Explore BrainMass

Numerical problem related to Michaelis-Menten Reactions

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

An enzyme catalyzes a reaction at 20&#956;mol/min when the substrate [s] is 0.01M. The Km for this substrate is 1 x 10-5M Assuming michaelis-menten kinetics are followed, what would the reaction velocity be when the concentration [s] is (a) 1 x 10-5 M (b) 1 x 10-6 M?

https://brainmass.com/biology/biological-chemistry/numerical-problem-related-michaelis-menten-reactions-181519

SOLUTION This solution is FREE courtesy of BrainMass!

Please see the file attached. Numerical problem related to Michaelis-Menten Reactions are solved in very simple and stepwise easy to understand way.

An enzyme catalyzes a reaction at 20μmol/min when the substrate [s] is 0.01M. The Km for this substrate is 1 x 10-5M Assuming michaelis-menten kinetics are followed, what would the reaction velocity be when the concentration [s] is (a) 1 x 10-5 M (b) 1 x 10-6 M?

The Michaelis-Menten equation is expressed here as:

where Km = Michaelis-Menten rate constant, [S] = substrate concentration, v = initial rate of production of the product, and Vmax = maximum initial rate of production of the product .

given v = 20x10-6 mol/min
[S] = 0.01M
Km = 1x10-5 M

Therefore, Vmax = v/[S] (Km + [S]) = 20x10-6 mol/min/0.01M (1x10-5 M + 0.01M)
= 2.002 x 10-5 mol/min

(a) [s] = 1x10-5 M

v = (2.002 x 10-5 mol/min x 1x10-5 M ) / (1x10-5 M + 1x10-5 M)
= 1.001x10-5 mol/min

(b) [S] = 1x10-6 M

v = (2.002 x 10-5 mol/min x 2x10-6 M ) / (1x10-5 M + 2x10-6 M)
= 3.33x10-6 mol/min

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!