An enzyme catalyzes a reaction at a rate of 20 micromolar/min.mg when the concentration of substrate (S) is at 0.01 M. The Km of the enzyme for this substrate is 1 x 10-5 M. Assuming that Michaelis-Menten kinetics are followed, what will the reaction rate be when the concentration of S is 1 x 10-6 M?

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Asked Oct 13, 2019
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An enzyme catalyzes a reaction at a rate of 20 micromolar/min.mg when the concentration of substrate (S) is at 0.01 M. The Kof the enzyme for this substrate is 1 x 10-5 M. Assuming that Michaelis-Menten kinetics are followed, what will the reaction rate be when the concentration of S is 1 x 10-6 M?

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Expert Answer

Step 1

The Michaelis-Menten equation of an enzyme catalysed reaction is given as:

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Vmax S] Vo KmS Where, Vo rate of an enzyme catalysed reaction maximal rate of that enzyme catalysed reaction Vmax S] substrate concentration Michaelis-Menten constant of that enzyme catalysed reaction Km

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Step 2

Given,

Rate of an enzyme catalysed reaction, Vo = 20 micromolar/min.mg

Concentration of substrate, [S] = 0.01 M

 Kof the enzyme for this substrate = 1 × 10-5 M

The maximal rate can be calculated using the Michaelis-Menten equation as :

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VmaxS Vo (KmS) Vmax S] Vo (Km [S]) O Vo (Km S) Vmax 20 micromolar/min.mg (1 x 10-5 M 0.01 M) Vmax 0.01 M 20 micromolar/min.mg (0.01001 M) Vmax 0.01 M Vmax 20.02 micromolar/min.mg

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Step 3

Now, the concentration of S, [S] is 1 × 10-6&n...

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Vmax [S] Vo KmS 20.02 micromolar/min.mg (1 x 10-6 M) Vo= (1 x 10-5 M) (1 x 10-6 M) 20.02 micromolar/min.mg (1 x 106 M) Vo= (1.1 x 10-5 M) Vo 1.82 micromolar/min.mg

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