# Numerical problem related to Michaelis-Menten Reactions

Please explain:

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?

Am I looking for Vo or Vmax or both? Please help!

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## 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 20Î¼mol/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

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