The below graph is the result of an experiment containing 0.10 µM enzyme and var

Question

The below graph is the result of an experiment containing 0.10 µM enzyme and various substrate concentrations. The activity of the enzyme was measured with each substrate concentration and plotted as Vo vs [S].

A. What is Vmax for this enzyme with this substrate?

10 µM/min                             30 µM/min                 0.80 µM/min

0.050 mM/min                        50 mM/min

B. What is the Km of this enzyme for this substrate?

1.5 µM                                    10 µM                         0.50 µM                     

1.0 mM                       1.5 mM

C. What is the kcat (turnover number) for this enzyme with this substrate?

5.0 x 102/s                         1.0 /min                      5.0 x 102 /min

5.0 x 103 /min                   1.0 x 102 /s

Explanation / Answer

(A)

From the graph we can see that Vmax = 50 uM/min = 0.050 mM/min

(B)

Michaelis Menten equation tells that:

V0 = Vmax[S] / ( Km + [S] )

Taking the data point , V0 = 25 at [S] = 1.5, and putting this value in the equation we get:

25 = 50*1.5/(Km + 1.5)

Thus, Km = 1.5 mM

(C)

Vmax = kcat * E0

50 = kcat * 0.1

So, kcat = 5*102 min-1

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