5.60 spring 2008 Lecture #35 1 Enzyme Catalysis Readings: SAB, pp. 745-752 Enzymes are biological catalysts, consisting of proteins, that greatly accelerate reaction rates and do so with exquisite specificity. That is, a chemical reaction involving one reactant goes much faster, but other chemical reactions involving the same reactant or the same chemical reaction performed on an analogous reactant are sometimes largely unaffected. (accelerated) SA ⎯E⎯ → PA (not accelerated) PC SB PB (not accelerated) The reactant in an enzymatic process is called a substrate. Examples: 1) invertase Sucrose + H2O(C12H22O11) ⎯⎯⎯ → glucose (C6H12O6) + fructose (C6H12O6). OH O 2) HO lactase lactose (a C12 milk sugar) ⎯⎯⎯⎯ → glucose + galactose (lactose intolerant: unable to produce lactose. LactAid and other products contain a form of lactase) 5.60 spring 2008 3) Lecture #35 H 2 H COO– H C COO– C fumarase +H 2 O ⎯ ⎯⎯⎯⎯ → C H –OOC (1) H –OOC fumarate Key features: C OH L-malate rate accelerated product specificity substrate selectivity [invertase does not work on “splenda” which is chlorinated sucrose.] stereospecificity Michaelis-Menten Mechanism One mechanism that describes the behavior of many enzymes is known as the Michaelis-Menten mechanism. 1 ⎯⎯ ⎯ → ES E + S← k ⎯ k −1 2 ES ⎯ ⎯ →E + P k (1) (2) S→P (overall) In the first step of this mechanism (1), enzyme binds substrate, creating a Michaelis complex (ES). This step is readily reversible. In the second step (2) the substrate is converted to product and is released from the enzyme. Frequently this step may be considered irreversible because the concentrations, [E] and [P], are quite small. 5.60 spring 2008 Lecture #35 3 Because [E] is so small and k2 is usually large, the Michaelis complex (ES) is generally present at low concentration. Treating ES as a steady-state intermediate, we can derive an expression for the reaction rate. d[ES] = k1[E][S] − k −1[ES] − k 2 [ES] = 0 . dt At this point it is convenient to also replace [E], because it is often difficult to measure accurately. This is done by expressing the total enzyme concentration [E]0 as a sum of free enzyme [E] and enzyme bound to substrate [ES]. [E]0 = [E] + [ES] [E] = [E]0 – [ES]. Although [ES] is small, the approximation [E] ≈ [E]0 is not valid, because [E]0 is usually very small as well. Substituting for [E], one obtains 0 = k1 ([E]0 − [ES])[S] − k −1[ES] + k 2 [ES] = 0 k1[E]0 [S] [E]0 [S] = k1[S] + k −1 + k 2 [S] + k −1 + k 2 k1 d[P] k 2 [E]0 [S] = k 2 [ES] = k + k2 dt [S] + −1 k1 [ES]SS = This expression is traditionally written v= k [E] [S] d[P] = cat 0 [S] + K m dt initial Michaelis-Menten equation where “v” stands for reaction “velocity”, k2 has been replaced by kcat, and k −1 + k 2 has been replaced by Km, the Michaelis constant. k1 One typically measures the initial value of v (rate of appearance of P) at a series of initial values of [S]. Two kinds of plots are convenient. 5.60 spring 2008 (i) v vs. [S], v = Lecture #35 4 d[P] dt initial k cat [E]0 = vmax “saturated” All E tied up in ES v vmax 2 initial slope = k cat [E]0 vmax = , when [S] << Km Km Km [S] = Km [S] The “initial slope” must not be confused with the “initial rate.” The initial slope is obtained by measuring the initial rate for several reaction mixtures at initial values of [S]. This illustrates two important limits: (i) [S] = K m → v ≈ (ii) [ S ] >> K m → v ≈ kcat [ E ]0 → constant with respect to [S] . k cat [E]0 [S] → linear in [S] Km vmax is the maximum reaction rate for a given quantity of enzyme [E]0. It is achieved at a saturating concentration of substrate, at which every enzyme molecule is “busy” carrying out a reaction (ES → E + P) and there are no free enzyme molecules “waiting” to find substrate. 5.60 spring 2008 Lecture #35 5 kcat is sometimes called the turnover number, because it represents the rate at which each enzyme molecule produces products under these saturating conditions. Maximum number of product molecules formed per unit time Number of enzyme molecules Maximum rate of product formation k max k cat [E]0 = = = = k cat Total concentration of enzyme [E]0 [E]0 Turnover number = (2) Lineweaver-Burk plot A second useful plot for enzyme kinetics is called the Lineweaver-Burk plot. It is useful because it is linear in 1/[S]. 1 [S] + K m 1 Km 1 = = + · v K cat [E]0 [S] k cat [E]0 k cat [E]0 [S] 1/v slope = • • –1 Km Km K = m k cat [E]0 vmax 1 1 = k cat [E]0 vmax 1/[S] Notice that with this type of plot, it is possible to obtain Km and vmax without measuring [E]0. However, in order to determine kcat = vmax/[E]0, one needs an accurate measurement of [E]0. To find accurate values of the slope and intercept, it is necessary to sample a wide range of values spaced roughly equally in 1/[S]. One usually sets up reactions with a variety of initial substrate concentrations and measures the initial rate for each to obtain the data for this plot.