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I have a question about replication in prokaryotes. I learned in school that:

  1. DNA polymerase needs 3'-OH to add a dNTP.
  2. The chromosomes of prokaryotes are usually circular.
  3. The primer in the leading strand (of course the primers in the lagging strand too) is removed by the 5'→3' exonuclease activity of DNA polymerase I.

It makes sense that on the lagging strand the primers are removed and then replaced by new dNTPs using the 3'-OH of the previous Okazaki fragment. However, for the primer on the leading strand, there is no Okazaki fragment upstream and thus no 3'OH that DNA polymerase can use for polymerisation.

How is the primer replaced by new dNTPs on leading strand?

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  • $\begingroup$ It seems you're asking about the leading strand, but you may have made a typo that made it unclear what you are actually asking. I tried to clarify your question, but please roll back the edit if it changed the meaning. $\endgroup$
    – canadianer
    Feb 10, 2018 at 0:14
  • $\begingroup$ @canadianer — I think you must be right. I have had to scrap my answer which assumed the lagging strand, because the question makes no sense for the leading strand where the primer is the growing DNA chain which is not removed. $\endgroup$
    – David
    Feb 10, 2018 at 9:44
  • $\begingroup$ @David Sorry for making your original answer obsolete. However, the leading strand does have a primer at the origin of replication that needs to be removed. The gap can be filled using the upstream Okazaki fragment as a primer. See this image for example. $\endgroup$
    – canadianer
    Feb 10, 2018 at 18:43
  • $\begingroup$ @canadianer — No problem about my original answer. And you're right, but I wasn't thinking about the ori. I doubt the poster was either, but I guess I should update my answer to cover that. That way it will be of some value. Tomorrow perhaps. $\endgroup$
    – David
    Feb 10, 2018 at 20:08
  • $\begingroup$ @canadianer — where was your figure from? I have now extended my answer and should acknowledge the source. $\endgroup$
    – David
    Feb 11, 2018 at 17:29

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The leading strand post-initiation

First consider the DNA replication fork after initiation has occurred.

The 3′-OH primer on the leading strand is the 3′-end of the strand of DNA being synthesized in the 5′ to 3′ direction (circled in the diagram below, modified from Berg, Biochemistry). It is not removed because there is no need to do so. (The primer on the lagging strand is only removed because it is RNA.) So the problem posed in the question does not arise.

Remember that the reason for Okazaki fragments is that there is no DNA 3'-OH primer for the lagging strand, so a temporary RNA 3′-OH primer is generated instead, RNA polymerase — unlike DNA polymerase — is able to copy DNA without a primer. There is no such problem for the leading strand.

Comparison of leading and lagging strands

The leading strand at the origin of replication

As @canadianer points out in a comment, the remarks above do not apply to the initiation of DNA replication, the single instance where the leading strand must have an RNA primer. In the case of prokaryotes such as Escherichia coli replication proceeds in both directions so that a short time after initiation one has a replication bubble as shown in the figure below (adapted from Russel, iGenetics) :

E.coli replication origin

It can be seen that, at the origin, the 3′-OH of the DNA from the lagging strand of the other direction of replication (boxed in red) can act as a primer for the DNA synthesis function of DNA polymerase I, after its exonuclease activity has removed a nucleotide from the start of the RNA primer on the leading strand.

References

Berg et al. Ch.27

The Cell, Ch. 5

Sandwalk: Blog page

Footnote

There was a typo in the original question which caused confusion about which strand the poster was concerned with. This answer assumes the leading strand.

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  • $\begingroup$ I am very sorry for my late response, and I guess someone already modified my question, but indeed my question was how the primer on the leading strand at the site of initiation could be replaced by a DNA fragment when there is no 3OH' (on the lagging strand, the 3OH' of an okazaki fragment just behind the primer can be used), but as you say, we have a 3OH' from the lagging strand which results from the replication in the opposite direction!!! Wow, that was very interesting. Thank you. $\endgroup$
    – Cochon noir
    Aug 17, 2018 at 9:15
  • $\begingroup$ However, I still have a small question if you allow me. I thought that the leading strand starting from the Ori would continue until the very end (when two forks meet) without any addition of another primer, so I'm confused about the first diagram. Does it mean that during the replication in procaryotes (my question has been always the case on procaryotes), on the leading strand, far from the initiation site, a few primers are added and in this case they are DNA primers? Or the explication with the first diagram is a general description of the replication in eucaryotes? Thank you so much. $\endgroup$
    – Cochon noir
    Aug 17, 2018 at 10:02

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