Question

1) You are a research scientist studying a novel enzyme X, and you want to characterize this new enzyme. You measure the velocity of the reaction with different substrate concentrations and get the following data:

[S] (mM)	Initial Velocity (mmol/min)
3.0	10.4
5.0	14.5
10.0	22.5
30.0	33.8
90.0	40.5

a) Graph the above data. From the graph, estimate KM and Vmax (Michaelis-Menten Plot)

b) Use Lineweaver-Burk plot to calculate Km and Vmax. Show all equations and calculations.

c) You decide to do this experiment again, but this time with only one third of the enzyme X concentration used in the first experiment. Draw a new graph on the same graph that you did the first graph on. Estimate Km and Vmax from the new graph.

Answer 1

1) Plot the data in excel. The Michelis-Menten equation can be written as

v = v_max[S]/K_m + [S] …..(1)

where v_max is the maximum reaction velocity (= maximum rate of reaction) and K_m is the Michelis-Menten constant.

Plot the values.

[S] (mM)	Initial velocity (mmol/min)
3.0	10.4
5.0	14.5
10.0	22.5
30.0	33.8
90.0	40.5

The plot is hyperbolic and it is difficult to obtain v_max and K_m from the Michelis-Menten plot. Hence, we move to Lineweaver-Burke plot, which is a linear relation between 1/v and 1/[S].

2) Take the reciprocal of equation (1) above.

1/v = K_m + [S]/v_max[S] = K_m/v_max.1/[S] + 1/v_max ……(2)

Equation (2) is of the form y = mx + c where y = 1/v and x = 1/[S]. The slope of the plot is K_m/v_max and the y-intercept is 1/v_max.

Obtain the requisite data from the table given.

[S] (mM)	V (mmole/min)	1/[S] (mM-1)	1/v (min/mmol)
3.0	10.4	0.333	0.096
5.0	14.5	0.200	0.069
10.0	22.5	0.100	0.044
30.0	33.8	0.033	0.029
90.0	40.5	0.011	0.025

Plot the graph.

Equation of the plot is shown alongside.

y-intercept = 0.0222 = 1/v_max

====> 1/v_max = 0.0222

====> v_max = 1/0.0222 = 45.045 ≈ 45.04 mmol/min (same unit as v)

Slope of the plot is 0.2242.

Slope = K_m/v_max

Slope will have the unit of (min/mmole)/(L/mmole).

Therefore,

0.2242 min/L = K_m/v_max = K_m/(45.04 mmole/min)

====> K_m = (0.2242 min/L)*(45.04 mmole/min) = 100.9796 mmole/L ≈ 100.98 mM (ans).