Enzyme Reaction Order

8/12/2019 Enzyme Reaction Order

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CHAPTER 2A Kinetics Primerfor the Tissue Analyst

A rudimentary working knowledge of enzyme kinetics is of greatvalue in the use of enzymes as analytical tools. An analysis may failif too little enzymeisused, because of incomplete reaction, but it mayalso fail if too much enzyme is used. This might not be true if absolutely pure, single-action, enzymes were available.In thiscase,atremendousexcessofenzymecouldbeemployed andno harm would bedone.Even crystalline enzymes, however, are frequently contaminated with interfering activities, or the enzyme itselfmayhavea disturbingside action. Fortunately,if justthe right amountis employed, enzyme preparations that are only partially purified canoften be used successfully. The right amount of enzyme (the least amountthat willdothejob) can berather exactly defined.Inordertodeterminehow much enzyme shouldbe used,kinetic information isneeded.Thesubject ofenzymekinetics often fills the neophyte with fear. It is thehope here to present simple kinetics in such a way that even the relatively inexperienced tissue analyst can proceed with confidence.

Nonenzyme KineticsBefore discussing enzymekinetics,it may beusefultopresent somefundamental concepts applicable to all chemical reactions.

Order of the ReactionChemical reactions are defined as first order, second order, orhigher order if there are one, two, or more reactants. The decay ofa radioactive element is a first-order reaction; the combination of an

alcohol with an acid to form an ester is a second-order reaction. Thehydrolysis of an ester is also second order, but because one of the reactants(H2O) is present in enormous excess, the reaction has the mathematical form of a first-order reaction. It is called pseudo first order orusually simply first order. There is also a (pseudo) zero order23


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24 Passonneau and Lowryreaction, which may sound impossible. Any reaction that proceedswith constant velocity over the period of observationfitshedefinitionof(pseudo)zero order. For example, the rate of decay of radium overa 24-h periodis,for practical purposes, zero order, and when measuring the activity of an enzyme it is desirable to make the reaction asnearly zero order as possible.Only reactions following zero-, first-, or second-order kinetics areof interest here. Of these we shall mainly emphasize those of firstorder.

First -Order R eac t ionsConsider the first-order reaction:

Its velocity (v) can be described by the equation:v = Jfc[A] [Eq. 2-1]

As the reaction proceeds, the velocity will decrease exactly as [A]decreases (Fig.1).When Aishalf-gone,thevelocity willbehalf of theinitial velocity, and so forth. [A] is said to diminish by a "die-away"curveor by"logarithmic decrement";kis a"first-orderrateconstant."Note its meaning. It is equal to the fraction of A, which would beconverted toBin unit timeifthe velocitydid not falloff.For example,if it=0.3/min,thevelocityatany instantisalways0.3[A]/permin (i.e.,30%/min of the amount ofApresent at that moment).Equation (2-1) describes the situation at any moment in Fig.1,butwe needanequation thatsums upthe total change in[A]in an intervalof time (f). This (integrated) equation is:

2.3 log[Ay[A] = kt [Eq. 2-2]where[A]0is the initial amount of A. W henA ishalf-gone, 2.3 lo g(l/0.5) =kt=0.7 (actually 0.69). The time is the "half-time" orti/2,i.e.:

tm=0J/k [Eq.2-3]


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Kinetics Primer 25

0.1250.062

1.42 2Time in half-times)

Fig.1.Plotof afirst-orderreaction.For example, ifk=0.3/min,tin= 2.3 min. Most people find thehalf-timeofareaction easier to visualize than the rate constant. (A somewhat more erudite term thanthehalf-timeis the"relaxation time,"twhichissimplythereciprocal ofthefirst-orderrateconstant [Fig.1].)Note that after 1, 2, 4, and 6half-times, A willbe 50,75,94, and98.5%gone (i.e.,asimpleirreversiblefirst-order reaction will becomplete,formost practicalpurposes, in5 or6ha lf-times).This appliestoboth true and pseudo-first-order reactions, andtoenzymaticaswell asnonenzymatic reactions.REVERSIBLE FIRST-ORDER REACTIONS

If the reactionisreversible,thesituationismore complicated:ABk2

[Re.2 - 2]


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26 Passonneau and LowryThe net velocity isthedifference betweenthetw o velocities (v1andV2)inthe forward and reverse direction. The velocity equationis:

V =Vj-v2 =kj[A] -fc2[B] [Eq.2 -4]Thus, the initial velocity (i.e., when [B]= 0) isthe same asforanirreversible reaction, but the net velocity w ill diminish m ore rapidlyand will become zero whenk\[A]=&2[B]. W hen this hap pens,thesystem isin equilibrium ([A ]=[A]eq, [B]=[B ]eq ). Itisclear that:

= K,, [Eq.2-5][BL, k j

T his describes the fundamental factthat theequilibrium constant isdetermined by the two rate constants. Since [B]=[A]Q- [A ], [B] canbe eliminated from Eq. (2-4)togiv e:V =(* ,+k 2)[A] -fc2[A ]0 [Eq.2-6]

This can b eintegrated, andbytaking advantage ofEq. (2-5) theintegrated eq uation can be put intoasim ple form:[A]0-[A ] . ,

2.3 log = (kj+k2)t [E q - 2 7 ][A ]- [ A ] ^

T his is exactly analogoustoEq. (2 -2). T he half-time ofapproachtoequilibrium is:0.7h a T [E*281

and the reaction w ill be 50, 75 , and 98.5% as far asitw ill ever goin1 ,2 , and 6 half-times. If, for exam ple,k\= 0.3/m in,andki- 0.075/min,K cq= 4 (from Eq. [2-5]),i.e.,equilibrium w ill occur w hen [A] hasfallento20%of [A]G .The half-tim e (i.e., when [A ]=60%of [A ]G ) w ill


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Kinetics Primer 27be0.7/0.375= 1.9 m in. Note thattheapproachtoequilibrium is fasterthan ifthereactionhadnotbeenreversibleand [A]had been approaching zero instead of20%.S e c o nd- O r de r Re a c t io ns

Most analytical reactions have the form offirst-or zero-orderreactions, or something in between. However, we will encounter someanalytical reactions that follow a second-order time-course,and,moreover, a brief presentation of certain aspects ofthesecond-order reaction may be helpful in understanding enzyme kinetics. Consider theirreversible reaction:

kA + B -> C [Re. 2-3]The velocity equation is:

V = A][B] [Eq. 2-9]Thus,the concentrations of both A and B affect the velocity, and theconstantkhas a morecomplex meaning than in the case ofafirst-orderreaction. Itisnumerically equalto theinstantaneous rateexpressed asthe fraction of one reactant, which is converted to productinunit timewhen the other reactantispresent atunitconcentration.For example,supposek=0.3 L/mol/min(0.3M"1min-1).Thiswould mean that if [A]is \M ,theinstantaneous velocitywillcorrespondto thedisappearanceofB at therate of 30%/min. If[A]is only 0.0 lAf,the rateof disappearance ofBwill be only0.3%/min.Thus, a second-order rate constanthas both a time and a concentration dimension.The integrated form of Eq. (2-9), which would describe the time-course ofasecond-order reaction, will not be presented. However, inthe enzyme section below, a graphic representation will be given forthe time-course when [A] is varied relative to [B].

IfRe.(2-3) is reversible:* ;A + B ^ C [Re. 2-4]k 2


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28 Passonneau and Lowryi.e., itissecond orderinone direction and first order in the other. Thevelocity equation is:

V=k i[A][B]-Jt2[C] [Eq.2-10]At equilibrium,fci[A][B]= fa[C] and:

[A][B] k2Kcq= = [Eq.2-11][C] kj

It is important to realize the difference between Eq. (2-11) andEq.(2-5).Kcqhas a concentration dimension, and the position of equilibrium is no longer independent of absolute concentrations. For exam ple,letk\= 10M-1m h r1andk2=0.1m hr1 . From Eq.(2-11),Keq=0.0 IM=10mM. At equilibrium, if[A]= [B] = 0.1M, [C] = IM, i.e.,91%ofthe total; if[A]= [B]=0.1mM, [C]=0.001mM, i.e., only1%of thetotal. (Ifthereaction isA + B-C + D, the equilibrium constant hasno concentration term, and the position of equilibrium is, of course,independent of absolute concentrations.)A shift in equilibrium with concentration occasionally has analytical significance. For example, glutamate can be oxidized enzymati-cally with NAD +to forma-ketoglutarateandNH3.Two componentsreact to form three; therefore, the equilibrium position is concentration-dependent.In themillimolar concentrationrange theequilibriumisunfavorable, butin themicromolar range, the reaction can easily bedriven to essential com pletion.

E n z y m e K i n e t i c sThe kinetics of catalyzed reactions are more complicated than thoseof spontaneous reactions, because the catalyst must of necessity enterinto the reaction, usually by forming a transient complex with the

reactant(s). The velocity ofanenzymatic reaction will be determinedby the amount of enzyme and by some function ofthereactant (substrate), but in contrast to a spontaneous reaction, the velocity willusually not be directly proportional to reactant concentration.


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Kinetics Primer100

L_ 1 1 I L0 2 4 6 8[S] multiples of Km

Fig . 2 . V elocity of an enzy m e reaction as a fraction of substrate concentration.T his describes the situation for an enzy m e w ith normal kinetics.

One-Step Reactions with One SubstrateIn simple uncomplicated cases, the initial velocity of an enzymereaction increases with substrate concentration in the manner showninFig.2.At low substrate concentrations,the raterises almost linearly

with concentration. At higher concentrations, the increase graduallylessens, until the reaction is hardly accelerated by increasing substrate. Under such conditions, the substrate is said to have "saturated"the enzyme, and the rate observed is the "maximum velocity" ormaxfor that concentration of enzyme. The actual chain of events in anenzyme reaction maybevery com plex, butinthe case ofasingle substrate, S, and an irreversible reaction that forms product, P, the situation can usually be explained for practical purposesbythree reactions:k j k 3E +S ^> ES -> P + E [Re. 2-5]k 2


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30 Passonneau and LowrywhereEis free enzym eand ESisatransient com p lex b etw een enzy m eand sub strate. (It w ou ld be equ ally satisfactory to write EP insteadof ES . ) From w hathasb een said earlier, it w ill be evident that of thethree reactions taking p lace, oneissecond orderandtw oarefirstorder.The third reaction (ES -> P) is the one ordinarily measured as theoverall enzy m e v elocity (represented b elow as either v or V3).Usu ally , at the start of such an enzy m e reaction, S greatly ex ceed sthe enzym e on a m olarb asis.In th is case, Re. (2 -5) can be consideredto take p lace in three ph ases. In the first phase, ES is form ed faster byreaction 1 than it breaks down by reactions 2 and 3; therefore, ESaccum ulates and E falls. T h is is the presteady-state ph ase, w h ichusually lasts only a matter of milliseconds. In the second phase, asteady state has been established in w h ich reaction1 is balanced byreactions2 and3 ,but S has notdim inished app reciably. E S is thereforeat its peak lev el and the net v elocity (decrease in S orincrease in P) iscalled initial velocity . Finally , in the third (and usua lly longest)p h ase, S fal ls tow ard z ero, during w h ich tim e a steady state is m aintained b etw een E , S, and E S .A t all tim es the net v elocity is proportional to [E S ], i.e.:

V =fcj[ES] [Eq. 2 -12 ]It w ill be seen from Re. ( 2 -5) that as [S ] is m ade larger and larger,it w ill finally dim inish[E] to anegligible v alue (i.e.,the totalenzym e, Ex,w il l beinthe formE S ,andtherefore, the velocity w ill be m axim al) thus:

W hen velocity is m axim al, it is independent of m odest ch anges inS and iscalled zero order inS. Itis,inreality , at leastasecond-orderreaction,andshou ld properlybecalled pseudo zero order. It is difficu ltto conceive of atruezero order reaction.)A t low er substrate lev el s, an equ ation is needed relating v to S andVmax.T he three separate v eloc ities of R e. (2 -5) are:

V l = *,[E][SLv2=^ [ E S ] , [ E q . 2 - 1 4 ]v3=3[ES]


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Kinetics Primer 31W hen the steady state is reached, Vi = V 2 +V3,or:

kj[E][S] =A:2[ES] +fc3[ES]=(ik2+^)[ES] [Eq.2 -15]Rearranging:

[E][S] k2^ [E ] Km= = / C m o r / ^ s ) or = gin[ES] k j [E S] [S ] W ^ ' JJ T m (the M ichaelis constant ) has t h e form of a n equilibrium constant(cf. Eq. [2-11]). Itisa true equilibrium constantif fa ism uch largerthanfa.Km isa readily determinable constant, w hereas,fa andf areusually very difficult t o evalu ate. From E q . (2 -17 ), it is seen that w hen[S ]=ATm, [ E ] = [ E S ] , i.e., half of the total enzym e w ill be present a s E S .Sincevis proportional to [ES ] (Eq. [ 2 -1 7] ) , w hen [S ]=Km , v ishalf-m axim al. Thu s,Km isnumerically equal tothe substrate concentration, which gives half-maximal velocity {see Fig.2) .

The desired equation can now be formulated. Rearranging Eq.(2 -1 7 ) , [ E S ] #m=[E][S] .Sub stituting [ E T ] - [ E S ] for [ E ] and rearrangingE T =total enzy m e):[EJfS][ES]= [Eq. 2-18][S]+*m

If both sides are m ultipliedby fa,this becom es (from E qs. [2 -12 ]and [2-13]):VJS] vmV = [S]+ K,,, 1+K J[S) [ E q- 2A9]

whichis the Michaelis-Menten equation. This describes th ehyp erbolic curve p lotted i n F i g . 2 . The reciprocal form of E q . (2 -19 ) is oftenm ore convenient becauseitdescribesastraight l ine:

_J_ J_ / ^m _ \ [Eq. 2-20a]


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32 Passonneau and Lowry

Cs]

Fig. 3 . Reciprocal L ineweaver-Burk plot of velocity against substrate concentration for an enzy m e w ith normal k inetics.

Equation(2-20a) is the basisofthewidely used Lineweaver-Burk plot,or reciprocal plot, of 1/v against 1/[S] (Fig. 3). The intercept on thevertical axis is\IVm,and the intercept on the horizontal axis is-\IKm.F irs t- O r der E nz y m e Re a c t io ns

When [S] is small compared toKm,Eq. (2-19) reduces to:v = Vm[S]/Km [Eq. 2-20b]

Since,withagivenlevelof enzyme,mandKmareboth constants, theycan be replaced by their ratio, which we can denote by kand call the"apparent first-order rate constant":k=VJKm [Eq.2-21a]

from which:v = *[S] [Eq.2-21b]


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Kinetics Primer 33Table 1Influence of Maximum Velocity and Michaelis Constanton Apparent First-Order Rate Constant and Reaction Times5

EnzymeAAB

V b'maxfxM/min70020050

^m\lM1000100050

kmiir 10.70.21

h amm13.50.7

'o.75nun271.4

*0.98nun5.6203.9

aThe reaction times apply to initial substrate levels well below theKnbA t the enzy m e concentration used.

Th is has the form of a first-order equ ation, and w e speak in this caseof the substrate concentration being at a "first-order level." Everything said above about nonenzymatic first-order reactions is applicable in this case. Equation (2-2) can be rewritten (S replacing A) as:2.3 [S]0t = l o g [ E q.2 -2 2 ]* [S]For the majority of analytical situations,k is more important thaneitherVmorKm alone. For example, in the hypothetical cases of Table1, enzy m e B at an activity Vmax)of 50 |iAf/min is more ac tive tow ardlow levels of substrate than enzyme A at an activity of 700 |iAf/min.If a substance is to be measured enzymatically, and the concentration is well below the Km, the timerequired as shown earlier, is 5 or

6 half-times. Thesame time would be requiredwhether [S]0was1/10thor1/1000ofKm.Notethat,in calculatingkyitisnecessarytoputmandKmintorationalcom patible term s. It is often conve nienttoexpressKma s\\M (|imol/L),and to expre ss Vmas jiM/min (i.e., (imol/L/m in). In this casekwill beexpressedas per minute.The advantage of this is that the interna tionalenzy m e unit (U) is 1 |imo l/L/m in, therefore, U/L = |iM/m in. S tockenzym e solutions can be conven iently expressed as mM /min (U/m L).M i x e d Z e ro - O r d er

a n d F i rs t - O r d e r E n z y m e R e a c t i o n sIf [S]0approachesKmor exceeds it, more time is required to complete the reaction than when [S] 0 is at a first-order level. The timerequired can be calculated from the integrated form of Eq. (2-19):


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23Km [ S ]0 [P]/ = log + [Eq .2 - 2 3 ]V m [S ] Vm

w hereP isthe product form ed ( i.e., S0 - S ) . I fkis again su b stitutedforV JK m :2 .3 [ S ]0 [P]t = log + [Eq. 2- 2 4 ]

k [SI VmTh is m ayb eseentodiffer from the equ ation fora first-order reaction(Eq . [ 2 -22 ] ) on ly b y th eaddition ofthe term [P]Vm .In other words, to complete a one-step enzyme reaction, with anylevel of substrate, requires the time calculated for a first-order reaction plus the time that wouldbe required if the entire reaction were toproceed at theVm.For the reaction to b e9 8% com p lete (and ignoring th e2% difference between [S]o and [P] ) , E q .(2 -23 ) becom es (with suf f icientlyclose approximation):

4 [ S ]0f0.98 ~ + [Eq. 2-25]k V

or sincek=0 .1Itm:[S ]0

' a 9 = 6 ' i / 2 + [ E * 2 - 2 6 1KmC onsider as an exam p le an enz ym e w ithKm =2 00 (lAf andVm(at the

lev el of enzym e used)=50 |iA f/min. Th is g ive sk=0.25/m in andtm = 3m in . To g ive9 8%conv ersion to product w ith initial substrate lev el so f1(lA f,10 0 (lAf, and 1m Af w ou ld require 18 m in, (1 8+ 2 )m in, and( 1 8+ 2 0 )m in, respectively . (SeeT able2 forthis and other examples.)

In a metabolite assay, th eproblem isu sual ly th econverse of theabove ( i . e. , h ow m uch enzym e shou ldb eusedtocom plete areaction


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Kinetics Primer 35Table 2Effect ofSubstrateConcentrationonReaction Time3

\xM/m in \xM \\M min 1 min m in50 200 1 0.25 3 1850 200 100 0.25 4 2050 200 1000 0.25 13 3810 10 1 1 0.7 410 10 100 1 5.7 1410 10 1000 1 51 104aT he half-tim es w ere calculated from Eq . (2 -23 ); for

exam p le, in the third line, tm = 0.1/k+ [P]/50 = 3 + 10 =13.T he98%tim es ( fo .9 s)w e r ecalculated in a sim ilar m anner(seetext).

in a give n t im e) . In th is case , it is s im pler to repla cekbyVm/Kmin E q.( 2 - 2 5 ) , wh ich then becomes :4 * m + [S]0 4Km + [S]0

*0.98 = or Vm = [Eq . 2-2 7]V m t().98

Fo r exa m ple , i t i s des i red to hav e a reac t ion (98 % ) com ple te in 10m in . The ATmis 10 | iM , and the high est conc entra t ion to be m ea sur edw il l be 100 |iAf in the assay sys tem . T he enz ym e stock is a 1% solu t ionwith ac t iv i ty of 8 U /m g ( i .e ., 80 | im ole s /m in /m L, or 80 ,00 0 ( iM/m in) .H ow m uch of th is s tock wi l l be needed? From Eq. (2 -27 ) ,Vm = (4 x 10+ 10 0) /1 0 = 14 ( iM/min . Therefo re , an enz ym e conc entra t ion of 1:5000should suff ice. Figure 4 provides a graphic calculat ion of the enzymerequirement wi th d i f feren t subst ra te leve ls and wi th enzymes hav ingdif feren t Michael is constan ts .

W h e n Km is substantial ly larger than [S]0 , Eq . (2-27) red uc es to Vm= 4KnJu Th us , if theKm i s 200 \iM to com ple te the reac t ion (98 % ) in20 min w ould re quire enz ym e ac t iv ity equal to 40 |iAf/min or 40 U /L.O n e - S t ep R e a c t i o n s w i t h T w o S u b s t r a te s

The major i ty of analyt ical react ions to be considered wil l involvetwo subst ra tes (A, B ) . In m any ca ses , how ever , on e of the reac tan ts (B)wil l be present in substantial excess and wil l therefore not be great lyreduc ed d ur ing the reac t ion . Th is m ean s tha t fo r p rac t ica l purp ose s th e


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8 12.units/liter -/j./V//mm

Fig. 4. E nzym e required for an enzym e reaction w ith normal k inetics to be 98%complete in 10m in.T he lines are calculated fromEq.(2 -27). For enzym es withKm 'sof 50, 100, 200, and 500\\M ,the minimal enzyme requirements for [S0] = 100\\Mw ould be 3 0, 50 ,9 0, and 21 0 U/L, respectively, and only 10 U/L less if [S ]0= 1\\M.

kinetic situation will reduce to that for one substrate, except that thelevel ofBwill influenceone or bothof the kinetic parameters for A(Vmand K A ) . (Forthecase in whichBas wellasA is substantially reducedduring the reaction,seethe next subsection.)The mutual effects of two or more cosubstrates can be complex;nevertheless, if[B] isheld constant,thereaction will usually obey Eq.(2-20) (and therefore Eqs. [2-22]-[26]), i.e.:

v = [Re.2-6]whereVm andKm only apply to the particular level ofB.In designingor modifying an analytical method, it is useful to know how the concentration ofBaffects the kinetics ofAin the analytical range, but itshould notbenecessaryto go intoa completeandperhaps complicatedkinetic analysis. Without any attempt to be comprehensive, some of


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K inetics Prim er 37the common types of kinetic interactions between cosubstrates can bementioned.CASE 1: "RANDOM ORDEROFADDITION"Inthesimplestcase,each substrateacts asthough it combines independently with the enzyme.

K

EAB - E + P [Re. 2-7]

B


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38 Passonneau and Lowrywhich is the same as the equation for a second-order nonenzymaticreaction (Eq. [2-11]). Note that as in the nonenzymatic case,fcabs asecond-orderrateconstant,andtherefore, h as bothatim eandconcentration dim ension(seeexam ples in Table 3).If [B ] is constant, E q. (2 -28) reduces to:

V I[A]v = [Re. 2-8][A]+KA

and Eqs. (2-29) and (2-30) reduce to v =k [A ] , wh ere:[B]

K = ka [ E q . 2 - 3 1 ][ B ] + BT able 3 g ives exam p les of the infuence of M ich aelis constants and

concentrations of sub strate B on the apparent first-order rate constantfor substrate A . T h is table w ou ld b e relevant for enzy m atic analy sis ofsubstrate A.C A SE 2: COOPERATIVE ADDITION

Inasecond tw o-substrate case, each sub strate influ encestheKm forthe other in a m anner that can b e represented b y:*A

K B

E A + B

B


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inetics Primer 9

3 sg | > COC/3 0) 33|2

en

iU8S I

id1-

=

tti^ i

w

ON O O O N ^. n ON (S in o TfCM C O

ON ^ O VO C M O N^ ' H d M" d o\ d

S o oo vo en o ooO T t ON CO 35 >CO Xl*12-8Is-&"*-* c*l?"J 8o a.S

S^ cs60 I s s .s JS * 1 ^S s i

5 w>

CA

oB.au pq H

c s-.a ooa - 8 1

0 8


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40 Passonneau and LowryKA/KB A

The velocity equation is:* X = * B


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Kinetics PrimerTable4

[B],m MInfinite1.000.100.01

V 'mM/min1.000.910.500.09

KA',mM0.100.180.550.92

ka,m in-110.05.00.90.1

CASE 3: "ORDERED ADDITION"A third type of two-sub strate case is in reality an extrem e e xam pleof the case just discussed in which KA B is zero and KB is infinitelylarge.

K A K B A k5E + A ^ E A + B ^ EAB - E + P t ^ -2 " 1 1

This is called "orde red add ition" of substrates, since it describes thesituation in which the second substrate can only combine with theenzyme if the first substrate is in place. For strictly ordered addition,"rapid equilibrium" kinetics, on which Eq. (2-33) is based, lead toerroneous results. Thus, according to rapid equilibrium kinetics, TA1becom es infinitely smallasthe conc entration of B approaches infinity,whereasKA in fact approaches K A . Th e reason for the discrepancy isthat the assumption of rapid equilibrium does not properly take intoaccoun t the contribution of5 to the kinetic situation. Further consideration of this special case is beyond the scope of this book.E n z y m e R e a c t i o n s w i t h S e c on d - O r d er K i n e t i cs

Th ere are a num ber of analytical situations involving tw o substratesin which both su bstrates are initially b elow their respective M ichaelisconstants and in which the auxiliary substrate B falls substantiallyduring the analytical reaction. This is often the case wh en B is N A D Hor N A D PH , and the disappearance is measured directly in the fluoro-meter. It is not poss ible to usealarge excess of the pyridine n ucleotide,becau se during the reaction, the percentage c hange w ould be too sm alltome asure accurately. In this situation, the reactionis nolonger pseud ofirst order. Th e practical consequ enc e is that the reaction veloc ity falls


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42 Passonneau and LowryI.Oi

0.8h

0.6hC D C HThe problem is to determine what discrepancy there may be betweenthe disappearance ofAand the appearance ofC(i.e., how fast and towhat extentBaccumulates [Fig. 6]).Since step2 bydefinitionisfirst order,Vi=fa[B],where, accordingto Eq. (2-21),ki =VmiJKn.Ultimately, a steady state will be established, at which time Vi = V2.W hen this occurs:

v l = M B ] m a x or rB ]m a x = y * 2 [Eq.2-38]

where [B]m axis the steady-state level of B.Thus,for the case ofazero order reaction followed byafirst-orderreaction, [B]maxwill be numerically equal to the velocity of the firstreaction divided by the rate constant for the second reaction. For exam ple,ifthevelocity ofthefirst reactionis 1\xMI mm andthe rateconstant forthe second reaction is3/min,[B]max will be1 |LLM/3= 0.33\\M.


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A - ^ B - ^ C

Lawl.44/, pk- ~2S2 K r n S^

\ Dieaway curve y>\ \ for step 2 AY

V^^V i i 7 ~

1

A

^Bmax^-^/ .K 12

0 1 2 3 4 5 6Time (half- times ofstep2 )

Fig. 6. Lag in a two-step reaction in which the first step is zero order, and thesecondstepisfirstorder.Thevertical distance betweenthelines equalstheamountofB accumulated, and the horizontal distance between the lines equals the time lagbetween disappearance ofAand appearance of C.

Sincek= 0.69/fi/2 (Eq.[2-3)],Eq. (2-38) can also be written as:

W =1 M

' l / 2v

l [Eq. 2-39]wheretm refersto thesecondstep.Thissaysthat [B]max represents theamount ofAconverted to B in 1.44 half-times for the second step.It is often more useful to consider the lag time between the disappearance ofAand the appearance ofC.The maximum lag time (Fig.6) is the time it takes to accumulate [B]maxor:[B ]lag max 1 Bmax = lM tm [Eq.2-39a]"m2

Notice that the lag time depends solelyonthe first-orderrateconstantofthesecondstep.


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K inetics Prim er 45Theamount ofBpresent beforethesteadystateis reachedisdescribedby the integrated equation:

[B] = ( v l / / : 2 )( l-0.5^i/2) [Eq.2-40]Accordingly, whentisequal to1,2,3,and so on, half-times (for step2), [B] will be 1/2, 3/4, 7/8, and so on, of the maximum; or whatamountsto thesamething,thelagtimewillbe1/2,3/4,7/8,andsoon,of the maximum lag time (Fig. 6).Example: V\=0.5 |iM/min,ti/2for the second step is 2 min. [B]max= 1.44 x2min x 0.5 |iAf/min=1.44\iM.At2,4 ,6 , and8min,[B]willbe 0.7, 1.1, 1.3, and 1.36 |iM, respectively.TWO-STEP REACTIONS WITH BOTH STEPS FIRST ORDER

k l k 2 [Eq. 2-41]A -> B - C ^This isacommon situation in metabolicassays.Itisusefultoknowthe rate of appearance ofCas a function ofk\andki.When the firststep is much slower than the second, the time lag curve is the same asthat given for the previous case (step1 zero order, step 2 first order).Departure from this simple situation is not serious until the rate of thefirst step approaches or exceeds the second. Therefore, in m ost analytical situations, the maximal time lag caused by the second step canbe taken as 1.5-2 half-times (for the second step).For those who may need more exact information, the followingmore detailed analysisispresented.Wewill approach this problem byregarding the time course ofthefirst step as primary and asking whattime lag in the overall reaction is caused by the second step. Thekinetic equations are more complicated than when the first step is ofzero order, but fortunately approximations and graphical representations will suffice for most purposes.The appearance ofCis the same as the disappearance of A+B.Theintegrated rate equation is :

| A 1 + [ B 1 . ? [ E q . 2.42](A] k,- k,


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Incubation time (hal f - tim es for step 2)

Fig . 7. L ag in a tw o-step reaction in w hich both steps are first-order. T im e isrecorded in m ultiples of thehalf-time for the secondstep .Ex amp les ofuse:L etkxlk2= 2 and the step 2 half-tim ebe3 m in. After incub ation for12 min(four tim es thehalftim e forstep2) , the lag in appearance of product is 7.5 m in (2.5 half-tim es for step 2 ).

The equation for the disappearance ofAis:[A]/[A]0 = e - V [Eq. 2-43]

This is the exponential form ofEq.(2-2).The time lag in appearance ofCcan be calculated from the difference between these two equations. For example, letk\=0.5/minand&2=1/min.It is calculated from Eq. (2-2a) that the half-time for disappearance of A would be 1.39 min. Itiscalculated from Eq. (42) thatthehalf-time for disappearance of (A+ B) is2.46min.Therefore, afteran elapsed time of 2.46 min, the lag is 2.46 - 1.39 = 1.07 min.Curves of time lags are given in Fig. 7 for fo held constant andk\varied from 0.2 to 5 timesfe.Whenk\is smaller thankithe lag approachesa finite fraction of the half-time for step 2 (Table5).Iftheratio ofk\tokiis madesuccessivelysmaller,thelag curveapproaches that obtainedwhen the first step is zero order. Thatis,the maximum lag approaches


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inetics Primer 7

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T t i n s q o q c s i n t ^ - ^ - ^ - ^ - ^^ ^ ,_; ^ c4 c4 ^

t ^c NON S T t m m r ^c s c N T t^ o in oq f-j - ^ so r*; oq oq cs -


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48 Passonneau and Lowry1.44 half-times (for step 2), and the lag is 1/2, 3/4, 7/8 , and so forth,of the maximum after 1, 2, 3, and so forth, half-times (of step 2).

Ifthehalf-time forstep 2is known,thehalf-time and eventhewholereaction curve for step1 can be calculated with the aid of Table5andFig.7 . As an example, suppose step2hasahalf-time of 2 min and theoverall reaction has a half-time of8min (four times longer). Table 5shows thatthefirst stepisabout 0.4times asfastas thesecondstep k\lki=0.4). If the firststep isknownto befirstorder, this would indicatea half-time of2.5min. To see ifthefirst step did in fact follow a first-order curve, its time curve could be calculated by deducting the lagtimes indicated inFig.7 along an interpolated curve fork\/k2=0.4.Inthe present example, readingsmadeat 2,4 , and6min (1,2 , and3half-times) would be correctedby0.75,1.3,and 1.4 half-times, or1.5,2.6,and 2.8 min. This procedure would give a close approximation to thetime curve for step 1, even if step1 turned out not to be first order.Another use for Table5and Fig. 7 would be to guide the choice ofassay conditions.Theyallow prediction of amounts of enzymes neededto complete a reaction in a specified time. For example, from Table 5it is easily seen that for the overall reaction to be 98% complete inabout10min the half-times forstep 1and step2could be (in m inutes)1.7 and0.17,1.5and 0.6, both 1.2,or0.17and 1.7. Note that the mostefficient use of the enzymes occurs when lhe half-times are aboutequal and that interchange of the two half-times does not affect theoverall time curve.

O t h e r F a c t o r s A f f e c ting E nz y m e K ine t i c sThe preceding has been devoted to the relationships of substrateconcentrations to reaction velocity. It should be stressed thatVm>Michaelis constants, and equilibrium constants are all subject to theinfluence of temperature,pH,ionicstrength, andthepresence of activatorsandinhibitors. A brief commentontemperatureand pH is inorder.TEMPERATURE

Ordinarilymisincreasedbyincreasing temperature until the pointof enzyme instability isreached.Temperature coefficientsvarygreatlyfrom one enzyme to another, and it is sometimes overlooked thatenzym e activity does not fall to zero in an ice bath. Usually themat0 is 3-10% of that at 38 and, in the case of some enzymes, may beas high as 20% or more.


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Kinetics Primer 49M ichaelis constants also w ould be exp ected to increase w ith tem perature. Th erefore, tem perature m ay h ave a less favorab le ef fect on

rates as lo w substrate lev els ( i.e., in the usu al situation encountered inanalytical w ork). In som e cases, rates may actually decrease w ith increasing tem perature.P H E F F E CT S

Theeffect of pH onVm is ,of course, extrem ely variable from enz ym eto enzy m e. For som e enzy m es, the pH optim um is extrem ely sharp,and for others very b road. T he pH for m axim um v elocity does notnecessarily coincide w ith the pH for m inim um M ichaelis constant.C onsequently, the pH optim um for an analytical reaction w ith lowsubstrate lev el m ay b e qu ite different from that for an enzym e assaycarried out at near-saturating substrate concentration.

Enzyme Reaction Order

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