DNA Replication in E.Coli
DNA polymerase III from E.coli catalyzes the stepwise addition of deoxyribonucleotides to the 3´-OH end of a DNA chain. The enzyme has the following requirements: all four dNTPs must be present to be used as precursors, Mg2+ is also required; a DNA template is essential, to be copied by the enzyme DNA polymerase III; a primer with a free 3´-OH that the enzyme can extend. The four dNTPs must be present in equally balanced levels otherwise one base may be more likely to be added in the wrong place.
DNA polymerases are template-directed enzymes, that is they recognize the next nucleotide on the DNA template and then add a complementary nucleotide to the 3´-OH end of the primer, creating a 3´5´phoshodiester bond, releasing pyrophoshate. DNA polymerase III catalyses the synthesis in the replicative fork.
DNA polymerase III also corrects mistakes in DNA by removing mismatched nucleotides. So, during polymerisation, if the nucleotide that has just been incorporated is incorrect, it is removed using a 3´5´exonuclease activity. This is very accurate, an error rate of less than 0.00000001 per base pair. DNA polymerase I also has 5´3´exonuclease activity. It can hydrolyse nucleic acid starting at the 5´end of the chain. This activity also has a role in removing the RNA primer used during replication. So DNA polymerase I has 3 catalytic activities.
When the bacterial chromosome is replicated it commences at a single origin. The double helix opens up and both strands serve as a template for DNA replication.
DNA synthesis then proceeds bi-directionally outward from the single origin , producing 2 daughter double strands each of which has one strand from the parent, thus it is semiconservative. The region of replicating DNA is called the replication bubble or eye and consists of two replication forks moving in opposite directions around the circle of DNA.
Double stranded DNA is anti-parallel; one strand running 5´3´, the other 3´5´. As the original double- stranded DNA opens up, new DNA is made against each template strand. However, all DNA polymerases make DNA only in the 5´3´direction, never in the other direction. What happens is that DNA is made continuously in the 5´3´direction (leading strand)and on the 3´5´strand, the lagging strand it is made by short (1kb) pieces at a time. The fragments are called Okazaki fragments. They are then joined together by DNA polymerase I. (DNA polymerase is believed to have some role in repair.)
DNA polymerase III (polIII) cannot start without a primer. Even each Okazaki fragment requires an RNA primer before synthesis can start. The primer used is a short piece of RNA ( about ) 5 nucleotides long, which is synthesized by an RNA polymerase called primase. Primase can make RNA directly onto the strand because, like all RNA polymerases, it can start synthesis without a primer. The RNA primer made by primase is then extended by pol III. Pol III makes DNA for both lagging and leading strand. After DNA synthesis by pol III, pol I uses its 5´3´exonuclease activity to remove the primer and fill the gaps between fragments with DNA. Pol III cannot do this because it doesn’t have the 5´3´exonuclease activity. Finally DNA ligase joins the ends together. In E.coli ligase uses NAD+ to donate an AMP group. The inability of replication to start without a primer increases fidelity because the first base pairs added are most likely to be inaccurate because of the cooperative nature of base pairing interactions. It doesn’t matter if the RNA is occasionally inaccurate here because it is replaced later by DNA under more accurate conditions when other base pairs are now present.
Pol I and III , ligase and primase are not the only proteins needed for replication of the bacterial chromosome. The DNA template is a double helix with each strand wound tightly round the other so that it needs to be unwound before replication can occur. DNA helicase is used to unwind DNA using ATP as an energy source and SSB (single-strand binding protein) prevents the unwound region base-pairing again because it has a high affinity for single stranded DNA but no base specificity. The energy release from SSB binding helps drive the strand separation to completion. When the strand is being read the SSB has to be taken away by the replicating machinery.
In principle, when the replication fork is moving along the coil needs to be unwound ahead of it. This would mean the helix behind it would rotate rapidly.
However in bacteria the chromosome is circular so it has no ends to rotate, so it would become overwound and supercoil, resisting separation. This problem is solved by topoisomerase I, which breaks a phosphodiester bond in one DNA strand (a single strand break) a small difference ahead of the fork, allowing the DNA to rotate freely around the other strand. The phosphodiester bond is then reformed by the topoisomerase. Topoisomerase I can only relax supercoiling, it cannot insert supercoiling. Actually in E.coli it can only relax negative supercoiling. The enzyme does not use ATP since the more relaxed form is at a lower energy level. This does not solve the winding problem since there is positive supercoiling ahead of the fork, caused by helicase unwinding. Topoisomerase II (gyrase) induces negative supercoiling and converts positively supercoiled DNA into negatively supercoiled DNA. When gyrase attaches to the strand it causes local positive supercoiling. Since no bonds have been broken there is no net supercoiling, so the rest of the strand is negatively supercoiled. The enzyme then cuts both strands of the DNA and uses ATP hydrolysis to drive the DNA through the gap formed. Then the gap is resealed on the other side. This converts the positive node in one step into a negative one, so negative supercoiling has been introduced, using ATP. However, how there is too much negative supercoiling so topoisomerase I relaxes it without change in free energy because it only transfers bonds, so it is freely reversible. This provides the right degree of supercoiling.
Gyrase can also separate the circles produced after bacterial chromosome replication has been completed, because an inevitable symptom of the semiconservative replication of a circular piece of DNA is that the two daughters are linked.
There is also another backstop mechanism operating in E.coli to correct mistakes in the base pairing. If an error has occurred there will be a slight deformation of the duplex due to imperfect base pairing. It must be the new strand that has the error, but the new strand is identical – except for one thing which gives it away: In E.coli wherever GATC appears the A is methylated. This takes some time to happen after the synthesis so the new strand will not have it yet. First MutS finds the distortion of the strand. Then MutH finds an unmethylated GATC and nicks the chain at this point. MutH is linked to MutS by MutL, over a distance of possibly several kb. Then helicase, SSB and an exonuclease cooperate to remove the strand from the nick to beyond the distortion. Then pol III replaces it with new DNA and ligase seals the replacement in.
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