Problem Statement: Consider the following enzymatic reaction mechanism where an enzyme E combines with a substrate molecule S to form an active complex ES, which then leads a product P. Both steps are reversible.

           k1          k2
  E + S <------> ES <------> E + P
           k-1         k-2

1. List the dynamic equations that describe the above elementary reaction steps for all species. (d[S]/dt=?, d[E]/dt=?, d[ES]/dt=?, d[P]/dt=?)

   Solution:

    dS/dt = -k1*E*S + k-1*ES
    dE/dt = -k1*E*S + k-1*ES + k2*ES - k-2*E*P
   dES/dt =  k1*E*S - k-1*ES - k2*ES + k-2*E*P
    dP/dt =  k2*ES  - k-2*E*P
   I.C.: S(0)=S0, E(0)=E0, ES(0)=0, P(0)=0.
   

2. Derive the rate expression v=dp/dt as a function of S, P, and E₀. Set up the equations; solve them when time permits. Clearly state your assumptions. The following is the eventual expression, provided here in case you think you need it to answer questions in the next steps.

   v=(vms*s-vmp*p)/(Km+s+Kp*p)
   

   Solution:

   Definition of rate:               dP/dt =  v = k2*ES  - k-2*E*P
   Conservation of enzyme:           E0 = E + ES
   Equil. or QSS assumption for ES:  dES/dt = 0 = k1*E*S - k-1*ES - k2*ES + k-2*E*P
   From these three equations, solve for three unknowns: E, ES, and v.
   

3. When does the rate expression derived in the last step ceases to be valid?

   Solution:

   The assumption from the last step is valid when E₀<<S₀.

4. Do you expect product to enhance or depress the reaction rate? Why?

   Solution:

   Since the last reversible step means that P competes with S to bind to E, we have competitive inhibition of the enzyme. The reaction rate is depressed.

5. What is the product yield in terms of the number of moles of product formed per mole of substrate consumed? What is the ratio of P/S when the system is allowed sufficient time for the reaction to proceed?

   Solution: From the overall stoichiometry, one S molecule is converted into one P molecule; thus, the product yield is 1. At equilibrium, P_eq/S_eq=K_eq=k₁*k₂/k_-1k_-2=v_ms/v_mp. The following Mathcad file contains the answer to these questions.

   • PDF Version (inhibit5.pdf)
   • Mathcad File (inhibit5.mcd)

6. For this part, assume that the enzyme deactivates with time in a first-order manner at a rate of k_d in a batch reactor and that the reactor is initially charged with an excess of substrate.

   General Solution:

   With an excess substrate, the rate expression does not depend on S or P (i.e., v=v_m=k*E). The following Mathcad file answers this part.

   • PDF Version (opt-temp.pdf)
   • Mathcad File (opt-temp.mcd)
   1. Find how enzyme activity changes with time.

      Solution:

      
      1st-order deactivation:  dE(t)/dt=-kd*E(t)
      Integrating the above equation with E(0)=E0 yields:
        E(t)=E0*exp(-kd*t)
      

   2. Find how product concentration changes with time in a batch reactor.

      Solution:

      v=dp/dt=k2*E(t)
      Integrating the above equation with E(t)=E0*exp(-kd*t) gives:
      p=k2/kd*E0*(1-exp(-kd*t))
      

   3. Do you expect the rate of reaction (in terms of the number of moles of product formed per unit time) to increase or decrease with temperature? Why?

      Solution:

      rate of reaction = v = k2*E
      Rate of reaction increases with k2, which increases with temperature.
      (All reaction rates, whether desirable or not, increase with temperature!)
      

   4. What is the product yield (in terms of the number of moles of product formed per mole of enzyme initially present) when the system is allowed indefinite time for the reaction to proceed?

      Solution:

      Substitute t=infinity to the expression of p in part ii) yields,
        pinfinity=k2/kd*E0
      Thus,
        product yield = pinfinity/E0=k2/kd
      

   5. Do you expect the product yield to increase or decrease with temperature? Why?

      Solution:

      p_infinity/E₀ decreases with increasing temperature, because as the deactivation term k_d in the denominator becomes more dominant with increasing temperature compared to the k₂ term in the numerator.

   6. Do you expect the product productivity (in terms of the number of moles of product formed per unit time reacted per unit enzyme initially present) to increase or decrease with temperature? Why?

      Solution:

      For a given reaction time, product productivity initially increases with temperature, reaches a maximum, then decreases with further increase in temperature; the temperature at which product productivity reaches a maximum shifts to a lower value as the reaction time increases. At a given temperature, the productivity decreases monotonically as reaction time increases because of deactivation.

   7. Of the last three measures (the rate of reaction, product yield, and product productivity), which one would you consider in choosing the operating temperature? If you choose none of the above, give an appropriate measure.

      Solution:

      None of the above three. In business practice, one would choose to maximize profit, more specifically the rate of return on investment. Never assume that high productivity means more profit, especially when high productivity is realized at a loss. (If it costs you $2 to crank out a $1 product, the faster you crank out your product, the more money you lose!)

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Biochemical Engineering -- Derivation of Enzymatic Reaction Rate /w Product Inhibition
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Nam Sun Wang Department of Chemical & Biomolecular Engineering University of Maryland College Park, MD 20742-2111 301-405-1910 (voice) 301-314-9126 (FAX) e-mail: nsw@umd.edu ©1996-2007 by Nam Sun Wang