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I have seen a paper in which deuterated water was used to follow dividing cells. The assumption was that deuterated water will be incorporated into newly synthesized DNA molecules. Is there any direct incorporation of water in the replication process? Is there any proton exchange?

The paper describing the method can be found here: http://www.pnas.org/content/105/16/6115.full.pdf

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  • $\begingroup$ Please add a reference for the method. I imagine that deuterium could be incorporated at several points in the pathway for de novo synthesis of dNTPs, either into the bases or the deooxyribose. $\endgroup$
    – Alan Boyd
    Jul 24, 2013 at 21:30
  • $\begingroup$ proton exchange always happens (i dont have anexact reference right now but i have read about it in spectroscopy texts) .. and hydrogen incorporation is also there as @AlanBoyd said.. $\endgroup$
    – WYSIWYG
    Jul 25, 2013 at 3:08
  • $\begingroup$ @WYSIWYG Exchanging protons wouldn't be useful to follow cell division as they exchange all the time. The deuterium would be incorporated into both new and old DNA, and would also be gone if you remove the D2O again. $\endgroup$
    – Mad Scientist
    Jul 25, 2013 at 7:26
  • $\begingroup$ Thank you all for the comments and answers. Just added the reference. The de novo synthesis pathway of dNTPs sounds like a reasonable place to start looking. Thanks, Alan. $\endgroup$
    – pankdora
    Jul 25, 2013 at 9:07

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I would assume that the labeling occurs in the reduction of NTPs to form dNTPS.

This process (catalyzed by ribonucleotide reductase) involves protonating the hydroxyl group on the 2' carbon, allowing it to leave as water, and then adding a hydride to the newly formed carbocation. The two hydrogen atoms (the proton and the hydride) come from two thiols on the enzyme, which in the process are oxidized to form a disulfide bond.

The crux is that, in the presence of heavy water, the two thiols would rapidly exchange their hydrogen for deuterium. That means that the hydride that gets added to the carbocation would be a deuterium, and the resulting dNTP would be deuterium labeled.

This is the only step that I can imagine the labeling working for, as a carbon-hydrogen bond is formed (which doesn't exchange rapidly with the solvent) using the hydrogen from a sulfer-hydrogen bond (which does exchange rapidly with the solvent).

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