For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, v (as a percentage of Vmax), observed at:a) S=Kmb) S = 0.5 Kmc) S = 0.1 Kmd) S = 2 Kme) S = 10 Km

a) 50% of the maximum velocity

b) 33.33% of the maximum velocity

c) 9.09% of the maximum velocity

d) 66.66% of the maximum velocity

e) 90.9% of the maximum velocity

Explanation:

The Michaelis-Menten kinetis is represented by

v = Vmax*S / (Km+S)

where

v = reaction rate

S = Substrate's concentration

Vmax = maximum rate of reaction

Km = constant

a) for S=Km

v = Vmax*Km / (2Km) = Vmax/2

v/Vmax = 1/2 = 50% of the maximum velocity

b) for S=Km/2

v = Vmax * (Km/2) / (3/2Km) = Vmax/3

v/Vmax = 1/3 = 33.33% of the maximum velocity

c) for S = 0.1*Km=Km/10

v = Vmax * (Km/10) / (11/10Km) = Vmax/11

v/Vmax = 1/11 = 9.09% of the maximum velocity

d) for S=2*Km

v = Vmax * (2*Km) / (3*Km) = (2/3) * Vmax

v/Vmax = 2/3 = 66.66% of the maximum velocity

d) for S=10*Km

v = Vmax * (10*Km) / (11*Km) = (10/11) * Vmax

v/Vmax = 10/11 = 90.9 % of the maximum velocity