Replicating the Genome Flashcards
Review the basic molecular mechanism of DNA replication Explain why cell cycle control is essential for genome maintenance Describe replication origins & their control Explain how cyclin dependent kinases control S and M-phases Describe the topological problems of DNA replication & their solutions Describe the problem of DNA end-replication & how telomerase acts
Describe the basic molecular mechanism of DNA replication
DNA synthesis is in S-phase by the replisome - a protein complex including DNA Polymerase which adds free nucleotides in the 5’ to 3’ direction to a RNA primer made by DNA primase. DNA Helicase unwinds the double stranded helix. The leading strand is the one replicated continuously, the lagging strand is replicated in Okazaki fragments made by multiple primers which are degraded and the fragments joined by DNA ligase.
Why are their controls in S and M phase and what do they regulate?
DNA Polymerase has high fidelity. The controls ensure that the genome is properly passed on, S and M phase are tightly controlled. The controls ensure that the chromosomes are copied once per S phase. M phase happens only when S phase is complete. S phase is always preceded by M phase.
What can happen if these controls fail?
If the controls fail, the cells can develop aneuploidy which means missing, extra, or rearranged chromosomes. In mitotic cells this compromises cell function/survival, which can lead to mutations, and then cancer. In meitotic cells, the gametes will be aneuploid, leading to sterility, or the offspring to have mental defects or other genetic diseases.
When is it acceptable for the cell cycle to be different? Give some examples.
In certain specialist cells/situations. Meiotic cells have 2 consecutive divisions with no S phase in order to half the chromosome number. Megakaryocytes have repeated S-phases in order to become polyploid and eventually rupture. Senescent cells go into G0 after certain signals. It can be reversible or permanent.
What are replication origins and what is their purpose in the genome?
For prokaryotic cells, DNA replication begins at a single point in the circular chromosome and two replication forks form which travel in opposite directions, duplicating the whole genome. To avoid taking 200 days to replicate, the mammalian genome has multiple replication origins - 1/70kb.
Explain origin licensing
Origin Recognition Complex has subunits 1-6 and serves as the landing pad for DNA synthesis proteins. Animals don’t have conserved sequences at origins unlike bacteria and simple eukaryotes. A Pre-Replicative complex is assembled in G1. Cdc6 and Cdt1 recruit the replicative helicases (inactive) to the ORC.
Explain origin activation
Precedes S-phase. Multiple initiation and replication proteins are recruited and phosphorylated (by Cyclin dependent kinase). Also, Cdk phosphorylates ORC, Cdc6 and Cdt1 in licensing. Helicases activated by Dbf4-dependent kinase (DDK). After S-phase, G2 is entered.
What happens if origin licensing and activation occur at the same time? How is this stopped?
There would be unlimited replication as each bubble would become two bubbles etc. Evidence shows that DNA re-replication -> impaired proliferation, DNA damage, cell death/oncogenesis/. Cdk inhibits ORC, Cdt1 and Cdc6 by phosphorylation. Having the same kinase activity both inhibit licensing and promote firing means that only one can happen at a time. In G2 and until M-phase, relicensing is suppressed by geminin (builds up in S and G2) which binds and inactivates Cdt1.
Explain Cdk and the two different types
They are serine-threonine protein kinases - transfer a phosphate from ATP onto ser or thr residues in their protein substrates. This either activates or inactivates the protein. They must be bound to cyclin to have any activity. The two different types are M-Cdk and S-Cdk. M-Cdk’s substrates are key mitotic proteins (involved in nuclear envelope breakdown, chromosome condensation).
What are the two methods of activity of S-Cdk and M-Cdk and which one is correct?
The qualitative model dictates that S and M cyclins accumulate in S and M phases. Therefore, M cyclins binding Cdks make them have substrate specificity for mitotic proteins (making them M-Cdks) and same for S-Cdk.
The quantitative model dictates that as the cyclin concentration increases, different substrates are acted on. As cyclin (and Cdk activity) accumulates, two thresholds are reached. First at G1/S boundary (S-Cdk) and the second one at G2/M boundary (M-Cdk). The models are not mutually exclusive.
How does origin licensing occur after origin activation if Cdks inhibit origin licensing?
The cyclin is proteolysed, reducing Cdk activity until it can no longer inhibit licensing. APC/C tags the cyclin with ubiquitin, marking them for degredation. In S, G2, early M phases, Cdk activity inhibits APC/C but in late mitosis, M-Cdk activity activates APC/C, degrading cyclin and cohesin (allows chromosome segregation) and geminin (origin licensing inhibitor). Cdk activity is inhibited to cause cell cycle arrest.
What are the topological problems associated with DNA replication?
DNA can’t be unwound indefinitely because it causes overwinding and places topological stress on the DNA ahead of the fork. To relieve this stress, supercoils form ahead of the fork and pre catenanes (intertwined sister chromatids) form behind.
How are the supercoils and precatenanes resolved?
Enzymes called Topoisomerases relieve topological stress. Type I Topoisomerase cleaves and reseals single strands in the supercoil, unwinding it.
Type II Topoisomerase cleave both strands of supercoils or precatenanes reducing the stress.
What are the effects of Top2 deletions and how can it be inhibited in a therapeutic manner?
Top2 deletions cause entangled sister chromatids at anaphase and abnormal chromosome segregation -> cell death. Top2 inhibitors cause double strand breaks (inhibiting its resealing ability). Used as a chemo as cancer cells are more sensitive to DNA damage than normal cells. Top2 inhibitors may be present in food at low concs as carcinogens.
Define and explain the end replication problem
The 5’ end of the lagging strand is not completely replicated because once the primers are degraded the gap at the end can’t be filled with DNA as it needs a primer to build from and the strand is finished so where would the primer go? The telomeres (end of chromosomes) become shorter with each S phase.
