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Enzyme Inhibitor Key

# Enzyme Inhibitor Key - (2 α = 1[I/K I K I = 10 mM/2.88 =...

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Additional Problem for Reversible Enzyme Inhibition for Practice Below are show two data sets, one for an enzyme-catalyzed reaction that occurs in the absence of an inhibitor, and one for the same enzyme in the presence of a non- competitive inhibitor (I know that I told you that you would never see a noncompetitive inhibitor in your life). -1.0 -0.8 -0.6 -0.4 -0.2 -0.0 0.2 0.4 0.6 0.8 1.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 No inibitor [I] = 10 mM 1/[S], mM -1 1/v, s/ μ M Calculate the Ki and/or Ki’ for the non-competitive inhibitor. Remember, the Ki = Ki’ in the case of a noncompetitive inhibitor. First, calculate the Km and the Vmax (1) Km = - 1/x-intercept (no inhibitor curve) = - 1/(-0.35 mM -1 ) = 2.86 mM (2) Vmax = 1/y-intercept(no inihibitor curve) = 1/(0.09 s/ μ M) = 11.1 μ M/s Then, calculate α from inhibitor curve: First calculate using the slope (1) slope inhibitor curve = α Km/Vmax α = 11.1 μ M/s x (0.5 s/ μ M/0.5 mM -1 )) = 3.88 2.86 μ

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Unformatted text preview: (2) α = 1 + [I]/K I K I = 10 mM/2.88 = 3.5 mM We could also determine KI’, the inhibition constant for inhibitor interaction with ES complex. This should be equal to the KI that was determined above. Calculate the α ' from the inhibitor curve, using the y-intercept (instead of the slope). y-intercept inhibitor = α / Vmax 0.3 = α ’ / 11.1 μ M/s α ' = 3.33 Calculate K I ’ from α α ’ = 1 + [I]/KI’ K I ’ = 10 mM/2.33 = 4.3 mM Note: Notice that the KI values are not apparently the same. This comes from the fact that I drew onto the graph lines extrapolated from the line created from the data points through the x and y axis. These lines are somewhat askew and give extrapolated points that are off a bit from the true extrapolated lines. In other words, I didn’t draw the extrapolated lines very well. The KI values, for all intents and purposes, the same....
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Enzyme Inhibitor Key - (2 α = 1[I/K I K I = 10 mM/2.88 =...

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