Biochemistry Online: An Approach Based on Chemical Logic

Biochemistry Online

CHAPTER 9 - SIGNAL TRANSDUCTION

E:  Metabolic Control Analysis and Systems Biology  

BIOCHEMISTRY - DR. JAKUBOWSKI

04/16/16

Learning Goals/Objectives for Chapter 9E:
After class and this reading, students will be able to

  • write mass balance expressions for Eo and use it to calculate the fraction of each enzyme form assuming rapid equilibrium;
  • draw "wire" diagrams showing activation and inhibition for coupled reactions;

E3. Metabolic Control Analysis and Simple Enzyme Inhibition


Biochemists model complex enzyme-catalyzed reactions in the presence and absence of modifiers (either activators or inhibitors) to develop mechanisms for the reactions. The following kinetic parameters are experimentally determined by fitting initial velocity (vo) vs substrate concentration ([S]) through nonlinear fitting algorithms.

 


An example of competitive inhibition illustrates a common type of analysis for such reactions.

comp_inhib_diagram

A modern pictorial depiction of a simple irreversible, enzyme-catalyzed reaction of substrate S going to product P with inhibition by the product and by an added inhibitor is shown below:

 

moderncompinhib

Consider the simple enzyme-catalyzed reaction for a reversible conversion of substrate S to product P that has 3 reversible steps.

rxderivation

If the forward (f) and reverse (r) chemical reaction steps (reactions 2 and -2) were irreversible and written separately, simple Michaelis-Menten equations could be written for each

 

rxderiv1

For the actual reversible reactions for step 2, the net forward rate cannot be found by simple subtraction of the two equations above as the differential equations describing the simple forward and reverse rates don’t account for the reverse steps

rxderv2

A simple derivation (assuming rapid equilibrium for both forward and reverse steps) can be made for the net forward reaction. Again consider the following enzyme catalyzed reaction:

rxderivation

You may remember that for the isolated E + S <----> ES and for E + P <----> EP reactions, the simple dissociation constants, KS and KP are given by

rxderiv3

The rapid equilibrium assumption states that the rate of dissociation of ES and EP, which are both physical steps, are much faster than the rate of the chemical conversion steps for each complex. Hence k-1 >> k2 and k3 >> k-2, so the relative amounts of ES and EP can be determined from the dissociation constants as shown above.

Mass conservation of enzyme gives

massconser

From this we can get the fractional amount of both ES and EP

rxeq4
Now we can derive the rate equation for the net forward reaction for the rapid equilibrium case:

rxeq5

Knowing that k2E0 and k-2E0 represent the maximal velocities, Vf and Vr, respectively, the equation becomes:

rxeq6

An equation of similar form can be derived from the steady state assumption. This equation is clearly different from the earlier equation (4) derived by intuitively assuming a simple subtraction of the irreversible forward and reverse rates which we have now shown to be invalid by comparison.

rxderv2

Programs like COPASI have many built in equations for velocities of many enzyme-catalyzed reactions that have similar forms. Two are shown below:

Reversible Michaelis-Menten: revMMeq

 

Competitive Inhibition Reversible: compinhreveq

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