1. An enzyme with a Km = 2 x 10-4 M for its substrate and a Vmax of 5.5 x 10-5 m

ID: 788321 • Letter: 1

Question

1. An enzyme with a Km = 2 x 10-4 M for its substrate and a Vmax of 5.5 x 10-5 moles per

liter-min was assayed for its catalytic rate with its substrate present at 2 x 10-4 M.

Calculate the anticipated initial rate under these conditions.

2. If the enzyme in problem #1 was incubated with a competitive inhibitor at 2.5 x 10-3 M

concentration and if the inhibitor demonstrated a KI = 2.5 x 10-3 M, what would be the

initial rate in the presence of the inhibitor? All other conditions are the same as those

presented in problem #1.

From Michealis Menton Kinetics,

v= vmax[S]/(Km+[S])

The given values are Km=2 x 10^-4 M, vmax= 5.5 x 10^-5 moles/L-min, [S]=2*10^-4 M

We find that Km=[S], So, v=vmax/2

So, v=(5.5x 10^-5)/2

=2.25 x 10^-5 moles/Liter-min

The velocity of the reaction is 2.25 x 10^-5 moles/L-min

2. For a competitve inhibition,

v=vmax[S}/Km* + [S]) where Km*=Km(1+ I/Ki)

Km*=2x10^-4(1+(2.5x10^-3/2.5x 10^-3)

=4*10^-4

v=5.5*10^-5*2*10^-4/(4+2)*10^-4

=11* 10^-9/6*10^-4

=1.833*10^-5 moles/L-min

The velocity of the reaction in the presence of a compettive inhibitor is 1.833*10^-5 moles/L-min

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