Lecture 5 - DNA Replication III Flashcards
How is repliation terminated in eukaryotes?
Torsional stress generated ahead of convergent replication forks must be removed
DNA ahead of the fork gains one positive supercoil for each turn DNA that is unwound. Increased supercoiling makes strand separation more difficult especially when two forks approach each other. Topoisomerases resolve the overwinding by transiently breaking DNA to remove supercoils allowing replication to be completely.
Termination of replication occurs at multiple sites in eukaryotes. When the replication forks converge the CMG complexes move past each other on the leading strand and CMG then transitions to encircling dsDNA. DNA polymerase δ, Fen1 and DNA ligase 1 recruited to complete maturation. The dsDNA remains entwined and this is resolved by a topoisomerase II.
What is the End-repilcation problem?
The problem with replicating a completely linear chromosome (eukaryotes) is that they have ends.
* RNA primer removal on the lagging strand leaves a gap that cannot be filled in
* Dissolution of the replication fork when the leading strand is finished before the lagging strand synthesis is complete - could lead to the loss of the terminal Okazaki fragment.
Both could lead to the erosion of chromosome end over several rounds of replication.
Telomeres protect the ends of linear chromosomes. Telomeres are simple sequence repeats found at the ends of chromosomes. Their length ranges depending on the species, they are G-rich and extends 5’-3’ towards the chromosome end and terminates in a ssDNA tail. A T-loop is formed by base pairing of the overhang with the double-stranded region.
Telomere repeats are binding sites for proteins that mark them as natural ends and distinguish them from DNA breaks - essential for chromosome stability.
What is the function of telomerases?
Telomerase solves the end-replication problem
Telomere sequences are maintained by telomerase, a specialized reverse transcriptase (DNA pol that uses ssRNA as a template to make DNA)
Telomerase is active in tissue-specific stem cells and germline cells. Telomerase synthesises one strand of the telomere - the G-rich strand 5’ to 3’ in a direction away from the centromere.
* Telomerase - has both protein and RNA components
The protein is: telomerase reverse transcriptase (TERT), well conserved in eukaryotes
* RNA component - template for synthesis of repeats and has a role in enzyme activity.
Telomerase elongates the 3’ end of telomeres
1. Telomerase binds to ss telomere DNA
2. 3’ end of DNA base-pairs with the RNA
3. RNA used as a template to synthesize DNA
4. Enzyme translocates and repeats synthesis to elongate the G-rich 3’end
5. The C-rich strand (other strand) is filled in by DNA polα
Somatic issues lack telomerase - telomere shortening occurs over successive cell divisions, therefore the is a finite number of replications. Short telomeres activate the DNA damage response. Telomerase expression re-activated in many cancers.
Why is regulation of initiation important in E.coli? and what are the methods of reguation
E.coli can divide as often as every 20 minutes in good conditions however it takes around 50 mins to replicate the bacterial genome. This is possible because in good growth conditions, a second round or replication is started before the first is complete. Re-replication is well regulated, and initiation at each oriC in a cell is always a power of two.
Regulatory inactivation of DnaA
Methylation status
DnaA sequestration
Recycling of DnaA-ADP to DnaA-ATP
How is regulatory inactivation of DnaA used to regulate initation of replication in E.coli?
- DnaA binds and multimerises at oriC when bound to ATP, and unwinds adjacent DNA
- Helicase/primase/sliding clamp initiate synthesis
- After initiation the sliding clamp stimulates hydrolysis of the DnaA-ATP to ADP
- DnaA-ADP dissociates and cannot reinitiate origin firing, preventing over-stimulation of replication initiation.
- This regulatory inactivation of DnaA (RIDA) blocks replication initiation. (Major regulatory step preventing initiation)
How is methylation status used to regulate initation of replication in E.coli
Methylation status
* 11 GATC sites in oriC overlap the DnaA boxes.
* DNA adenine methylase aka Dam methylase methylates GATC sequences in E.coli genome
* Before replication GATC sequences are fully methylated
* DnaA binds DnaA boxes at oriC when GATC fully methylated (1)
* After replication DnaA boxes are transiently hemi-methylated (2)
* SeqA binds to hemi-methylated GATC sites (3)
§ Prevents DnaA rebinding DnaA box
§ Prevents Dam methylase methylating at GATC
* SeqA binding causes temporary block
* SeqA dissociates – may be replaced by another SeqA , or dam methylase may bind and methylate at GATC (blocking further SeqA binding).
Origin is usually fully methylated 10 mins after initiation. The fully methylated chromosome is ready for DnaA binding.
GATC Sites in oriC: oriC is the origin of replication in the E. coli genome, and it contains specific sequences called DnaA boxes where the DnaA protein binds to initiate replication. There are 11 GATC sites in oriC that overlap with the DnaA boxes.
DNA Adenine Methylase (Dam Methylase): This enzyme, also known as Dam methylase, is responsible for methylating adenine bases within the GATC sequences in the E. coli genome.
Methylation Status Before Replication: Before replication begins, the GATC sequences in the genome are fully methylated by Dam methylase.
DnaA Binding: DnaA binds to the DnaA boxes at oriC when the GATC sequences are fully methylated. This binding is essential for initiating the replication process.
Hemi-Methylation After Replication: After replication, the newly synthesized DNA strands are hemi-methylated, meaning that only one strand of the DNA double helix is methylated while the other is not.
SeqA Binding to Hemi-Methylated GATC Sites: SeqA is a protein that specifically binds to hemi-methylated GATC sites. This binding serves several purposes:
It prevents DnaA from rebinding to the DnaA boxes, thereby inhibiting premature re-initiation of replication.
It prevents Dam methylase from methylating the GATC sites, maintaining the hemi-methylation status.
Temporary Block: SeqA binding causes a temporary block in the replication process, allowing time for the replication machinery to complete the replication of the entire chromosome.
SeqA Dissociation and Methylation Restoration: SeqA eventually dissociates from the hemi-methylated GATC sites. This can occur spontaneously, or it may be replaced by another SeqA protein. Alternatively, Dam methylase may bind to the hemi-methylated sites and methylate the adenine bases, restoring full methylation status.
Fully Methylated Chromosome: Approximately 10 minutes after replication initiation, the fully methylated chromosome is ready for DnaA binding again, initiating a new round of replication.
How can DnaA sequestration help regualte initiation of replication in E.coli?
- DnaA can bind to the ~ 1kbp datA (DnaA titrator) DNA region, adjacent to oriC
- This regions sequesters about 25% of free DnaA in the cell
- Just after replication, the datA region is duplicated, sequestering twice as much DnaA, making it unavailable to rebind to oriC
- After segregation of chromosomes, there is one copy of datA so there is again enough DnaA available to bind to oriC and promote initiation
How does recyling of Recycling of DnaA-ADP to DnaA-ATP aid in the regulation of replication initiation in E.coli?
- DnaA-ATP must be provided for the next round of initiation
- DnaA-ATP is newly synthesized, and DnaA-ADP is recycled to DnaA-ATP
- DnaA binds to DARS regions of DNA (DnaA Reactivating Sequences)
- DARS serves as cofactors to stimulate exchange of ATP for ADP
Why is regulation of initation important in eukaryotes?
DNA replication is tightly controlled in eukaryotes. Chromosomal DNA replication occurs only during S phase of cell cycle. During S-phase all the DNA in the cell must be duplicated exactly once.
* Incomplete replication leads to inappropriate links between daughter chromosomes
* Segregation of linked chromosomes causes breakage or loss
* Re-replication can also have severe consequences.
How is eukaryotic chromosome replication controled?
What is origin timing?
Control of eukaryotic chromosome replication
* Eukaryotic chromosomes are replicated from multiple origins
* Origin firing in S requires MCM loading to establish a pre-RC in G1
* Not every pre-RC will fire in every cell cycle – controlled by origin efficiency and origin timing
* Origin timing relates to when an origin fires ‘early’ or ‘late’
* Origin efficiency is the probability that an origin will fire
* Most origin firing is stochastic – dependent on replication protein availability.
* Replication of unused origins removes preRC
* MCM proteins cannot load in S
Origins are only licensed in G1 and helicases are only activated to unwind after cells transition to S phase
How does proteolysis regulate origin licensing in mammalian cells?
- Origins are selected in G1 (‘origin licensing’) by
○ (1) ORC binding
○ (2) Cdt1/Cdc6 mediated loading of MCM DNA helicase to form pre-RC- Following loading of MCM DNA helicase, Cdc6 and Cdt1 proteins are degraded
- In S phase, helicases activate to initiate origin unwinding – once fired, origins cannot be reused until the next G1
- Cell cycle kinases are involved in restricting initiation steps and origin firing to the appropriate timing
Different eukaryotes have different mechanisms to prevent helicase loading outside of G1
Yeast - Regulation is achieved by exporting proteins from the nucleus – also transcription inhibition and proteolysis for some proteins
Metazoa, the main mechanism is proteolysis of proteins required to form pre-replication complex in S-phase
Geminin provides an additional control layer – binds to Cdt1 and prevents it loading MCM during S, G2, and M phases.
What is requried for replication of chromosomes?
Chromosome replication requires duplication of both the DNA and protein components of chromatin
* DNA is packaged into chromatin by wrapping around nucleosomes (~146 bp DNA wrapped around a histone octamer)
* Nucleosome structure must be faithfully replicated on the two daughter DNA molecules
* Histone modifications must be maintained – epigenetic inheritance
* Synthesis of main histone subunits H2A, H2B, H3 and H4 is tightly coupled to DNA synthesis – replication-dependent histones or S phase histones
How are histone proteins reused?
Parental histones are recycles to newly replicated DNA
* Nucleosomes are disrupted at replication fork by CMG helicase
* Parental histones are actively transferred to daughter DNA – parental histone segregation
* Provides a guide to reconstruction of daughter chromatin with same modifications as were present in parental chromatin.
* Basis of epigenetic inheritance
* Parental H3 and H4 are distributed equally between daughters as tetramers – half go to one daughter, half to the other
* Parental H2A and H2B are disassembled and reassembled on daughters as dimers
* Nucleosome disassembly and reassembly is mediated by histone chaperones
Histone chaperones interact with replication fork proteins to coordinated nucleosome disassembly and reassembly
How are nuclososomes synthesised and modified?
Synthesis and modification of new nucleosomes
* Segregation of parental nucleosomes to daughters results in each daughter having half the required number of nucleosomes
* New histones are synthesized in S-phase
* New H3 and H4 histones are acetylated on specific lysine residues in N-terminal tail
* ASF1 and CAF1 promotes assembly of new H3 and H4 histones into a tetramer and loading onto DNA (via interaction with PCNA) to form a half nucleosome
* NAP1 and FACT add newly synthesized H2A/H2B to loaded half-nucleosomes to form complete nucleosomes
* Histones H3 and H4 are deacetylated
* Parental histones recruit histone modification enzymes to modify new histones (guided by modification on recycled old nucleosomes)
