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BCHM218 > Mod 2 > Flashcards

Mod 2 Flashcards

1
Q

What is the eukaryotic cell cycle

A
  • G1 Phase is the first growth phase not part of the cell cycle
  • S phase is the synthesis phase where the cell replicates the entire genome
  • The M phase which is he mitosis and the actual cell division
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2
Q

What is a brief overview on Arthur Kornberg for readout of polymerization activity and specific activity

A
  • Radout ofr polymerization activity
  • Uses radioactive fluroescent dNTPs
  • Specific activity - Understanding what protein does rpelication
  • Fractionation seperates then used exchange chromatography at 280nm and this is he wavelength aino acids absorb ligt anf he noted which proteins were catalyzing the reaction
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3
Q

What are the requirements for Polymerase activity

A
  • dCTP, dGTP, dATP
  • Mg
  • DNA
  • Without any of these factors polymerase cannot work - there was a complete loss of radiolabelled dTTP incorporatoon
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4
Q

What are the 6 key principles for DNA replication

A
  • It is semi conservative
  • There are also other theories
  • Conservative - Parents stay together and the yare fully copied
  • anf dispersive was that the parent was broken apart and combined with pther parts to make a patchwork of different DNA copies

2 Replication is initiated at specific sites
* Called the origin of replication
* Eukaryotic DNA has multiple sites of replication

3 Replication is bidrection but always in the 5-3’ direction
* Will always move away from the origin of replication but can only build in 5-3’ directtion

4 Replication is semi discontinous
* Discontinous on the lagging strand
* Due to the anti parallel nature of DNA the lagging strand must be synthesied in fragments
* leading strand is continppusly synthesized

5 RNA primers are needed to start replication
* Need RNA primers to start
* Lay down the primer strands which are made of DNA
* They must be complementary to the template RNA and have a free 3’ OH
* Neded to incorporate incoming dNTPs to synthesize new DNA off of them

6 - Nuclease, polymerase and ligases replace RNA primers with DNA and seal the nicks
* RNA must be removed becuase it is unstable and it will contaminante the stable DNA molecule
* DNA polymerase I will fill the gaps with dNTPs\Polymerase I will degrade the RNA primer with exonnuclease domain and it inserts the dNTP
* DNA ligases will then seal the gap

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5
Q

What is the structure of DNA polymerase I

A
  • Has 3 major Domains
  • Fingers domain where the dNTP enter
  • Thumb which holds the DNA in place as its being synthesized
  • Palm domain is where the active site is

Conformations
* Open and losed conformation and it altnerates between
* There is an open 3’ OH un the active ste in the open conformation there is a dNTP in the finger domain
* In closed conformation it will brinh the dNTP to the active site at the free 3’ OH and allows a phosphodiester bond to form

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6
Q

How does DNTP selection involve shape recognition

A
  • AT GC bases have shapes that fit well into closed active site of the polymerase
  • Relies on H bonds, VDWF and ionic interactions ensures that there is he correct base pairs
  • If they are incorrect base pairs it will not enzymatically favour and it is significantly slower
  • This reduces the stability from the base stacking incorporating correct bases creates more stable stacking
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7
Q

How does the polymerase active site work

A
  • When correct nucleotide is added
  • Reaction proceedes using 2 negatively charged acidic residues that coordinate with 2 Mg 2+
  • First Mg will deproteate the 3’ I on the growung strands
  • Generates he 3’ O nucleophile
  • The nucleophile attacks the alpha pjosphate and it breaks the bonds between the aloha nad the betas
  • The secon dMg binds negatively charged phosphate group and it removes it from tne enzyme. It slingshots out of tyhe actuve site
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8
Q

What is the DNA polymerase reaction

A
  • Polymerase needs a primer strnad and dNTPS and the free 3’ OH
  • After incorporation of dNTP it moves forward so the 3’ OH is not in the active site and in the post insertion site
  • After incorporation of dNTP it moves forward so the 3’ OH is not in the active site and in the post insertion site
  • Allows post insertion site and then allows the translocation of the enzyme
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9
Q

What are the effects of mispairing bases

A
  • Wrong base pair is incorporated and does not fit the active site
  • The polymerase will detect this based on forces
  • They are removed by exonuclease domain it can fray the DNA
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10
Q

What does the 3-5’ exon nuclease do on the active site

A
  • Works in a similar 2 metal system
  • One Mg Deprotenates a water molecules - happens on the exonuclease
  • Forms an OH nucleophile that attacks the phosphate of the incorporated dNTP then leaves he free 3’ OH which is facilliated by the 2nd Mg
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11
Q

What is the DNA base excision rate

A
  • Polymerase is unique
  • has 5-3’ nuclease DNA it can excuse un the direction of polymerase movement
  • The 3-5 does it in the opposite direction
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12
Q

What is nick translation by Pol 5’3

A

Pol I degrades RNA primer in the 5-3 releases dNMPs and extends the 3’ terminus with the dNTP in the same direction
Nick moves the 5-3’ direction along DNA until all RNA is removed
DNA ligases seal the fragments

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13
Q

What are the 5 polymerases of E coli

A
  • Polymerase 2 - Not well understood involved in DNA repair, has 3-5 exnonuclease
  • Polymerase 3 - Has proofreasing exonnuclease and does most of the work called the replicase has the 3-5’ exon
  • Polymerase 4,5 - Lack exonucleases and incorporate a lot of error they are used to incorporate a base pair to continue replication
  • IIf replication fork is halted it would be better to get an incorrect pair instead of having DNA that is not replicated
  • Polymerase I does Okazaki fragments and DNA repair has both sides exonnucleases
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14
Q

What are the stages of Ecoli replication

A
  • Intiation of replication
  • Elongation by replisome
  • Termination
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15
Q

Why does E coli have A=T rich 13mer and DNAA 9Mer Sites - initiation

A
  • Has a A=T repeats that ahve the same directionality
  • Only 2 H bonds which base pair and are easier to disrupt
  • easier to unwind
  • DNAA 9mer site 4 copies of the 9 base pair sites where the bacterial initiator protein binds
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16
Q

What is the first step of generating the open complex

A
  • DNAA binds to specific sites in the oriC
  • Initiates replication
  • Part of the AAA+ Protein famiky becasue it saves ATPase activity
  • DNA binding at the oRiC it will oligomerize and cause many different complexes to appear and there is a bunch of different strands on the proteins
  • It will destabilize and form the ssDNA bubble
  • The Use of DNAAtpase activity here
  • Formatio of the bubble faciliated by histone like protein DNAA ATPase complex forms its bubble att he cmplex and generates the open complex and opens the bubble
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17
Q

What is the second step - the activation of replication in E Coli

A
  • DNA is an open complex
  • Now ssDNA have been opened
  • The 2 exposed areas there is 2 hexamers of DnaB called helicase sometimes
  • It can bind to each strand
  • DNAC is required to load the DNAB onto the single stranded DNA at the start of these replication forks
  • What it does it pry open the DnaB and it shoves the single strand on it
  • It will then bind and stay in the closed conformation
  • The DNA bubble at he OriC is called the pre priming complex
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18
Q

Step 3 - assembly of the e coli replication forks

A
  • Now we need primers yto start the replication
  • DnaB binds ATP which allows it to translocate
  • Opens the replication bubble
  • This will open up the replicatioj bubble and make it bidirectional
  • Topisomerase removes stress
  • DNA woll also be coated with ss binding proteins to prevent enzymes from coming and degrading the DNA
  • After replication bubble and grown to 200bp in size it will stop
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19
Q

WHat is the Ecoli replisome and what does it do

A
  • After primers are made, te primer allows for the direction of the loading and assembly of the beta clamp holoenzyme
  • Reach polymerase III enzyme are the yellow things 2 at each fork follow the replication bubble
  • Produces more ssDNA
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20
Q

What is a basic overview of the Replisome

A
  • Multi protein complex
  • Promotes DNA synthesis at the replication fork, each fork going biidirectional
  • Primase - lays down the primers
  • DnaB Helicase
  • Polymerase III holoenzyme
  • 3 Polymerase IIII cores
  • 3 beta clamps
  • Beta loading clamp
  • DnaB helicase unwinds DNA
  • Primase adds RNA primers
  • DnaB helicase connects to the III core with the green loaders - if it does not connect to polymerase core directly it would be much slower having a direct connection of unwinding happens much faster
  • 1 polymerase core with leading strand
  • 2 polymerase cores with lagging strand
21
Q

What is DNA polymerase III Holoenzyme

A
  • Everything is connected the beta clamp curcles the DN nad they hold the Polymerase on the DNA
22
Q

What is the point point of having the Polymerase III core connected to the clamp

A
  • Creates a high processvity
  • No 5-3 Exonuclease
  • Palm is the active site
  • Fingers at the dNTPs
  • Thumb holds everything and how it connects to the beta clamp
  • Betaclamp is homodimer
  • Beta clamp make whole replication faster
  • 3 Polymerases in the holoenzyme allows for the coordinated synthesis of leading and lagging strands
  • 1 beta clamp on leading and 2 beta on the lagging
  • There is no covalent bonds there is only transient bonds and H bonds and VDWF
23
Q

What experiment showed that the beta clamp improves pricessivity

A
  • They had 2 plasmids - Larger with beta clamo anf the smaller without
  • After reaction progressed it was ran on a gel
  • In both cases the original donor plasmid
  • The plasmid with beta clamp see larger replication at an earlier time
  • On reaction B we see how it takes onger Polymerase III can hop between clamps
24
Q

What is the E coli clamp loader

A
  • Beta clamp cannot put itself on the clamp
  • It is the tao complex
  • Uses Energy of ATP binding to open up the clamp
  • Then loads the clamp on a primed region of DNA
  • ATP hydrolysis will then conform to a shape that cannot bind to the loader so it can start replicating DNA
  • The whole idea of the clamp loader works the same way DnaB is loaded onto DnaC
  • ATP is hydolyzed to get clamo then it changes
25
Q

What is the trombone model

A
  • Descrubes how the lagging strand is synthesized
  • Lagging strand extends the 3’ terminus that is opposite to fork movemrnt
  • When lagging strand reaches the end of the region the clamop will become part of the strand and a new clamp will bind and it will continue working
26
Q

Hoe are clamps reccled by DNA polymerase I and Ligase

A
  • Clamp attracts to Pol I to removed RNA PRIMER
  • Clamp attracts ligase to seal ssDNA break
  • Clamp is opened and removed
  • Clamps that are left behind on the okazaki fragments are recycled once the Pol III core disassciates it can attract a polymerase I whhich removes the RNA primer and the nick translation using the 5-3’ Exonuclease
  • Leaves a single stranded break behind
  • the ligase will seal the nick
  • The clamp is then unloaded and recycled back into the reaction of a new okazaki fragment
27
Q

How is replication terminated

A
  • When replication starts it cannot stop
  • Stope when replication forks meet each other
  • Only stop DnaB Helicase if it moving in direction opposite to it
  • They prevent 1 replication site from doing more work than the otehr even if they are at different s;ees
  • Active replication bacteria metabolize at the asme time - in eukaryotes this does not happen just straight replication but the baceria is making RNA shit also which can cause stalling and shit
  • The ter Tus sites prevent collisions from the proteins on other sides so the collisiosn can only be co directional
28
Q

How is eukaryotic terminatuon the end replication a problem

A
  • After leading strands have been completed they have no continuitu and the lagging strands has a gaop that needs to be filled in
  • When the RNA primer ends up being removed at the extreme end there iis nowhere for the polymerase to extend off of so there s this gap that would progressivly cause the gene to shorten
29
Q

What are the steps of eukaryotic end termination

A
  • Telomere anneals to RNA nucleotides in telomerase
  • Telomerase extends 3’ end of ssDNA
  • After adding 6nt repeats, telomerase synthesizes many telomere repeats
  • The telomerase extended 3’ ssDNA terminus is converted duplex DNA with the help of primase and Pol III
  • 3’ Terminus of the new telomere still has ssDNA and as a result these are bound and protected by the telomere DNA
30
Q

What are the explanation notes of end of replication

A
  • Telomeres are repeats of a sequence
    • Telomerase is an enzyme that extends on the 3; end of the ssDNA 6 nt repeats
    • This all occurs in the S phase
    • At the 3’ end of linear DNA - 3 nt in the telomerase will allign with the telomere and it will extend it by 1 telomere repeat 6 nt
    • After adding the 6nt repeat it seperates the DNA RNA hybrid and sets itself up for the next repeat
    • It then extends 6 more nt and extends the 3’ end of the ssDNA
    • It is then converted to duplex DNA
    • Telomerase extended the 3’ end so no info is lost when the primase are taken out and the telomeres are special and not subjec to DNA repair mechanisms because there is this overhand that ssDNA can bind to and prevent degredation
    • Since there was an extension of a sequence that does not code for anything with every replication it will be continued to increase but it does not matter because it does not have any useful information on it
      Telomerase is a DNA polymerase but carries it own DNA template
31
Q

What are notable features of a cancer karotype

A
  • They are mixed and matched
  • Chromosomal translocation is he smattering of dfferent colours
  • They could be insertions nad deletions
  • Some are longer or shorter or there are many different kinds of shapes and sizes
32
Q

What is a DNA lesion ( DSB)

A
  • Deadliest and double stranded break
  • Bad because it prevents replication from happening
  • UV exposure, gamma rays cause DSBs
33
Q

What is a noteable factor about Deinococcus radiodurans

A
  • Can repair its own DNA rapidly and reassemble its genome before the next cycle of gene repair
  • It always has copies of its DNA that it can pool from
  • The overlapping DNA fragments allow it to repair DNA very quickly
34
Q

What is Recombinational DNA repair

A
  • enzymes can repair double stranded breaks
  • Not as efficient as the bacteria deinococcus
  • Requires the presence of undaamaged homologous Chromosome
  • Or a sister hcromatid which is present after replication to act as template
  • After having double stranded break the breaking ends are cleaved back with nucleases. They degrade the 5’ end
  • There will be a 3’ overhand they will be protected by single stranded binding proteins
  • They then invade healthy chromosome and displace it and bind to the healthy
  • Binding reaction is the invadin strand by DNA polumerase and causes the DNA polymerase to synthesize new DNA
  • Across the double stranded break
  • Using homology of chromosome there is a template that can be synthesized that is identical
  • Any info lost on the oyher side of the break can be regained
35
Q

How is recombinational DNA repair completed

A
  • 2 Possible pathways
    Synthesis dependent strand annealing
  • Lengethened strands are extended by polymerase and they are sealed by ligase

**Double stranded break repair **
* This is ligation when the strands are linked and ligated called holliday intermediates
* Specialized endonucleases resolve these intermediates

36
Q

How are Holiday Junctions resolved

A
  • 2 outcomes crossover/non crossover
  • If both intermediates are cleaved in same direction there is no crossover
  • One at X and one at Y
  • Then tehre is crossover
  • It is called crossover because material outside the repair site is exchanged in the feet and ends of the chromosome
37
Q

What is DNA recombination

A
  • Originally a repair proccess but evolved to be a form of adaptation
  • Homologous recombination between 2 DNA molecules of similar sequence such as homologous chromosomes or chromosomes with slight differences
  • This happens in all cells happens in meiosis and mitosis for eukaryotes
38
Q

What is meiosis and when does DNA recombination happen in it

A
  • To produce gametes that are both haploid and genetic different
  • Duplicate DNA in S phase
  • we then have 4n
  • Then meiosis I and meiosis II
  • In meiosis I such as metaphase when they are all lined up and pulled apart in anaphase thats when they are alligned and crossover - This increases the genetic diveristy in DNA replication
  • In mitosis they recombination happens with DSB
  • In meiosis they are pre programmed outcomes
  • Allowing the alligning of molecules at the genetic diversity
  • Chiasmata is the site where the chromosome break and where they crossover
39
Q

How is meiotic recombination initiated

A
  • AS the cell enters meiosis DSB are set intentionally along areas
  • They arent predetermined but there are hotspost where they are likely to happen
40
Q

What is Spo11 and what does it do

A
  • Is a dimer uses active site tyrosine to attack phosphodiester bonds
  • Replaces the linkage with 5’ phosphoyltyrosyl linkage similar to the linkage in topisomerase
  • Each subunit of protein then cleaves one strand of the protein
  • Last steos of the protein are removing the spo11
  • MR11 with bond to 3’ side of spo11 and cut the DNA with the spo11 in it
  • Sae2 will then degrade the remaining DNA left on the spo and acts like an endonuclease
  • Whole purpose is to create a double stranded break during Meiosis
  • Sgs1 acts like helicase
41
Q

What are D loops of Holliday junctions

A
  • Now that the 3’ overhands are present
  • Then they bind to the template DNA
  • Bound by protein RPA
  • DMC1 and Rad51 are loaded onto the 3’ extension of either DSB
  • These overhands gets repeated cycles of binding and sampling and cycling until it fidns the correct spot to bind to
  • Because the first invasion of the D loop might not be the correct homologous pair it is binding to
  • Recombinases will then attack the DNA together
42
Q

What proteins are involved in Resolving Holliday Junctions

A
  • They nick and ligase
  • RUVAb complex processes Holliday intemrediates
  • The tetramers. The RuvA keeps it in the single stranded state RuvB is a translocase and it unqinds shit and moves it along
  • RuvC is the resolvase and once it its stabilized by the RuvAB complex it will place itself on the RuvA tetramer and it nicks strands with the same polatity and ligates them
43
Q

Using the holliday map how are they resolved

A
  • Vertical cuts are on the left and right
  • Horizontal cuts are on the top and bottom
44
Q

What is site specific recombination

A
  • Precise and predictable
  • Involbed DNA and RNA being arranged between specific sequences
  • Site specific recombination can alter gene order - homologous recombination canntot do this
  • Only happens in some areas and shares the info
45
Q

What are LoxP and FRT sites

A
  • LoxP and FRT are not naturally occuring they are artifically placed
  • Cre and FLP are recombinase enzymes that recognize the artifical site and cleave them
  • They will cleave or invert the sequences
  • Something to notice is that they are palendromic seuqneces of each other that are seperated by a core sequence that is not palendromic
  • Can have sites that are inverted or parallel in the same direction
  • Recombinases will flip sequences
  • Deletion and insertion are also possible
  • if there is foreign DNA that contain a region that is homolgous but is also matching the loxP site it can also be inserted in the DNA and can be used to insert
46
Q

What are ZFN and talens

A
  • Zinc figer nuclease protein DO,ain has a zinc atom connected to 2 histidine or 2 cysteine residues that form this structure.
  • ALpha helix portion contacts 3 nucleotides and intermingles. Caan be added on to create domain thats are just 3 nucleotides but longer also
  • Can also be used to be cut DNA such as FOKI which allows it to cut DNA at sites it is intended to
  • TALEN uses tail domains or transcrtiption activator domains they can recognize single base pairs, they cut DNA as well both of these medthods are good since they create a double stranded breaks
  • A huge limitation is how much editing they require. If you want to edut another protein has to be re-engineered
47
Q

What is the CRISPR-CAS system

A
  • Origunally a bacteria immunity system but now used in gene editing
  • SOlves problem with having to re edit enzymes
  • More robust approach viral DNA is seized by bacterial DNA and it is incorproated into the CRISPR regions which are the scripy repeats
  • There is also new spacer from the brisu
  • They create a short hairpin RNAs called guide RNAs
  • Then when anotehr viral infections come around it cna be recognize the virla RNA and cleave it
48
Q

What is CRISPR/Cas9 In bacteria

A
  • Bypasses the troublesome by using guice RNA and tracer RNA guide the sequence
  • Dark green is the guide RNA
  • is synethsized by the cripsr spacer
  • The tracer RNA creates the cas system to cleave it
49
Q

What is CRISPR

A
  • CAS9 has 2 seperate nuclease domains
  • One is to cleave the DNA paired with the RNA and the other is to cleave the opposite strand
  • If you inactivate one it results in a DSB
  • Can use homology directed repair to incorporate them into sequence
  • We can also introduce a DSB which repaired by non homologous end jouning - very error prone and can result in genes bing functionless

BCHM218 (3 decks)

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