The rate of the hydrolysis of sucrose to glucose is quite slow in the absence of

Question

The rate of the hydrolysis of sucrose to glucose is quite slow in the absence ofa catalyst.

Sucrose + water ---> glucose + fructose

c. Determine the velocity of the uncatalyzed reaction when the concentration ofsucrose is 0.050M. The rate law is: rate = k [sucrose]. The rate constant for theuncatalyzed reaction is 5.0 x 10-11 s-1.

d. Determine the velocity of the catalyzed reaction when the concentration ofsucrose is 0.050M. The rate law is: rate = k [sucrose]. The rate constant for thecatalyzed reaction is 1.0 x 10^4 s-1.

e. The enzyme catalyzed reaction has a KM of 0.135 mM and a Vmax of 65?molmin-1. What is the reaction velocity when the concentration of sucrose is 1.0 mM?

**I know the answers, I need to understand the steps in obtaining the answers. For C and D, my understanding is that the velocity would be (k)(constant). But (0.050M)(5.0*10^-11 s-1) does not give me the correct answer. This is the same for D. For E, I think the equation would be Velocity=(Vmax)(s)/(Km+s). But I am unable to obtain the correst answer using this equation.Thanks!

Explanation / Answer

c) Using the given rate law:

rate = k[sucrose]

rate = 5.0 x 10-11 s-1 x 0.05 = 2.5 x 10^-12 M/s

d) rate = k[sucrose]

rate = 1.0 x 10^4 s-1 x 0.05 = 5 x 10^2 M/s

e) Using the Michaelis Menton eqn.

V = Vmax(s)/(Km +S)

V = 0.065 x 1 / (1+ 0.135)

= 5.72 x 10^-2 mmol/min = 5.72 x 10^-5 mol/s

NOTE: What you did in parts c and d are absolutely correct. Hence I feel that the only possibility here is that whoever has calculated the answers that you have mentioned before have calculated by using the concentration value as 0.5 M instead of 0.05 M. That factor of 1/10 is the only change between your answers and the ones that you have been told!

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