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Mouse action summary

First explore the enzyme-DNA complex in the original ribbon and backbone mode. (To get the original presentation, just click the "select/resize/color" button, the reset button under the jmol window, or hit reload on your browser.)

The form of the enzyme that was crystallized has two domains, a 3' to 5' exonuclease domain and the polymerization domain. The polymerization domain has three named regions: fingers. palm, and thumb. Use the buttons to highlight each.

Now look at the substrates: DNA, the incoming nuleotide, and 2 magnesium ions. Identify the template and primer strands. When you restrict to just the substrates, there is an "extra" base hanging away from the double-helical portion of the DNA. What is it? (Try to answer, then click here .)

How far is the polymerization active site from the exonuclease domain? You can get distances as follows: From the standard buttons below the Jmol window select "distance" under "mouse click." Now distances are given between atoms clicked in succession. When you are done reset "mouse click" to indentify."

Obviously, a major rearrangement of the path of the primer strand through the enzyme is required for editing.

Let's investigate the active site. Reset the image and then click a "retrict" button under "Substrates: DNA and dNTP", zoom in a bit. Notice how the incoming nucleotide has positioned itself so that it is in the right position to become B-DNA. Once you are oriented show all of the atoms. If you lose sight of the incoming dNTP, color its carbons white.

Now use the "zoom" button to see the details of the active site. Color the template and primer, if you need to orient yourself. Observe how the 3'-OH of the primer is situated so that it can attack the phosphorous of the alpha-phosphate of the incoming dNTP.

The reaction that creates the new phosphoester bond is an SN1 displcement on the phosphorous of the alpha-phosphate. The 3'-OH is the nucleophile and the beta phosphate is the leaving group. The two Mg++ ions aid this reaction.
Ion A coordinates the 3'-OH to position it and to fascilitate the dissociation of its hydrogen. It also neutralizes the oxygens of the alph-phosphate, so that they will readily reside at the basal positions of the trigonal bipyramid that forms during the transitions state. Ion B positions the beta- and gamma-phosphates, which will leave as pyrophosphate.

As mentioned above, to be an effective nucleophile, the 3'-OH lose its hydrogen. Is there a suitable group around to accept this hydrogen? I'm not certain , but my guess is or one of two waters coordinated to Mg++ A. . The Mg++ may cause one of these to dissociate to OH-. Remeber, hydrogens aren't seen in X-ray structures such as these.

Similarly, to make the beta phosphate a better leaving group, it should be protonated. Which residue might donate such a proton? My choice is .

If the image gets too messed up, just click the select whole complex button or reload.

CREDITS