Example CD7-4
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Solve for the concentrations of the complexes (IE) and (IES) in terms of the free enzyme concentration, (E), and substrate concentration, (S), in a manner similar to that for (ES); substitute these expressions in the total enzyme balance and then obtain the rate law for noncompetitive inhibition. (Let Ks, KI,,and represent the equilibrium constants for reactions 1 through 4, respectively.) | |||
Solution | |||
For reaction 2, | |||
(EI) =KI (I)(E) |
(CDE7-4.1) | ||
We can use either Equation (CD7-19), | |||
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(CDE7-4.2) | ||
or (CD7-20), | |||
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(CDE7-4.3) | ||
to express IES in terms of IE,
and S. The mole balance on the total amount of enzyme, bound and free (E), is |
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Et = E + ES + IE + IES |
(CDE7-4.4) | ||
Substituting Equations (CD7-24), (CDE7-4.1), and (CDE7-4.2) into (CDE7-4.4) and rearranging gives us | |||
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(CDE7-4.5) | ||
The rate of formation of the product for a noncompetitive inhibition of an enzymatic reaction is | |||
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(CDE7-4.6) | ||