12/8/2011 Updated: 8 December 2011 CEE 670 Kinetics Lecture #9 1 Print version CEE 670 TRANSPORT PROCESSES IN ENVIRONMENTAL AND WATER RESOURCES ENGINEERING Kinetics Lecture #9 Enzyme Kinetics: Basic Models Kinetic Theory: Encounter & Transition Model & IS Effects Clark, 9.6 Brezonik, pp. 130-158 Enzyme Kinetics David A. Reckhow Enzymatic Reactions 2 Many ways of illustrating the steps Substrate(s) bond to active site Product(s) form via transition state Product(s) are released CEE 670 Kinetics Lecture #9 David A. Reckhow 1
12/8/2011 Note that some Basic Enzyme Kinetics references use k 2 for k -1 , and k 3 for k 2 3 Irreversible k 1 k 2 → E + S ES E + P ← k- 1 Single intermediate The overall rate is determined by the RLS, k 2 d [ S ] d [ P ] r k 2 ES [ ] dt dt But we don’t know [ES], so we can get it by the SS mass balance d [ ES ] 0 k [ E ][ S ] k [ ES ] k [ ES ] 1 1 2 dt Again, we only know [E o ] or [E tot ], not free [E], so: 0 k [ E ] [ ES ] [ S ] k [ ES ] k [ ES ] 1 o 1 2 CEE 670 Kinetics Lecture #9 David A. Reckhow Reactants, products and Intermediates 4 Simple Progression of components for simple single intermediate enzyme reaction Shaded block shows steady state intermediates Assumes [S]>>[E] t From Segel, 1975; Enzyme Kinetics CEE 670 Kinetics Lecture #9 David A. Reckhow 2
12/8/2011 Basic Enzyme Kinetics II 5 And solving for [ES], k [ ES ][ S ] k [ ES ] k [ ES ] k [ E ][ S ] 1 1 2 1 o k [ E ][ S ] 1 o [ ES ] k [ S ] k k 1 1 2 [ E ][ S ] o [ ] ES k k [ S ] 1 2 k 1 CEE 670 Kinetics Lecture #9 David A. Reckhow Michaelis-Menten 6 Irreversible k 1 k 2 → E + S ES E + P ← k- 1 Single intermediate d [ P ] [ ] r k 2 ES dt [ E ][ S ] o [ ES ] k k [ S ] 1 2 k 1 d [ P ] k [ E ][ S ] r [ S ] 2 o max r k k dt K [ S ] [ S ] 1 2 k s 1 CEE 670 Kinetics Lecture #9 David A. Reckhow 3
12/8/2011 Michaelis Menten Kinetics 7 Classical substrate plot rmax 100 80 Reaction Rate 60 0.5r max 40 K s d [ P ] r [ S ] max r s dt K [ S ] 20 0 0 20 40 60 80 100 120 CEE 670 Kinetics Lecture #9 David A. Reckhow Substrate Concentration Substrate and growth 8 d [ P ] d [ S ] 1 dX If we consider Y r dt dt Y dt We can define a microorganism-specific substrate utilization rate, U dX r dt U YX X Y And the maximum rates are then max U k max Y 1 d [ S ] k [ S ] 1 d [ X ] [ S ] max U and s s X dt K [ S ] X dt K [ S ] CEE 670 Kinetics Lecture #9 David A. Reckhow 4
12/8/2011 Linearizations 9 Lineweaver-Burke Double reciprocal plot Wikipedia version Voet & Voet version CEE 670 Kinetics Lecture #9 David A. Reckhow 10 das CEE 670 Kinetics Lecture #9 David A. Reckhow 5
12/8/2011 3 types 11 Lineweaver Burk Hanes Eadie-Hofstee CEE 670 Kinetics Lecture #9 David A. Reckhow Compare predictions 12 ad CEE 670 Kinetics Lecture #9 David A. Reckhow 6
12/8/2011 k 1 k 2 k 3 → E + S ES EP 2 2E + P 2 Multi-step ← k- 1 P 1 13 Double intermediate Also gives: d [ P ] r [ S ] max r s dt K [ S ] But now: k k [ E ] k k k 2 3 o 3 1 2 r K s 1 max k k k k k 2 3 2 3 Note what happens when: k 3 >> k 2 CEE 670 Kinetics Lecture #9 David A. Reckhow Activation Energy 14 Activation Energy must always be positive Unlike ∆ H, which may be positive or negative Differing reaction rates Activated Complex Activated Complex E a E a Energy Energy reactants reactants E H E H f f products products Reaction Coordinate Reaction Coordinate CEE 670 Kinetics Lecture #9 David A. Reckhow 7
12/8/2011 Encounter Theory I 15 Uncharged Solutes Nature of diffusion in water Encounter within a solvent cage Random diffusion occurs through elementary jumps of distance Molecular Molecular 2 r diameter radius Average time between jumps 2 For a continuous medium: 2 2 or D 2 D For a semi-crystalline structure: 6 2 2 or D 6 D For water, D ~ 1x10 -5 cm 2 s -1 , and � = 4x10 -8 cm, so ~ 2.5x10 -11 s If time between vibrations is ~ 1.5x10 -13 s, then the average water molecule vibrates 150 times (2.5x10 -11 /1.5x10 -13 ) in its solvent cage before jumping to the next one. CEE 670 Kinetics Lecture #9 David A. Reckhow Encounter Theory II 16 Probability of Encounter If A and B are the same size as water They will have 12 nearest neighbors Probability that “A” will encounter “B” in a solvent cage of 12 neighbors is: Proportional to the mole fraction of “B” P 6 X With each new jump, “A’ has 6 new neighbors B A B Where: n # molecules of “B” per cm 3 B X B 1 3 # molecules of solvent per cm 3 Molecular volume (cm 3 ) Geometric packing factor CEE 670 Kinetics Lecture #9 David A. Reckhow 8
12/8/2011 Encounter Theory III 17 And combining the rate of movement with the probability of encountering “B”, we get an expression for the rate of encounter with “B” 1 6 D P 2 A B AB Then substituting in for the probability 3 6 D ( 6 n ) 1 B 2 AB 36 n D B For water, =0.74, and the effective diffusion coefficient, D AB = D A + D B , and � =r AB , the sum of the molecular radii Then we get: 1 25 r D n AB AB B AB # of encounters/sec for each molecule of “A ” CEE 670 Kinetics Lecture #9 David A. Reckhow Encounter Theory IV 18 Now the total # of encounters between “A” and “B” per cm 3 per second is: n 25 A r D n n AB AB A B AB In terms of moles of encounters (encounter frequency) this becomes: 3 3 cm cm 1000 n 1000 L A L Z 25 r D n n e , AB AB AB A B N molecules N molecules o Mole o Mole 25 r D [ A ] n AB AB B n B =[B]/N 0 /1000 L/cm 3 cm 2 s -1 M -1 s -1 cm #/Mole 2 Z 2 . 5 x 10 r D N [ A ][ B ] e , AB AB AB 0 CEE 670 Kinetics Lecture #9 David A. Reckhow 9
12/8/2011 Encounter Theory V 19 Frequency Factor 2 Z 2 . 5 x 10 r D N [ A ][ B ] e , AB AB AB 0 A When E a = 0, k=A Activated Complex E a E a / RT k Ae Energy reactants products Reaction Coordinate CEE 670 Kinetics Lecture #9 David A. Reckhow Transition State Theory I 20 Consider the simple bimolecular reaction k A B C Even though it is an elementary reaction, we can break it down into two steps k A B AB C “Activated Complex” Where the first “equilibrium” is: [ AB ] K [ A ][ B ] [ AB ] K [ A ][ B ] Activated So the forward rate is: Complex E a d [ C ] k [ AB ] k K [ A ][ B ] Energy reactants dt products CEE 670 Kinetics Lecture #9 David A. Reckhow Reaction Coordinate 10
12/8/2011 Transition State Theory II 21 Now the transition state is just one bond vibration away from conversion to products Frequency of vibration (s -1 ) E vib Planks Law: h vibrational Planck’s constant (6.62 x 10 -27 ergs·s) energy Bond energy must be in the thermal region: Temperature (ºK) E bond Bond kT energy Boltzman constant (1.3807×10 − 16 ergs ºK -1 ) So equating, we get: kT h kT h And since conversion occurs on the next vibration: kT k k K K h CEE 670 Kinetics Lecture #9 David A. Reckhow Transition State Theory III 22 Now from basic thermodynamics: G o G o RT ln K or K e RT And also G o H T S S e So: H K e R RT kT And combining: S H k e R e RT h Recall: E H P V H kT E S a And substituting back in: k e R e RT h A CEE 670 Kinetics Lecture #9 David A. Reckhow 11
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