In class, we derived the following form for the Michaels-Menten rate law for a s

Question

In class, we derived the following form for the Michaels-Menten rate law for a simple enzymatic reaction: v = V_max[S]_0/K_M + [S]_0 We then rearranged it to show that a plot of 1/v vs. 1/[s]_0should give a line with intercept 1/V_max and slope KM/V_max State how the parameters K_M and V_max are obtained from a plot of v vs. [S]. Demonstrate that a plot of v vs.v/[S]_0 should give a line and indicate how K_M and V_max could be obtained from such a plot. Demonstrate that a plot of [S]/v vs. [S] should give a line and indicate how K_M and V_max could be obtained from such a plot.

Explanation / Answer

the Michaelis-Menten kinetics is V= Vmax*S/(KM+S)

so a plot of V vs 1/S is hyperbolic. When V=Vmax/2,   Vmax/2= Vmax*S/(KM+S)

1/2 = S/(KM+S), KM+S= 2S or KM=S. So at V=Vmax/2 , KM can be located. When the enzyme is saturated with substrate gives Vmax

The evaluation is shown in the plot

2. V= Vmax*S/ KM+S)

V= VmaxS/ (KM+S)

V*(KM+S)= VmaxS

V/S*(KM+S)= Vmax

(V/S)*KM+ V= Vmax

V= Vmax- (V/S)*KM

So a plot of V vs V/S gives a straight line whose slope is -KM and intercept if Vmax.

3.   V= Vmax*S/(KM+S)

1/V= (KM+S)/VmaxS

multiplyig with

S/V= KM+S)/Vmax

S/V= KM/Vmax +S/Vmax

So a plot of S/Vmax vs S gives a straight line whose slope is 1/Vmax and intercept is KM/Vmax

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