9.2.2 Mechanisms

In developing some of the elementary principles of the kinetics of enzyme reactions, we shall discuss an enzymatic reaction that has been suggested by Levine and LaCourse as part of a system that would reduce the size of an artificial kidney.8 The desired result is the production of an artificial kidney that could be worn by the patient and would incorporate a replaceable unit for the elimination of the body’s nitrogenous waste products, such as uric acid and creatinine. In the microencapsulation scheme proposed by Levine and LaCourse, the enzyme urease would be used in the removal of urea from the bloodstream. Here, the catalytic action of urease would cause urea to decompose into ammonia and carbon dioxide. The mechanism of the reaction is believed to proceed by the following sequence of elementary reactions:

  1. The enzyme urease (E) reacts with the substrate urea (S) to form an enzyme–substrate complex (E · S):

    9-13

    image

    The reaction mechanism

    image

  2. This complex (E · S) can decompose back to urea (S) and urease (E):

    9-14

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  3. Or it can react with water (W) to give the products (P) ammonia and carbon dioxide, and recover the enzyme urease (E):

    9-15

    image

Symbolically, the overall reaction is written as

image

We see that some of the enzyme added to the solution binds to the urea, and some of the enzyme remains unbound. Although we can easily measure the total concentration of enzyme, (Et), it is difficult to measure either the concentration of free enzyme, (E), or the concentration of the bound enzyme (E · S).

Letting E, S, W, E · S, and P represent the enzyme, substrate, water, the enzyme–substrate complex, and the reaction products, respectively, we can write Reactions (9-13), (9-14), and (9-15) symbolically in the forms

9-16

image

9-17

image

9-18

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Here P = 2NH3 + CO2.

The corresponding rate laws for Reactions (9-16), (9-17), and (9-18) are

9-16A

image

9-17A

image

9-18A

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where the specific reaction rates are defined with respect to (E · S). The net rate of formation of product, rP, is

9-19

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For the overall reaction

image

we know –rS = rP.

This rate law (Equation 9-19) is of not much use to us in making reaction engineering calculations because we cannot measure the concentration of the enzyme substrate complex (E · S). We will use the PSSH to express (E · S) in terms of measured variables.

The net rate of formation of the enzyme-substrate complex is

rE·S = r1E·S + r2E·S + r3E·S

Substituting the rate laws, we obtain

9-20

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Using the PSSH, rE·S = 0, we can now solve Equation (9-20) for (E · S)

9-21

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and substitute for (E · S) into [Equation (9-19)]

9-22

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We still cannot use this rate law because we cannot measure the unbound enzyme concentration (E); however, we can measure the total enzyme concentration, Et.

We need to replace unbound enzyme concentration (E) in the rate law.

In the absence of enzyme denaturation, the total concentration of the enzyme in the system, (Et), is constant and equal to the sum of the concentrations of the free or unbounded enzyme, (E), and the enzyme–substrate complex, (E · S):

Total enzyme concentration = Bound + Free enzyme concentration.

9-23

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Substituting for (E · S)

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solving for (E)

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substituting for (E) in Equation (9-22), the rate law for substrate consumption is

9-24

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Note: Throughout the following text, Et ≡ (Et) = total concentration of enzyme with typical units such as (kmol/m3) or (g/dm3).

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