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Kinetics: Reaction Order

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 .

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Kinetics: Reaction Order

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  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 to 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 (t½) 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. t½ = ln2/k = 0.693/k

  4. Enzyme Kinetics E + S  ES  P + E • E- enzyme, S- substrate, ES- enzyme-substrate complex, P-product • k1 and k-1 are the forward and reverse rate constants for formation of ES • k2 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”.

  5. Enzyme Reaction Velocity For the complex enzymatic reaction: v = d[P]/dt = k2 [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 = k1 [E][S] - k-1[ES] - k2[ES]

  6. Simplifying Assumptions • The velocity equation is too complex to be directly evaluated, so assumptions must be made. • Equilibrium. If k-1 >> k2, 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]

  7. 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 = k1[E][S] = k-1[ES] + k2 [ES] By defining the Michaelis Constant (KM) = (k-1 + k2)/k1 then rearranging and substituting, [ES] can be expressed as: [ES] = [E]T [S]/(KM + [S]) The initial velocity (vo) = k2 [ES] = k2 [E ]T [S]/(KM + [S]) The key point is that [E]T and [S] are measurable.

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

  9. Maximal Velocity When the enzyme is saturated with substrate, the reaction is progressing at its maximal velocity, Vmax. At saturation [E]T = [ES], and the equation for reaction velocity simplies to Vmax = k2 [E]T Combing with the expression for vo leads to the Michaelis-Menten Equation of enzyme kinetics: [Fig. 12-3] Vo = Vmax [S] / (KM + [S])

  10. Implications of M-M Equation • Vo = Vmax [S] / (KM + [S]) • 1. From the M-M equation, when [S] = KM, vo = Vmax/2. • This means that low values of KM imply the enzyme achieves maximal catalytic efficiency at low [S]. • 2. The catalytic constant, kcat = Vmax / [E]T • kcat 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.

  11. Implications of M-M Equation (cont.) • 3. When [S] << KM, very little ES is formed. Under these conditions: vo ~ (kcat/Km)[E][S] • The kcat/Km term is a measure of the enzyme’s catalytic efficiency. • The upper limit to kcat/Km is k1, I.e. the decomposition of ES to E + P can occur no more frequently than ES is formed. • The most efficient enzymes have kcat/Km values near the diffusion-controlled limit of 108-109 M-1 s-1. They will catalyze a reaction almost every time a substrate is bound in the active site- catalytic perfection!!!

  12. Practical Summary- Analysis of Kinetics • Using Fig. 12-3, the plot of vo vs. [S] • When [S] << KM, the reaction increases linearly with [S]; I.e. vo = (Vmax / KM ) [S] • When [S] = KM, vo = Vmax /2 (half maximal velocity); this is a definition of KM: the concentration of substrate which gives ½ of Vmax. • When [S] >> Km, vo = Vmax

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

  14. 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

  15. 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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