Current LectureMCB 250 1st Edition Lecture 12Outline of Current Lecture 1. Replisome machine2. Starting3. Regulation4. Control5. DNA replication in Eukaryotes DNA replication6. End Replication in Eukaryotes7. TelomeresCurrent Lecture1. Replisome machine:- Pol III holoenzyme has 3 core polymeraseso One leading strand synthesiso 2 others trade off on the lagging strand to increase the speed of the synthesis of the lagging strando New primers, clamp loader loads camp, core binds to it and synthesizes the Okazaki fragment- Replisome travels around the cell very fast: 1000 nucleotides per second.- If you grand two strands of the kite string and unwind it: the front will become overwound and have knots (+ supercoiled) at the front of the Replisome and the knots. -> Topoisomerases prevent positive supercoil from building up in front of the Replisome. Without it, the force of the supercoil will stop DNA replication. - Type I and II can act. In E. Coli, gyrase relaxes- Topoisomerase/ gyrase is not part of the replication machine that’s traveling down DNA. Supercoils build up at front, gyrase acts all over cell/DNA. It relieves and acts constently the postitive supercoil2. Starting- E. Coli is single, covalently closed circle. Single origin of replication that allows us tobuild two replication forks and two machines- Single initiation site = oriC. It is recognized by DnaA- DNA A must be bound to ATP, the ATP form bound to DNA a acts as an initiator protein to start the process.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- DNA a + ATP starts replication that is regulated- OriC is a sequence stretch of DNA in the E. Coli chromosome ~250bp long- 9-mer sites are specific sequences: for DNA-a- AT rich region next to the 9-mer site- Dam methylation sites in DNA: DNA methylation in the region- DNA-A is going to bind to the particular site. Causes AT rich DNA to unwind. - DNA binds and gets wrapped around DNA complex3. Regulation- Control DNA replication so there is at least 2 chromosomes to divide to daughter cells- Want the oriC to initiate replication at controlled circumstances. - DAM!o Recognizes GATCo Methylates Ao It will recognize any GATC site in the chromosomeo You are most likely to find GATC in every 256 bpo Methylation is slower than replicationo After replication fork to pass, DAM methylates new DNA aftero Old DNA is fully methylated, but newly synthesized DNA is only hemi-methylated (old is methylated, but new is not) DAM hasn’t gotten around to it yeto Tells you it is a new piece of DNA that was recently synthesized and tells you which one is the old/template strand. o If polymerase did make a mistake and there is a mismatch, how do you know which one to throw out? The newly piece should be thrown out and keep the oldone.4. Control- Rate limiting step in initiation of DNA replication is the binding of DnaA-ATP to OriC- The concentration of DnaA-ATP is tightly controlledo Amount of DnaA is proportional to cell mass: old cells have more of it, new cells have less of it. New cells are just replicated and you may not want to do it againo As replication fork/helicase passes, kicks DnaA off of DNA due to the hydrolysis ofATP to ADPo Now have DnaA-ADP must lose and bind to new ATP, takes a while to recycle DnaA to DnaA-ATP- There are 11 GATC sites in oriC (250 bp)o On random would find GATC maybe once in the origin; instead, there is 11 of there. There is enrichment of GATC DAM methylation sites in this sequenceo Dnaa-ATP only recognizes fully methylated replication siteso Once the fork passes, there is hemi-methylated DNA (one old and one new). Thatis not a target for DnaA-ATP.o SeqA protein binds the same site as DNA but only hemimethylated siteso SeqA binds the orgin only immediately after fork passes. This “sequestures” the origin away from the DAM and DNA-Ao Blocks DAM from coming in to methylate the DNA; keeps it hemimethylatedo SeqA isn’t bound covalently to the DNA it comes on and off. There is competition between DAM and SeqA bindingo At some frequency, DAM is able to come in and methylate it, but it is a slow processo If that happens, seqA wont recognize it (because it recognizes hemi-methylated)o If there is enough DnaA-ATP you can rebind and reinitiate synthesis (40 minutes passed and synthesized chromosome)o DnaA-ATP levels are controlled, converts DnaA-ADP to bind to new ATP again. Slowly methylated and slowly build up Dna-ATP again and initiate another round of replication5. DNA replication in Eukaryotes- Replisome in eukaryotes is functionally equivalent to E. coli- Eukaryotes are more complicated- All the proteins have a different name!- Differences: Replisome is much more slowly (e. coli is thousands nucleotide/sec) (eukaryotes 30-50 n/s)6. End Replication in Eukaryotes- Leading strand synthesis runs right off. The lagging strand contains the okazaki fragmentsand the RNA primers are changed to DNA using the 3’OH of previous Okazaki fragments- However, the last okazaki fragment does not have a 3’OH group to fix it after the RNA is removed. You end up with a 3’ overhang- Replication of that chromosome becomes shorter- There is something called telomeres that help this7. Telomeres- The end of eukaryotic chromosomes is called telomeres. They contain repeats of TG rich sequences (10 kb)- They protect the ends of the linear chromosomes from degradation- Lots of proteins realize the structure and grabs ahold of it and protects the end so nothing happens- Telomerase: reverse transcriptase: makes DNA using a RNA template- Telomerase is a ribonucleoprotein: made up of RNA and protein. Protein carries its own RNA and uses it as a template- Add dNTPS from 3’ end of a primer but the primer is DNA and template is