Explanation: Michaelis-Menten MechanismExplanation: Michaelis-Menten Mechanism

E is free enzyme and ES is an enzyme-substrate complexE is free enzyme and ES is an enzyme-substrate complex

All of the reactions in the M-M Scheme are All of the reactions in the M-M Scheme are elementary kinetic stepselementary kinetic steps..

VVss is found to be directly proportional to the total enzyme is found to be directly proportional to the total enzyme

concentration (Econcentration (Eoo) )

VVss ~ (E ~ (Eoo) )

(S) concentration required to reach half maximum initial velocity(S) concentration required to reach half maximum initial velocity (V(Vi i = V= VSS/2) found to be independent of (E/2) found to be independent of (Eoo). ). (S)(S)1/21/2 = K = KMM

E + S ES E + PE + S ES E + Pk 1⏐ → ⏐ ⏐

k −1← ⏐ ⏐ ⏐ k 2⏐ → ⏐ ⏐

It may generally be assumed that (S) >> (E) since E are so efficientIt may generally be assumed that (S) >> (E) since E are so efficient they catalyze reaction at very small concentration.they catalyze reaction at very small concentration.

(E(Eoo) = (E) + (ES) mass balance [(E) = (E) + (ES) mass balance [(Eoo) is the total ) is the total

enzyme concentration]enzyme concentration]

Steady State assumption:Steady State assumption:

0 = k0 = k11(E) (S) -k(E) (S) -k-1-1 (ES) - k (ES) - k22 (ES), but (E (ES), but (E00) = (E)+(ES)) = (E)+(ES)

= kk11 (E) (S) - k (E) (S) - k-1-1 (ES) - k (ES) - k22 (ES) (ES)d(ES)

dt

0 =d(ES)

dt

0 = k0 = k11 {(E {(E00) - (ES)} (S) - (k) - (ES)} (S) - (k-1-1 + k + k22) (ES)) (ES)

(ES) =k1(E0 )(S)

k−1 +k2 +k1(S)

Rate =d(P)dt

=k2 (ES)

R =d(P)dt

=k1k 2(E0 )

k1 +k−1 +k2

(S)

MechanismMechanism kk11

E + S E + S ES ES Step 1Step 1 kk-1-1

ES ES E + S E + S Step 2Step 2 kk22

ES ES P P Step 3Step 3All are elementary kinetic steps.All are elementary kinetic steps.

Step 1Step 1 Step 2Step 2 Step 3Step 3

Divide top and bottom by kDivide top and bottom by k11 to get to get

= k= k11EE00/[k/[k11+(k+(k-1-1+k+k22)/(S)])/(S)]

dP

dt=

k2(E0 )

1 + K m

(S)Here we define KHere we define Kmm = =

k−1 +k2

k1

dP

dt=

k2(E0 )

1 + K m

(S)

dP

dt ⎛ ⎝

⎞ ⎠max

=lims→ ∞

=k2 (E 0 )=VsCase I : Case I : (max rate) (max rate)

Case II :Case II : Half Maximum RateHalf Maximum RatedP

dt=

Vs

2=

k2

2(E 0 )⇒

Bonus * Bonus * Bonus * Bonus * Bonus * BonusBonus * Bonus * Bonus * Bonus * Bonus * Bonus

{V{VSS proportional proportional

to (Eto (E00)})}

What is (S) when dP/dt = (VWhat is (S) when dP/dt = (VSS)/2? )/2?

Depends linearly on [S] in region of low substrate concentration.Depends linearly on [S] in region of low substrate concentration.

k2

2(E0 ) =

k2 (E0 )

1 + K m

(S)12

1+K m

(S)12

=2 (S)12

=K m

Case III : Case III : Km

(S)>>1

ie. Kie. Kmm >> (S) [small (S)] >> (S) [small (S)]

dP

dt=

k2(E0 )

1 + K m

(S)

dP

dt=

k2 (E 0)K m

(S)

(S)(S)1/21/2 is the substrate concentration when the is the substrate concentration when the

initial rate reaches half its maximum value.initial rate reaches half its maximum value.

VVSS/2/2

dP

dt=V =

Vmax

1 + K m

(S)1

V=

1Vmax

+Km

Vmax

1S

Plot1

Vversus

1

(S)

gives straight line of slope gives straight line of slope and intercept [V and intercept [Vmaxmax]]-1-1Km

Vmax

KKmm is rate at which ES decomposes by two mechanisms (k is rate at which ES decomposes by two mechanisms (k-1-1 or k or k22))

divided by rate of formation ES.divided by rate of formation ES.

Large KLarge Kmm weak binding of E to S weak binding of E to S

Small KSmall Kmm strong binding of E to S strong binding of E to S

{Lineweaver-Burke Plot}

KKmm = = k−1 +k2

k1

MechanismMechanism kk11

E + S E + S ES ES Step 1Step 1 kk-1-1

ES ES E + S E + S Step 2Step 2 kk22

ES ES P P Step 3Step 3

Invert thisInvert thisequation toequation toget---get---

QuickTime™ and aAnimation decompressor

are needed to see this picture.

Application of chemical kinetics to Application of chemical kinetics to ecologicalecological and and toxicologicaltoxicological problems problems

While extremely beneficial for protection of crops, While extremely beneficial for protection of crops, pesticides can have serious environmental impact pesticides can have serious environmental impact

Possible deleterious consequences include seepage of Possible deleterious consequences include seepage of these otherwise these otherwise helpfulhelpful chemicals into soil and ground water chemicals into soil and ground water

I. Application of enzyme kinetics:I. Application of enzyme kinetics: Degradation of organophosphate pesticidesDegradation of organophosphate pesticides

Enzymes can be used to catalyze degradation of pesticidesEnzymes can be used to catalyze degradation of pesticides

Case study: organophosphate pesticides Case study: organophosphate pesticides

abundantabundant highly toxichighly toxic “ “neutralized” via reactionneutralized” via reaction

The The samesame enzyme catalysts which can neutralize these pesticides enzyme catalysts which can neutralize these pesticidescan also be used to can also be used to detoxifydetoxify chemical nerve agents. chemical nerve agents.

nerve agents are structurally similar to nerve agents are structurally similar to organophosphate pesticidesorganophosphate pesticides

both contain organophosphate estersboth contain organophosphate esters

paraoxonparaoxon

Nerve gas used Nerve gas used in Japanesein Japanesesubway attack.subway attack.

Degradation Degradation product producedproduct producedin body from parathionin body from parathion

GeneralGeneralformula forformula forpesticidespesticides

Recent studies have shown that enzymes, which effectively degrade Recent studies have shown that enzymes, which effectively degrade organophosphates, can be incorporated into polymers -- organophosphates, can be incorporated into polymers -- specifically, specifically, foamsfoams -- in order to aid in their practical application. -- in order to aid in their practical application.

One enzyme under current investigation is One enzyme under current investigation is organophosphorus organophosphorus hydrolasehydrolase (OPH) a.k.a. (OPH) a.k.a. phosphotriesterasephosphotriesterase..

[derived from [derived from Escherichia coliEscherichia coli]]

The degradation reaction of organophosphates works via hydrolysis.The degradation reaction of organophosphates works via hydrolysis.

Attack by waterAttack by water

(yellow in solution)(yellow in solution)

Here, ENZ = enzyme = OPH (a.k.a. phosphotriesterase)Here, ENZ = enzyme = OPH (a.k.a. phosphotriesterase)

The kinetics can be monitored by following the production of The kinetics can be monitored by following the production of p-nitrophenol, a yellow colored chemical, spectroscopically.p-nitrophenol, a yellow colored chemical, spectroscopically.

Enzyme kinetics can be used to determine the efficiency of OPHEnzyme kinetics can be used to determine the efficiency of OPHwhen it is integrated into a polymeric when it is integrated into a polymeric foamfoam compared to OPH’s compared to OPH’s efficiency in efficiency in solutionsolution..

evaluate Kevaluate KMM and V and Vmaxmax and compare and compare evaluate Kevaluate KMM and V and Vmaxmax and compare and compare

M.M. Benning, J.M. Kuo, F.M. Raushel and H.M. Holden, Biochemistry, Vol. 33 pp. 15001-15007 (1994).

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

V = rate = kV = rate = k22[ES][ES]

To obtain VTo obtain Vmaxmax and K and KMM from this data: from this data:

Method IMethod I extrapolate plot above to find Vextrapolate plot above to find Vmaxmax

find Vfind Vmaxmax/2/2

find [substrate] corresponding to the value at Vfind [substrate] corresponding to the value at Vmaxmax/2/2

This [substrate] is KThis [substrate] is KMM

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 0.5 1 1.5 2 2.5

rate

rate (mM/min)

[substrate] (mM)Ap

pea

ran

ce o

f p

-nit

rop

hen

ol p

rod

uct

V=

First, let’s look First, let’s look at the activity of at the activity of OPH in solutionOPH in solution

OPH hydrolysis of paraoxonOPH hydrolysis of paraoxon

(mM = millimolar)

Here, VHere, Vmaxmax = 0.063 mM/min and K = 0.063 mM/min and KMM = 0.08 mM = 0.08 mM

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 0.5 1 1.5 2 2.5

rate (mM/min)

[substrate] (mM)

Vmax

Km

Vmax/2

Ap

pea

ran

ce o

f p

-nit

rop

hen

ol

V=

OPH hydrolysis of paraoxonOPH hydrolysis of paraoxon

Extrapolate plot to find VExtrapolate plot to find Vmaxmax

find Vfind Vmaxmax/2/2

find [substrate] corresponding to find [substrate] corresponding to the value at Vthe value at Vmaxmax/2/2

This [substrate] is KThis [substrate] is KMM

This yields VThis yields Vmaxmax = 1/[14.3(min/mM)] = 1/[14.3(min/mM)]-1-1 = 0.07 mM/min, = 0.07 mM/min,

and Kand KMM = [1.4 (min)] = [1.4 (min)] •• V Vmax max = 0.098 mM= 0.098 mM

10

20

30

40

50

60

0 5 10 15 20 25 30

1/rate (min/mM)

1/[substrate] (1/mM)

The line that fits the data has the following form:The line that fits the data has the following form:y = 14.3 min/mM + 1.4 min y = 14.3 min/mM + 1.4 min •• x x

V

Method II: plot Method II: plot 1/V versus 1/[substrate]1/V versus 1/[substrate]

(Lineweaver-Burke Plot)(Lineweaver-Burke Plot)

vs (0.08)vs (0.08)vs (0.063)vs (0.063)

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0 0.5 1 1.5 2 2.5

rate (mM/min)

[substrate] (mM)

Ap

pea

ran

ce o

f p

-nit

rop

hen

ol

V=

Again, Again, VVmaxmax can be found from the asymptote of the curve, and can be found from the asymptote of the curve, and KKMM is is

that substrate concentration which corresponds to that substrate concentration which corresponds to VVmaxmax/2/2..

Now, let’s see how well OPH hydrolyzes paraoxon when it is Now, let’s see how well OPH hydrolyzes paraoxon when it is encapsulated in foam.encapsulated in foam.

VVmaxmax

VVmaxmax for OPH in foam = 0.035 mM/min, while K for OPH in foam = 0.035 mM/min, while KMM = 0.19 mM. = 0.19 mM.

VVmaxmax/2/2

KKMM

We get similar results from a Lineweaver-Burke plot:We get similar results from a Lineweaver-Burke plot:

0

50

100

150

200

250

0 5 10 15 20 25 30

1/rate (min/mM)

1/[substrate] (1/mM)

from the y-intercept, Vfrom the y-intercept, Vmaxmax = 0.040 mM/min; = 0.040 mM/min;

from the slope, Kfrom the slope, KMM = 0.24 mM = 0.24 mM

Now, we can compare these parameters for the cases whereNow, we can compare these parameters for the cases whereOPH is in solution and integrated into foam:OPH is in solution and integrated into foam:

This comes from:This comes from:V

(vs 0.19)(vs 0.19)(vs 0.035)(vs 0.035)

VVmaxmax((mMmM//minmin))

KKMM(mM)(mM)

in solutionin solution in foamin foam 0.0630.063 0.0350.035 0.080.08 0.190.19

VVmaxmax decreasesdecreases by a factor of 2 when the enzyme OPH is by a factor of 2 when the enzyme OPH is

encapsulated in foam and Kencapsulated in foam and KMM increasesincreases by approximately by approximately

the same factor.the same factor.

What does this mean????What does this mean????

The lower the value for The lower the value for KKMM, the , the betterbetter the enzyme works the enzyme works

(tighter binding) as a catalyst for the hydrolysis of paraoxon.(tighter binding) as a catalyst for the hydrolysis of paraoxon.

OPH is a tighter binding (more efficient) catalyst in solutionOPH is a tighter binding (more efficient) catalyst in solution than in foam!!!than in foam!!!

V = rate =k 2[OPH 0]

1 +k – 1 + k 2

k 1

1[paraoxon]

KM =k –1 + k2

k1

KKMM(mM)(mM)in solutionin solution in foamin foam 0.080.08 0.190.19

OPH + paraoxon <k 1

k – 1> [OPH...paraoxon]

k 2

H2O> OPH + p–nitrophenol + product

Why is OPH less efficient as a catalyst (has smaller VWhy is OPH less efficient as a catalyst (has smaller Vmaxmax) )

when used in a foam?when used in a foam?

Let’s look at Let’s look at VVmaxmax for some clues: for some clues:

RecallRecall

where where [OPH[OPH00]] is the amount of enzyme present. is the amount of enzyme present.

In order to get the enzyme into foam, certain groups of atomsIn order to get the enzyme into foam, certain groups of atoms(called functional groups) within the enzyme (e.g. NH(called functional groups) within the enzyme (e.g. NH22 and OH) and OH)

react with the foam, immobilizing a percentage of the total enzyme.react with the foam, immobilizing a percentage of the total enzyme.

V

max= lim

[paraoxon] →∞

k 2 [OPH0]

1 +KM

[paraoxon]

= k 2[OPH 0]

There are 2 possible effects at work here:There are 2 possible effects at work here:

1. “Inhibition” effect:1. “Inhibition” effect:

That is, the foam is NOT inert!That is, the foam is NOT inert!

Not exactly ...Not exactly ... for use on pesticides, delivery of catalysts like OPH via for use on pesticides, delivery of catalysts like OPH via foams is likely not the best means to degrade organo- foams is likely not the best means to degrade organo- phosphates -- reduced efficiencyphosphates -- reduced efficiency

Is the decrease in catalytic efficiency when OPH is Is the decrease in catalytic efficiency when OPH is used in a foam a negative result?used in a foam a negative result?

2. Change in the rate constant k2. Change in the rate constant k22..

A decrease in kA decrease in k22 may occur when OPH is used in foam, leading may occur when OPH is used in foam, leading

to a smaller Vto a smaller Vmaxmax as compared to that parameter measured when as compared to that parameter measured when OPH is in solution.OPH is in solution.

for use on other toxins (e.g. nerve agents) delivery for use on other toxins (e.g. nerve agents) delivery in foams is extremely beneficialin foams is extremely beneficial

OPH is still highly effective and delivery viaOPH is still highly effective and delivery viafoams is more convenientfoams is more convenient

ReferencesReferences

K.E. LeJeune and A. J. Russell, Biotechnology and K.E. LeJeune and A. J. Russell, Biotechnology and Bioengineering, Vol. 51, pp. 450-457 (1996)Bioengineering, Vol. 51, pp. 450-457 (1996)

K.E. LeJeune and J.R. Wild, Nature, Vol. 395, pp. 27-28K.E. LeJeune and J.R. Wild, Nature, Vol. 395, pp. 27-28 (1998)(1998)

K.E. LeJeune, A.J. Mesiano, S.B. Bower, J.K. Grimsley, J.R. K.E. LeJeune, A.J. Mesiano, S.B. Bower, J.K. Grimsley, J.R. Wild and A.J. Russell, Biotechnology and Bioengineering, Wild and A.J. Russell, Biotechnology and Bioengineering,

Vol. 54 pp. 105-114 (1997)Vol. 54 pp. 105-114 (1997)

S.E. Manahan, Environmental Chemistry (5S.E. Manahan, Environmental Chemistry (5thth ed), Lewis ed), Lewis Publishers: Chelsea, MI (1991)Publishers: Chelsea, MI (1991)

D.J. Hanson, Chemical and Engineering News, pp. 20-22 D.J. Hanson, Chemical and Engineering News, pp. 20-22 (Sept. 28,1998)(Sept. 28,1998)

M.M. Benning, J.M. Kuo, F.M. Raushel and H.M. M.M. Benning, J.M. Kuo, F.M. Raushel and H.M. Holden, Biochemistry, Vol. 33 pp. 15001-15007 (1994).Holden, Biochemistry, Vol. 33 pp. 15001-15007 (1994).