7 29-10enzymeskinetics-coloso-110715062024-phpapp01[1]


Published on

Published in: Business, Technology
1 Like
  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

7 29-10enzymeskinetics-coloso-110715062024-phpapp01[1]

  1. 1. Enzymes Pt 2: KineticsRelicardo M. Coloso, Ph. D.College of MedicineCentral Philippine University
  2. 2. Kinetics of enzyme action Michaelis-Menten model An enzyme converts one chemical (the substrate), into another (the product). A graph of product concentration vs. time follows three phases as shown in the following graph.At very early time points, the rate of product accumulation increases over time.Special techniques are needed to study the early kinetics of enzymeaction, since this transient phase usually lasts less than a second (the figuregreatly exaggerates the first phase).
  3. 3. Enzyme velocity as a function of substrate concentration If you measure enzyme velocity at many different concentrations of substrate, the graph generally looks like this: Enzyme velocity as a function of substrate concentration often follows the Michaelis-Menten equation: Where KM –Michaelis-Menten constant Vmax – maximum velocity of the reaction
  4. 4. Vmax is the limiting velocity as substrate concentrations get very large.Vmax (and V) are expressed in units of product formed per time. If you know the molar concentration of enzyme, you can divide the observedvelocity by the concentration of enzyme sites in the assay, and express Vmax as units of moles of product formed per second per mole ofenzyme sites. This is the turnover number, the number of molecules of substrate converted to product by one enzyme site per second. Indefining enzyme concentration, distinguish the concentration of enzyme molecules and concentration of enzyme sites (if the enzyme is a dimerwith two active sites, the molar concentration of sites is twice the molar concentration of enzyme). KM is expressed in units of concentration, usually in Molar units. KM is the concentration of substrate that leads to half-maximal velocity. Toprove this, set [S] equal to KM in the equation above. Cancel terms and youll see that V=Vmax/2.
  5. 5. Note that KM is not a binding constant that measures the strength of binding between the enzyme and substrate. Its value includes the affinity of substrate for enzyme, but also the rate at which the substrate bound to the enzyme is converted to product.The Michaelis-Menten model is too simple for many purposes. The Briggs-Haldane model has proven more useful:Under the Briggs-Haldane model, the graph of enzyme velocity vs.substrate looks the same as under the Michaelis-Menten model, butKM is defined as a combination of all five of the rate constants in themodel.
  6. 6. Significance of KM of an enzyme Example: Hexokinase – enzyme that phophorylates glucose Glucose + ATP Glucose – 6-P + ADP + H+ Rates of phosphorylation of glucose and fructose in the brain Properties of brain Sugar concn in Calculated rate Sugar brain cell hexokinase of phosphorylation Vmax KM In vivo Glucose 17 10-5 10-5 8.5 Fructose 25 10-3 10-6 10-2Units: Vmax – micromol/min/g; KM – Molar; Sugar concn – Molar; rate of phosphorylation – micromol/min/gKM value tells us whether or not the enzyme is physiologically important.It also gives information on the affinity of the enzyme for its substrate
  7. 7. Assumptions of enzyme kinetic analyses Standard analyses of enzyme kinetics (the only kind discussed here) assume: The production of product is linear with time during time interval is used •• •The concentration of substrate vastly exceeds the concentration of enzyme. This means that the free concentration of substrate is very close to the concentration you added, and that substrate concentration is constant throughout the assay. • A single enzyme forms the product. • There is negligible spontaneous creation of product without enzyme • No cooperativity. Binding of substrate to one enzyme binding site doesnt influence the affinity or activity of an adjacent site. • Neither substrate nor product acts as an allosteric modulator to alter the enzyme velocity.
  8. 8. Linear form of the Michaelis-Menten equationTransform the curved data into a straight line, so they could beanalyzed with linear regressionOne way to do this is with a Lineweaver-Burk plot.Take the inverse of the Michaelis-Menten equation and simplify:Ignoring experimental error, a plot of 1/V vs. 1/S will be linear, with a Y-interceptof 1/Vmax and a slope equal to Km/Vmax. The X-intercept equals 1/Km.
  9. 9. Example of double reciprocal plot to solve for KM and Vmax
  10. 10. Enzymes can be affected by inhibitory compounds or inhibitors•most clinical drug therapy is based on inhibiting the activity of enzymes,•analysis of enzyme reactions using the tools described above has beenfundamental to the modern design of pharmaceuticals.Enzyme inhibitors fall into two broad classes:1) those causing irreversible inactivation of enzymes and2) those whose inhibitory effects can be reversed.Irreversible inhibitors•These inhibitors usually cause an inactivating, covalent modification of enzymestructure. Examples: many poisons, such as cyanide, carbon monoxide and polychlorinated biphenols (PCBs) Cyanide is a classic example of an irreversible enzyme inhibitor: by covalently binding mitochondrial cytochrome oxidase, it inhibits all the reactions associated with electron transport.• are usually considered to be poisons and are generally unsuitable fortherapeutic purposes.
  11. 11. Reversible inhibitors can be divided into two main categories;competitive inhibitors and noncompetitive inhibitors, with a thirdcategory, uncompetitive inhibitors, rarely encountered. Binding Site on Enzyme Kinetic effectInhibitor Type Specifically at the catalytic site, where it Vmax is unchanged; Km, as defined 1) Competitive competes with substrate for binding in a by [S] required for ½ maximal Inhibitor dynamic equilibrium- like process. Inhibition activity, is increased. is reversible by substrate. Binds E or ES complex other than at the Km appears unaltered; Vmax is2)Noncompetitive catalytic site. Substrate binding unaltered, but decreased proportionately to Inhibitor ESI complex cannot form products. Inhibition inhibitor concentration. cannot be reversed by substrate. Binds only to ES complexes at locations other than the catalytic site. Substrate binding Apparent Vmax decreased; Km, as3) Uncompetitive modifies enzyme structure, making inhibitor- defined by [S] required for ½ Inhibitor binding site available. Inhibition cannot be maximal activity, is decreased. reversed by substrate.The hallmark of all the reversible inhibitors is that when the inhibitor concentrationdrops, enzyme activity is regenerated. Usually these inhibitors bind to enzymes bynon-covalent forces and the inhibitor maintains a reversible equilibrium with theenzyme.
  12. 12. Michaelis–Menten curves for enzyme with or without inhibitor
  13. 13. Example: Given the enzyme Q which A ---------> BConverts substrate A to product B Enz (Q) Tube A Tube B Tube C Tube D [S], or 4.8 mM 1.2 mM 0.6 mM 0.3 mM Conc of A 1/[S] 0.21 0.83 1.67 3.33 Δ OD540 (Vi) or 0.081 0.048 0.035 0.020 Rate of reaction 1/Vi 12.3 20.8 31.7 50.0 Making a Lineweaver-Burk plot of these results shows (red line in graph) that 1/Vmax = 10, so Vmax = 0.10 −1/Km = − 0.8, so Km = 1.25 mM (In other words, when [S] is 1.25 mM, 1/Vi = 20, and Vi = 0.05 or one-half of Vmax.)
  14. 14. With Non competitive inhibitor Competitive inhibitorLineweaver-Burk Plot
  15. 15. The table below summarizes the results with competitive inhibitor Tube A Tube B Tube C Tube D [S] 4.8 mM 1.2 mM 0.6 mM 0.3 mM 1/[S] 0.21 0.83 1.67 3.33 ΔOD540 0.060 0.032 0.019 0.011 (Vi) 1/Vi 16.7 31.3 52.6 90.9The Lineweaver-Burk plot of these results is shown above (green line in graph). 1/Vmax = 10, so Vmax remains 0.10. Now, however, −1/Km = − 0.4, so Km = 2.50 mM(In other words, it now takes a substrate concentration [S] of 2.50 mM, to achieve one-half of Vmax.)
  16. 16. With Non competitive inhibitor Competitive inhibitorLineweaver-Burk Plot
  17. 17. The table below summarizes the results with non competitive inhibitor Tube A Tube B Tube C Tube D [S] 4.8 mM 1.2 mM 0.6 mM 0.3 mM 1/[S] 0.21 0.83 1.67 3.33 ΔOD540 0.040 0.024 0.016 0.010 (Vi) 1/Vi 25 41 62 102 The Lineweaver-Burk plot of these results is shown above( blue line in graph). Now 1/Vmax = 20, so Vmax = 0.05. But −1/Km = − 0.8, so Km = 1.25 mM as it was in the first experiment.So once again it only takes a substrate concentration,[S], of 1.25 mM to achieve one- half of Vmax.
  18. 18. With Non competitive inhibitor Competitive inhibitorLineweaver-Burk Plot
  19. 19. Random bi bi mechanismOrdered bi bi mechanism Enzyme reaction Intersecting LB plots mechanisms
  20. 20. ACP,acyl carrier protein; HSL,homoserine lactone; SAM,S-adenosylmethionine; MTA,5′-methylthioadenosineOrdered bi bi mechanism inacyl homoserine lactone synthase
  21. 21. Adenylate kinase (myokinase) is a phosphotransferase enzyme that catalyzesthe interconversion of adenine nucleotides, and plays an important role in cellular energy homeostasis. The reaction catalyzed is: 2 ADP ATP + AMPThe reaction is randombi bi mechanism
  22. 22. Ping pong mechanism Parallel LB plots
  23. 23. A friendly animated ping pong game
  24. 24. Ping pong mechanism for uridylyltransferase
  25. 25. Thus,1. Substrates may add in a random order (either substrate may combine first with the enzyme) or in a compulsory order (substrate A must bind before substrate B).2. In ping-pong reactions, one or more products are released from the enzyme before all the substrates have added.
  26. 26. Keep youreyeon theball!