vi Enzyme Kinetics ing is a plot of an experimental result for determination of

Question

vi Enzyme Kinetics ing is a plot of an experimental result for determination of kat, Kin and Vs of an enzyme that catalyze the transformation of its substrate (S). The reaction measured in unit of ??/min and concentration of substrate is in mM. for each parameter is clearly shown (Spt each) rates were Make sure the unit 120.0 100.0 2 80.0 60.0 40.0 20.0 0.0 30 40 50 60 10 20 s] mM a. Estimate Va b. estimate Km c. If the concentration of the enzyme in the enzymatic reation mixture is 60 nM, Calculate kat d, what would be the initial reaction rate if the concentration of substrate is 100 ?M

Explanation / Answer

a. The value of Vmax is the maximum of the curve at very high [S]. After estimated extrapolation, the value of Vmax = 120 uM/min.

b. Km is the value of concentration of substrate at half of the Vmax. From graph, Km = 4 mM = 4000 uM.

c. [E] = 60 nM.

Vmax = Kcat x [E] => Kcat = Vmax / [E] = 120 uM/min / 60 nM = 2000 min.

d. [S] = 100 uM.

v = Vmax [S]/ (Km+[S]) = 120 x 100/(4000 + 100) = 2.92 uM/min.

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