Question

An enzymatic reaction was analyzed for rate versus substrate concentration. Roughly estimate the Vmax and Km from these measurements. (Hint: Slope = rise/run and remember what Km represents!)

v^o(nM/s)           [S] (nM)

30                           100,000

29.5                        10,000

29                           1,000

28                           200                        

25                           150

20                           100

10                           50

5                              20

a. Vmax = 10 nM/s, Km = 20 nM

b. Vmax = 20 nM/s , Km = 50 nM

c. Vmax = 40 nM/s , Km = 150 nM

d. Vmax = 70 nM/s, Km = 250 nM

ANSWER is (c), how?

Answer 1

The Michaelis–Menten kinetics,

v=V_max[s]/K_m+[s]

Here,

V_max is the the maximum rate at saturated substrate concentration. K_m is the Michaelis constant, [s] is the substrate concentration

At high concentration of substrate, [s] >>>K_m, the rate of reaction is V_max.v=V_max i.e rate of reaction is  [s] independent.

At low concentration of substrate, [s] <<K_m, the rate of reaction is v=V_max/K_m

In the given data, at very high   [s] rate is independent.

So consider the data between concnetration of 20 to 1000 nm and plot the graph.

In the given data, the highest rate occurs at 30 so a value above the curve is considered, in the options nearest values to the curve is 40.

Km occurs at 0.5 V max, so at 20 nM/s the substrate concentration is 150 nm , so K m is 150 nM