Isn't helicase always free floating in bacterial cells, and the DNA without a nuclear membrane and uncoiled and freefloating and so why doesn't the helicase keep breaking the double helix of DNA? Also, since other DNA synthesizing enzymes like polymerase and RNA primer and other replication enzymes also free floating, so why doesn't the DNA in bacterial cells keep replicating at all times?


1 Answer 1


Bacterial DNA replication is initiated at the oriC by DnaA in E. coli. Think about ways in which DnaA binding or activity can be regulated in a way that inhibits or permits DNA replication. In recently replicated bacterial DNA, the DNA is hemimethylated (parental strand has a methyl group, daughter strand doesn't): An inhibiting protein binds to hemimethylated DNA, that can block DnaA activity at the oriC. A DAM methylase adds methyl groups after a while to the newly synthesized DNA, in which case DnaA may bind. Due to the lag between replication and methylation, DNA replication can't be continuously initiated! This is one of many examples (for example, DnaA must bind ATP for it's mechanism of action), but the take home point is DNA replication must be initiated and if it can't be, then it doesn't take place!


Breaking that down: (note, this is in E. coli)

DnaA is a protein with DNA-binding activity. When many DnaA bind at the oriC, it causes the DNA to bend open and loop. At this point, DnaB would have access to the DNA, and this is the helicase that unwinds everything for replication (you'll also see that DnaB must be loaded onto the DNA by DnaC, a helicase loader). So blunty enough, oriC is an origin of replication, everything starts here. There are repeats of AT rich regions denoted 13mer and 9mer in the oriC (number of bases in the region if you will). We know that this is beneficial because AT rich regions are easiest to pull apart due to the number of hydrogen bonds between A and T.

In the example I provided, the assumption is that the DNA of E. coli is methylated on adenines on both strands. When replication occurs, the new DNA doesn't have this methyl group on it's adenines like the parental strand. A protein that recognizes and binds so-called hemimethylated (half methylated) DNA, SeqA, binds where it can, including in the oriC. It blocks DnaA subsequently for a short time, and falls off the DNA on its own. The DAM methylase adds the methyl groups when it can, and once it does the SeqA can no longer bind, but the replication-initiation complex thus can.

There are other modulators of DNA replication, and the concentration of the DnaA itself is a big factor because you need many units of DnaA to bind to facilitate initiation. So then you have to consider is DnaA being expressed strongly enough? And this has it's own pathway outside the scope of this question.

  • $\begingroup$ Ok, I get some of it. Could you explain it at a little lower level (keeping in mind that I am still a highschool student but I have done extended research so you can use some jargon). What is DnaA? and oriC? So when the DNA is methylated, it can replicate? then what does the inhibiting protein does (what does blocking DnaA activity do?)? So according to the lag point, the DNA, methylation activates replication? I get the core point though :P $\endgroup$ Commented Dec 6, 2015 at 8:09
  • $\begingroup$ @SarthakGarg I made some edits that explain it a little more detailed. Always keep note that this is in E. coli and other families may adopt different strategies, but the take-home point is there (you need the initiation complex). $\endgroup$
    – CKM
    Commented Dec 7, 2015 at 20:57

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