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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!
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).
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 
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:
Both the sentence can be understood from the below explanation.
In a newly synthesizing DNA strand, the DNA polymerase forms extensive hydrogen bonds with the newly added base everytime a new nucleotide is added. (First sentence) Polymerase also forms bonds with the DNA backbone of phosphate and sugar. (Second sentence)
The first sentence talks of how polymerase will form bonds via the minor groove of the newly synthesized DNA. Second sentence talks about the changing of an amino acid from the DNA polymerase (which is important for forming bonds with DNA backbone). This changing of amino acid will cause error in the nucleotide added (as the Polymerase is not positioned properly which you will understand in the following explaination)
These bonds of polymerase and bases along with the bonds between the DNA backbone and polymerase position the DNA polymerase.
When a newly added nucleotide is correctly base paired, the structure of the paired bases will be such that the bonds formed between polymerase and base will allow DNA polymerase to be positioned for the addition of next new nucleotide. If a incorrect nucleotide was added the DNA polymerase positions such that it's exonuclease site is at a position to catalyse exonuclease activity.
Thus, the hydrogen bonds between the base and Polymerase is needed for correct positioning of polymerase to either allow the next nucleotide to be added (if correct base was added) or allow exonuclease activity of Polymerase (if an incorrect base was added). This will help in proofreading and thus provide high fidelity.
