# Michaelis-Menton equation

(See attached file for diagrams)

1. For a Michaelis-Menten reaction, k1 = 5 x 107 M-1 s-1, k-1 = 2 x 104 s-1, and k2 = 4 x 102 s-1. Calculate Ks and Km for this reaction. Does substrate binding achieve equilibrium or the steady state?

2. At what substrate concentration will an enzyme having a kcat of 30 s-1 and a Km of 0.005 M show one quarter of its maximum rate?

3. If a classic noncompetitive inhibitor (KI = 5 x 10-7 M) is added to a concentration of

1 x 10-5 M in a reaction containing the enzyme in problem #2 and 5 x 10-3 M substrate, what percentage of enzyme activity will remain?

4. If the addition of a competitive inhibitor to a concentration of 1 x 10-5 M in a reaction containing the enzyme in problem #2 yields an apparent Km = 1.5 x 10-2 M, what is the KI for the inhibitor?

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#### Solution Preview

Please see the attached file.

1. For a Michaelis-Menten reaction, k1 = 5 x 107 M-1 s-1, k-1 = 2 x 104 s-1, and k2 = 4 x 102 s-1. Calculate Ks and Km for this reaction. Does substrate binding achieve equilibrium or the steady state?

k2 + k-1

As Km = -------------------

k1

therefore putting the values for k1, k-1 and k2,

we get,

(4 X 102) + (2 x 104)

Km = -------------------------- = 4.08 X 10-4 M s

5 x 107

2. At what substrate concentration will an enzyme having a kcat of 30 s-1 ...

#### Solution Summary

Solution provides detailed answers for 4 questions regarding the Michaelis-Menten reaction.