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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
