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Lec 4Ch13-BactReplicCellCycle-2011-b

Lec 4Ch13-BactReplicCellCycle-2011-b - Chapter 13 Bacterial...

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Unformatted text preview: Chapter 13 Bacterial Replication Is Connected to the Cell Cycle Sections 13.12, 13.7, 13.10, 13.12 Bacterial replication is connected to the cell cycle 1. two links between replication and bacterial cell growth: a. Frequency of initiation adjusted to fit rate cell is growing. b. The completion of a replication cycle is connected to cell division Connecting bacterial replication to the cell cycle 2. Doubling time: time needed for cell numbers to double. In E. coli they range from 18 min to slower than 180 min. (average in lab is 60 min) Replication Is Connected to the Cell Cycle s The doubling time of E. coli can vary over a 10 range, depending on growth conditions. It requires 40 minutes to replicate the bacterial chromosome (at normal temperature). Known as "C". Completion of a replication cycle triggers a bacterial division 20 minutes later. Known as "D". C+D= 60 min. This is the total time between the initiation of DNA replication and cell division. s s s s If the doubling time is <60 minutes, a replication cycle is initiated before the division resulting from the previous replication cycle. Fast rates of growth therefore produce multiforked chromosomes. This initiation is for the next generation. s Connecting bacterial replication to the cell cycle Example: cell divides every 35 minutes There is a fixed interval of 60 minutes between initiation of replication and cell division. Note: E. coli is able to divide every 18 minutes max.. How is this possible? At 10 min. initiation occurs at both origins on the partially replicated chromosome= multifork chromosome. Note: there is an origin for both of the daughter cells. Overlap of replication cycles Initiation of replication 60 min Cell division 25 min Doubling time In this example, the doubling time is 25 min. Connecting bacterial replication to the cell cycle 3. Constants in replication cycle: a. C; fixed time (40 min) to replicate entire chromosome. Replication fork moves at ca. 50,000 bp/min) b. D; fixed time (20 min) between end of replication and cell division. c. C+D = 60 min. These values apply for doubling times between 18 and 60 min. Connecting bacterial replication to the cell cycle 3. Constants in replication cycle (continued): d. When doubling time is less than 60 min, initiation of replication must occur before the end of previous division cycle. How does the cell know when to initiate the replication cycle? Rapidly growing cells are larger and possess a greater number of origins (multiforked chromosomes). Initiation occurs at a constant ratio of cell mass to the number of chromosome origins. a. Bacterium unit cell= an entity 1.7 m long. One origin per unit cell. A rapidly growing cell with two origins will be 1.7-3.4 m long. b. At 10 min. after division, the cell mass increases sufficiently to support an initiation at both origins s A replication cycle is initiated at a constant ratio of mass/number of chromosome origins. There is one origin per unit cell of 1.7 m in length. s How is cell mass titrated? 1. Current model suggests that initiation is controlled by the accumulation of a positive acting factor. Accumulation of a critical amount would trigger initiation. (This factor is dilute in newly divided cells). DnaA protein is likely candidate. 2. Alternative model suggests that initiation is controlled by the accumulation of a negative acting factor (an inhibitor). Inhibitor may be synthesized to a fixed level (no initiation) and diluted below an effective level in larger cells (initiation). How many replication origins are active in an E. coli cell that is 2.1 m in size? 1. 2. 3. 4. One Two Three Not enough information A cell with a length A cell with a length between 1.7-3.4 m between 1.7-3.4 m would have two origins. would have two origins. The oriC and termination regions of the chromosome have specific locations during the cell cycle. The separation of the origins reduces recombination between the daughter genomes before cell division. If recombination occurs, a genome dimer is formed that must be resolved by a site-specific recombination (a second recombination). The Xer site-specific recombination system to resolve a dimeric genome into monomer genomes after replication. Xer-directed recombination occurs at the dif site which must be within ~30 kb from the septum that is forming between the daughter cells. Plasmid Incompatibility Is Determined by the Replicon s Plasmids in a single compatibility group have origins that are regulated by a common control system. Example shows single-copy plasmids Note: plasmids Note: plasmids in the same in the same compatibility compatibility group are not group are not able to coexist able to coexist in the same in the same cell. cell. The ColE1 Compatibility System Is Controlled by an RNA Regulator s Replication of pColE1 requires transcription to pass through the origin. There the transcript is cleaved by RNAase H to generate a primer end. s The regulator RNA I is a short antisense RNA that: pairs with the transcript (primer) prevents the cleavage that generates the priming end s The Rom protein enhances pairing between RNA I and the transcript. Control of Plasmid Replication Plasmid incompatibility is connected to the regulation of copy number and segregation 1. Compatibility group: a set of plasmids that are unable to coexist in the same bacterial cell. a) Both plasmids have the same type of origin. . . b) For the ColE1 plasmid, copy number is controlled by a repressor (RNA I) that "measures" the concentration of origins. 2. ColE1 plasmid employs negative control for its copy number and incompatibility system. (ca. 20 copies per cell). Mechanism for controlling copy number: regulate activity of the primer Primer RNA II: 555 bases, starts upstream of the ori and extends into the ori. a. cleaved by RNase H (cuts RNA:DNA hybrids) b. 3' OH of RNA serves as primer to initiate DNA synthesis. c. only cut by RNase H if it is not duplexed with antisense RNA (RNA I) Replication of ColE1 plasmid DNA-multicopy control; Replication starts with transcription RNA II is required for priming DNA synthesis at the origin. Positive regulation. -555 RNA polymerase P-II RNA II (primer RNA) -20 RNA I P-I origin P-I and P-II are promoters for the regulatory transcripts. Anti-sense RNA I (108 b) acts as a negative regulator RNA I is complementary to the 5'-terminal region of RNA primer II Replication of ColE1 plasmid DNA-multicopy control -555 -265 -20 P-II P-I RNA-loop structure is required to stabilize the RNA:DNA persistent hybrid . origin 3' RNA II 5' ds stem RNaseH creates the 3'-terminus if loops are present. ss loop Replication of ColE1 plasmid DNAmulticopy control -555 -265 -20 P-II P-I RNA II serves as primer for DNA synthesis for replication. origin 3' RNA II (primer RNA) 5' Replicase Replication of ColE1 plasmid DNAmulticopy control -555 -265 -20 P-II P-I RNA I origin RNA II RNA I hybridizes to the 5' terminal region of RNA II and disrupts the RNA loops. Replication of ColE1 plasmid DNAmulticopy control -555 -20 P-II P-I RNA I 5' origin 3' RNA II Rom Without loops, RNA II is not able to form a persistent hybrid at the origin. No replication but transcription continues DNA synthesis can not proceed without a stable primer Plasmid incompatibility is connected with copy number 1. RNA I antisense inhibitory transcript forms duplex with Primer RNA II. 2. The Primer RNA:RNA I duplex molecule is not cut at ori by RNase H, and the persistent hybrid at the origin is not formed. DNA synthesis is not initiated. New incompatibility groups generated by mutations 4. Mutations in Primer RNA/RNA I interaction region may result in formation of a new compatibility group. RNA:RNA duplex can not be formed between the mutated RNA primer and the original RNA I. The mutated and original replicons no longer regulate each other. Original & mutant plasmids behave as members of different compatibility groups. Q: How does RNA I negative regulator "count" the copy number? A: P-I promoter regulates the expression level of RNA I regulatory transcript. 1) At low levels of RNA I- replication occurs. 2) At higher levels of RNA I- replication shut down. Q: How many levels of regulation? A: Positive- Primer RNA II enhances replication by providing 3'OH end B: Negative- RNA I repressor shuts down replication, but allows continued transcription of RNA II . C: Negative- Rom protein helps to shut down replication because it enhances RNA primer/RNA I duplex formation The End of chapter 13 ...
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