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Kinetics: Reaction Order Reaction Order: the number of reactant molecules that need to come together to generate a product. A unimolecular S P reaction is first order. A bimolecular 2S P reaction is second order. A bimolecular S1 + S2 P is second order, first order in S1 and first order in S2. The reaction velocity for bimolecular reactions are not linearly dependent on S. 2S P v = k [S] 2 S1 + S2 P v = k [S1] [S2]

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First Order Rate Equations Rate equation: describes the progress of a reaction over time. For a unimolecular reaction: v = d[P]/dt = -d[S]/dt= k[S] rearranging gives: d[S]/[S] = d ln[S] = -k dt Integrating from the starting concentration [S] o at time t o to the concentration [S] at a future time t gives: d ln[S] = -k dt ln[S] = ln[S] o –kt This allows the calculation of [S] at any time t from the starting concentration and the reaction rate constant.

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First Order Rate Equations (cont.) ln[S] = ln[S] o –kt This equation for a first order reaction is linear A plot of ln[S] vs. t is a straight line: [Fig. 12-1] The intercept is the starting concentration [S] o Slope is the negative of the rate constant (k) Half-life of the reaction (t ½ ) is the time required for half of S to be used up. Since the rate is intrinsic to the reaction in a first order reaction, the slope of the line in the plot never changes and the half-life is the same regardless of the starting concentration. t ½ = ln2/k = 0.693/k

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Enzyme Kinetics E + S ES P + E E- enzyme, S- substrate, ES- enzyme-substrate complex, P-product k 1 and k- 1 are the forward and reverse rate constants for formation of ES k 2 is the rate constant for the decomposition of ES to form P, and we assume that the reverse reaction is negligible. When [S] is high enough, all of the enzyme is in the ES form and the second step is rate limiting. Above this [S], the progress of the reaction is insensitive to the amount of S present, i.e. “the reaction is zero order in substrate”.

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Enzyme Reaction Velocity For the complex enzymatic reaction: v = d[P]/dt = k 2 [ES] The overall rate of the production of ES is given by rate of production from E and S minus the rate of decomposition of ES either via the reverse reaction or formation of product: d[ES]/dt = k 1 [E][S] - k -1 [ES] - k 2 [ES]

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Simplifying Assumptions The velocity equation is too complex to be directly evaluated, so assumptions must be made. 1.Equilibrium. If k -1 >> k 2, the first step of the reaction reaches equilibrium. Unfortunately, this situation rarely holds. 2. Steady State. If [S] >> [E], [ES] remains constant because the large excess of S molecules rapidly fill all enzyme active sites: d[ES]/dt = 0. This holds true until all of S is used up. [Fig. 12-2]

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Expressing With Measurable Quantities The total concentration of enzyme [E] T can be determined, but it is very difficult to measure [E] and [ES]. Steady state rate equation: v = k 1 [E][S] = k -1 [ES] + k 2 [ES] By defining the Michaelis Constant (K M ) = (k -1 + k 2 )/k 1 then rearranging and substituting, [ES] can be expressed as: [ES] = [E] T [S]/(K M + [S]) The initial velocity (v o ) = k 2 [ES] = k 2 [E ] T [S]/(K M + [S]) The key point is that [E] T and [S] are measurable.

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Importance of Initial Velocity In reality, v o is velocity measured before 10% of S is consumed. Before [ES] has built up- can assume reverse reaction is negligible. Working with v o minimizes complications Reverse reactions Inhibition of reaction by product

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Maximal Velocity When the enzyme is saturated with substrate, the reaction is progressing at its maximal velocity, V max. At saturation [E] T = [ES], and the equation for reaction velocity simplies to V max = k 2 [E] T Combing with the expression for v o leads to the Michaelis-Menten Equation of enzyme kinetics: [Fig. 12-3] V o = V max [S] / (K M + [S])

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Implications of M-M Equation V o = V max [S] / (K M + [S]) 1. From the M-M equation, when [S] = K M, v o = V max /2. This means that low values of K M imply the enzyme achieves maximal catalytic efficiency at low [S]. 2. The catalytic constant, k cat = V max / [E] T k cat is also called the turnover number of the enzyme, i.e. the number of reaction processes (turn-overs) that each active site catalyzes per unit time. Turnover numbers vary over many orders of magnitude for different enzymes in accord with need.

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Implications of M-M Equation (cont.) 3. When [S] << K M, very little ES is formed. Under these conditions: v o ~ (k cat /K m )[E][S] The k cat /K m term is a measure of the enzyme’s catalytic efficiency. The upper limit to k cat /K m is k 1, I.e. the decomposition of ES to E + P can occur no more frequently than ES is formed. The most efficient enzymes have k cat /K m values near the diffusion-controlled limit of 10 8 -10 9 M -1 s -1. They will catalyze a reaction almost every time a substrate is bound in the active site- catalytic perfection!!!

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Practical Summary- Analysis of Kinetics Using Fig. 12-3, the plot of v o vs. [S] When [S] << K M, the reaction increases linearly with [S]; I.e. v o = (V max / K M ) [S] When [S] = K M, v o = V max /2 (half maximal velocity); this is a definition of K M : the concentration of substrate which gives ½ of V max. When [S] >> K m, v o = V max

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Practical Summary - Practical Summary - V max and K m K M gives an idea of the range of [S] at which a reaction will occur. The larger the K M, the WEAKER the binding affinity of enzyme for substrate. V max gives an idea of how fast the reaction can occur under ideal circumstances. K cat / K M gives a practical idea of the catalytic efficiency, i.e. how often a molecule of substrate that is bound reacts to give product.

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Inhibition of Enzymes Enzyme Inhibitor: a molecule that reduces the effectiveness of an enzyme. Mechanisms for enzyme inhibition Competitive inhibition- substrate analogs bind in the active site reducing availability of free enzyme Uncompetitive inhibition- inhibitor binds to the substrate-enzyme complex and presumably distorts the active site making the enzyme less active Mixed or non-competitive inhibition- combination of competitive and uncompetitive inhibition Inactivator- irriversible reaction with enzyme

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Enzyme Inhibitors in Medicine Many current pharmaceuticals are enzyme inhibitors (e.g. HIV protease inhibitors for treatment of AIDS- see pages 336-337). An example: Ethanol is used as a competitive inhibitor to treat methanol poisoning. Methanol is metabolized by the enzyme alcohol dehydrogenase producing highly toxic formaldehyde. Ethanol competes for the same enzyme. Administration of ethanol occupies the enzyme thereby delaying methanol metabolism long enough for clearance through the kidneys.

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Rules for deriving rate laws for simple systems 1.Write reactions involved in forming P from S 2. Write the conservation equation expressing the distribution.

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