Rae laws, Reacion Orders The rae or velociy of a chemical reacion is loss of reacan or appearance of produc in concenraion unis, per uni ime d[p] d[s] The rae law for a reacion is of he form Rae d[p] k[a] n a []n b [C]n c... The order in each reacan is is exponen, n i The overall order of he reacion is he sum of he n i Reacion Order Moleculariy The order of a reacion is deermined by he number of paricipans ha mus come ogeher before he rae-limiing-sep (RLS) of he reacion I is deermined by he mechanism, no soichiomery of a reacion I mus be deermined (measured) experimenally Deermining Reacion Order Iniial Rae Mehod Choose iniial concenraions of reacans Measure Iniial Rae, before concenraions of reacans change appreciably (angens) [A] 8 6 4 Iniial rae Disappearance of A Course of reacion 5 5 5 3 35 4 45 5 ime Deermining Reacion Order Iniial Rae Mehod Choose iniial concenraions of reacans Measure Iniial Rae, before concenraions of reacans change appreciably Change iniial concenraions, and measure iniial rae again Example: For A + C, assume Rae k[a] na [] nb [A]o []o Rae if hen R R R n a R R R n a R 3 R 4R n a R 3 R n b, ec. Once you know he n i, plug in and solve for k Deermining Reacion Order Inegraed Rae Equaion Mehod Allow he rx o proceed and measure change in concenraion of reacan or produc. y rial and error, deermine which equaion fis he daa 8 Disappearance of [A] Inegraed Rae Equaions If our rae law is d[p] k[a] n a []n b [C]n c... or d[a] k[a] n a []n b [C]n c... [A] 6 4 3 4 5 ime We will have o inegrae i in order o use our experimenally-deermined values of [P], [A], ec., o make a plo like he one on he previous slide.
Deermining Reacion Order Inegraed Rae Equaion, Zero Order Choose iniial concenraions of reacans Measure Rae over a subsanial porion of he reacion The produc and reacan concenraions change auomaically Fi o heory Example: zero order reacion A d[] d[a] The rae law is k[a] k Inegraing from [A] a o [A] a [ A] d[a] k [ A] o [A] [A] k So, a plo of [A] - [A] vs is linear, wih a slope of k, IFF i is zero order Inegraed Rae Equaion Firs Order Reacion A The rae law is Rewrie as d[] d[a] [A] k d[a] k[a] Inegrae from [A] a o [A] a ln[a] ln[a] ln( A A ) k d[a] k [A] A plo of ln (A /A ) vs is linear, wih a slope of k If so, he reacion is s Order in A and zero order in all oher componens or [ A] [ A] o ln( A A ) k Inegraed Rae Equaion Second Order Reacion A The rae law is Rewrie as d[] d[a] k[a] Inegrae from [A] a o [A] a [A] d[a] [A] k [A] k or [A] d[a] k So, a plo of (/[A] ) vs is linear, wih a slope of -k If so, he reacion is nd Order in A and zero order in all oher componens [A] [ A] [ A] o [A] k Deermining Reacion Order Muliple reacans, firs order in each A + C d[c] d[a] d[] k[a][] Use soichiomery o relae all reacan concenraions o heir iniial values and o [A] If [A] [] and he soichiomery is, hen he inegraed rae equaion is he same as for nd order in A If [A] [] and Soichiomery is n (A + n C) Then subsiue [] [] -([A] -[A]) before inegraing Remember, soichiomery doesn deermine reacion order, bu i may be useful in guessing wha he order migh be Unis of Rae Consans If rae d[a], i mus have unis of (concenraion) (ime) And he producs on he righ side of he equaion mus have he same unis, so he unis of k mus be differen Zero order reacions Rae (M min - ) k (M min - ) Firs order reacions Rae (M min - ) k (min - ) [A](M) Second order reacions Rae (M min - ) k (M - min - ) [A] (M ) Third order reacions Rae (M min - ) k (M - min - ) [A] 3 (M 3 ) Someimes iniial rae and inegraed rae measuremens disagree - Why? The inegraed rae equaion gives he concenraions of componens whose concenraions change during he reacion consider: A + C Rae k [A][] The reacion is second order overall, firs order in each reacan Rae decreases as [A] decreases and as [] decreases. Wha if [] remains consan during a run? because i is presen in large excess (e.g., solven), because is concenraion is buffered (e.g., H +, OH - ), or because i is a caalys, and is regeneraed. Then he reacion is pseudo-zero order in, and pesudo-firs order overall: Rae k apparen [A], where k apparen k []
Model an Enzyme Reacion The enzyme and subsrae(s) ge ogeher Transformaions occur The produc(s) is (are) released, regeneraing enzyme k k E + S " "" E S " "" E + P k We have o worry abou four concenraions [E], [S], [E S], and [P] and four reacions, some probably s order, some nd k [E][S], k [E S], k - [E S] and k - [E][P] Too complicaed! k Simplifying Assumpions. Measure iniial raes (<% of reacion) hen [P] so we can neglec he reverse reacion, k - [E][P] and we can ignore produc inhibiion (laer) and [S] + [E S]. Use caalyic amouns of enzyme, [E] << [S] Then [S] + 3. Enzyme exiss in only wo forms, E and E S, which sum o 4. [E S] reaches a consan level rapidly and remains consan during he measuremen Could happen in wo ways k k k - >> k, rapid equilibrium E + S E S E + P k - k, seady sae k - 5. k [E S] Simplifying Assumpions seady sae k k E + S E S E + P d[e S] or k [E][S] k - [E S] k [E S] k [E][S] [E S]{k - + k } u we also know ha [S] and [E]+[E S], so ha [E] - [E S], and k [E S] (assumes he caalyic sep is he RLS) So, le s subsiue for free enzyme [E] and solve for [E S] k { - [E S]} [E S]{k - + k } Collec erms in [E S] (nex slide) k - Seady Sae k k - k E + S E S E + P k - [E S]{k + (k - + k )} k [E S] solve for [E S] k + (k - + k ) muliply by k k k [E S] k k + (k - + k ) k + (k V - + k ) max (k + k ) - k +[S] k k Simplifying Assumpions rapid equilibrium k k E + S E S E + P [E S] is conrolled by he equilibrium and is slowly bled off oward produc [E] [E S] [E S] [E] k - [E S] { - [E S]} [E S]{+ } K [E S] S E + [S] T + Comparison seady sae vs rapid equilibrium Seady Sae k + k k ime unis: concenraion ime - conc k Rapid Equilibrium unis: ime concenraion ime - conc ecause a equilibrium k [E] k [E S] k so k [E] k [E S] k [E S] k + + The same equaion So, hey really only differ in he relaive magniude of k - and k k k - k E + S E S E + P 3
Michaelis-Menen Equaion Michaelis-Menen Equaion Vmax (unis of velociy) Could also be wrien + Which migh be useful if we have more han one subsrae Vo 5 Or, as [S] + K M + [S] Which is uniless (unis of concenraion) 4 6 8 4 6 [S]o Fracion of Fracion of Subsrae bound Wha Daa are Required? If is oo high poor esimaion of - looks zero order If is oo low poor esimaion of - no evidence of sauraion poor esimaion of - looks firs order Need daa wih in he viciniy of, Need daa wih boh above and below Order of he Reacion Vo >> V [S] max V [S] max X + Mixed order zero order 5 Vo 5 4 6 8 4 6 [S]o So how can i be boh firs order and zero order? << V [S] max V [S] max + [S] X 4 6 8 4 6 [S]o firs order More Order Zero Order Region d[p] d[s] d[s] [S] >> [S ] Firs Order Region << d[s] [S ] So, a plo of -[S] or [P] vs is linear, wih slope [S ] [S] d[s] [S] [S] ln [S] So, he semilog plo vs is linear, wih slope / In beween -can simplify inver before inegraing: V o and [S] o Relaions [S] + + [S] So, if V. is % of, + [S] and. so, S.5K M..5 Or, if 5, how close is he rae o? 5K M.83 + 5 ln [S] + {[S] } The sum of he limiing cases How abou? K M.9 + 4
Complicaions - Subsrae Inhibiion I s acually hard o ge o A, you observe only 9% of The example acually had And, someimes subsrae binds incorrecly and inhibis Never reaches Vo vs [S]o Can ge 75 Lineweaver-urk Inver boh sides: y - inercep + + slope Vo 5 5 4 6 8 [S]o x - inercep - Lineweaver-urk Double-reciprocal plo / vs / has linear region, even wih subsrae inhibiion Eadie-Hofsee Plo alernaive linear ransform + +. + + /Vo -5 5 5 5 /[S]o - Nowadays, you can use nonlinear leas squares fiing Measure and several and pu hem ino V [S] max + To calculae. Deermine he difference beween he measured and calculaed (he residuals, (calc obs )) Square he residuals for all values of, and add hem ogeher Adjus and unil he sum of he squares of he residuals is a a minimum. Which mehod o use? Lineweaver-urk, Eadie-Hofsee and non-linear leas-squares give he same answers, because hey are differen forms of he same equaion u, if here is noise in he daa, he noise has differen effecs, use weighed leas squares for L- plo, because he lowes raes have he mos error, and end o underesimae. 5
/( + ) >> zero order Non-Michaelis-Menen ehavior cooperaiviy Mixed order Michaelis-Menen Negaive Cooperaiviy Vo 5 V o Posiive Cooperaiviy << firs order 5 5 [S] o [S] o Induced Fi Enzymes and Ideas are Flexible Conformaional Change Hexokinase is a monomer, and no cooperaive, bu i was an early demonsraion of induced fi Cooperaiviy S S K d + S K d + S Hexokinase Hexokinase ound Glc (Glc no shown) S S Higher [S] always shifs he equilibrium o he righ (from cubes o cylinders) If K d > K, cooperaiviy is posiive. d If K d < K, cooperaiviy is negaive. d How Cooperaive is i? Koshland s Cooperaiviy facor: For normal MM kineics R S giving V.9 giving V. V [S].9 V [S] 9. + + [S] K 9 M R S 9 /9 8 For posiive cooperaiviy, R s < 8 For negaive cooperaiviy, R s > 8 Dependence of V o on k Unis of are velociy, e.g., concenraion/min u So, i s no an absolue quaniy Mus be expressed as per mg proein for comparison Like specific aciviy Turnover number V Δ[ S] Δ[ P] max ime or ime [E] [E] ime Molecules S P per enzyme molecule/ime Also k, migh call i k ca if i is acually for caalysis (no, e.g., produc release) 6
Dependence of V o on k, and, always A low, / A high, So, V o, always, always, always Dependence of V o on excepions - irreversible inhibior The irreversible inhibior inacivaes he enzyme soichiomerically When you have added more enzyme han here is inhibior, he reacion rae is proporional o Excep If an irreversible inhibior is presen If a second subsrae is limiing If he assay sysem becomes limiing If an Enzyme-cofacor or oher equilibrium is involved normal behavior Enzyme is inacivied unil irreversible inhibior is used up Dependence of V o on excepions - second subsrae becomes limiing Example: glucose oxidase α-d-glucose + O + H O g δ-gluconolacone + H O u O is no very soluble in waer (.4 x -4 M @ 5 C) So, rae can coninue o increase wih forever. normal behavior [O ] sars o become limiing, here Dependence of V o on excepions - coupled enzyme assay becomes limiing A common way o assay an enzyme is o use a second enzyme ha uses he produc of he firs enzyme o make somehing ha s easy o deec, such as NADH, which absorbs a 34nm A C E E normal behavior NAD + NADH + H + Coupling enzyme can keep up More han one Subsrae Excep for isomerases, mos enzymes have more han one subsrae How do we deal wih his fac? A + E D E A + is awkward noaion A beer noaion is due o W. W. Cleland (Madison,WI): Rewrie normal MM sequence E + S D E S E + P as S P E E S D E P E Then, more complex reacions can be wrien simply Noe also ha [S] + [S] [S] + Models wo subsraes on, wo producs released E + A + D P + Q + E Subsraes could add ogeher, one afer he oher A P E E A E A D E P Q EQ E This is an ordered bi bi reacion Kineic equaion can be generaed as before We ll assume rapid equilibrium Q 7
Remember Single Subsrae eqn? rapid equilibrium k k E + S E S E + P Three Main Poins in he derivaion. Enzyme is pariioned ino wo forms [E] + [E S]. Dissociaion consans can be defined for complexes [E] [E S] k - 3. k k ca rae consan for RLS, so k ca [E S] Ordered bi bi assume rapid equilibrium E + A D EA + K E EA k ca g E + P + Q k ca [EA] [E] + [EA] + [EA] [A][E] [EA] K [][EA] [EA]. Subsiue [E] - [EA] - [EA] ino eq for. Solve for [EA]; subsiue ha ino eq for K 3. Solve for [EA] 4. Muliply by k ca : V max + K [] + K [A][] Random bi bi eiher subsrae can add firs, eiher produc can leave firs E A E A P Q E Q E E A D E P Q E P A Q P inding of A and may or may no be independen E K Random bi bi E + A D EA + E αka + E αk E + A D EA g E + P + Q k ca Same seps - jus more of hem solving subsiuing muliply by k ca k ca [EA] [E] + [EA] + [E] + [EA] [A][E] [EA] K [][E] [E] α [A][E] [EA] αk [][EA] [EA] + α [A] + αk [] + αk [A][] Ping-Pong One produc leaves before las subsrae adds A P E E A D F P F F D EQ E Q Double reciprocal plos Vary one subsrae a differen fixed levels of oher subsrae 4 " K % $ ' #[]&[A] + Ordered, vary [A] " + K % $ ' # []& + K [] + [A] /V 35 3 5 5 [] 5-4 6 8-5 /[A] 8
,, K? Noe ha he inercep! + K $ # & " []% K + [] Double reciprocal plos ordered bi bi, coninued " K % $ ' #[]&[A] + " + K % $ ' # []& So, plo he inercep versus /[], and obain / from he inercep of he replo exrapolaing [] o infiniy The apparen / is apparen K [] + So, use a replo, as above. (For K of he varied subsrae (A), he inersecion of he lines is a (-/ ), / ) Inercep / -apparen /[] /[] Noe ha plo of same daa vs second subsrae looks differen; rearrange o see why (collec erms in /[], and pu in slopeinercep form wih x /[]) /V /V vs /[] a fixed [A] [A] 5 5-3 4 5 6-5 /[] Random bi α 4 35 3 5 αk # A + K & % ( $ []'[A] + [] Random, vary [A] # + αk & % ( $ [] ' α.5 4 35 3 5 Random, vary [A] [],, K?! + αk $ Noe ha he inercep # & + αk " [] % [] So, plo he inercep versus /[], and obain / from he inercep of he replo exrapolaing [] o infiniy Inercep /[] /V 5 /V 5 Various replos can be used. 5 5-4 6 8-4 6 8-5 -5 /[A] /[A] (For K of he varied subsrae, he inersecion of he lines is a (-/ ), (-α)/ ) Double reciprocal plos ping pong Consan slope Decreasing inercep [A] + /V 3 8 3 8 " + K % $ ' # []& Ping Pong, vary [A] [],, K? Slope K [] + In K [] + So, use an inercep replo o find, and K Use a slope replo for /, and solve for. 3 - - 4 6 8 /[A] 9
Quick Example Aldehyde Dehydrogenase Acealdehyde + NAD + + CoA è NADH + AcCoA + H + As we will see, he kineic mechanism is i Uni Uni Uni Ping Pong (ing ang walla walla bing bang) Vary acealdehyde a several fixed concenraions of NAD + and CoA (a a fixed raio) Vary NAD + a several fixed concenraions of acealdehyde and CoA (a a fixed raio) Vary CoA a several fixed concenraions of acealdehyde and NAD + (a a fixed raio) Superficial analysis Since here are hree subsraes, he sysem is fairly complex, and we should do more experimens u. The L plo of acealdehyde as he varied subsrae a differen fixed levels of NAD + and CoA (a a fixed raio) looks fairly normal, which suggess binding of acealdehyde and he nex subsrae is random (α ). Wih NAD + as he varied subsrae, i looks like he random case, wih α >, suggesing ha one of he fixed subsraes alers is binding (bu no acealdehyde) 3. Wih CoA as he varied subsrae, i looks like Ping Pong, and he subsrae inhibiion suggess ha i migh inerfere wih NAD + binding, consisen wih. Kineic Mechanism consisen wih hese resuls