Explanation: Michaelis-Menten Mechanism Vs is found to be directly proportional to the total enzyme concentration (Eo) Vs ~ (Eo) (S) concentration required to reach half maximum initial velocity (Vi = VS/2) found to be independent of (Eo). (S)1/2 = KM Explanation: Michaelis-Menten Mechanism E + S ES E + P E is free enzyme and ES is an enzyme-substrate complex All of the reactions in the M-M Scheme are elementary kinetic steps. It may generally be assumed that (S) >> (E) since E are so efficient they catalyze reaction at very small concentration. (Eo) = (E) + (ES) mass balance [(Eo) is the total enzyme concentration]

Steady State assumption: = k1 (E) (S) - k-1 (ES) - k2 (ES) Step 1 Step 2 Step 3 Steady State assumption: 0 = k1(E) (S) -k-1 (ES) - k2 (ES), but (E0) = (E)+(ES) 0 = k1 {(E0) - (ES)} (S) - (k-1 + k2) (ES) = k1E0/[k1+(k-1+k2)/(S)] Mechanism k1 E + S  ES Step 1 k-1 ES  E + S Step 2 k2 ES  P Step 3 All are elementary kinetic steps. Divide top and bottom by k1 to get 

Bonus * Bonus * Bonus * Bonus * Bonus * Bonus Here we define Km = Case I : (max rate) {VS proportional to (E0)} Case II : Half Maximum Rate What is (S) when dP/dt = (VS)/2? 

(S)1/2 is the substrate concentration when the initial rate reaches half its maximum value. VS/2 Case III : ie. Km >> (S) [small (S)] Depends linearly on [S] in region of low substrate concentration.

Invert this equation to get--- Mechanism k1 E + S  ES Step 1 k-1 ES  E + S Step 2 k2 ES  P Step 3 {Lineweaver-Burke Plot} gives straight line of slope and intercept [Vmax]-1 Km is rate at which ES decomposes by two mechanisms (k-1 or k2) divided by rate of formation ES. Km = Large Km  weak binding of E to S Small Km  strong binding of E to S

Application of chemical kinetics to ecological and toxicological problems I. Application of enzyme kinetics:  Degradation of organophosphate pesticides Enzymes can be used to catalyze degradation of pesticides While extremely beneficial for protection of crops, pesticides can have serious environmental impact Possible deleterious consequences include seepage of these otherwise helpful chemicals into soil and ground water Case study: organophosphate pesticides  abundant  highly toxic  “neutralized” via reaction

The same enzyme catalysts which can neutralize these pesticides can also be used to detoxify chemical nerve agents.  nerve agents are structurally similar to organophosphate pesticides Nerve gas used in Japanese subway attack. General formula for pesticides  both contain organophosphate esters Degradation product produced in body from parathion paraoxon

Recent studies have shown that enzymes, which effectively degrade organophosphates, can be incorporated into polymers -- specifically, foams -- in order to aid in their practical application. One enzyme under current investigation is organophosphorus hydrolase (OPH) a.k.a. phosphotriesterase. [derived from Escherichia coli] Attack by water The degradation reaction of organophosphates works via hydrolysis. (yellow in solution)

Here, ENZ = enzyme = OPH (a.k.a. phosphotriesterase) M.M. Benning, J.M. Kuo, F.M. Raushel and H.M. Holden, Biochemistry, Vol. 33 pp. 15001-15007 (1994). The kinetics can be monitored by following the production of p-nitrophenol, a yellow colored chemical, spectroscopically. Enzyme kinetics can be used to determine the efficiency of OPH when it is integrated into a polymeric foam compared to OPH’s efficiency in solution.  evaluate KM and Vmax and compare

Let’s review how these parameters are derived in enzyme kinetics:

V = rate = k2[ES]

 OPH hydrolysis of paraoxon First, let’s look at the activity of Appearance of p-nitrophenol product V= First, let’s look at the activity of OPH in solution  (mM = millimolar) To obtain Vmax and KM from this data: Method I  extrapolate plot above to find Vmax  find Vmax/2  find [substrate] corresponding to the value at Vmax/2  This [substrate] is KM

OPH hydrolysis of paraoxon  Extrapolate plot to find Vmax Appearance of p-nitrophenol V=  find Vmax/2  find [substrate] corresponding to the value at Vmax/2  This [substrate] is KM Here, Vmax = 0.063 mM/min and KM = 0.08 mM

1/V versus 1/[substrate] (Lineweaver-Burke Plot) Method II: plot 1/V versus 1/[substrate] (Lineweaver-Burke Plot) The line that fits the data has the following form: y = 14.3 min/mM + 1.4 min • x V This yields Vmax = 1/[14.3(min/mM)]-1 = 0.07 mM/min, and KM = [1.4 (min)] • Vmax = 0.098 mM vs (0.063) vs (0.08)

Now, let’s see how well OPH hydrolyzes paraoxon when it is encapsulated in foam. Appearance of p-nitrophenol V= Vmax Vmax/2 KM Again, Vmax can be found from the asymptote of the curve, and KM is that substrate concentration which corresponds to Vmax/2. Vmax for OPH in foam = 0.035 mM/min, while KM = 0.19 mM.

We get similar results from a Lineweaver-Burke plot: from the y-intercept, Vmax = 0.040 mM/min; from the slope, KM = 0.24 mM (vs 0.035) (vs 0.19) This comes from: V Now, we can compare these parameters for the cases where OPH is in solution and integrated into foam:

What does this mean???? Vmax(mM/min) KM(mM) in solution in foam 0.063 0.035 0.08 0.19 Vmax decreases by a factor of 2 when the enzyme OPH is encapsulated in foam and KM increases by approximately the same factor. What does this mean????

The lower the value for KM, the better the enzyme works (tighter binding) as a catalyst for the hydrolysis of paraoxon. KM(mM) in solution in foam 0.08 0.19 OPH is a tighter binding (more efficient) catalyst in solution than in foam!!!

Why is OPH less efficient as a catalyst (has smaller Vmax) when used in a foam? Let’s look at Vmax for some clues: Recall where [OPH0] is the amount of enzyme present. There are 2 possible effects at work here: 1. “Inhibition” effect: In order to get the enzyme into foam, certain groups of atoms (called functional groups) within the enzyme (e.g. NH2 and OH) react with the foam, immobilizing a percentage of the total enzyme. That is, the foam is NOT inert!

2. Change in the rate constant k2. A decrease in k2 may occur when OPH is used in foam, leading to a smaller Vmax as compared to that parameter measured when OPH is in solution. Is the decrease in catalytic efficiency when OPH is used in a foam a negative result? Not exactly ...  for use on pesticides, delivery of catalysts like OPH via foams is likely not the best means to degrade organo- phosphates -- reduced efficiency  for use on other toxins (e.g. nerve agents) delivery in foams is extremely beneficial OPH is still highly effective and delivery via foams is more convenient

References K.E. LeJeune and A. J. Russell, Biotechnology and Bioengineering, Vol. 51, pp. 450-457 (1996) K.E. LeJeune and J.R. Wild, Nature, Vol. 395, pp. 27-28 (1998) K.E. LeJeune, A.J. Mesiano, S.B. Bower, J.K. Grimsley, J.R. Wild and A.J. Russell, Biotechnology and Bioengineering, Vol. 54 pp. 105-114 (1997) S.E. Manahan, Environmental Chemistry (5th ed), Lewis Publishers: Chelsea, MI (1991) D.J. Hanson, Chemical and Engineering News, pp. 20-22 (Sept. 28,1998) M.M. Benning, J.M. Kuo, F.M. Raushel and H.M. Holden, Biochemistry, Vol. 33 pp. 15001-15007 (1994).