8.) (2 pts) For a competitive inhibitor. If the ] 1/2K, and IS- 3K.m. What is th

Question

8.) (2 pts) For a competitive inhibitor. If the ] 1/2K, and IS- 3K.m. What is the reaction velt (expressed as a percentage of Vmax) 9.) (2 pts) For a noncompetitive inhi (expressed as a percentage of V nux) bitor. Ifthe [1] and [S]=4K, what is the reaction velocity, v 2K or an uncompetitive inhibitor.If the 3K, and [S] - 5K.m, What is the reaction velocity, v iti (expressed as a percentage of Vmax ) 11.) (4 pts) The Ki for a noncompetitive inhibitor of a particular enzyme is 52.5 M. How much inhibitor would be required to reach 90% inhibition of the Vmax for the enzyme catalyzed reaction?

Explanation / Answer

8) For competitive inhibition KI = KM (1 + [I] / Ki )
[I] =1/2Ki so KI = KM ( 1 + 1/2Ki / Ki ) = 1.5 KM
it is also given that [S] = 3KM
but v = Vmax [S] / KI + [S] , sunstitute all the above into the Michaelis Menten equation
v = Vmax * 3KM / ( 1.5KM + 3KM) = Vmax (3/4)
v = 3/4 * Vmax ; so the reaction velocity v is 75% of Vmax (3/4 * 100)

9) For non-competitve inhibition,
v = ( Vmax / 1 + [I] / Ki ) [S] / (KM + [S] )
put [I] = 2Ki and [S] = 4KM into the above equation,
v = ( Vmax / 1 + 2Ki / Ki ) *4KM / (KM + 4KM ) = 4/15 * Vmax
v = 0.2667 Vmax = 26.7 % of Vmax

10) For uncompetitive inhibition,
  1/v = ( KM / Vmax [S] ) + (1 + [I] / Ki ) / Vmax
put [I] = 3Ki and [S] = 5KM
1/v = ( KM / Vmax *5KM ) + (1 + 3Ki/ Ki ) / Vmax
1/v = 1/5Vmax + 4/Vmax = 1/Vmax * [ 1/5 + 4] = 4.2 / Vmax
v = Vmax / 4.2 = 0.2381 Vmax = 23.81 % of Vmax

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