Rate laws, Reaction Orders The rate or velocity of a chemical reaction is loss of reactant or appearance of product in concentration units, per unit time d[p] = d[s] The rate law for a reaction is of the form Rate = d[p] = k[a] n a []n b [C]n c... The order in each reactant is its exponent, n i The overall order of the reaction is the sum of the n i Reaction Order Molecularity The order of a reaction is determined by the number of participants that must come together before the rate-limiting-step (RLS) of the reaction It is determined by the mechanism, not stoichiometry of a reaction It must be determined (measured) experimentally Determining Reaction Order Initial Rate Method Choose initial concentrations of reactants Measure Initial Rate, before concentrations of reactants change appreciably (tangents) 2 Disappearance of A 8 Initial rate Course of reaction [A] 6 4 2 5 5 2 25 3 35 4 45 5 time
Determining Reaction Order Initial Rate Method Choose initial concentrations of reactants Measure Initial Rate, before concentrations of reactants change appreciably Change initial concentrations, and measure initial rate again Example: For A + C, assume Rate = k[a] na [] nb [A]o []o Rate if then R R 2 = R n a = 2 R 2 R 2 = 2R n a = 2 R 3 R 2 = 4R n a = 2 R 3 = R n b =, etc. Once you know the n i, plug in and solve for k Determining Reaction Order Integrated Rate Equation Method Allow the rx to proceed and measure change in concentration of reactant or product. y trial and error, determine which equation fits the data Disappearance of [A] 2 8 [A] 6 4 2 2 3 4 5 time Integrated Rate Equations If our rate 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... We will have to integrate it in order to use our experimentally-determined values of [P], [A], etc., to make a plot like the one on the previous slide. 2
Determining Reaction Order Integrated Rate Equation, Zero Order Choose initial concentrations of reactants Measure Rate over a substantial portion of the reaction The product and reactant concentrations change automatically Fit to theory Example: zero order reaction A d[] = d[a] The rate law is = k[a] = k Integrating from [A] at t = to [A] t at t = t [ A]t d[a] = k [ A] o t [A] t [A] = kt So, a plot of [A] t - [A] vs t is linear, with a slope of k, IFF it is zero order Integrated Rate Equation First Order Reaction A The rate law is Rewrite as d[] d[a] [A] = k = d[a] = k[a] Integrate from [A] at t = to [A] t at t = t t d[a] = k [A] [ A]t [ A] o ln[a] t ln[a] = ln( A t A ) = kt ln( A A t ) = kt A plot of ln (A /A t ) vs t is linear, with a slope of k If so, the reaction is st Order in A and zero order in all other components or Integrated Rate Equation Second Order Reaction A The rate law is Rewrite as d[] = d[a] = k[a] 2 Integrate from [A] at t = to [A] t at t = t [A]t d[a] [A] 2 = k [A] = kt or [A] 2 d[a] = k So, a plot of (/[A] t ) vs t is linear, with a slope of -k If so, the reaction is 2nd Order in A and zero order in all other components [A]t [ A]t [ A] o = [A] kt t 3
Determining Reaction Order Multiple reactants, first order in each A + C d[c] = d[a] = d[] = k[a][] Use stoichiometry to relate all reactant concentrations to their initial values and to [A] If [A] = [] and the stoichiometry is, then the integrated rate equation is the same as for 2 nd order in A If [A] [] and Stoichiometry is n (A + n C) Then substitute [] = [] -2([A] -[A]) before integrating Remember, stoichiometry doesn t determine reaction order, but it may be useful in guessing what the order might be Units of Rate Constants If rate = d[a], it must have units of (concentration) (time) And the products on the right side of the equation must have the same units, so the units of k must be different Zero order reactions Rate (M min - ) = k (M min - ) First order reactions Rate (M min - ) = k (min - ) [A](M) Second order reactions Rate (M min - ) = k (M - min - ) [A] 2 (M 2 ) Third order reactions Rate (M min - ) = k (M -2 min - ) [A] 3 (M 3 ) Sometimes initial rate and integrated rate measurements disagree - Why? The integrated rate equation gives the concentrations of components whose concentrations change during the reaction consider: A + C Rate = k 2 [A][] The reaction is second order overall, first order in each reactant Rate decreases as [A] decreases and as [] decreases. What if [] remains constant during a run? because it is present in large excess (e.g., solvent), because its concentration is buffered (e.g., H +, OH - ), or because it is a catalyst, and is regenerated. Then the reaction is pseudo-zero order in, and pesudo-first order overall: Rate = k apparent [A], where k apparent = k 2 [] 4
Model an Enzyme Reaction The enzyme and substrate(s) get together Transformations occur The product(s) is (are) released, regenerating enzyme k k E + S " 2 "" E S " "" E + P k k 2 We have to worry about four concentrations [E], [S], [E S], and [P] and four reactions, some probably st order, some 2 nd k [E][S], k 2 [E S], k - [E S] and k -2 [E][P] Too complicated! Simplifying Assumptions. Measure initial rates (<% of reaction) then [P] = so we can neglect the reverse reaction, k -2 [E][P] and we can ignore product inhibition (later) and = [S] + [E S] 2. Use catalytic amounts of enzyme, [E] << [S] Then = [S] + 3. Enzyme exists in only two forms, E and E S, which sum to 4. [E S] reaches a constant level rapidly and remains constant during the measurement Could happen in two ways k k 2 k - >> k 2, rapid equilibrium E + S E S E + P k - k 2, steady state k - 5. = k 2 [E S] Simplifying Assumptions steady state k k E + S E S 2 E + P d[e S] or = = k [E][S] k - [E S] k 2 [E S] = k [E][S] [E S]{k - + k 2 } ut we also know that [S] = and = [E]+[E S], so that [E] = - [E S], and = k 2 [E S] (assumes the catalytic step is the RLS) So, let s substitute for free enzyme [E] and solve for [E S] = k { - [E S]} [E S]{k - + k 2 } Collect terms in [E S] (next slide) k - 5
Steady State k k 2 E + S E S E + P = k - [E S]{k + (k - + k 2 )} k - solve for [E S] k [E S] = k + (k - + k 2 ) multiply by k 2 k = k 2 [E S] = k 2 k + (k - + k 2 ) k = 2 + (k + k ) V = max = (k + k ) - 2 k - 2 +[S] k = k 2 Simplifying Assumptions rapid equilibrium k k E + S E S 2 E + P [E S] is controlled by the equilibrium and is slowly bled off toward product k - K S = [E] [E S] [E S] = [E] K S [E S] = { - [E S]} K S [E S]{+ } = K [E S] = S E K S K S + [S] = T K S + K S = k 2 [E S] = k 2 K S + = K S + The same equation Comparison steady state vs rapid equilibrium Steady State = k + k 2 k time units: = concentration time - conc K S = k Rapid Equilibrium units: time = concentration time - conc ecause at equilibrium k [E] = k [E S] k so k = [E] k [E S] So, they really only differ in the relative magnitude of k - and k 2 k k - k 2 E + S E S E + P 6
Michaelis-Menten Equation Vmax (units of velocity) Vo 5 (units of concentration) 2 4 6 8 2 4 6 [S]o Michaelis-Menten Equation Could also be written = + Which might be useful if we have more than one substrate Or, as [S] = + K = M + [S] Which is unitless Fraction of Fraction of Substrate bound What Data are Required? If is too high poor estimation of - looks zero order If is too low poor estimation of - no evidence of saturation poor estimation of - looks first order Need data with in the vicinity of, Need data with both above and below Vo 5 2 4 6 8 2 4 6 [S]o So how can it be both first order and zero order? 7
Order of the Reaction >> = V [S] max V [S] max = X + Mixed order zero order Vo 5 << = V [S] max V [S] max = + [S] X 2 4 6 8 2 4 6 [S]o first order More Order Zero Order Region = d[p] = d[s] = >> [S ]t d[s] = [S ] t [S] = t So, a plot of -[S] or [P] vs t is linear, with slope = First Order Region << = d[s] = [S] [S ] [S] d[s] [S] = t ln [S] = t So, the semilog plot vs t is linear, with slope = / In between -can t simplify invert before integrating: t = ln [S] + {[S] } The sum of the limiting cases V o and [S] o Relations = [S] = + + [S] So, if V =.2 = is 2% of, + [S] and =.2 so, S =.25K M.2 =.25 Or, if = 5, how close is the rate to? 5K = M =.83 + 5 How about =? K = M =.9 + 8
Complications - Substrate Inhibition It s actually hard to get to At =, you observe only 9% of The example actually had = And, sometimes substrate binds incorrectly and inhibits Never reaches Vo vs [S]o Can t get 75 Vo 5 25 2 4 6 8 2 [S]o Lineweaver-urk Invert both sides: = K + [S] M = + y - intercept = slope = x - intercept = - Lineweaver-urk Double-reciprocal plot / vs / has linear region, even with substrate inhibition. /Vo -5 5 5 25 /[S]o 9
Eadie-Hofstee Plot alternative linear transform = + + = V V + = V + = max [S] [S] = - Nowadays, you can use nonlinear least squares fitting Measure and several and put them into = V [S] max + To calculate. Determine the difference between the measured and calculated (the residuals, (calc obs )) Square the residuals for all values of, and add them together Adjust and until the sum of the squares of the residuals is at a minimum. Which method to use? Lineweaver-urk, Eadie-Hoftsee and non-linear least-squares give the same answers, because they are different forms of the same equation ut, if there is noise in the data, the noise has different effects, use weighted least squares for L- plot, because the lowest rates have the most error, and tend to underestimate.
= /( + ) >> zero order Mixed order Vo 5 << first order 5 5 [S] o Non-Michaelis-Menten ehavior cooperativity Michaelis-Menten Negative Cooperativity V o Positive Cooperativity [S] o Induced Fit Enzymes and Ideas are Flexible Conformational Change Hexokinase is a monomer, and not cooperative, but it was an early demonstration of induced fit Hexokinase Hexokinase ound Glc (Glc not shown)
Cooperativity S S K d + S K d2 + S S S Higher [S] always shifts the equilibrium to the right (from cubes to cylinders) If K d > K, cooperativity is positive. d2 If K d < K, cooperativity is negative. d2 How Cooperative is it? Koshland s Cooperativity factor: R S = giving V =.9 giving V =. For normal MM kinetics V [S] =.9 = V [S] = 9 =. = + + = 9 R S = 9 /9 = 8 For positive cooperativity, R s < 8 For negative cooperativity, R s > 8 Dependence of V o on = k 2 Units of are velocity, e.g., concentration/min ut So, it s not an absolute quantity Must be expressed as per mg protein for comparison Like specific activity Turnover number = V Δ[ S] Δ[ P] max time or time = [E] [E] time Molecules S P per enzyme molecule/time Also = k 2, might call it k cat if it is actually for catalysis (not, e.g., product release) 2
Dependence of V o on = k 2, and, always At low, = / At high, = So, V o, always, always, always Except If an irreversible inhibitor is present If a second substrate is limiting If the assay system becomes limiting If an Enzyme-cofactor or other equilibrium is involved Dependence of V o on exceptions - irreversible inhibitor The irreversible inhibitor inactivates the enzyme stoichiometrically When you have added more enzyme than there is inhibitor, the reaction rate is proportional to normal behavior Enzyme is inactivited until irreversible inhibitor is used up Dependence of V o on exceptions - second substrate becomes limiting Example: glucose oxidase α-d-glucose + O 2 + H 2 O g δ-gluconolactone + H 2 O 2 ut O 2 is not very soluble in water (2.4 x -4 M @ 25 C) So, rate can t continue to increase with forever. normal behavior [O 2 ] starts to become limiting, here 3
Dependence of V o on exceptions - coupled enzyme assay becomes limiting A common way to assay an enzyme is to use a second enzyme that uses the product of the first enzyme to make something that s easy to detect, such as NADH, which absorbs at 34nm NAD + NADH + H + A C E E 2 normal behavior Coupling enzyme can t keep up More than one Substrate Except for isomerases, most enzymes have more than one substrate How do we deal with this fact? A + E D E A + is awkward notation A better notation is due to W. W. Cleland (Madison,WI): Rewrite normal MM sequence E + S D E S E + P as S P E E S D E P E Then, more complex reactions can be written simply Note also that = V [S] max + [S] = [S] + Models two substrates on, two products released E + A + D P + Q + E Substrates could add together, one after the other A P E E A E A D E P Q EQ E This is an ordered bi bi reaction Kinetic equation can be generated as before We ll assume rapid equilibrium Q 4
Remember Single Substrate eqn? rapid equilibrium k k E + S E S 2 E + P Three Main Points in the derivation. Enzyme is partitioned into two forms = [E] + [E S] 2. Dissociation constants can be defined for complexes K S = [E] [E S] k - 3. k 2 = k cat = rate constant for RLS, so = k cat [E S] Ordered bi bi assume rapid equilibrium K A E + A D EA + K E EA k cat g E + P + Q = k cat [EA] = [E] + [EA] + [EA] K A = [A][E] [EA] K = [][EA] [EA]. Substitute [E] = - [EA] - [EA] into eq for K A 2. Solve for [EA]; substitute that into eq for K 3. Solve for [EA] 4. Multiply by k cat : V = max + K [] + K AK [A][] Random bi bi either substrate can add first, either product can leave first A E A P Q E Q E E E A D E P Q E P A Q P E inding of A and may or may not be independent 5
K Random bi bi K A E + A D EA + E αka + E αk E + A D EA g E + P + Q k cat Same steps - just more of them solving substituting multiply by k cat = k cat [EA] = [E] + [EA] + [E] + [EA] = K A = [A][E] [EA] K = [][E] [E] αk A = [A][E] [EA] αk = [][EA] [EA] + αk A [A] + αk [] + αk AK [A][] Ping-Pong One product leaves before last substrate adds A P E E A D F P F F D EQ E Q = + K [] + K A [A] Double reciprocal plots Vary one substrate at different fixed levels of other substrate = K A " K % $ ' #[]&[A] + " + K % $ ' # []& 4 Ordered, vary [A] 35 3 [] 25 /V 2 5 5-2 2 4 6 8-5 /[A] 6
, K A, K? Note that the intercept =! + K $ # & = " []% K + [] So, plot the intercept versus /[], and obtain / from the intercept of the replot extrapolating [] to infinity The apparent /K A is = K A apparent K A K [] + So, use a replot, as above. (For K of the varied substrate (A), the intersection of the lines is at (-/K A ), / ) K A Intercept /K A -apparent /[] /[] Double reciprocal plots ordered bi bi, continued = K A " K % $ ' #[]&[A] + " + K % $ ' # []& /V vs /[] at fixed [A] Note that plot of same data vs second substrate looks different; rearrange to see why (collect terms in /[], and put in slopeintercept form with x = /[]) /V 2 [A] 5 5-2 3 4 5 6-5 /[] Random bi = αk # A + K & % ( $ []'[A] + # + αk & % ( $ [] ' α = 4 Random, vary [A] α =.5 4 Random, vary [A] 35 3 [] 35 3 [] 25 25 /V 2 5 /V 2 5 5 5-2 2 4 6 8-2 2 4 6 8-5 -5 /[A] /[A] 7
, K A, K? =! + αk $ Note that the intercept # & = + αk " [] % [] So, plot the intercept versus /[], and obtain / from the intercept of the replot extrapolating [] to infinity Various replots can be used. (For K of the varied substrate, the intersection of the lines is at (-/ ), (-α)/ ) Intercept /[] Double reciprocal plots ping pong = K A [A] + " + K % $ ' # []& Ping Pong, vary [A] Constant slope Decreasing intercept /V 23 8 3 [] 8 3-2 -2 2 4 6 8 /[A], K A, K? Slope = K [] + K A Int = K [] + K A So, use an intercept replot to find K A, and K Use a slope replot for K A /, and solve for. 8
Quick Example Aldehyde Dehydrogenase Acetaldehyde + NAD + + CoA è NADH + AcCoA + H + As we will see, the kinetic mechanism is i Uni Uni Uni Ping Pong (ting tang walla walla bing bang) Vary acetaldehyde at several fixed concentrations of NAD + and CoA (at a fixed ratio) Vary NAD + at several fixed concentrations of acetaldehyde and CoA (at a fixed ratio) 9
Vary CoA at several fixed concentrations of acetaldehyde and NAD + (at a fixed ratio) Superficial analysis Since there are three substrates, the system is fairly complex, and we should do more experiments ut. The L plot of acetaldehyde as the varied substrate at different fixed levels of NAD + and CoA (at a fixed ratio) looks fairly normal, which suggests binding of acetaldehyde and the next substrate is random (α = ) 2. With NAD + as the varied substrate, it looks like the random case, with α >, suggesting that one of the fixed substrates alters its binding (but not acetaldehyde) 3. With CoA as the varied substrate, it looks like Ping Pong, and the substrate inhibition suggests that it might interfere with NAD + binding, consistent with 2. Kinetic Mechanism consistent with these results 2