Michaelis-Menten Theory of Enzyme Action

Enzyme kinetics Michaelis-Menten theory of enzyme action 2 Enzyme action due to enzyme-substrate complex Enzyme action due to enzyme-substrate complex An enzyme, E, catalyzes conversion of substrate, S, to product, P. The binding is fast: E + S ⇄ ES 1. Substrate binds to the enzyme The conversion to product and its release is slow: ES → E + P 2. Substrate is converted to product The rate of formation of product is the rate of the second, slow step 3. Product is released from the enzyme 3 4 1 Enzymes catalyze biochemical reactions Enzymes catalyze biochemical reactions Typical enzyme concentrations, [E], are nmol/L Carbonic anhydrase, E, catalyzes hydration of CO2(aq), S, to HCO3−(aq), P. Typical substrate concentrations, [S], are much larger, mmol/L. Sketch how you imagine initial rate, V0, depends on the initial substrate concentration, [S]0. By limiting measurements to initial rate, V0, of formation of product, substrate concentration nearly constant at its initial value, [S]0. 5 Enzymes catalyze biochemical reactions [S]0 mmol/L V0 mmol/(L s) [S]0 mmol/L V0 mmol/(L s) 0.100 0.00248 5.00 0.083 0.250 0.00610 10.0 0.125 0.500 0.0119 15.0 0.150 1.00 0.0227 20.0 0.167 2.50 0.0500 25.0 0.179 6 Enzymes catalyze biochemical reactions The red dots are the initial rate, V0, of formation of bicarbonate, in mmol/L/s, versus initial concentration, [S]0, of CO2, in mmol/L, at pH = 7.1, 273.5 K, and an enzyme concentration of 2.3 nmol/L. Plot initial rate, V0, versus initial substrate concentration, [S]0. 7 8 2 Enzymes catalyze biochemical reactions Enzymes catalyze biochemical reactions The green line shows that for very small [CO2] , the rate of formation of bicarbonate is first order in [CO2]. As [CO2] becomes larger, however, the rate of formation of bicarbonate progressively slows. 9 Enzymes catalyze biochemical reactions 10 Enzymes catalyze biochemical reactions The blue curve is a fit to the observed data. The dashed line is the asymptote of the fit. It is the maximum rate, Vmax, possible rate of formation of bicarbonate for the given enzyme concentration. Add a curve showing how you imagine initial rate, V0, changes when the initial enzyme concentration is lowered. 11 12 3 Enzymes catalyze biochemical reactions Michaelis-Menten mechanism of enzyme action Repeating the measurements for 40% of the original enzyme concentration shows that both the initial slope and the asymptote are proportional to the enzyme concentration. Enzyme catalysis shows these characteristic features: 13 Leonor Michaelis and Maud Menten, ~ 1913 14 Leonor Michaelis and Maud Menten , ~ 1945 In 1913 Leonor Michaelis and Maud Menten provided the general theory of enzyme action that accounts for these features of enzyme catalysis 15 16 4 Michaelis-Menten mechanism of enzyme action Michaelis-Menten mechanism of enzyme action The Michaelis-Menten (MM) mechanism accounts for enzyme action in two steps. Then, ES transforms the substrate into product much more slowly, First, the enzyme, E, binds to the substrate, S, to form an enzyme-substrate complex, ES, (1) E + S ⇄ ES (2) ES → E + P rate2,for = k2 [ES] Since we are concerned just with initial rates … [P] remains so small that we can ignore the reverse process, rate1,for = k1 [E] [S]0 E + P → ES rate1,rev = k−1 [ES] Therefore, the rate of formation of product is the rate of step (2), Typical enzyme concentrations are nmol/L and typical substrate concentrations are mmol/L. For initial rates this means … V0 = k2 [ES] [S] changes only a little from its initial value, [S]0, 17 18 Michaelis-Menten mechanism of enzyme action Michaelis-Menten mechanism of enzyme action The initial rate V0 = k2 [ES] depends on the concentration of the enzymesubstrate complex. ES is formed by forward reaction (1), Because not all of the enzyme is bound in ES, let’s instead express the total enzyme concentration as [Et] = [E] + [ES] ( d[ES]/dt )formed = k1 [E] [S]0 = k1 ( [Et] – [ES] ) [S]0 ES is depleted by the reverse of reaction (1) and by forward reaction (2). ( d[ES]/dt )depleted = k−1 [ES] + k2 [ES] We can then express the unbound enzyme concentration as = ( k−1 + k2 ) [ES] [E] = [Et] – [ES] 19 20 5 Michaelis-Menten mechanism of enzyme action Michaelis-Menten mechanism of enzyme action Since reaction (2) is so slow, we assume a steady state [ES] is achieved, Determine the units of the Michaelis constant Km = ( k−1 + k2 ) / k1 ( d[ES]/dt )formed = ( d[ES]/dt )depleted Hint: E + S ⇄ ES, k1 ( [Et] – [ES] ) [S]0 = ( k−1 + k2 ) [ES] rate1,for = k1 [E] [S]0 , rate1,rev = k−1 [ES] We can rearrange this steady-state equality to express [ES] as ES → E + P, [ES] = k1 [Et] [S]0 / ( k1 [S]0 + ( k−1 + k2 ) ) = [Et] [S]0 / ( [S]0 + ( k−1 + k2 ) / k1 ) rate2,for = k2 [ES] = [Et] [S]0 / ( [S]0 + Km ) In terms of the Michaelis constant Km = ( k−1 + k2 ) / k1 21 22 Michaelis-Menten mechanism of enzyme action Michaelis-Menten mechanism of enzyme action Since Using our expression of [ES], rate1,for = k1 [E] [S]0, and so k1 ~ 1/(M s) [ES] = [Et] [S]0 / ( [S]0 + Km ) rate1,rev = k−1 [ES], and so k−1 ~ 1/s we can write the rate of formation of product as the Michaelis-Menten equation rate2,for = k2 [ES], and so k2 ~ 1/s V0 = k2 [Et] [S]0 / ( [S]0 + Km ) the units of the Michaelis constant are … Km = ( k−1 + k2 ) / k1 ~ (1/s + 1/s) / (1/(M s)) ~M 23 24 6 Michaelis-Menten mechanism of enzyme action Michaelis-Menten mechanism of enzyme action For values of [S]0 very large relative to Km, For values of [S]0 very small relative to Km, [S]0 + Km ≈ [S]0 [S]0 + Km ≈ Km Therefore, the Michaelis-Menten equation simplifies to Therefore, the Michaelis-Menten equation simplifies to V0 = k2 [Et] = Vmax V0 = k2 [Et] [S]0 / Km = ( Vmax / Km ) [S]0 That is, Vmax is the plateau, that the rate approaches in proportional to the total enzyme concentration. That is, for very small values of [S]0, the rate of formation of product is linear in the substrate concentration [S]0, with slope Vmax/Km. This means we can express the Michaelis-Menten equation as V0 = Vmax [S]0 / ( [S]0 + Km ) 25 26 Getting Vmax and Km from kinetic data Getting Vmax and Km from kinetic data Lineweaver-Burk plot Rearrange the Michaelis-Menten equation V0 = Vmax [S]0 / ( [S]0 + Km ) to sketch how 1/V0 depends on 1/[S]0 27 28 7 Getting Vmax and Km from kinetic data Getting Vmax and Km from kinetic data The equation for the Lineweaver-Burk plot is Here is data for catalysis of the hydrolysis of aqueous carbon dioxide at pH = 7.1, 273.5 K, and carbonic anhydrase concentration 2.3 nmol/L is 1/V0 = (Km / Vmax ) 1/[S]0 + 1 / Vmax The plateau rate, Vmax, is the reciprocal the y intercept. The Michaelis constant, Km, is Vmax times the slope. CO2 / (nmol/L) 1.25 2.5 5 20 V0 / (nmol/(L s)) 2.78 × 10−2 5.00 × 10−2 8.33 × 10−2 1.67 × 10−1 Construct the Lineweaver-Burk plot for this catalysis reaction and use it to determine that Vmax = 0.250 nmol/(L s) Km = 10.0 nmol/L 29 30 Getting Vmax and Km from kinetic data Getting Vmax and Km from kinetic data As a check, plot the Michaelis-Menten equation, As a check, plot the Michaelis-Menten equation, V0 = Vmax [S]0 / ( [S]0 + Km ) V0 = Vmax [S]0 / ( [S]0 + Km ) using Vmax = 0.250 nmol/(L s) and Km = 10.0 nmol/L. using Vmax = 0.250 nmol/(L s) and Km = 10.0 nmol/L. Do this by calculating V0 for values of [S]0 equal to 0.10, 0.50, 1, 1.25, 2.5, 5, 10, 15, 20, and 25. Note that your values for [S]0 equal to 1.25, 2.5, 5, and 20 should reproduce the measured values V0. Finally, add to your plot the line 31 V0 = ( Vmax / Km ) [S]0 To see that it is tangent to the Michaelis-Menten curve near [S]0 ≈ 0. 32 8 Getting Vmax and Km from kinetic data As a check, plot the Michaelis-Menten equation, V0 = Vmax [S]0 / ( [S]0 + Km ) using Vmax = 0.250 nmol/(L s) and Km = 10.0 nmol/L. Here is what your results should look like. 33 9