At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme

Question

At room temperature sucrose is hydrolyzed by the catalytic action of the enzyme sucrase as follows: Starting with a sucrose concentration C_A0 =1.0 milli-mole/liter and an enzyme concentration C_E 0 = 0.01 milli-mole/liter, the following data are obtained in a batch reactor (concentration calculated from optical rotation measurements): Using the integral method, determine whether these data can reasonably fitted by kinetic equation of the Michaelis-Menton type, or: -r_A = k_3 C_A C_E 0/C_A + M Where M = Michaelis Constant And if so, evaluate the constants "k_3" and "M".

Explanation / Answer

Given –rA= KCEOCA/ ( CA+M)

Let K1= K3CEO/M    and K2=1/M

Then -rA= K1CA/ (1+K2CA)

-dCA/dt= K1CA/ (1+ K2CA)

{(1+K2CA)/ CA}dCA= -K1dt

When integrated noting that at t=0 , CA= CAO and t=t , CA=CA

Ln (CAO/CA)+ K2*(CAO-CA)= K1t

Ln (CAO/CA)/ (CAO-CA)+ K2= K1t/(CAO-CA)

Ln (CAO/CA)/ (CAO-CA)= - K2 + K1t/(CAO-CA)

So a plot of ln (CAO/CA)/ (CAO-CA) vs t/(CAO-CA) gives slope of   K1 and intercept of   -K2.

The plot is shown below

Intercept = -5.096= -K2 and hence K2= 5.096

M= 1/K2= 1/5.096= 0.197

Slope = 1.01 = K3CEO/M or K3= 1.01*0.197/0.01=19.897

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