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LS3 > DNA Replication > Flashcards

DNA Replication Flashcards

1
Q

When is DNA replicated?

A

replicated once every time a cell divides

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2
Q

In which direction does DNA synthesis occur?

A

5’ to 3’ direction by DNA polymerase

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3
Q

What are the 3 hypotheses of DNA replication?

A
  1. conservative
  2. semiconservative
  3. dispersive
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4
Q

Conservative Model of DNA replication

A

In this model the two parental DNA strands are back together after replication has occurred. That is, one daughter molecule contains both parental DNA strands, and the other daughter molecule contains DNA strands of all newly-synthesized material

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5
Q

Semiconservative Model of DNA replication

A

In this model the two parental DNA strands separate and each of those strands then serves as a template for the synthesis of a new DNA strand. The result is two DNA double helices, both of which consist of one parental and one new strand.

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6
Q

Dispersive Model of DNA replication

A

In this model the parental double helix is broken into double-stranded DNA segments that, as for the Conservative Model, act as templates for the synthesis of new double helix molecules. The segments then reassemble into complete DNA double helices, each with parental and progeny DNA segments interspersed.

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7
Q

Dispersive Model of DNA replication

A

In this model the parental double helix is broken into double-stranded DNA segments that, as for the Conservative Model, act as templates for the synthesis of new double helix molecules. The segments then reassemble into complete DNA double helices, each with parental and progeny DNA segments interspersed.

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8
Q

Who did what experiment to define the nature of DNA replication?

A

Meselson and Stahl Experiment
-used heavy isotopes (more dense) N15 that can incorporate itself into DNA without affecting DNA function
(DNA is normally made of N14)
*when bacteria are grown in a medium containing N15 their DNA is denser than bacteria that grow under normal conditions with N14
-DNA of different densities can be separated through centrifugation in a heavy salt (CsCl) density gradient. Heavy DNA towards the outer end> 14N-15N hybrid DNA> light DNA on the inner end

Setup:

  1. Grow E.coli cells in a medium with 15N as sole source of nitrogen. Collect sample and purify DNA. (15N-15N heavy DNA)
  2. Transfer cells to medium containing 14N. After cells divide once, collect sample and purify DNA. (15N-14N hybrid DNA)
  3. After cells have divided a second time in 14N medium, collect sample and purify DNA. (15N-14N hybrid DNA and 14N-14N light DNA)
  4. Centrifuge the three samples and compare the locations of the DNA bands.
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9
Q

Who did what experiment to define the nature of DNA replication?

A

Meselson and Stahl Experiment
-used heavy isotopes (more dense) N15 that can incorporate itself into DNA without affecting DNA function
(DNA is normally made of N14)
*when bacteria are grown in a medium containing N15 their DNA is denser than bacteria that grow under normal conditions with N14
-DNA of different densities can be separated through centrifugation in a heavy salt (CsCl) density gradient. Heavy DNA towards the outer end> 14N-15N hybrid DNA> light DNA on the inner end

Setup:

  1. Grow E.coli cells in a medium with 15N as sole source of nitrogen. Collect sample and purify DNA. (15N-15N heavy DNA)
  2. Transfer cells to medium containing 14N. After cells divide once, collect sample and purify DNA. (15N-14N hybrid DNA)
  3. After cells have divided a second time in 14N medium, collect sample and purify DNA. (15N-14N hybrid DNA and 14N-14N light DNA)
  4. Centrifuge the three samples and compare the locations of the DNA bands.
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10
Q

How to analyze the results of the Meselson & Stahl experiment.

A

Meselson and Stahl tested the hypothesis of DNA replication. They cultured bacteria in a 15N medium. 15N is a heavy isotope of nitrogen so the DNA synthesized is of heavy density. They then shifted the bacteria to a 14N medium, DNA was isolated at different times corresponding to replication cycles 0, 1, and 2. After one replication cycle, the DNA was all of intermediate density. This rules out the conservative replication model, which predicts that both heavy density DNA and light density DNA will be present, but none of intermediate density will be present. This result is consistent with the semiconservative replication model, which predicts that all DNA molecules will consist of one 15N-labeled DNA strand and one 14N-labeled DNA strand. The result does not rule out the dispersive replication model, which also predicts that all DNA will be of intermediate density, consisting of interspersed double-stranded 15N-labeled and 14N-labeled segments.

After two replication cycles, two bands of DNA were seen, one of intermediate density and one of light density. This result is exactly what the semiconservative model predicts: half should be 15N-14N intermediate density DNA and half should be 14N-14N light density DNA. This result rules out the dispersive replication model, which predicts that after replication cycle 1, the DNA density of all DNA molecules will gradually become lower, so no intermediate density DNA should remain at replication cycle 2. The semiconservative model is correct.

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11
Q

How to analyze the results of the Meselson & Stahl experiment.

A

Meselson and Stahl tested the hypothesis of DNA replication. They cultured bacteria in a 15N medium. 15N is a heavy isotope of nitrogen so the DNA synthesized is of heavy density. They then shifted the bacteria to a 14N medium, DNA was isolated at different times corresponding to replication cycles 0, 1, and 2. After one replication cycle, the DNA was all of intermediate density. This rules out the conservative replication model, which predicts that both heavy density DNA and light density DNA will be present, but none of intermediate density will be present. This result is consistent with the semiconservative replication model, which predicts that all DNA molecules will consist of one 15N-labeled DNA strand and one 14N-labeled DNA strand. The result does not rule out the dispersive replication model, which also predicts that all DNA will be of intermediate density, consisting of interspersed double-stranded 15N-labeled and 14N-labeled segments.

After two replication cycles, two bands of DNA were seen, one of intermediate density and one of light density. This result is exactly what the semiconservative model predicts: half should be 15N-14N intermediate density DNA and half should be 14N-14N light density DNA. This result rules out the dispersive replication model, which predicts that after replication cycle 1, the DNA density of all DNA molecules will gradually become lower, so no intermediate density DNA should remain at replication cycle 2. The semiconservative model is correct.

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12
Q

What substrates are used for DNA replication?

A
  1. dNTPs: dGTP, dCTP, dATP, dTTP

2. Primer: a short sequence of nucleotides that is bp to the template with a 3’OH end of growing DNA

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13
Q

In a dNTP, which bond is high in energy?

A

the bond between the alpha and beta phosphate group. it breaks through hydrolysis which makes it an energetically favorable reaction

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14
Q

What is the chemistry of DNA synthesis?

A

A dNTP comes in to fill the spot at the end of the primer on the 3’OH side. The 3’OH of the primer bonds with the alpha phosphate group of the incoming dNTP and the dNTP base pairs with the template. Two phosphate groups of the dNTP are removed by PYROPHOSPHATES.

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15
Q

What is the chemistry of DNA synthesis?

A

A dNTP comes in to fill the spot at the end of the primer on the 3’OH side. The 3’OH of the primer bonds with the alpha phosphate group of the incoming dNTP and the dNTP base pairs with the template. Two phosphate groups of the dNTP are removed by PYROPHOSPHATES.

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16
Q

DNA Polymerase

A
  • resembles a partially closed right hand
  • DNA sits in the cleft of it
  • has 3 domains:
    1. palm (cleft where DNA sits)
    2. thumb
    3. fingers
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17
Q

How is DNA polymerase flexible but specific?

A
  • specific: active site is used to catalyze 4 different reactions (addition of the 4 dNTPs)
  • flexible: geometry of A:T, T:A, G:C, and C:G base pairs are similar, but very different from any mismatched base pairs. So what prevents the wrong base pair from being added? When a correct base pair is formed between the template and incoming dNTP, the 3’OH of the primer is positioned properly to attack the incoming alpha-phosphate (forming a new phosphodiester bond). When an incorrect base pair forms, the alpha phosphate is displaced from the site of catalysis, strongly reducing the rate of catalysis up to 10,000 fold. “kinetic proofreading”
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18
Q

How does DNA polymerase distinguish between rNTPs (uracil) and dNTPs (thymine)?

A

The dNTP’s alpha phosphate is in the catalytic center, ready for catalysis. The rNTP’s alpha phosphate is displaced from the catalytic center.

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19
Q

What substrates are used for DNA replication?

A
  1. dNTPs: dGTP, dCTP, dATP, dTTP

2. RNA Primers: a short sequence of nucleotides that is bp to the template with a 3’OH end of growing DNA

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20
Q

How does DNA polymerase distinguish between rNTPs (uracil) and dNTPs (thymine)?

A

The dNTP’s alpha phosphate is in the catalytic center, ready for catalysis. The rNTP’s alpha phosphate is displaced from the catalytic center.

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21
Q

How are two metal ions (Mg2+) in the active site of DNA Polymerase, critical for catalysis?

A
  1. one generates a 3’O that is ready for catalysis by reducing the affinity of the 3’OH for its H
  2. one neutralizes the negative charges on the triphosphate and stabilizes the pyrophosphate generated from catalysis
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22
Q

T/F. Polymerase closes and reopens its grip with each nucleotide added.

A

true

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23
Q

The replication fork

A

The junction between the unreplicated DNA duplex and the newly separated template strands

  • both strands of DNA are replicated at the same time and must be separated to serve as templates for DNA synthesis
  • leading strand is in the same direction as the replication fork 5’-3’
  • lagging strand is in the opposite direction of the replication fork 5’-3’
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24
Q

Leading strand synthesis

A

-synthesized in 5’-3’ direction which is the same direction as the direction of the replication fork movement

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25
Q

Lagging strand synthesis

A
  • synthesized in the 5’-3’ which is in the opposite direction of the replication fork
  • synthesis is discontinuous, making short “Okazaki” fragments are more lagging strand template is exposed: this is transient and non-covalently attached
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26
Q

DNA Helicase

A

Unwind duplex DNA at the replication fork

  • uses the energy from ATP hydrolysis to catalyze separation of two DNA strands at the replication fork
  • hexameric “ring”- encircles one of the two single strands
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27
Q

What kind of primers are needed for DNA synthesis?

A

RNA primers

-All DNA polymerases require a primer with a fee 3’OH (DNA or RNA)

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28
Q

Primase

A

a special type of RNA polymerase that generates short 5-10nt long RNA primers on a ssDNA template

29
Q

How do Okazaki fragments work?

A

They are not joined together initially. Once the RNA primer gets replaced with DNA the Okazaki fragments are joined by DNA ligase.

30
Q

Where are promises located at the replication fork?

A

They associate with Helicase at the replication fork. Primes activity dramatically increases when it associates with DNA helices at the replication fork. As helices unwinds, the DNA

31
Q

Where are promises located at the replication fork?

A

They associate with Helicase at the replication fork. Primase activity dramatically increases when it associates with DNA helicase at the replication fork. As helicase unwinds the DNA and makes ssDNA template, primase can make a new primer for the next Okazaki fragment

32
Q

Removal of RNA primers from DNA

A
  1. RNase H (hybrid): is an endonuclease that degrades RNA in RNA: DNA hybrid; only cleaves a bond between 2 ribonucleotides, leaving behind an end of the primer
  2. 5’ exonuclease cuts out the part of the primer that was left behind by RNase H
  3. DNA polymerase fills in this gap with DNA and 2 separate Okazaki fragments are now covalently linked/continuouse
  4. DNA ligase seals the niche
33
Q

single stranded binding proteins (SSBs)

A

rapidly coat and stabilize the newly exposed sand template, preventing base pairing

34
Q

Topoisomerases

A

unwinding of the double helix by HELICASE introduces POSITIVE supercoils in the unreplicated sDNA in front of the replication fork because LK changes with covalent breaks in the backbone
-topoisomerases remove positive supercoils ahead of the replication fork

35
Q

Which topoisomerase exist in E.coli?

A
  • Topo I

- Gyrase

36
Q

Enzymes that work at the replication fork are conserved. T/F

A

True.

But proteins that act at the replication fork interact tightly in a sequence-independent manner with DNA unlike enzymes.

37
Q

DNA Polymerase also has ______ Exonuclease activity.

A

3’—>5’
This is a separate active site on the same polypeptide as the polymerase active site. It detects and removes a mistakenly added nucleotide

38
Q

How accurate is DNA replication?

A
  • incorporates one incorrect nucleotide for every 10^5 nucleotide added (kinetic proofreading)
  • proofreading exonuclease decreases this number to 1 in every 10^7 nucleotides added (this is the 3’ exonuclease activity of the DNA polymerase)

*final error rate: 1 in every 10^10 due to post replication repair mechanisms

(don’t have to know the numbers)

39
Q

How accurate is DNA replication?

A
  • incorporates one incorrect nucleotide for every 10^5 nucleotide added (kinetic proofreading)
  • proofreading exonuclease decreases this number to 1 in every 10^7 nucleotides added (this is the 3’ exonuclease activity of the DNA polymerase)

*final error rate: 1 in every 10^10 due to post replication repair mechanisms

(don’t have to know the numbers)

40
Q

What is the rate-limiting step for DNA synthesis?

A

the initial binding of polymerase

~1 sec

41
Q

How fast can DNA polymerase add nucleotides?

A

1,000 nucleotides per second
*processivity can increase overall DNA synthesis 1,000-fold (consecutive binding without its release from the strands)- so the longer it stays on, the more productive it is

42
Q

On its own, DNA polymerases at the replication fork will only synthesize ________ bp before dissociating from the template DNA.

A

20-100bps

43
Q

What helps DNA polymerase stay on and not release itself from the template?

A

Sliding DNA clamps located behind the DNA polymerase

44
Q

Sliding DNA clamps

A
  • tetramer of 3 identical subunits
  • opened and placed on DNA by enzymes called CLAMP LOADERS in a process that requires ATP binding and hydrolysis
  • occurs at primer:template junctions (dsDNA)

*so DNA polymerase still starts to dissociate from the template every 20-100bp and because it binds tightly to the sliding clamp, polymerase rebinds to the same template and continues DNA synthesis

45
Q

Why does polymerase get released when synthesis is complete?

A

The conformation of DNA polymerase changes when it transitions from having a ssDNA:sDNA junction to sDNA in its active site.
Polymerase bound to dsDNA in its active site has a low affinity for the sliding clamp and the DNA dissociates

46
Q

Polymerase I

A

prokaryotes

  1. RNA primer removal
  2. DNA repair
47
Q

Polymerase III holoenzyme

A

prokaryotes

1. chromosome replication: highly processive

48
Q

Polymerase alpha

A

eukaryotic

1. primer synthesis during DNA replication

49
Q

Polymerase delta

A

eukaryotic

  1. Lagging-strand DNA synthesis
  2. nucleotide and base excision repair
50
Q

Polymerase epsilon

A

eukaryotic

  1. Leading-strand DNA synthesis
  2. nucleotide and base excision repair
51
Q

Explain eukaryotic DNA polymerase Switching.

A

DNA pol alpha immediately starts DNA replication following primer synthesis by primes, but has very low processivity. DNA Pol delta and epsilon are highly protective and replace Pol alpha and extend the DNA product up to 10,000 nucleotides

52
Q

How do replication fork proteins stimulate each other’s activities?

A
  1. DNA sliding clamp- increases the proclivity of polymerase
  2. Helicase- stimulates rate of primase synthesis
  3. DNA pol III holoenzyme- stimulates helicase rate of movement
53
Q

How do replication fork proteins stimulate each other’s activities?

A
  1. DNA sliding clamp- increases the proclivity of polymerase
  2. Helicase- stimulates rate of primase synthesis
  3. DNA pol III holoenzyme- stimulates helicase rate of movement
54
Q

E. coli cells have only one origin of replication for their one circular chromosome. T/F

A

True

55
Q

Replication of Eukaryotic Linear DNA

A
  • initiated at many sites along the DNA
  • multiple initiation is a means of reducing total replication time
  • bidirectional
  • origins are 30kb apart, so large human chromosomes may have thousands
56
Q

replicator

A

the cis-acting DNA sequence, required to direct DNA replication initiation

57
Q

initiator

A

a trans-acting DNA protein that recognizes a DNA sequence in the replicator and activates replication initiation (recruits additional factors)

58
Q

oriC

A

the single E.coli replicator

  1. 9-mer motif: binding site for the DnaA (e.coli initiator protein); 5 repeats
  2. 13-mer motif: AT-rich DNA that is the initial site of DNA winding into ssDNA; 3 repeats
59
Q

DnaA

A

E.coli initiator protein

  1. binds to oriC via 2 separate DNA binding domains. one is used to bind the 9-mer double stranded DNA elements; sequence specific
    * DnaA must be bound to ATP
  2. second domain used to recognize 13-mer repeats. its for ssDNA, resulting in DNA unwinding/strand separation at the 13-mer repeats
60
Q

Recruitment of the replication machinery

A
  1. DNA helicase (DnaB) and the helices loader (DnaC): recruitment of these proteins is mediated by their interactions both with ssDNA and DnaA at the oriC
  2. Helicase recruits primase
  3. DNA pol III comes to the oriC via interactions with helices and the presence of a primer: template junction
61
Q

Recruitment of the replication machinery

A
  1. DNA helicase (DnaB) and the helices loader (DnaC): recruitment of these proteins is mediated by their interactions both with ssDNA and DnaA at the oriC
  2. Helicase recruits primase
  3. DNA pol III comes to the oriC via interactions with helices and the presence of a primer: template junction
62
Q

Completing replication of circular chromosomes

A
  • two daughter DNA moelcules are linked as catenanes

- Type II topoisomerases decatenate/ segregate the two circular DNA chromosomes

63
Q

The end replication problem on linear chromosomes

A

shorter chromosomes formed after the repair of Okazaki fragments

64
Q

Telomeres

A

A specialized DNA polymerase
-RIBONUCLEOPROTEIN: RNA and protein component

  • ends of eukaryotic chromosomes
  • composed of TG-rich DNA
  • include 3’ overhands
  • these special structures recruit Telomerase

Extends the 3’end of a DNA substrate but does not need exogenous template; the RNA component of the enzyme serves as template for DNA synthesis at the telomeres. The newly synthesized DNA (at the 3’end) is single-stranded.

65
Q

TER: Telomerase RNA

A

150-1,300 bases; contains a complementary sequene to the telomere, allowing telomerase to anneal to the sDNA at the 3’telomeric end (partially)

66
Q

TERT: telomerase reverse transcriptase

A

reverse transcribes RNA (from template, TER) into DNA on the chromosome

67
Q

What about the 5’end?

A

okazaki fragment can be placed but telomere extension still results in a 3’ overhang

68
Q

Could telomere length be the cause of these limitations on cell division?

A

–Telomere lengths on chromosomes shorten as people age
–‘normal’ cells have limited telomerase activity
–Stem & cancer cells have high telomerase activity
–Expressing telomerase in normal cells increases their proliferative capacity

69
Q

Can cells only divide a limited number of times?

A

the # of times is characteristic to that cell type

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