7.29.10 enzymes (kinetics) coloso


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Lecture: Enzyme02

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7.29.10 enzymes (kinetics) coloso

  1. 1. Enzymes Pt 2: Kinetics Relicardo M. Coloso, Ph. D. College of Medicine Central Philippine University
  2. 2. 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 enzyme action, since this transient phase usually lasts less than a second (the figure greatly exaggerates the first phase). Kinetics of enzyme action Michaelis-Menten model
  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 K M –Michaelis-Menten constant V max – 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 observed velocity by the concentration of enzyme sites in the assay, and express Vmax as units of moles of product formed per second per mole of enzyme sites . This is the turnover number , the number of molecules of substrate converted to product by one enzyme site per second. In defining enzyme concentration, distinguish the concentration of enzyme molecules and concentration of enzyme sites (if the enzyme is a dimer with two active sites, the molar concentration of sites is twice the molar concentration of enzyme). K M is expressed in units of concentration, usually in Molar units. K M is the concentration of substrate that leads to half-maximal velocity. To prove this, set [S] equal to K M in the equation above. Cancel terms and you'll see that V=Vmax/2.
  5. 5. Note that K M 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, but K M is defined as a combination of all five of the rate constants in the model.
  6. 6. Significance of K M 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 Sugar Properties of brain hexokinase Sugar concn in brain cell Calculated rate of phosphorylation In vivo Vmax K M Glucose 17 10 -5 10 -5 8.5 Fructose 25 10 -3 10 -6 10 -2 Units: Vmax – micromol/min/g; K M – Molar; Sugar concn – Molar; rate of phosphorylation – micromol/min/g K M 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 •    <ul><li>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. </li></ul>•    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 doesn't 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. One way to do this is with a Lineweaver-Burk plot . Take the inverse of the Michaelis-Menten equation and simplify: Linear form of the Michaelis-Menten equation Transform the curved data into a straight line, so they could be analyzed with linear regression Ignoring experimental error, a plot of 1/V vs. 1/S will be linear, with a Y-intercept of 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 K M and V max
  10. 10. Enzymes can be affected by inhibitory compounds or inhibitors <ul><li>most clinical drug therapy is based on inhibiting the activity of enzymes, </li></ul><ul><li>analysis of enzyme reactions using the tools described above has been fundamental to the modern design of pharmaceuticals. </li></ul><ul><li>Enzyme inhibitors fall into two broad classes: </li></ul><ul><li>those causing irreversible inactivation of enzymes and </li></ul><ul><li>those whose inhibitory effects can be reversed. </li></ul><ul><li>These inhibitors usually cause an inactivating, covalent modification of enzyme structure. </li></ul>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. Irreversible inhibitors <ul><li>are usually considered to be poisons and are generally unsuitable for therapeutic purposes. </li></ul>
  11. 11. Reversible inhibitors can be divided into two main categories; competitive inhibitors and noncompetitive inhibitors , with a third category, uncompetitive inhibitors , rarely encountered. encountered. The hallmark of all the reversible inhibitors is that when the inhibitor concentration drops, enzyme activity is regenerated. Usually these inhibitors bind to enzymes by non-covalent forces and the inhibitor maintains a reversible equilibrium with the enzyme. Inhibitor Type Binding Site on Enzyme Kinetic effect 1) Competitive Inhibitor Specifically at the catalytic site , where it competes with substrate for binding in a dynamic equilibrium- like process. Inhibition is reversible by substrate. V max is unchanged; K m , as defined by [S] required for ½ maximal activity, is increased. 2)Noncompetitive Inhibitor Binds E or ES complex other than at the catalytic site . Substrate binding unaltered, but ESI complex cannot form products. Inhibition cannot be reversed by substrate. K m appears unaltered; V max is decreased proportionately to inhibitor concentration. 3) Uncompetitive Inhibitor Binds only to ES complexes at locations other than the catalytic site . Substrate binding modifies enzyme structure, making inhibitor- binding site available. Inhibition cannot be reversed by substrate. Apparent V max decreased; K m , as defined by [S] required for ½ maximal activity, is decreased.
  12. 12. Michaelis–Menten curves for enzyme with or without inhibitor
  13. 16. A ---------> B Enz (Q) Example: Given the enzyme Q which Converts substrate A to product B 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 .)   Tube A Tube B Tube C Tube D [S], or Conc of A 4.8 mM 1.2 mM 0.6 mM 0.3 mM 1/[S] 0.21 0.83 1.67 3.33 Δ OD 540 (V i ) or Rate of reaction 0.081 0.048 0.035 0.020 1/V i 12.3 20.8 31.7 50.0
  14. 17. Competitive inhibitor With Non competitive inhibitor Lineweaver-Burk Plot
  15. 18. The table below summarizes the results with competitive inhibitor The 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 .)   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 ΔOD 540 (V i ) 0.060 0.032 0.019 0.011 1/V i 16.7 31.3 52.6 90.9
  16. 19. Competitive inhibitor With Non competitive inhibitor Lineweaver-Burk Plot
  17. 20. The table below summarizes the results with non competitive inhibitor 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 .   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 ΔOD 540 (V i ) 0.040 0.024 0.016 0.010 1/V i 25 41 62 102
  18. 21. Competitive inhibitor With Non competitive inhibitor Lineweaver-Burk Plot
  19. 22. Random bi bi mechanism Ordered bi bi mechanism Enzyme reaction mechanisms Intersecting LB plots
  20. 23. Ordered bi bi mechanism in acyl homoserine lactone synthase ACP,acyl carrier protein; HSL,homoserine lactone; SAM,S-adenosylmethionine; MTA,5′-methylthioadenosine
  21. 24. Adenylate kinase ( myokinase ) is a phosphotransferase enzyme that catalyzes the interconversion of adenine nucleotides, and plays an important role in cellular energy homeostasis . The reaction catalyzed is: 2 ADP ATP + AMP The reaction is random bi bi mechanism
  22. 25. Ping pong mechanism Parallel LB plots
  23. 26. A friendly animated ping pong game
  24. 27. Ping pong mechanism for uridylyltransferase
  25. 28. <ul><li>Thus, </li></ul><ul><li>Substrates may add in a random order (either substrate may combine </li></ul><ul><li>first with the enzyme) or in a compulsory order (substrate A must bind </li></ul><ul><li>before substrate B). </li></ul><ul><li>In ping-pong reactions , one or more products are released from the </li></ul><ul><li>enzyme before all the substrates have added. </li></ul>
  26. 29. Thank you! Keep your eye on the ball!