LECTURE 2 (The molecular basis of mutation 2) Flashcards

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1
Q

What are spontaneous mutations? what’s their frequency?

A

they arise in the absence of a mutagen and have a frequency of 10^-4 in retrovirus, 10^-9 in euk

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2
Q

Are spontaneous mutations due to external stimuli or due to a pop?

A

Fluctuation test: this was addressed by luria & delburk, they won a nobel price. E.coli were plated in the presence of phage T1 (infects and kills the bact but some bact can survive) they let 20 independent cultures to grow for a few days until they were 20^8 and then analyze them and also 10 aliquots from a large culture also let them grow until they were 10^8 . the result show that in the independent cultures the numbers of T1 mutants vary so much whereas in the large cultures all the aliquots had more or less the same amount of T1 mutants, this means that the mutations were not due to the T! phage otherwise all the cultures would have the same amount of mutants because all were exposed to the same amount of phages.

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3
Q

How do we get the mirror image bt plates in replica plating?

A

by using a velvet covered stamp

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4
Q

Can the existence of mutants in a population before selection be demonstrated directly?

A

yes through replica plating: Lederberg
we plate colonies in a master plate (non-selective medium) therefore no phage and all colonies grow. Then we imprint with a velvet stamp and stamp on different plates w/ phage (same amount), we can compare the plates bc they’re mirror images and we realize that some colonies in the 1st rich medium were resistant and all the replica plates show the same pattern. This suggests that mutations occurred before exposure to T1 phage otherwise different patterns would be observed.
This can be used to screen auxotrophic mutants.

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5
Q

How can spontaneous mutations occur?

A

By errors in DNA replication or spontaneous lesions

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6
Q

What are the most common spontaneous lesions?

A
  • deamination and depurination and oxidative damage
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7
Q

What is deamination? give examples

A

It’s a way of causing an spontaneous lesion. loss of an amine.

e. g deamination of cytosine leads to uracil so instead of a GC pair we get a AU pair (AT at the level of DNA)
e. g deamination of 5’methylcytosine generates thymine (5’methyluracil) again GC to AT pairing.

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8
Q

How can a deamination be observed at the level of DNA?

A

because if there’s a U it shouldn’t be there. so uracil glycosilates initiate the base excision repair mechanism

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9
Q

What is depurination?

A

The loss of a purine this occurs quite often in mammals 10000npurines are lost every 20h and this is quite critical. sometimes another base is inserted in the apurinic site but if it’s the wrong one then it’s a mutation

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10
Q

What are oxidative damaged lesions?

A

this occurs due to oxygen e.g the 8-oxodG produced due to radiation (G pairing with A) these oxidative lesions cause many humans diseases.

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11
Q

What are errors in DNA duplication?

A
  • base substitutions in already present bases or new bases. tautomeric shift can cause misfiring in DNA replication.
  • indels (leading to frameshift)
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12
Q

How are indels caused in DNA replication?

A

via a mechanism called replication slippage. indels arise when loops in ss regions are stabilized by slipped mispairing of repeated seq

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13
Q

Give an example of replication slippage in nature.

A

in the LacI gene of E.coli there are CTGG sequences repeated mutations can occur here if there’s an addition or loss of CTGG sequences. if there’s an addition DNAP needs to complete that addition plus the rest of the molecule therefore you end up with a much larger seq. the majority of the mutations in this lacI gene result from the addition of ctgg sites rather than the deletion

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14
Q

Appart from replication slippage, can indels appear through another mechanism?

A

yes, they can appear due to offset homologous recombination between copies of the repeats.

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15
Q

What are the main DNA repair mechanisms?

A
  • direct reversal of damage
  • base excision repair (BER)
  • nucleotide excision repair
  • mismatch repair
  • non-homologous end joining
  • homologous recombination
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16
Q

Give examples of direct reversal of damage.

A

e. g pyrimidine photodimers are formed due to uv light, this can be reversed by using photolyase which is activated by white light and separates the dimers in two
e. g alyltransferases remove alkyl group from the 0-6 position ensuring the correct pair. however this removal inactivates the protein and thus this system can be saturated if theres a high level of alkylation.

17
Q

Describe base excision repair.

A

It is to excise a base thats been damaged. it’s carried by DNA glycosylases which hydrolyze the sugar bond surrounding the damaged base which is excised leaving a apurinic or apyrimidinic site.

From here there are two pathways:

  • Single nucleotide BER: uses lyases which cleave the 3’ site of DNA. betalyase leaves a 3’deoxiribophosphate and deltabetalyase leaves a 3’ and 5’ phosphate. then a AP endonuclease eliminates the 3dRP and DNA fills the gap
  • Long patch BER: AP endonuclease cleaves the 5’ end of the a basic site 3’oh is the substrate of DNAP which can continue the synthesis and a few extra bases are added this displaces the strand which is removed by flap endonuclease.
18
Q

What is the function of DNA glycosylases?

A

to hydrolyze the sugar bond surrounding a damaged base which is then excised

19
Q

What’s the function of AP endonucleases?

A

They cut the sugar phosphate backbone

20
Q

What is the function of lyases?

A

They cleave the DNA at the 5’ end

21
Q

Describe the NER mechanism.

A

There are two ways to get to a common pathway:
- TC NER:
RNAP stops at the lesion and CSA & CSB recruit the TFIIH complex (10 S.U) in which XPD & XPB are found

  • GG NER
    XPC-HR23B complex stops at the lesion, then it’s degraded into XPC and HR23B. XPC recruits the TFIIH

common pathway
the TFIIH comes in and XPA & CAK are recruited. XAP recruits RPA which coats the undamaged DNA strand (complementary to damaged RNA) CAK is released and XPF & XPG carry the incision of 5’ and 3’ in damaged finally DNAP fills the gap and ligases end this.

22
Q

Describe the mismatch repair system.

A

It takes advantage of the methylation of A in the old strain. the new one won’t be methylated until the end. MutS recognizes the damage and recruits Mut L and MutH which cuts the section with wrong bases then DNAP resynthesizes.

23
Q

Why are methylated A imp in repair systems?

A

Because only the old strand of DNA has methylated A until the synthesis of the strand finishes so A can be methylated. this allows to identify which one is the new strand and see the error here.

24
Q

explain non.homologus end joining

A

DNA dependent kinase and Ku80-70 complex are recruited in the dsbreak
the artremis endonuclease leaves 3’ and 5’ overhangs and DNAP finishes the synthesis

25
Q

Describe homologous recombination

A

To repair dsbreaks using the sister chromatid
MRN complex finds the dsbreak and Ctbp produces 3’ends overhangs, BRCA very imp bc it’s in charge of finding whether the overhangs have homology with the sister chromatid
RPA stabilizes 3’end overhangs DNAP finishes the polymerization