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Km and Vmax dependence on enzyme concentration

  1. Sep 15, 2009 #1
    We know that Vmax depends on enzyme concentration since Vmax = k2[E]

    However, what I have trouble grasp is that why Km does not depend on enzyme concentration if Km is the substrate concentration where V = 1/2 Vmax. If you increase Vmax, shouldn't Km increase as well since it is dependent on it?

    I asked my biochem prof today and he gave some explaination about substrate concentration which I didn't catch. I am hoping someone can clarify it here. thanks!
  2. jcsd
  3. Sep 15, 2009 #2


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    If [tex]K_M = [/tex] then
    [tex]v_0 = v_{max}\frac{}{K_M+}=v_{max}\frac{}{2}=\frac{v_{max}}{2}[/tex]

    Note though, that this doesn't mean [tex]K_M[/tex] is dependent on . [tex]K_M[/tex] is defined as a measure of the equilibrium constant for E + S <-> ES. Which is not dependent on any concentration, as long as the reaction is first order, which Michaelis-Menten kinetics assumes. (it also assumes a steady-state where [ES] is near-constant)

    I see http://en.wikipedia.org/wiki/Michaelis–Menten_kinetics" [Broken] has a thorough derivation of Michaelis-Menten kinetics, btw.
    Last edited by a moderator: May 4, 2017
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