Lecture 2 - Exam 2 Flashcards

1
Q

Discuss the vital prokaryotic DNA replication enzymes.

A

DNA pol III: Main enzyme in DNA replication. Synthesizes new strands by adding nucleotides to the 3’ - OH group of the RNA primer. 3’ - 5’ exonuclease activity allows proofreading of DNA strand. 1000 nt/s.
DNA pol I: Removes the RNA primer from Okazaki fragments by 5’-3’ exonuclease activity and fills the gap by adding nucleotides to the lagging strand fragments… It has both 5’-3’ and 3’-5’ exonuclease activity.
DNA pol II: Its main role is in repair and also a backup of DNA polymerase III. It has 3’-5’ exonuclease activity.

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2
Q

As replication in bacteria is bidirectional, the replication forks will….?

A

As replication in bacteria is bidirectional, at some point the two replication forks will meet (about 180 degrees from oriC).

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3
Q

As replication in bacteria is bidirectional, at some point the two replication forks will meet (about 180 degrees from oriC). This phase of replication is unique to?

A

This phase of replication is unique to termination and is defined as replication fork “convergence.”

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4
Q

Termination occurs in the…?
What does this place contain?

A

“Termination zone,” which contains multiple Ter sites and bind to Tus proteins on the replisome.

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5
Q

Replication forks can meet…?

A

Anywhere within the termination zone, either at a Ter site or between Ter sites.

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6
Q

Eventually, the two replicating forks will meet somewhere 180 degrees from the oriC, and will ____________ replication.

A

Terminate

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7
Q

There are multiple _________ sites (______ sites) on the bacterial chromosome. E. coli has how many?

A

Termination ; Ter
E. coli has 6.

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8
Q

What are Ter sites?

A

22-bp sequences that binds to specific proteins called TUS proteins (Termination Utilization Substance). Ter sites serve as binding sites for Tus protein. Ter sites only allow passing of the replication fork in one direction.

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9
Q

Ter sites are divided into?

A

Two groups of three (or more) termination sequences:
terA, terD, and terE prevent counterclockwise replication.
terC, terB, and terF prevent clockwise rotation.

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10
Q

These termination recognizing sequences (Ter sites) create a “_____” that the replication fork can enter, but can’t leave.

A

Trap

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11
Q

The one way travel is imposed at Ter sites by?

A

The Tus protein which bind to Ter sites and inhibit the helicase (DnaB) activity.
When Tus binds to Ter site, a complex is formed that locks the DNA strands together and prevents DnaB helicase from further separating the DNA.

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12
Q

What happens now after replication has been terminated? How is the replication of chromosomes completes? How are the two pieces of replicated DNA joined?

A

Completion of replication involves:
1) A short-term over-replication of the region where the replication forks converge.
2) The excess regions are then cut by RecBCD proteins, and the strands are joined together with DNA ligase.

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13
Q

Completion of replication therefore requires _____________ needed to repair _____________, part of the SOS response that is also used to repair DNA damage.

A

Several proteins (RecBCD) ;
double-strand DNA breaks

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14
Q

What are SbcC, SbcD, and Exol?

A

Highly conserved nucleases required for genomic stability-enzyme process DNA intermediates at sites where replication forks converge.

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15
Q

What does the process of overreplication of replication forks look like?

A
  1. Replication forks continue past their meeting point, creating a palindromic substrate.
  2. SbcCD-Exol cleave and process the overreplicated intermediate.
  3. RecBCD-mediated resection and joining of the DNA ends completes replication.
  4. In vitro and proposed in vivo substrates for SbcCD, respectively.
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16
Q

What are the models of DNA replication?

A

Factory model: DNA polymerase is stationary in the cell and the DNA is pulled through it (pulled through the polymerase).
Train on track model: DNA polymerase moves along the DNA strands like a train moving on the track.

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17
Q

More recent data tends to support which model of DNA replication?

A

Train on the track.

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18
Q

Under optimal growth conditions for E. coli, cell division occurs more rapidly than…?
What does this require?

A

Genome replication.
This requires a new round of genome replication to begin in the newly replicated chromosome before the previous round has been completed.

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19
Q

Under optimal growth conditions for E. coli, cell division occurs more rapidly than genome replication.
This requires a new round of genome replication to begin in the newly replicated chromosome before the previous round has been completed.
However, there must be only one…?

A

There must be only one replication per cell cycle.

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20
Q

To ensure that initiation at an oriC occurs only once per cell cycle, what happens?

A

Specific mechanisms exist to control chromosomal replication. This control is primarily mediated by DNA methylation.

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21
Q

DNA methylation is a _____________ that acts by the ______________ from an S-adenosyl-methionine molecule to an ______ or a ______ in the DNA.

A

Base modification system ;
Addition of a methyl group ;
Adenine ;
Cytosine

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22
Q

DNA methylation plays vital roles in…?

A

DNA replication and repair.

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23
Q

Enzymes responsible for DNA methylation are called?

A

DNA-methyltransferases (MTases)

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24
Q

“Mature” DNA is methylated (primarily) at the?

A

N6 positions in adenines found in the sequence 5’ GATC 3’.

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25
Q

Is there a time where DNA is not methylated?

A

Yes, there is a short amount of time when the newly synthesized DNA is NOT methylated. During this period, the DNA is referred to as hemimethylated, because the parent strand is methylated and the daughter strand is not.

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26
Q

There are ______ GATC sequences in the oriC locus, so this area initially has?

A

Multiple ;
This area initially has a lot of unmethylated DNA following replication.

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27
Q

What is the protein called that binds to the hemimethylated oriC DNA?

A

A protein called SeqA binds to this hemimethylated oriC DNA and “sequesters” it, delaying DNA methylation and recognition and binding of DnaA.

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28
Q

DNA damage is a common occurrence in all cells. A bacterial cell growing in an aerobic environment will suffer 3000-5000 DNA lesions per cell per generation. Most of this damage is repaired by…?

A

DNA repair mechanisms and do no impact replication.
This is the reason why the replication forks are not meeting perfectly at 180 degrees from oriC and meeting at the same time, because they will break down during replication, as DNA damage is common in cells.

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29
Q

When a DNA polymerase encounters an unrepaired DNA lesion, it is unable to do what?

A

It is unable to bypass the DNA damage and the DNA Pol breaks down and the replication fork stalls. Under normal cellular growth conditions, nearly every bacterial replication fork will suffer this fate.

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30
Q

What does a stalled replication fork trigger?
What is this process called?

A

An elaborate enzymatic response, ultimately restoring an active replication fork without introducing a genetic mutation. This process is called homologous recombination.

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31
Q

Homologous recombination repair mechanisms involve _____ enzymes and is _____.

A

Multiple ;
Complex

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32
Q

In a nutshell, what is homologous recombination?

A

Homologous recombination (HR) repair uses a complex “cross-over” mechanism to faithfully repair the damaged DNA. It uses the methylated strand to know where to go in and repair the DNA without a mutation.

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33
Q

Typically, homologous recombination yields…?

A

Two complete chromosome monomers when replication is terminated.

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34
Q

Sometimes there are unequal numbers of crossover events. What happens?

A

Sometimes, when there are unequal numbers of crossover events, homologous recombination causes the two replicated chromosomes to be connected, resulting in a dimeric genome (covalent chromosome dimers)…
At the end of replication, you won’t have two separate chromosomes, will have a big ass chromosome dimers because of unequal crossovers.

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35
Q

What is chromosome separation?

A

Detachment of the newly replicated chromosomes from one another.

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36
Q

What is chromosome partitioning?

A

The segregation of sister chromosomes to opposite halves of the cell prior to cell division.

37
Q

Once replication has stopped at the Ter region, the two daughter chromosomes are…?

A

The two daughter chromosomes are entangled and must be separated or there is no cell division.

38
Q

How do cells solve the problem of the two daughter chromosomes being entangled once replication has stopped at the Ter region?

A

Cells use topoisomerase and the XerCD system to separate sister chromosomes.

39
Q

What are the two ways that chromosome separation can take place?

A
  1. If the linkage between the daughter cells is noncovalent (catenane) then then the cells are separated using topoisomerase IV.
  2. If an unequal number of recombinations have occurred between sister chromosomes, the sister chromosomes are covalently linked, and a site-specific recombination must occur to separate the chromosomes.
40
Q

What are catenanes?

A

Interlinked duplex DNA molecules formed during each round of replication.

41
Q

What are covalent chromosome dimers?

A

Formed by HR-dependent (homologous recombination dependent) crossover between sister chromosomes.

42
Q

Chromosome dimers can result from…?
How are they resolved?

A

Chromosome dimers can result from crossing over during homology-directed repair and are resolved by site-specific recombinase XerCD acting at the dif site, also linked with cell division.
Note: This type of linkage is less common.

43
Q

What are catenated chromosomes?

A

After replication, the two completed chromosomes are non-covalently linked together as two independent connected circles.
These must be separated for each chromosome to move to a daughter cell.

44
Q

What is decatenation?

A

The process of separating catenanes.
Decatenation separates the chromosome using topoisomerases.

45
Q

What are topoisomerases?

A

Enzymes that can break DNA strands (remember DNA gyrase that affects DNA supercoiling)

46
Q

Which topoisomerase is used for decatenation?

A

Topoisomerase IV.
Topoisomerase IV breaks the dsDNA of one chromosome and allows the other chromosome to pass through the break, then seals the break, separating the two chromosomes.

47
Q

Where is Topo IV brought to for decatenation and how?

A

Topo IV is brought to the decatenation site (a 28 bp locus called dif site, within the Ter region) by interaction with two recombinase enzymes, XerC and XerD, which interacts with the cell FtsZ ring (remember, FtsZ is the tubulin-like polymer ring that constricts to divide cells).

48
Q

During covalent chromosome dimer separation, what are the covalent chromosome dimers separated by?

A

Topoisomerase IV will not be able to separate the two chromosomes. Covalent chromosome dimers must be separated by site-specific recombination using enzymes called recombinases.
The recombination occurs at a 28 bp locus called dif within the Ter region.

49
Q

Covalent chromosome dimer site-specific recombination is catalyzed by…?

A

Two recombinases, XerC and XerD.

50
Q

Discuss the roles of XerC and XerD.

A

The XerC and XerD recombinases bind cooperatively at a 28 bp recombination site dif which is located close to the replication terminus.
XerC and XerD recombinases cut DNA at the dif site to resolve the chromosomal dimer into two chromosomal monomers.
XerC and XerD also act as “chaperones” to bring topoisomerase IV to the dif site during decatenation.
Note: XerC and XerD is not involved in cutting the DNA, it is responsible for drawing topoisomerase to the dif site to cute catenanes.

51
Q

Chromosomal partitioning and replication are _________.

A

Concurrent.

52
Q

In bacteria, there is no ________ apparatus involved in chromosomal partitioning as occurs in eukaryotes.

A

Spindle.

53
Q

If bacteria do not have a spindle apparatus, what enables the sister chromosomes to move to the opposite poles of the cell?

A

Not really a great idea, but maybe the Min system.

54
Q

Describe the chromosome partitioning process in bacteria.

A

Entropic repulsive forces move chromosomes apart, while the Min system creates an oscillating gradient of chromosome tethering sites. This oscillation keeps the two pieces of chromosomes pushed towards the sides of the cells.

55
Q

Previously, the hypothesis for chromosome segregation was that something was ________ the chromosomes toward opposite poles.

A

Pushing

56
Q

The researchers that demonstrated the Train on Track model also found what?

A

Also found that not only did the two replisomes move independently, but also that the movement of the newly replicated DNA appeared to be driven by redistribution of DNA mass from strongly confined towards less confined space (the opposite pole of the elongating cell)

57
Q

Timing of DNA replication: Under conditions when nutrients are lower, there is…?

A

There is only one round of DNA replication occurring at one time. This could be seen at stationary and lag phases.

58
Q

When nutrients levels are high and the environment is conducive to growth, there is…?

A

There is multiple rounds of DNA replication that will occur simultaneously. This could be seen at exponential phase.
Cells grow rapidly, always replicating their DNA.

59
Q

What are some of the different mutations that can arise?

A

Point mutations, insertions and deletions, inversion: flipping, and reversion.

60
Q

What is a missense point mutation?
What is a nonsense point mutation?

A

Missense point mutation: Put wrong amino acid there, not going to stop replication. Introduces a mutation into DNA.
Nonsense point mutation: Introduces a stop codon. Going to stop the transcription of the gene.

61
Q

What is an insertion frameshift?

A

Insert two bases into a codon that shifts the reading frame and changes all the downstream amino acids. Everything downstream will be the wrong amino acids.

62
Q

What is a deletion frameshift?

A

Removes two nucleotides from the codon, causing a frameshift. It changes everything downstream (since codons are read in threes).

63
Q

What is an inversion mutation?

A

Flips a DNA sequence. The 3’ codon will be put into the 5’ strand AND it will be backwards. Same with the 5’ strand codons.

64
Q

What are some of the mechanisms of mutation?

A

Spontaneous mutation: errors in the replication of DNA
-Tautomeric forms of bases (amino vs imino and keto vs enol)
-Change in bonding properties

65
Q

The newly synthesized DNA must be a _______ copy of the template strand (THE MOTHER STRAND!!!)?

A

Precise

66
Q

The newly synthesized DNA must be a precise copy of the template strand. What does this involve?

A

This involves the specificity of the DNA polymerase in selecting nucleotides that are correctly aligned. Essentially, you are relying on the polymerase to pick the right nucleotide to pair to the base.

67
Q

The fidelity of replication is enhanced by…?

A

A second exonuclease function of DNA polymerases: The 3’ to 5’ exonuclease activity, which is able to remove the nucleotide at the growing end (3’ end) of the DNA chain.

68
Q

The DNA polymerase will only extend the DNA chain by adding nucleotides to the 3’ end, but is contingent on…?

A

Will only extend the DNA chain if the last base at the 3’ end is correctly paired with the template strand. If it is not correct, polymerization will stop and the 3’ to 5’ exonuclease function will remove the incorrect nucleotide, allowing a further attempt to be made.

69
Q

What are the benefits and drawbacks of proofreading?

A

Benefits: Reducing the rate of spontaneous mutation, increasing fidelity.
Drawbacks: Extensive error-checking slows down the rate of replication.

70
Q

The balance between the rate of replication and the extent of error-checking will be determined by..?

A

The nature of the DNA polymerase itself.
-Some DNA polymerases do not show efficient proofreading and therefore result in a much higher degree of spontaneous errors.
-The rate of spontaneous mutation shown by an organism is therefore (at least in part) a genetic characteristic that is subject to evolutionary pressure.

71
Q

The fidelity of replication is further enhanced by…?

A

DNA repair mechanisms.

72
Q

Mutagens can cause?

A

Mutations

73
Q

What are some examples of mutagens?

A

Chemical agents:
Base analogs, base modifiers, and intercalators
Electromagnetic radiation:
X-rays & gamma rays -> break the DNA
and ultraviolet rays -> form pyrimidine dimers

74
Q

What is an example of a strong mutagen?

A

Ethidium Bromide
It slides in between the bases of the DNA and allows you to visualize them BUT causes issues in women of reproductive age.

75
Q

What is DNA repair?

A

A collection of processes by which a cell identifies and corrects damage to the DNA molecules that encode its genome:
1. Methyl-directed mismatch repair
2. Nucleotide excision repair
3. Recombination repair
4. SOS repair

76
Q

What is methyl-directed mismatch repair based on?

A

It is based on recognition of the methylation pattern in DNA bases.

77
Q

In E. coli, most of the ____ positions in adenines found in the sequence ________ are ________ by DNA adenine methyltransferase.

A

N6 ; GATC ; methylated

78
Q

Will newly synthesized DNA be methylated?

A

NO

79
Q

Methyl-directed mismatch repair uses…?

A

Uses methylation of the parental strand to discriminate from newly replicated DNA.
The premise is that the parental strand will contain the proper DNA sequence.

80
Q

E. coli MMR requires multiple protein components. What are they?

A

MutS, MutL, MutH.
MutS, MutL, and MutH initiate MMR and play specialized biological roles in MMR in E. coli.

81
Q

Describe the Methyl-directed mismatch repair (MMR) process briefly.

A
  1. Base mismatches causing DNA distortion
    -MutS, the “mismatch recognition protein” bind the mismatched DNA and recruits MutL and MutH to the site.
  2. MutL binds the adjacent unmethylated GATC site.
  3. The MutHLS complex causes a loop to form in the DNA.
    -MutH cleaves the unmethylated strand 5’ to the GATC.
  4. Exonuclease then removes the damaged DNA strand.
  5. DNA polymerase synthesizes a new strand.
82
Q

Ultraviolet radiation causes what?
What does the product of this ultraviolet radiation activate?
Describe what happens during this process.

A

The formation of pyrimidine dimers that distort the shape of the double helix.
The distortion activates nucleotide excision repair, initiated by an endonuclease.
The endonuclease cuts the DNA strand on either side of the damage.
This cut exposes a 3’ OH group ; this can be used as a primer by DNA polymerase I to replace the short region of DNA between the nicked sites.
The final step is the joining of the newly repaired strand to the existing DNA, by DNA ligase.

83
Q

DNA damage that interferes with base-pairing will prevent replication, due to the requirement of the DNA polymerase for an accurately paired 3’ end. The replication fork pauses. It is possible to restart replication beyond the lesion, thus _________ in the newly synthesized strand.

A

Leaving a gap

84
Q

With wrong base-pairing mistakes, a gap can be left in the newly synthesized strand. What is this gap filled with? What is the process that mediates this?

A

The gap can be filled using a portion of DNA from the other pair of strands, by a recombination process mediated by RecA (RecA binds ssDNA and mediates homologous recombination).
The original damage can now be repaired by nucleotide excision repair, while the gap in the other DNA molecule can be filled in by DNA polymerase I and DNA ligase.

85
Q

What is the SOS repair induced by?

A

Extensive DNA damage.

86
Q

Describe the SOS repair mechanism.

A

RecA co-protease activity stimulates autodigestion of the LexA repressor.
Expression of many DNA repair enzymes.
-Among them, 2 “sloppy” DNA polymerases that lack proofreading activity.
However, the cell has no other option: “Mutate or die”

87
Q

A protein called _____ acts as a _____ for the SOS system – this keeps the system off unless needed.

A

LexA ; repressor

88
Q

_____ is an activator of the SOS system and a ______amount of it is always produced by the cell.

A

RecA ; small

89
Q

Discuss the activation of the SOS response system.

A

Activation: DNA damage causes accumulation of ssDNA.
RecA binds to ssDNA and activates the protease function that degrades the SOS repressor, LexA.
-Degradation of the repressor initiates expression of the SOS genes.
SOS proteins inhibit cell division, maintain (low fidelity) replication, and function in nucleotide excision repair.