An enzyme catalyzes the reaction S → P. The table shows how the reaction rate (v)...

An enzyme catalyzes the reaction S → P. The table shows how the reaction rate (v) of the enzyme varies with the substrate concentration [S].

0,0500	0,04545
0,0250	0,03448
0,0167	0,02778
0,0125	0,02326
0,0100	0,02000

a) Create a Lineweaver-Burk plot of the data given in the table above.
b) Calculate Km, Vm, and Vm / Kmfor the enzyme.
c) Also calculate kcat and kcat / Km (the specificity constant) when given that
[E] = 10 nM.d) How saturated will an enzyme be if the activity is measured at a substrate concentration 2x greater than Km? Give the answer as a percentage of Vm.

Solutions

Expert Solution

a) According to the Michaelis menten equation, when rate of an enzymatic reaction is plotted with respect to the varying substrate concentration, a hyperbolic curve is obtained. The equation is-

vo = Vmax*[S]/ Km+[S]

When this equation is reversed, we get Lineweaver Burk plot, which is a straight line.

Slope	0.6094 ± 0.07610
Y-intercept when X=0.0	0.01627 ± 0.002059
X-intercept when Y=0.0	-0.02670

b) 1/vo =Km/Vmax (1/[S]) + 1/Vmax

The y intercept is = 1/Vmax

x intercept = -1/Km

Using the equation of the line-

According to the graph,

the y-intercept= 0.0163

Thus Vmax= 1/0.0163= 61.35 (unit of concentration/time)

The x-intercept= -0.0267

Thus the Km= 1/0.0267= 37.45 (unit of concentration)

Km/Vmax= slope of the line= 0.61 (per time)

c) Kcat is calculated as= Vmax/[E]= 61.35/10 = 6.135 per unit time (since [E]= 10nM)

Kcat/ Km = catalytic efficiency = 6.135/37.45 = 0.164

gladiator answered 1 year ago

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