Replication has an error rate of less than 1 in 100 million. DNA polymerase forms H-bond with the H-bond acceptor atoms in the minor groove. <-- enhance fidelity here?

Binding of the triphosphate group to the active site of DNA polymerase triggers a conformational change. Changing a conserved Tyr residue increases the error rate by 40 fold.

I don't quite understand the above two statements. Can anyone explain in detail to me? Thanks!

  • $\begingroup$ Where do these statements come from? Can you post a link to the source? $\endgroup$ – user137 Oct 17 '14 at 14:19
  • $\begingroup$ it comes from my lecture note. Maybe similar statements can be found from here. The lecture quotes this from this reference book link $\endgroup$ – Dexter Wong Oct 17 '14 at 14:20
  • $\begingroup$ There's a few probs with this question one is h-bond acceptor- what is that? Then w the tyr residue are you talking about the polymerase? I have a feeling your instructor is using this mediocre example to explain to you: the structure of dna, and the fact that aa sequence determines over protein structure which determines protein function. I'm gonna post some info on high fidelity pol but it's not going to completely answer your question cause I believe, respectfully, there could be some errors in it. $\endgroup$ – rhill45 Oct 17 '14 at 20:44

DNA polymerase must catalyse the addition of 4 different nucleotides to the growing strand. This means that it cannot directly determine which base to incorporate at a specific point (how would it 'know' which base to incorporate and how it would it change its specificity for different bases). This means that the specificity for which base pair to incorporate is dependent on the template DNA strand.

Correct Watson-Crick base pairing (that is, hydrogen bonding) between the template strand and the nucleotide to be incorporated triggers the closing of the finger domain of DNAP around the primer-template junction and positions the latter in the optimal position for catalysis (with the $\ce{\alpha-PO_4}$ of the incoming nucleotide near the $\ce{3'-OH}$ of the primer for a nucleophilic attack catalysed by two $\ce{Mg^{2+}}$ ions). This is where the conserved tyrosine residue you mentioned comes into play. An incorrectly paired nucleotide will not trigger this conformational change and will not be positioned optimally, thus catalysis is less likely.

Furthermore, the DNA polymerase makes contacts with the minor groove of the primer-template junction through hydrogen bonds. This interaction is not base-specific (all Watson-Crick base pairs have he same pattern of hydrogen-bond acceptors in the minor groove) but only occurs when the correct nucleotide is incorporated, thus stabilising the complex.

Finally, discrimination between ribonucleotides and deoxyribonucleotides is done by steric exclusion of the $\ce{2'-OH}$ by amino acid residues in the binding pocket.

These factors can be thought of as kinetic proofreading as they simply slow down the reaction rate and provide time for an incorrectly paired nucleotide to dissociate. However, they can still be incorporated and many DNA polymerases have a $\ce{3'->5'}$ proofreading exonuclease that can remove incorrectly paired nucleotides. This proofreading is again mediated by interactions between the DNAP and the primer-template junction (ie hydrogen bonding with the minor groove). A weakened interaction due to an incorrectly paired base reduces the affinity between DNA and the catalytic site and increases the affinity between DNA and the proofreading site (because it has a preference for cleaving ssDNA from the 3' end).

  • $\begingroup$ This is really helpful and detail. Now I understand! $\endgroup$ – Dexter Wong Oct 19 '14 at 6:40
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    $\begingroup$ I would upvote this over my own , good info $\endgroup$ – rhill45 Oct 20 '14 at 20:11

High-fidelity DNA polymerases have several safeguards to protect against both making and propagating mistakes while copying DNA.

Such enzymes have a significant binding preference for the correct versus the incorrect nucleoside triphosphate during polymerization.

If an incorrect nucleotide does bind in the polymerase active site, incorporation is slowed due to the sub-optimal architecture of the active site complex. This lag time increases the opportunity for the incorrect nucleotide to dissociate before polymerase progression, thereby allowing the process to start again, with a correct nucleoside triphosphate (1,2).

If an incorrect nucleotide is inserted, proofreading DNA polymerases have an extra line of defense dna mismatch

The perturbation caused by the mispaired bases is detected, and the polymerase moves the 3´ end of the growing DNA chain into a proofreading 3´→5´ exonuclease domain. There, the incorrect nucleotide is removed by the 3´→5´ exonuclease activity, whereupon the chain is moved back into the polymerase domain, where polymerization can continue.

From: New England biolabs:



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