Hochkoeppler, Alejandro

Almae Matris Studiorum Campus

Prof. Alejandro Hochkoeppler Department of Pharmaceutical Sciences and Biotechnology University of Bologna E-mail: a.hochkoeppler@unibo.it KINETICS KINETICS € A + B k−1k1↔ C + D Enzymes do not shift the equilibrium Conversion of  “A” and “B”  into “C” and “D”  is reversible: both k1 and k-1 are larger than the kinetic constants under non-catalysed conditions The molecular order is related to the reaction mechanism The kinetics order defines the time course of a reaction KINETICS Let us consider the time-course of 3 different reactions: TIME (s) [P]   (µM) [P]10   (µM) [P]16000 (µM) 1 3.83 2.57    (0.257) 2.72   (0.170 nM) 2 7.66 5.04    (0.504) 5.20   (0.325 nM) 3 11.49 7.44    (0.744) 7.55   (0.472 nM) 4 15.32 9.75    (0.975) 9.78   (0.611 nM) 5 19.15 11.98  (1.198) 11.86  (0.741 nM) 6 22.98 14.13  (1.413) 13.81  (0.863 nM) 7 26.81 16.22  (1.622) 15.66  (0.979 nM) 8 30.64 18.23  (1.823) 17.41  (1.088 nM) 9 34.47 20.17  (2.017) 19.07  (1.192 nM) 10 38.30 22.04  (2.204) 20.64  (1.290 nM) KINETICS € Zero - order kinetics :v = k       The equation does not contain variablesZero-order: slope is v First-order (squares): non-linear Second-order (triangles): slower Velocity is constant when  kinetics is zero-order KINETICS € First - order kinetics :v = k • A[ ]          The equation contains a single variable€ S0[ ]S0[ ] − P[ ]=S0[ ]S[ ]−∂ S[ ]∂t = k • S[ ]        −∂ S[ ]S[ ]= k •∂t      − ∂ S[ ]S[ ]S0S∫ = k ∂t     t0t∫ lnS0[ ]S[ ]= k • tSlope is k KINETICS €   ln S0[ ]S[ ]= k • tTIME (s) [S]   (µM) ln(S) ln(S0/S) 1 7.143 1.966 0.034 2 6.896 1.931 0.069 3 6.656 1.896 0.104 4 6.425 1.860 0.140 5 6.202 1.825 0.175 6 5.997 1.790 0.210 7 5.778 1.754 0.246 8 5.577 1.719 0.281 9 5.383 1.683 0.317 10 5.196 1.648 0.352 First-order kinetics S0 = 7.4 µM   ln(S0) = 2 KINETICS € Second - order kinetics :v = k • [A] [B]     The equation contains two variables€ −∂ S[ ]∂t = k • S[ ]2       − ∂ S[ ]S[ ]2= k •∂t−∂ S[ ]S[ ]2= k ∂t       1S[ ]= k • t + 1S0[ ]t0t∫S0S∫A and B: same concentration Slope is k, intercept is 1/[S0] KINETICS €   ln S0[ ]S[ ]= ln( S0[ ] • k • t +1)TIME (s) [S]   (nM) ln(S0/S) ln(t) ln([S0]kt + 1) 1 4.830 0.034 0.000 0.034 2 4.675 0.067 0.639 0.067 3 4.528 0.099 1.099 0.099 4 4.389 0.130 1.386 0.130 5 4.259 0.160 1.609 0.160 6 4.137 0.189 1.792 0.189 7 4.021 0.218 1.946 0.218 8 3.912 0.245 2.079 0.245 9 3.808 0.272 2.197 0.272 10 3.710 0.298 2.303 0.298 Second-order kinetics S0 = 5 nM   ln(S0) = 1.609 k = 0.00695 nM-1s-1 KINETICS One-to-one stoichiometry (lactose hydrolyzed, O consumed)                Discontinuous and laborious method [O2] = 0.23 mM (liquid), 8.9 mM (gas). Conversion from P to V.  Lactase assay: •  100 mM lactose (buffered) •  addition of lactase •  20 min at 28 °C •  heating at 100 °C, 3 min •  100 mM acetate, pH 5.3 •  transfer to a Warburg vial:   the side arm contains P.   notatum glucose oxidase   (notatine) KINETICS Lactose: galactose + glucose Glucose-oxidase: glucose + O2            gluconate +  H2O2 Determination of [O2]  Lactase was purified from Escherichia coli protein extracts o-nitrophenyl-β-D-galactopyranoside:        galactose + o-nitrophenol(ate) o-nitrophenolate absorbs at 420 nm ε = 4860 M-1 cm-1 (pH 8.0) KINETICS                 Continuous and sensitive method ε of PNP (PNPGal) = 12900 M-1 cm-1 (pH 8.0). More sensitive? Lactase assay: •  Joshua Lederberg (1950) •  Biochemistry of LacZ (Monod) •  1 mM ONPGal (buffered) •  Addition of enzyme •  Determination of absorbance increase    at 420 nm as a function of time •  The method can be miniaturized (96-    microwells plates) •  [P] calculated using ε KINETICS Zero-order First-order Second-order TIME (s) ABS 420 [P] (µM) ABS 420 [P] (µM) ABS 420  [P] (nM) 1 0.0186 3.83 0.00125 0.257 0.8310-6 0.170 2 0.0372 7.66 0.00244 0.504 1.6010-6 0.325 3 0.0558 11.49 0.00362 0.744 2.2910-6 0.472 4 0.0744 15,3 0.00474 0.975 2.9710-6 0.611 5 0.0930 19.15 0.00582 1.198 3.6010-6 0.741 6 0.1116 22.98 0.00687 1.413 4.1910-6 0.863 7 0.1302 26.81 0.00788 1.622 4.7610-6 0.979 8 0.1488 30.64 0.00886 1.823 5.2910-6 1.088 9 0.1674 34.47 0.00980 2.017 5.7910-6 1.192 10 0.1860 38.30 0.01071 2.204 6.2710-6 1.290 Time-course of ONPGal hydrolysis KINETICS € v = k • [A] [B]     v = k • [E] [S]Reaction rate depends on both enzyme and substrate. Specificity should be tested under second-order conditions. Velocity at the expense of different substrates (equal concentration). Importance of second-order kinetics: •  In 10 seconds the second-order reaction generates a modest absorbance   increase (0.00000627). •  Longer absorbance measurements? Velocity decreases.  t0 ABS = 0. After 10 seconds ABS = 6.2710-6. KINETICS Michaelis and Menten model: € E + S  k−1k1↔ES  kcat⎯ → ⎯   E + Pk1 E[ ] S[ ] = k−1 ES[ ]     k−1k1=E[ ] S[ ]ES[ ]= Kmv = kcat ES[ ]      ES[ ] =E[ ] S[ ]Km    Et[ ] = E[ ] + ES[ ]ES[ ] = S[ ]Km• ( Et[ ] − ES[ ])         Et[ ]ES[ ]=Km + S[ ]S[ ]v = kcat Et[ ] S[ ]Km + S[ ]          v = Vmax S[ ]Km + S[ ]kcat << k1 and k-1 Approximation of the first step as a rapid equilibrium Velocity is determined by kcat and [ES] Km represents the KD of ES complex Vmax depends on [Et] KINETICS € v = Vmax • S[ ]Km + S[ ]=kcat Et[ ] S[ ]Km + S[ ]When S[ ] 〉〉  Km[ ]     v = kcat Et[ ]When S[ ] 〈〈  Km[ ]     v =kcatKmEt[ ] S[ ]When S[ ] 〈〈〈  Km[ ]     v =kcatKmE[ ] S[ ]For E. coli β-galactosidase: kcat = 765 s-1 Km (ONPGal) = 110 µM kcat/Km = 6.95106 M-1 s-1 When [Et] is 5 nM: kcat[Et] = 3.825 µM/s kcat[Et]/Km = 0.03477 s-1 Calculate velocities for [S] equal to 1 mM, 7.5 µM and 5 nM KINETICS When [S] ≅ [Km] reaction is neither zero- nor first-order € −∂ S[ ]∂t =kcat Et[ ] S[ ]Km + S[ ]=kcat Et[ ]1+ Km /  S[ ]( )−∂ S[ ] 1+ KmS[ ]⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = kcat Et[ ]∂t− ∂ S[ ]S0S∫  −  Km∂ S[ ]S[ ]S0S∫ =  kcat Et[ ] ∂tt0t∫S0[ ] − S[ ] + Km • lnS0[ ]S[ ]= kcat Et[ ]tIntegration of Michaelis-Menten equation When [S0] is large (>> Km): ln[S0]/[S] is small compared to [S0] - [S] Reaction is zero-order  When [S0] is small (<< Km): ln[S0]/[S] is large compared to [S0] - [S] Reaction is first-order When [S] ≅	 Km both ln[S0]/[S] and [S0] - [S] are relevant   KINETICS Velocities as a function of [S] according to Michaelis-Menten: Zero-order (Vmax) First-order [Et] = 5 nM Km is equal to the [S] at which v is half of Vmax € 0.5 Vmax =Vmax S0.5[ ]Km + S0.5[ ]    0.5(Km + S0.5[ ]) = S0.5[ ]Km = S0.5[ ]KINETICS Kinetics must  be  zero-order Velocity depends on enzyme concentration Velocity as a function of [Et]: linear dependence Kinetics must  be  first-order Velocity depends on substrate concentration Velocity as a function of [S]: linear dependence [Et] to be determined: [S] to be determined: 

Kinetics: reaction order, Michaelis-Menten Model

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Kinetics: reaction order, Michaelis-Menten Model

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