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

i think so it’s will be b yes i believe its

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