For a competitive inhibitor. If the [I] = 2K_ and [S] = 2K_m. What is the reacti

Question

For a competitive inhibitor. If the [I] = 2K_ and [S] = 2K_m. What is the reaction velocity, v (expressed as a percentage of V_max) For a noncompetitive inhibitor. If the [I] = 2K_ and [S] = 2K_m. What is the reaction velocity, v (expressed as a percentage of V_max) For an uncompetitive inhibitor. If the [I] = 2K_ and [S] = 2K_m. What is the reaction velocity, v (expressed as a percentage of V_max) The K_i for a noncompetitive inhibitor of a particular enzyme is 32.5 mu M. How much inhibitor would be required to reach 90% inhibition of the V_max for the enzyme catalyzed reaction?

Explanation / Answer

34) Competitive Inhibition: the inhibitor competes with the substrate for binding to the enzyme.
Ki represents the dissociation constant for the inhibitor; Km represents the dissociation constant for the substrate.

Ki = [E][I]/[EI]
Km = [E][S]/[ES]

These equations can be used to arrive at modified Michaelis-Mentin equation:
v = (Vmax*[S])/(Km*(1 + [I]/Ki) + [S])

Substituting the given expressions for [I] and [S] in the above equation, we get:
v = Vmax*(2/5)
v = 0.4Vmax
Expressing as a percentage, v = 40% of Vmax

35) Non-competitive Inhibition: This occurs when I binds to both E and ES. For this case it is assumed that dissociation constants for both the reactions are same. The modified Michaelis-Menten for this is given below:

v = (Vmax*[S])/((Km + [S])*(1 + [I]/Ki))

Substituting the given expressions for [I] and [S] in the above equation, we get:
v = Vmax*(2/9)
v = 0.222Vmax
Expressing as a percentage, v = 22.2% of Vmax

36)  Uncompetitive Inhibition: this type of inhibitor kinetics is occasionally observed, primarily with multisubstrate enzymes. It occurs when the inhibitor only binds to the ES complex. The modified Michaelis-Menten for this is given below:
  v = (Vmax*[S])/(Km + (1 + [I]/Ki)*[S])
Substituting the given expressions for [I] and [S] in the above equation, we get:
v = Vmax*(2/7)
v = 0.286Vmax
Expressing as a percentage, v = 28.6% of Vmax

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