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H/D exchange

  1. Mar 21, 2005 #1


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    For a hydrogen/deuterium exchange experiment, the protein of interest is first incubated at pD = 7, after which it is quenched at pH = 2-3.

    Why it the exchange done at a neutral pD and why has the exchange of hydrogen atoms a minimum at pH 3?
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  3. Mar 21, 2005 #2


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    This site pretty much explains the gist of it, there's probably a better explanation... nevertheless, here's the kinetic isotope effect


    The lower the pH or pD the lower the relative concentration. According to the site, the activation energy for the breakage of a deuterium bond will be higher than the corresponding bond with hydrogen...due to the differences in mass of each, deuterium being heavier. Kinetic collision theory basically states that atoms collide, and this collision supplies the energy for the breakage of bonds (among other things), you can imagine that it'll be a bit more difficult to break the deterium bond.

    Activation energy relates to the rate, the higher the energy, the slower the rate. If you wish to replace deuterium with hydrogen you'll need to increase the rate of formation of hydrogen; note that deuterium is not exactly stuck, it's in equilibrium.

    The site also mentions subtle differences in electronegativity between the two in accounting for the kinetic isotope effect.

    Probably not the best explanation, but hope it helps.
  4. Mar 21, 2005 #3


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    the hydrogen analog (compounds), in replacing deuterium with hydrogen
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