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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.

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Presentation on theme: "Kinetics: Reaction Order Reaction Order: the number of reactant molecules that need to come together to generate a product. A unimolecular S  P reaction."— Presentation transcript:

1 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]

2 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.

3 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


5 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”.

6 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]

7 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]


9 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.

10 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

11 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])


13 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.

14 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 M -1 s -1. They will catalyze a reaction almost every time a substrate is bound in the active site- catalytic perfection!!!

15 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

16 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.

17 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

18 Enzyme Inhibitors in Medicine Many current pharmaceuticals are enzyme inhibitors (e.g. HIV protease inhibitors for treatment of AIDS- see pages ). 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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