DNA strand is replaced with a sequence complementary to the other strand resulting in aberrant gamete ratios.

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32
Q

How does DNA recombination aid recovery of problems that arise during DNA replication?

A

Partner chromosomes are found and single strands exchanged to form DNA branched structures. They often offer a template to Copt any genetic information that might be lost.

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33
Q

What is the holiday model?

A

Homologous chromosomes are nicked at identical locations, the strand strands from one side of the nick invades the homologous chromosomes base pairing with complementary strands.
- invading strands are covalently linked to the original strands strands ar the nicked points. This forms a holiday junction. The holiday junction migrates away from the nick points in a process known as branch migration. As it does so DNA is swapped between chromosomes. This creates heteroduplex regions on both chromosomes where minor base sequence differences between homologous chromosomes results in a region of DNA with low percentage of mismatched base pairs.

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34
Q

How and what are the results of the 2 different cleavage of Holliday junctions?

A
  • if crossed strands are cleaved by an endonuclease then after ligation within the chromosome there will be 2 non-recombinant chromosomes.
  • if one rotates 180 degrees a process called isomer action it’s easier to visualise how uncrossed stands are broken, this process results in recombinant chromosomes with short heteroduplex regions.
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35
Q

What is the difference between hetero/homoduplex?

A 
Homo = a DNA molecule composed of 2 chains with each derived from the same parent molecule
Hetero = a DNA molecule composted of 2 chains etched derived from a different parent molecule.
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36
Q

Explain the process of homologous dependant double strand break (DSB) repair. (SDSA)

A

Duplex is broken with double strand break.

  • first the DNA undergoes nuclease degradation forming DNA duplexes with 3’ ended single stranded tails.
  • one of the single stranded tails invades the non-broken duplex at a region of homology. Forming a region of heteroduplex.
  • the displaced pink strand which is of the same polarity of the invading grey strand forms a loop the D loop (the displacement loop)
  • the 3’ end of the invading strand acts as a primer for DNA synthesis the complementary strand serves as a template.
  • a small bubble is formed consisting of the template, a small region of Newley synthesised DNA and the displacement strand.
  • this continues until the new DNA complimentary to the double stranded break in the grey stand is synthesised.
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37
Q

What are the pre-synaptic stages of recombination?

A

Events take pales that are needed for the initiation of of recombination; this stage includes the formation of gaps, single or double strand breaks and the exposure of single stranded regions.

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38
Q

What are the synopsis stages of recombination?

A

The homologous strands are paired and strand exchange occurs between them to produce joint-molecule intermediates.

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39
Q

What are the Post-synapsic stages of recombination?

A

The joint molecule intermediates are processed to mature recombinant products.

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40
Q

What is branch migration?

A

It moves the joint along the DNA resolution: involves strand cleavage at the joint to separate the linked DNA molecules.

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41
Q

What is RecBCD involved in?

A

Is involved in end processing.

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42
Q

What are RecB/D

A

Helicases that travel along each of the strands at a DNA break.

RecB moves in the 3’ to 5’ direction, whilst recD move 5’ to 3’ on the opposite strand. Thus they unwind together as bipolar helicases.

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43
Q

What does the nuclease domain of RecB do?

A

Cuts both DNA strands as it passes through the RecBCD complex.

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44
Q

What occurs at a chi site?

A

Cutting of 3’ strand ceases, but continues on the 5’ strand, RecA can load onto the exposed ssDNA overhang

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45
Q

Explain the repair of a DSB with relation to chi sites.

A

8Bp chi sites are NOT randomly positioned throughout the E.coli chromosome. They are over represented and most are orientated away from the Ori.

  • this promotes recovery close to a break at a replication fork.
  • recombination at such breaks allows replication to restart mediated by PriA.
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46
Q

What is the purpose and function of RecA protein in E.coli?

A

Binds to ssDNA.

  • polymerises on the ssDNA to form a helical nucleoprotein filament that can be seen my elctromicroscopy.
  • it is the RecA nucleoprotein that catalysts the homologous DNA pairing and strand exchange states of recombination.
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47
Q

How does strand exchange by RecA occur in vitro?

A

Reactions require ATP and MG2+

  • RecA polymerises at the ssDNA gap and initiates pairing with the 332P-labels linear duplex.
  • strand exchange can be followed by removing proteins and analysis on agarose gel.
  • strand exchange generates a holiday junction intermediate, but continues to the end of the duplex to generate complete strand exchange products.
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48
Q

LOOK UP IN BOOK RECA

A

Look

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49
Q

What is the function of RecQ?

A

RecQ is a helicase that functions at the at the interface between DNA replication and recombination to repair damaged replication forks.

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50
Q

What is the structure of RecQ?

A

Has a highly conserved helicase domain that comprises 400 AA.
- the RQC domain in unique to the RecQ family and consists of a zinc binding domain and a winged helix involved in DNA recognition.

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51
Q

Why do we use model organisms?

A

Life is diverse so studying a few organisms is easier
It allows an understanding of how proteins function together to produce life.
Organisms tend to be selected for ease of manipulation and experimentation.

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52
Q

How can RecQ aid replication?

A
  • can be impeded by DNA secondary structure including hairpins or G-graduplexes
  • RecQ helicases act to disrupt these structures and allow smooth replies one progression.
  • RecQ helicase unwinds DNA with 3’ to 5’ polarity
  • It typically catalysts Holiday junction branch migration, fork regression and also will unwind D-loops, triple helicase, hairpins and G-guadruplex DNA.
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53
Q

How are topoisomerases used to relieve torsional stress arising from DNA supercoiling?

A

By transient lay breaking the DNA backbone in one (tyope 1) or both (type 2) strands then passing intact DNA through the opening before resealing the break.

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54
Q

How does holiday junction resolution occur by dual strand scisson?

A

Once the 4-stranded holiday junction has formed it can branch migrate over long distances as the two paired DNA molecules are homologous.
- resolution occurs by nicking an opposing pair of strands at the point of the crossover.

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55
Q

What is Lex A involved in? What is the response?

A

Lex A regulates the expression of multiple genes involved in DNA damage repair, including RecA RuvAB and uvrAB.

This is the sos response

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56
Q

How does Lex A conduct the sos response?

A

It binds genes to promoters preventing access of RNA polymerase but DNA damage is signalled by RecA binding to ssDNA which stimulates auto cleavage of lexA.

57
Q

How does branch migration occur by RuvAB?

A

A complex of RuvA and B unwind holiday junctions. RuvA assembles 2 helicase rings.

58
Q

How does RuvC act to resolve holiday junctions?

A

Each active site consists of four acidic residues.
Efficient cleavage requires a 4 bp target. (Point of strand crossover must branch migrate to the target sequence before cleavage can occur)

59
Q

Briefly summarise E.coli’s RuvABC system.

A
  1. RuvA and RuvC directly targets X junctions.
  2. RuvAB and RuvBC interact and stabilise the complex.
  3. RuvC introduces paired cleavages across the junction
  4. RuvAB move the junction to preferred RuvC cut sites
  5. RuvC functions as part of a complex with RuvAB in vivo.
60
Q

What are the consequences of two vertical or 2 horizontal cuts in double stranded DNA resolution?

A

Will give non-crossover.

61
Q

What is the result of one verticals and one horizontal cut in double stranded DNA recombination?

A

Will get crossover regions.

62
Q

What is another homology dependant process for double strand break repairs?

A

Synthesis-dependant strand annealing

63
Q

How does synthesis-dependant strand annealing (SDSA) differ from RuvABC system?

A

This process does not involve the formation of a holiday junction, although the D loop structure is still formed.
Instead the 3’ strand that invades the homologous partner chromosome is used to prime DNA synthesis.
- this D loop is then unwound by a helicase (e.g. UvrD) allowing RecA to reassemble and locate the other end of the original break with the Newley synthesised partner.
- SDSA always results in no crossover.

64
Q

What is extended synthesis-dependant strand annealing (ESDSA)?

A
  • chromosomes are shattered by radiation
  • broken ends are trimmed and paired with homologous DNA fragments
  • – these prime DNA synthesis and further annealing the recombination putting all the pieces back together.
65
Q

What is DNA mutation?

A

Any alteration to the genetic material (DNA or RNA) that produces a heritable change in the nucleotide sequence.

66
Q

What is DNA mutation not?

A

It is not chemical damage or modifications that cause temporary changes in genes or gene function.

67
Q

Difference between wild type and mutant

A

Wild type = standard type of gene or organism

Mutant = the altered gene or organism produced by mutation

68
Q

What is a forward mutation and what is reversion?

A

A process that converts a wild- type to mutant.

Reversion is the opposite

69
Q

List 3 global change mutation types

A

Chromosome aberrations
Genome rearrangements
Changes in chromosome number

70
Q

List 4 chromosomal aberrations

A
  • deletions
  • insert actions
  • duplications
  • inversions
71
Q

What are genome rearrangements?

A
  • Redistribution of genetic material between chromosomes (translocation)
  • often arises from chromosome breakage.
72
Q

How can there be changes in chromosome number?

A

e.g. Trisomy of chromosome 21 usually arises from mistakes in chromosome segregation at cell division

73
Q

What are the 6 types of localised change induced mutations?

A 
Base substitutions 
- single base changes (point mutation)
Deletions/ insert actions
- frame shift mutation (loss or gain of bases usually 2 or more)
Duplications
- a sequence is repeated
Inversion 
Translocation
-movement of a piece of DNA from one location to another
Base pair substitutions
74
Q

What are the two types of base pair substitutions? And what’s the differences?

A

Transition mutation- changes a purine to purine or prymadine to prymadine
Transfers ion mutation- Changes a purine to prymadine and prymadine to purine

75
Q

What is the major start codon?

What are the 2 minor start codons?

A

ATG

GTG TTG

76
Q

What are the 3 main stop codons?

A

TAA TAG TGA

77
Q

Base pair mutations can have 4 effects what are they and what do they do?

A
  1. Silence or same sense = no effect on AA sequence may code for same AA.
  2. Missense= results in AA substitution
  3. Nonsense= changes an AA for a translation termination STOP codon
  4. Readthough= Changes a stop codon to an amino acid so translation continues
78
Q

Why do addition or deletions cause frame shift mutations?

A

Induce a shift in the reading frame.

This changes the protein sequence downstream of the change and often results in a truncated protein.

79
Q

What is a mutagen?

A

A chemical or physical agent that causes mutation

80
Q

What is spontaneous DNA damage?

A
  • changes in nucleotide sequence that appear to have no known cause
    >loss of bases
    > loss of amine groups from bases
  • metabolic products (e.g. Reactive oxygen species)
  • Misincorporation of bases or errors by DNA polymerase (e.g. Mismatches) and DNA repair. Often occur during replication
81
Q

What causes induced DNA damage?

A

-result from influence of extraneous factors
- misreported of damage caused by radiation
> UV light, X- Ray’s, Yrays
- Mehtylation of bases
- cross linking reagents
- intercalated molecules

82
Q

What is depurination in terms of spontaneous damage and breakdown?

A

A chemical reaction of purine deoxyribonucleosides/adenosine/guanosine and …. In which the B-N-glycosidic Bond is hydrolytically cleaved releasing a nucleic base, adenine or guanine respectively. The second product is a sugar

83
Q

What is dreaming tigon in respect to spontaneous DNA damage and breakdown?

A

Amine groups on the rings of the bases are susceptible to spontaneous oxidation to aldehyde groups, a process known as defamation.
- delaminates not alters the pairing properties of the bases e.g. Cytosine delaminates to uracil which can base pair with adenine

84
Q

What are taut imperil shifts in relation to spontaneous DNA damage and breakdown?

A

Bases in DNA can occur in several forms (tautomers) which differ in the positions of their atoms and in the bonds between the atoms.
- DNA bases can isomeraze, with the different isomers having different base pairing properties, these changes are a significant source of spontaneous mutation.
Guanine undergoes a toutomeric shift to its rare enrol form (g*) just prior to replication its enrol form pairs with thymine but post replication it reverts back to its normal keto form.

Another round of replication results in a GC to AT transition mutation in one of the progeny.

85
Q

How does Endogenous oxidative damage cause induced DNA damage?

A
  • mostly due to the production of oxygen radicals- uncharged molecules with a single unpaired e-.
  • there are multiple sites of mitochondrial superoxide anion (O2._)
  • This causes radical damage
  • thymine glycol is formed by hydroxyl radical attack of thymine.
  • thymine glycol blocks DNA replication.
86
Q

What are common antioxidants and what is their function?

A
  • prevention is better than repair.

> catalysts, peroxidases and superoxide dis mutates minimise the accumulation of these two oxdants.

87
Q

What are the indirect effects of radiation?

A

Particles interact with the other molecules which then interact with the DNA.
> e.g. Cause the radiolysis of water to give hydroxyl radicals (which generate damaged bases)
> The production of clustered lesions can also lead to chain breaks

88
Q

What are the indirect effects of radiation?

A
  • Particle imparts its energy directly to the DNA molecule

- Breakage of the bonds that hold the sugar phosphate backbone together.

89
Q

What are the main products of UV radiation?

A

Cyclone tank pyramiding diners (CPD) and 6-4 photo products are the main products of UV irradiation.

90
Q

How are CPD’s formed?

A

At adjacent prymadines resulting in covalently fused diners.
The lesion inter fears with normal base spring and can yield base substitutions.
- however the double helix is distorted and a 4-members cyclobutane ring is for me between 2 adjacent pyramidines.

91
Q

What is a 6-4 photo product?

A

The C6 Carbon of one pyramidines covalently links with the C4 Carbon of an adjacent pyrimidine.

  • As with CPD, the DNA backbone is distorted resulting in a bulky lesion 5’ -methyl cytosine found in higher organisms never forms 6-4 lesions.
92
Q

What do alkylating agents do?

A
  • Add methyl or ethyl groups to bases
    > the addition of methyl or ethyl groups to aromatic rings can alter the pairing properties and so are often called miscoding lesions.
    > some even block replication
93
Q

What causes DNA cross linking?

A
  • Pyrimidine diners are an example of intra strand cross linking
  • Certain chemicals are capable of covalently joining two bases in complementary strands to form an inter strand cross link.
  • DNA cross links prevent separation of the two strands and hence block the transcription of DNA replication.
94
Q

What do intercalating agents do?

A

DNA double helix is partly held together by stacking interactions.
- some chemicals are capable of intercalating the stacked bases which can cause single nucleotide insert actions and deletions as well as blocking replication and transcription.

95
Q

What is AAF (hint. It is a mutagen produced by metabolic activation)

A

An aromatic amine originally used as an insecticide that is converted to an alkylating agent by estero fixation. It is a mutagen that requires bio activation to electorphiles that bind covalently to DNA.

96
Q

What is direct repair? What are its benefits?

A

Direct reversal of DNA damage, it is simple energetically advantageous and safe.

  • it minimises the chance of mistakes introduced by the repair process itself. (However limited scope)
  • Direct repair does not require the removal of damaged bases it can instead repair the nucleotide directly.
97
Q

What are the 3 known examples of direct repair?

A

A) repair of single-strand breaks
B) repair of pyrimidine dimers
c) repair of methyl groups

98
Q

Direct repair: how do single stranded breaks occur?

Why are single strand breaks a problem?

A

DNA replication, recombination and DNA repair
Exposure to X- Ray’s and y-radiation
- single-strand breaks generally do not present much of a problem for non-dividing cells although they could inter fear with transcription.
- Breaks that occur close to each other in opposing strands or replication of a single strand break can generate the potentially more hazardous double-strand break.

99
Q

Direct repair: How does DNA ligase act to repair single-stranded DNA breaks?

A

Requires 3’ OH and 5’ P groups, larger gaps cannot be repaired.
- Bacterial DNA ligase use NAD, bacteriophages and eukaryotic DNA ligases use ATP as an energy co-factor.
_DNA ligases cannot joint free ssDNA molecules
- Phage T4 DNA ligase can join blunt-ended dsDNA molecules useful for DNA cloning.

100
Q

What is DNA photolyase?

A

Found in viruses, bacteria, yeast, plants and more.

  • nearly all cells contain the photoreactivating enzyme DNA photolyase.
  • the enzyme uses light energy to form an excited state that cleaves the diner into its component bases.
  • the excision of modified bases is an example of BER
101
Q

Direct repair: how can DNA photolyase repair cyclobutane pyrimidine dimers?

A

The binding of the DNA photolyase enzyme to damaged DNA flips the affected base out of the DNA double helix and into the active site of the enzyme.

102
Q

Direct repair: list the steps of DNA photolyase repair.

A

The enzyme acts as a glycol are cleaving the glycosidic bond to release the damaged base.
At this stage the DNA backbone is in tact but a base is missing (AP site)
- An AP endonuclease recognises the defect and nicks the backbone adjecent to the missing base.
- Deoxyribose phosphorites tease excises the residual deoxyribose phosphate unit and DNA pol 1 inserts and unmanaged nucleotide.
- Repaired strand is re-sealed by DNA ligase.

103
Q

Direct repair: in 7 steps explain the process of repair of a prymadine dimer:

A
  1. Unmanaged DNA
  2. Pyrimidine dimer in DNA induced by UV (220-310nm)
  3. Complex of DNA with DNA photolyase enzyme
  4. Folate harvests light energy and transfer it to flavin
  5. Absorption of blue light (300-500nm)
  6. Flavin (FADH) breaks apart the dimer by donating an electron
  7. Release of enzyme to restore native DNA
104
Q

what can chemical agents to to DNA, which can then be fixed by direct repair?

A

Can induce the methylation or alkylation of guanine residues in the DNA at the oxygen on the 6th position.

105
Q

Direct repair: What enzyme is used to repair O6- methy guanine ?

A

O6-methyl guanine-DNA-Methyltransferases. It removes alkyl groups only from position 6 of guanine, restoring the correct structure to DNA preventing mutations from occurring.

106
Q

How does O6-Methyguanine-methyl transferase work?

A

The damage is caused by O6 methyl guanine mispairing with thymine.

  • Methyltransferase recognises distortion in DNA backbone.
  • works best on dsDNA
  • also known as suicide Methyltransferase.
107
Q

What is the function of E.coli’s Ada protein?

A

Regulates a set of genes involved in repairing alkylating damage. The N- terminal half of Ada switches on this ‘adaptive response’ once it has been methylated.

108
Q

How does E. Coli’s Ada protein work to repair alkylating damage?

A

Removal of methyl groups is achieved by a single step, methyltransferase reaction, where Ada accepts the adducts from the modified oxygen molecule, onto internal cystine residues, directly restoring the DNA damage and inactivating the protein. - once modified the inactivated protein is targeted for degradation.

109
Q

Why are some miss paired bases not recognised?

A

Despite proof reading mismatches can still occur. They present a particular problem for removal as they cannot be recognised in the same way as damaged bases as they appear ‘normal’.

110
Q

How are miss-match mutations identified?

A
  • the major repair system in bacteria uses the MutHLS protein and depends on DNA Mehtylation to distinguish between template and Newley synthesised strands.
111
Q

How does Dam methylase have a key function in miss-match repair? (MMR)

A

Dam methylase adds a methyl (CH3) group to adenine in the sequence 5’- GATC - 3’.
- the short delay in methylation of the newly-synthesised strand allows enzymes to identify new DNA and target repair to this strand.

112
Q

MMR: What is the function of MutS?

A

Binds DNA mismatch, clamps on to DNA and scans for mismatches until it finds a distortion in the DNA backbone cased by the mispaired bases.

113
Q

MMR: what is the function of MutH?

A

Cuts unmethylated strand at GATC sites. It assembled with MutL to form a complex.

114
Q

MMR: What is the function of MutL?

A

Stimulates MutS and MutH.

115
Q

Briefly explain the MutLHS system.

A

Methylation by DAM does not occur until several minutes after the strand is made. Providing a distinguishing feature between the new daughter strand.

  • MutS binds to mismatch sites and forms complies with MutL, formation of complex recruits MutH which binds to hemimethylated site.
  • The MutS/L complex interact with MutH with DNA looping mechanism.
  • MutH makes a cut in the non-methylated strand and an exonuclease digests non-methylated strand from cleavage site to just beyond the mismatch site.
  • the gap in the daughter strand is filled by DNA Pol 3 and sealed by ligase. Hence the mismatch is replaced.
116
Q

What does the BER pathway specialised to repair?

A

Protects cells from the damaging effects of oxidation, alkylation and deamination.

  • Fixing single-base lesions containing small chemical modifications.
  • The repair pathway is initiated by removal of the damaged base followed by incision at the site of the mission base.
117
Q

What are the 2 key enzymes involved in BER what are their functions?

A

DNA glycosylase > recognises the type of damaged base and cleaves the bond between the base and sugar backbone.
AP endonuclease > recognises the basic site and cleaves the bond between phospodiester backbone on either the 3’ or 5’ side.

118
Q

Explain the structure of DNA glycosylase.

A

Relatively small protein.

  • target dsDNA with the exception of uracil DNA glycosylase which also works on ssDNA.
  • E.coli has 6 different types.
  • can be mono functional or bi functional.
119
Q

How can uracil get into DNA? (2 ways)

A

A) Deamination of cytosine to uracil.
- this reaction is 4000 faster in ssDNA making it significant event in actively transcribed genes and at replication forks.
B) Misincorporation of dUTP during DNA synthesis
- Bacteria have an enzyme called Dut which has dUTPase activity degrading dUTP to dUDP thus reducing the pool of avalible dUTP.

120
Q

Explain the structure and function of AP endonucleases.

A

E.coli has 4.
- exonuclease III (Xth) is mono functional and cleaves by hydrolysis on the 5’ side of AP sites. It has the capacity to repair various blocking groups on the 3’ termini of DNA which includes the products of AP leases (sometimes called 3’ repair).

  • exonuclease IV (Nth) is bifunctional and has DNA glycosylase activity on Urea and thymine glycol. It cleaves on the 3’ side of AP sites. Bifunctional AP endonuclease cleave by a b-elimination reaction.
121
Q

Briefly outline the BER pathway.

A
  • DNA glycosylase cuts the damaged base. By hydrolysis of N-glycosidic bond between the deoxyribose and the base.
  • AP endonuclease cuts on either the 3’ or 5’ side of abasic site.
  • the backbone is processed ( removal of dR-P deoxyribose phosphate) to leave a single nucleotide gap that is filled by DNA polymerase and the Nick sealed by DNA ligase.
122
Q

What systems do humans have in place instead of DNA photolyase for repair of pyrimidine dimers?

A

NER. It involves the removal of the damaged base, or bases, as part of a short oligonucleotide.

123
Q

Why are interastrand cross links so detrimental?

A
  • seporation of the 2 stranded double helix is critical for replication and transcription the therefore intrastrand cross links (ICL’s) are extremely toxic.
124
Q

What are error prone DNA polymerases?

A

They are capable of bypassing lesions by inserting ‘any old’ nucleotide at the damaged site, they polymerases are called Pol IV and Pol V.
- both are tightly regulated by the SOS response so this type of DNA synthesis only occurs when cells are so damaged that managing to replicate and transcribe takes precedence over the risk of generating lots of mutations.

125
Q

What proteins are used in NER?

A

The Uvr ABCD system

126
Q

What does UvrA do?

A

Binds UV-irradiated dsDNA; can dimerise and complex with UvrB; has 2 ATP binding sites and 2 zinc fingers.

127
Q

What is the function of UvrB?

A

Binds UvrA
5’ to 3’ DNA helicase activity
UvrA2B2 complex
1 ATP binding site

128
Q

What is the function or UvrC?

A

Interacts with UvrB-DNA complex
2 endonuclease active sites
Makes 5’ to 3’ incisions

129
Q

What is the function of UvrD?

A

3’ to 5’ DNA helicase
Binds ATP and DNA
Releases (unwinds) oligonuclotide containing damaged bases

130
Q

What is the function of Mfd?

A

Specificity for transcribed strand repair by UvrABC
Displaces RNA polymerase
Interacts with UvrA

131
Q

NER: what is the function of PolA?

A

DNA polymerase 1

Fills in the ssDNA gap left by release of the excised oligonucleotide

132
Q

NER: what is the function of LigA?

A

Seals the Nick (reforms phosphodiester bond) to complete the repair process.

133
Q

What is stage 1 of NER?

A

Process begins with loading do OD protein trimmer onto the DNA double helix (consisting of 2 UvrA and 1 UvrB)
When the trimmer encounters DNA damage such as a thymine dimer it stops.
The UvrB protein remains at the damaged site UvrA is released.
UvrC binds to damage site and makes 2 cuts on the damaged strand both cuts are several bases away from the damaged site.

134
Q

What happens after UrvC cuts around damaged sites. AKA what is stage 2 of NER?

A

UvrD a helicase binds to the site and separates the segment of damaged strand from the rest of the DNA molecule.

  • UvrB/C/D are then all released from the damaged region
  • this leaves a gap, DNA polymerase fills the gap and nicks are sealed by DNA ligase.
  • Damaged DNA has been released with the correct bases.
135
Q

Explain how the UvrABC proteins work together to perform NER.

A

A dimer of Uvr A recognises distortion in the backbone induced by the DNA lesion.

  • UvrA2B form a complex which can then bind to damaged DNA
  • the DNA is kinked by UvrB to give a 130 degree bend.
  • UvrA is released leaving a UvrB-DNA complex.
  • UvrC bind to to the UvrB-DNA complex and makes the first 3’ incision.
  • his causes the DNA to straighten and UvrC makes a second incision at the 5’ side.
  • UvrD releases the 12-13 oligonuclotide and UvrC.
  • DNA pol I fils in the gap and releases UvrB
  • DNA ligase seals Nick
136
Q

What is the purpose of transcription-coupled repair (TCR)?

A

Describes the preferential repair of transcribed strands.

137
Q

How can lesions that impede DNA and RNA polymerases kill cells?

A
  1. By interfering with DNA replication in actively growing and dividing cells
  2. Blocking transcription and thus depriving cells of an essential active genes.
    - cells have a system that targets DNA repair enzymes to transcription ally active genes.
    - in bacteria RNA polymerase stalled at a lesion is recognised by an enzyme called Mfd also known as Transcription repair coupling factor (TRCF)
138
Q

What is the advantage TCR process?

A

Damage is repaired more quickly than it would be in non-transcribed regions that are repaired by the global genomic repair pathway provided by UvrABC alone.

139
Q

In 10 steps outline TCR.

A
  1. RNA polymerase backtracks at lesions in transcribed strand.
  2. Mfd recognises the stalled RNA polymerase.
  3. ATP hydrolysis mediates forward translocation.
  4. RNA polymerase released, unmasking UvrA binding surface.
  5. UvrAB recruited
  6. Damage verification and formation of a tight pre-incision complex
  7. UvrC recruitment by UvrB and dual strand incision by UvrC
  8. damage-containing oligonuclotide removal by UvrD
  9. Gap filling and ligation by DNA Pol I and ligase
  10. Restoration of genomic integrity

Genetics (5 decks)

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