Lecture 19: DNA Replication and Repair Flashcards

1
Q

What are the 6 problems that need to be solved during DNA replication?

A

(1) Strand polarity
(2) Unzip DNA
(3) Processivity
(4) Untangle
(5) One copy
(6) Accurate copy

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2
Q

Replication

A
  • Is semi-conservative
  • > Means that each DNA strand is used as a template for the synthesis of a complementary strand
  • Damaged DNA, if unrepaired, can persist through cell divisions.
  • Errors in copying, if unrepaired, are propagated through cell divisions
  • Need to make sure that there’s no errors in any strands so these don’t happen
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3
Q

Replication #2

A
  • Occurs in the 5’ to 3’ direction
  • New nucleotides are added at the 3’ end.
  • Chain growth occurs in the 5’ to 3’ direction.
  • Addition of deoxynucleotides requires a primer (3’ hydroxyl group).
  • This mode of replication results in the antiparallel double helix structure
  • Since the DNA is being synthesized in the 5’ to 3’ direction, this means that the DNA is being read in the 3’ to 5’ direction
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4
Q

DNA polymerase

A
  • What DNA replication is catalyzed by
  • An enzyme with fingers (grips the DNA), palm (where catalysis happens) and thumb (holds everything in place) domains
  • All nucleic acid polymerases have a similar structure
    5’ to 3’ polymerization occurs with an error rate of 1 in 105 nucleotides. Other processes reduce the error rate further
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5
Q

Solution to strand polarity

A
  • All synthesis is 5’ to 3’
  • Because DNA is antiparallel, one strand can be synthesized continuously (leading strand) while the other is synthesized in Okazaki fragments (lagging strand)
  • Okazaki fragments are synthesized on the lagging strand.
  • DNA Ligase seals gap between successive fragments
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6
Q

Solution to unzip DNA

A
  • A hexameric (6 identical subunits) complex called DNA helicase unzips the DNA.
  • It uses ATP and acts as a rotary engine.
  • Unzipped DNA is stabilized by single-stranded DNA binding protein
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7
Q

Solution to processivity

A
  • A sliding clamp holds the DNA polymerase in place
  • The clamp is loaded on DNA by a clamp loader that uses ATP hydrolysis to lock the clamp around DNA
  • Helicase is attached to the lagging strand, not the leading strand
  • Lagging strand is bent around so both molecules of DNA polymerase are synthesizing in the same direction
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8
Q

Rapid rotation of DNA is needed ahead of the replication fork

A
  • The rapid rotation of the DNA introduces torsional stress on the DNA molecules
  • Built up torsional stress leads to supercoils
  • Topoisomerases relieve torsional stress
  • Solution to Untangle
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9
Q

Nick and swivel mechanism

A
  • Topoisomerase I has a tyrosine in its active site which contains a hydroxyl group
  • The hydroxyl group performs a nucleophilic attack on a nucleotide and replaces the hydroxyl group on the nucleotide, which breaks the phosphodiester bond between the adjacent nucleotide, creating a nick
  • The strand with the nick can now swivel around the opposite strand, which relieves torsional stress
  • Once the torsional stress is relieved, the 3’ hydroxyl group next the nick attacks the benzene ring on the tyrosine, which causes Topoisomerase I to detach from the DNA
  • The phosphodiester bond then spontaneously reforms and the DNA is restored
  • Since bond energies can just be transferred back an forth, Topoisomerase I does not require ATP to relieve strain
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10
Q

Gating mechanism

A
  • Topoisomerase II (mirror-image enzyme) requires ATP to untangle DNA
  • The enzyme attaches covalently to both strands of one DNA helix, creating a gap in the DNA helix.
  • The other DNA helix is then passed through the gap and the broken DNA helix is then reattached
  • There is a loss of energy in the reversal of the covalent attachment, so ATP is needed in order to restore it
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11
Q

Where does DNA replication begin?

A
  • Begins at replication origins
  • A replication origin is a region of very high AT content (2 hydrogen bonds vs 3 for GC bonds, so easier to break)
  • Both directions are synthesized simultaneously, so both directions have a leading strand and a lagging strand
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12
Q

Replication origins in prokaryotes

A
  • For prokaryotes, there a replication origin and initiator proteins bind to the replication origin first, which allows the loading of the DNA helicase
  • DNA helicase will then start unwinding the DNA in one direction and another DNA helicase will be attached to the DNA and start unwinding in the other direction
  • It is important to have a single round of replication every time the cell divides in order to prevent the accumulation of DNA
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13
Q

Prokaryotic origins have a refractory period

A
  • This means that once replication is initiated at the origin, it can’t be initiated again for a little while
  • Methylation of adenines on the parent strand at the replication origin regulates DNA replication in prokaryotes
  • Once the daughter strands are finished synthesizing, they too are methylated, and once they’re methylated, then the cell can start replication again
  • Methylation is mediated by the Dam methylase in prokaryotes
  • Solution to One copy, but this only works for prokaryotes
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14
Q

When do eukaryotes replicate DNA?

A
  • During S-phase of cell cycle
  • Eukaryotic DNA replication is more complex than in prokaryotes
  • Eukaryotes have large genomes that need to be replicated
  • Eukaryotes have many origins of replication
  • Replication of DNA needs to occur once & only once per cell cycle
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15
Q

Not all eukaryotic origins are used in all S phases

A
  • Green replication origins are used most of the time, while red replication origins are used less of the time
  • Multiple origins of replication allows the genome to be replicated more rapidly
  • ORC-binding site: binds to the origin of replication complex
  • Unwinding region: where the DNA helicase binds
  • Auxiliary protein binding site: where additional proteins bind
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16
Q

One origin fires once during each cell cycle

A
  • A pre-replicative complex, which includes a helicase, assembles on the origin in the G1 phase (prior to S phase)
  • Pre-RC is held inactive by proteins until phosphorylation by a cell division kinase (cdk) during S phase
  • Solution to One copy, but only works for eukaryotes
17
Q

One origin fires once during each cell cycle #2

A
  • Cdc6 is phosphorylated which causes it degradation and then Cdt1 is released
  • Activation by Cdk during the S phase eventually results in the phosphorylation of the Origin Recognition Complex and activation of the helicase
  • Once replication has completed, the phosphorylated ORC keeps the origin inactive until after the M phase
18
Q

Eukaryotes need to replicate histones

A
  • H2A and H2B are completely removed from the DNA during replication
  • H3-H4 tetramers are randomly assorted between the two strands that are being replicated
  • This means that the new strands get half of the original H3-H4 and the H2A-H2B are reloaded in the new strands
19
Q

How are the histones added to the new strands?

A
  • H2A-H2B dimers and new H3-H4 tetramers are loaded using histone chaperones
  • CAF-1 loads newly synthesized H3-H4 dimers
  • NAP-1 loads the old H2A-H2B dimers
20
Q

When are histone modifications reestablished?

A
  • After DNA replication
  • The reader-writer complex modifies the nucleosomes that aren’t modified that are attached to the new strands of DNA to match the parental modifications
  • Re-establishment of histone marks after cell division is crucial for the maintenance of gene expression programs
21
Q

What can the “end-replication” problem lead to?

A
  • Can lead to progressive DNA loss from the ends of linear chromosomes
  • The 3’ end of the lagging strand has a gap after replication before modification because there’s no RNA primer for the DNA polymerase to attach to in order to replicate the 3’ end
22
Q

Telomerase

A
  • Prevents linear DNA ends from being lost during replication
  • The ends of linear chromosomes have repetitive sequences complementary to telomerase RNA
  • Telomerase uses its own RNA as a template the replicate the 3’ end of the lagging strand of DNA
  • It’s a DNA polymerase, so it has the same fingers, palm, and thumb structure
23
Q

What 3 main mechanisms do cells use to achieve high fidelity of DNA replication?

A
  • 5’ to 3’ polymerization (error of 1 in 10^5 bases)
  • 3’ to 5’ exonucleolytic proofreading (error of 1 in 10^2 bases)
  • Strand-directed mismatch repair (error of 1 in 10^3 bases)
  • Total error is 1 in 10^10 bases
  • Solution to Accuracy
24
Q

DNA polymerase can proofread

A
  • Lack of base-pairing by a mismatched nucleotide prevents extension by the polymerase.
  • This dependence of perfect base-pairing of the -1 base is one reason why DNA polymerase can only extend a base-paired primer
25
Q

Strand-directed mismatch repair

A
  • Can fix replication errors
  • The newly synthesized strand is recognized by the presence of an unsealed nick (recognized by MutL)
  • Re-synthesis after elimination of a section of the mismatched strand fixes the error
26
Q

What are the 3 types of ways that somatic DNA can be damaged?

A
  • Oxidation
  • Hydrolysis
  • Methylation
27
Q

Depurination

A
  • Accidental removal of a base
  • One of the most common forms of DNA damage
  • Hydrolysis can cause this
28
Q

Deamination

A
  • Removal of an amine group
  • One of the most common forms of DNA damage
  • Hydrolysis can cause this
29
Q

DNA glycosylases

A
  • Recognize individual damaged bases by base flipping
  • Can remove unnatural DNA bases
  • 5-methyl cytosine is about 3% of cytosines in vertebrate DNA
30
Q

Nucleotide Excision Repair

A
  • UV-irradiation can generate pyrimidine dimers that can be fixed by this
31
Q

Base Excision Repair

A
  • Results in the surgical removal of the damaged base followed by DNA synthesis and ligation
32
Q

Translesion polymerase

A
  • Can “replicate” damaged DNA
  • There are 7 in humans
  • They are recruited when replicative DNA polymerase encounters a DNA lesion.
  • They can typically add only one to a few nucleotides.
  • The added nucleotides do not follow Watson-Crick base pairing and are just good guesses
33
Q

Double-stranded breaks can also be repaired

A
  • 2 ways that double-stranded breaks can be repaired
  • Nonhomologous end joining: Can occur at anytime
  • Homologous recombination: Can only occur after DNA replication but before cell division