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Enzymes II

Enzymes II. Andy Howard Introductory Biochemistry 21 October 2010. Enzymes Enzyme kinetics Michaelis-Menten kinetics: overview Kinetic Constants Kinetic Mechanisms Induced Fit. Enzymes (concluded) Bisubstrate reactions Measurements and calculational tools Inhibition Why study it?

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Enzymes II

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  1. Enzymes II Andy HowardIntroductory Biochemistry 21 October 2010 Biochem: Enzymes II

  2. Enzymes Enzyme kinetics Michaelis-Menten kinetics: overview Kinetic Constants Kinetic Mechanisms Induced Fit Enzymes (concluded) Bisubstrate reactions Measurements and calculational tools Inhibition Why study it? The concept Types of inhibitors What we’ll discuss Biochem: Enzymes II

  3. [B] Kinetics, continued t • In most situations more product will be produced per unit time if A0 is large than if it is small, and in fact the rate will be linear with the concentration at any given time: • d[B]/dt = v = k[A] • where v is the velocity of the reaction and k is a constant known as the forward rate constant. • Here, since [A] has dimensions of concentration and d[B]/dt has dimensions of concentration / time, the dimensions of k will be those of inverse time, e.g. sec-1. Biochem: Enzymes II

  4. More complex cases • More complicated than this if >1 reactant involved or if a catalyst whose concentration influences the production of species B is present. • If >1 reactant required for making B, then usually the reaction will be linear in the concentration of the scarcest reactant and nearly independent of the concentration of the more plentiful reactants. • In fact, many enzymes operate by converting a second-order reaction into a pair of first-order reactions! Biochem: Enzymes II

  5. Bimolecular reaction • If in the reactionA + D  Bthe initial concentrations of [A] and [D] are comparable, then the reaction rate will be linear in both [A] and [D]: • d[B]/dt = v = k[A][D] = k[A]1[D]1 • i.e. the reaction is first-order in both A and D, and it’s second-order overall Biochem: Enzymes II

  6. Forward and backward • Rate of reverse reaction may not be the same as the rate at which the forward reaction occurs. • If the forward reaction rate of reaction 1 is designated as k1,the backward rate typically designated as k-1. Biochem: Enzymes II

  7. Multi-step reactions • In complex reactions, we may need to keep track of rates in the forward and reverse directions of multiple reactions. • Thus in the conversion A  B  Cwe can write rate constantsk1, k-1, k2, and k-2as the rate constants associated with converting A to B, converting B to A, converting B to C, and converting C to B. Biochem: Enzymes II

  8. [ES] Michaelis-Menten kinetics t • A very common situation is one in which for some portion of the time in which a reaction is being monitored, the concentration of the enzyme-substrate complex is nearly constant. Thus in the general reaction • E + S  ES  E + P • where E is the enzyme, S is the substrate, ES is the enzyme-substrate complex (or "enzyme-intermediate complex"), and P is the product • We find that [ES] is nearly constant for a considerable stretch of time. Biochem: Enzymes II

  9. Michaelis-Menten rates • Rate at which new ES molecules are being produced in the first forward reaction is equal to the rate at which ES molecules are being converted to (E and P) and (E and S). • Formation of ES is first-order in both S and available [E] • Therefore: rate of formation of ES from left =vf = k1([E]tot - [ES])[S]because the enzyme that is already substrate-bound is unavailable! Biochem: Enzymes II

  10. Equating the rates • We started with the statement that the rate of formation of ES and the rate of destruction of it are equal • Rate of disappearance of ES on right and left isvd = k-1[ES] + k2[ES] = (k-1+ k2)[ES] • This rate of disappearance should be equal to the rate of appearance • Under these conditions vf = vd. Biochem: Enzymes II

  11. Derivation, continued • Thus since vf = vd by assumption, k1([E]tot - [ES])[S] = (k-1+ k2)[ES] • Km (k-1+ k2)/k1 = ([E]tot - [ES])[S] / [ES] • [ES] = [E]tot [S] / (Km + [S]) • But the rate-limiting reaction is the formation of product: v0 = k2[ES] • Thus v0 = k2[E]tot [S] / (Km + [S]) Biochem: Enzymes II

  12. Maximum velocity • What conditions would produce the maximum velocity? • Answer: very high substrate concentration ([S] >> [E]tot),for which all the enzyme would be bound up with substrate. Thus under those conditions we getVmax = v0 = k2[ES] = k2[E]tot Biochem: Enzymes II

  13. Using Vmax inM-M kinetics • Thus sinceVmax = k2[E]tot, • v0 = Vmax [S] / (Km+[S]) • That’s the famous Michaelis-Mentenequation Biochem: Enzymes II

  14. Graphical interpretation Biochem: Enzymes II

  15. Physical meaning of Km • As we can see from the plot, the velocity is half-maximal when [S] = Km • Trivially derivable: if [S] = Km, thenv0 = Vmax[S] / ([S]+[S]) = Vmax /2 • We can turn that around and say that the Km is defined as the concentration resulting in half-maximal velocity • Km is a property associated with binding of S to E, not a property of turnover Michaelis Constant:British Christian hiphop band Biochem: Enzymes II

  16. kcat • We’ve already discussed what Vmax is; but it will be larger for high [E]tot than otherwise. • A quantity we often want is the maximum velocity independent of how much enzyme we originally dumped in • That would be kcat =Vmax / [E]tot • Oh wait: that’s just the rate of our rate-limiting step, i.e. kcat = k2 Biochem: Enzymes II

  17. Physical meaning of kcat • Describes turnover of substrate to product:Number of product molecules produced per sec per molecule of enzyme • More complex reactions may not have kcat = k2, but we can often approximate them that way anyway • Some enzymes very efficient:kcat > 106 s-1 Biochem: Enzymes II

  18. Specificity constant, kcat/Km • kcat/Km measures affinity of enzyme for a specific substrate: we call it the specificity constant or the molecular activity for the enzyme for that particular substrate • Useful in comparing primary substrate to other substrates (e.g. ethanol vs. propanol in alcohol dehydrogenase) Biochem: Enzymes II

  19. Dimensions • Km must have dimensions of concentration (remember it corresponds to the concentration of substrate that produces half-maximal velocity) • Vmax must have dimensions of concentration over time (d[A]/dt) • kcat must have dimensions of inverse time • kcat / Km must have dimensions of inverse time divided by concentration, i.e.inverse time * inverse concentration Biochem: Enzymes II

  20. Typical units for kinetic parameters • Remember the distinction between dimensions and units! • Km typically measured in mM or µM • Vmax typically measured in mMs-1 or µMs-1 • kcat typically measured in s-1 • kcat / Km typically measured in s-1M-1 Biochem: Enzymes II

  21. Kinetic Mechanisms • If a reaction involves >1 reactant or >1 product, there may be variations in kinetics that occur as a result of the order in which substrates are bound or products are released. • Examine eqns. 13.48, 13.49, 13.50, and the unnumbered eqn. on p. 430 in G&G, which depict bisubstrate reactions of various sorts. As you can see, the possibilities enumerated include sequential, random, and ping-pong mechanisms. Biochem: Enzymes II

  22. Historical thought • Biochemists, 1935 - 1970 examined effect on reaction rates of changing [reactants] and [enzymes], and deducing the mechanistic realities from kinetic data. • In recent years other tools have become available for deriving the same information, including static and dynamic structural studies that provide us with slide-shows or even movies of reaction sequences. • But diagrams like these still help! Biochem: Enzymes II

  23. Sequential, ordered reactions W.W.Cleland • Substrates, products must bind in specific order for reaction to complete A B P Q_____________________________E EA (EAB) (EPQ) EQ E Biochem: Enzymes II

  24. Sequential, random reactions • Substrates can come in in either order, and products can be released in either order A B P Q EA EQ__E (EAB)(EPQ) E EB EP B A Q P Biochem: Enzymes II

  25. Ping-pong mechanism • First substrate enters, is altered, is released, with change in enzyme • Then second substrate reacts with altered enzyme, is altered, is released • Enzyme restored to original state A P B QE EA FA F FB FQ E Biochem: Enzymes II

  26. Induced fit Daniel Koshland • Conformations of enzymes don't change enormously when they bind substrates, but they do change to some extent. An instance where the changes are fairly substantial is the binding of substrates to kinases. Cartoon from textbookofbacteriology.net Biochem: Enzymes II

  27. Kinase reactions • unwanted reactionATP + H-O-H ⇒ ADP + Pi • will compete with the desired reactionATP + R-O-H ⇒ ADP + R-O-P • Kinases minimize the likelihood of this unproductive activity by changing conformation upon binding substrate so that hydrolysis of ATP cannot occur until the binding happens. • Illustrates the importance of the order in which things happen in enzyme function Biochem: Enzymes II

  28. iClicker quiz, question 1 • The Michaelis constant Km has dimensions of • (a) concentration per unit time • (b) inverse concentration per unit time • (c) concentration • (d) inverse concentration • (e) none of the above Biochem: Enzymes II

  29. iClicker quiz question 2 • kcat is a measure of • (a) substrate binding • (b) turnover • (c) inhibition potential • (d) none of the above Biochem: Enzymes II

  30. Hexokinase conformational changes G&G Fig. 13.28 Biochem: Enzymes II

  31. Measurements and calculations • The standard Michaelis-Menten formulation is v0=f([S]), but it’s not linear in [S]. We seek linearizations of the equation so that we can find Km and kcat, and so that we can understand how various changes affect the reaction. Biochem: Enzymes II

  32. Lineweaver-Burk Dean Burk • Simple linearization of Michaelis-Menten: • v0 = Vmax[S]/(Km+[S]). Take reciprocals: • 1/v0 = (Km +[S])/(Vmax[S])= Km /(Vmax[S]) + [S]/(Vmax[S])1/v0 =(Km/Vmax)*1/[S] + 1/Vmax • Thus a plot of 1/[S] as the independent variable vs. 1/v0 as the dependent variable will be linear with Y-intercept = 1/Vmax and slope Km/Vmax Hans Lineweaver Biochem: Enzymes II

  33. How to use this • Y-intercept is useful directly:computeVmax = 1/(Y-intercept) • We can get Km/Vmax from slope and then use our knowledge of Vmax to get Km; or • X intercept = -1/ Km… that gets it for us directly! Biochem: Enzymes II

  34. Demonstration that the X-intercept is at -1/Km • X-intercept means Y = 0 • In Lineweaver-Burk plot, • 0 = (Km/Vmax)*1/[S] + 1/Vmax • For nonzero 1/Vmax we divide through: • 0 = Km /[S] + 1, -1 = Km/[S], [S] = -Km. • But the axis is for 1/[S], so the intercept is at 1/[S] = -1/ Km. Biochem: Enzymes II

  35. Graphical form of L-B 1/v0, s L mol-1 1/Vmax,s L mol-1 Slope=Km/Vmax 1/[S], M-1 -1/Km, L mol-1 Biochem: Enzymes II

  36. Are those values to the left of 1/[S] = 0 physical? • No. It doesn’t make sense to talk about negative substrate concentrations or infinite substrate concentrations. • But if we can curve-fit, we can still use these extrapolations to derive the kinetic parameters. Biochem: Enzymes II

  37. Advantages and disadvantages of L-B plots • Easy conceptual reading of Km and Vmax(but remember to take the reciprocals!) • Suboptimal error analysis • [S] and v0 values have errors • Error propagation can lead to significant uncertainty in Km (and Vmax) • Other linearizations available(see homework) • Better ways of getting Km and Vmax available Biochem: Enzymes II

  38. Don’t fall into the trap! • When you’re calculating Km and Vmax from Lineweaver-Burk plots, remember that you need the reciprocal of the values at the intercepts • If the X-intercept is -5000 M-1, thenKm = -1/(X-intercept) =(-)(-1/5000 M-1) = 2*10-4M • Remember that the X intercept is negative, but Km is positive! Biochem: Enzymes II

  39. Sanity checks • Sanity check #1:typically 10-7M < Km < 10-2M (table 13.3) • Typically kcat ~ 0.5 to 107 s-1 (table 13.4),so for typical [E]tot =10-7M,Vmax = [E]totkcat = 10-6 Ms-1 to 1 Ms-1 • If you get Vmax or Km values outside of these ranges, you’ve probably done something wrong Biochem: Enzymes II

  40. iClicker quiz: question 3 • The hexokinase reaction just described probably operates according to a • (a) sequential, random mechanism • (b) sequential, ordered mechanism • (c) ping-pong mechanism • (d) none of the above. Biochem: Enzymes II

  41. iClicker quiz #4 • If we alter the kinetics of a reaction by increasing Km but leaving Vmax alone, how will the L-B plot change? Biochem: Enzymes II

  42. iClicker question 5 • Enzyme E has a tenfold stronger affinity for substrate A than for substrate B. Which of the following is true? • (a) Km(A) = 10 * Km(B) • (b) Km(A) = 0.1 * Km(B) • (c) Vmax(A) = 10 * Vmax(B) • (d) Vmax(A) = 0.1 * Vmax(B) • (e) None of the above. Biochem: Enzymes II

  43. Another physical significance of Km • Years of experience have led biochemists to a general conclusion: • For its preferred substrate, the Km value of an enzyme is usually within a factor of 50 of the steady-state concentration of that substrate. • So if we find that Km = 0.2 mM for the primary substrate of an enzyme, then we expect that the steady-state concentration of that substrate is between 4 µM and 10 mM. Biochem: Enzymes II

  44. Example:hexokinase isozymes Mutant human type I hexokinase PDB 1DGK, 2.8Å110 kDa monomer • Hexokinase catalyzeshexose + ATP  hexose-6-P + ADP • Most isozymes of hexokinase prefer glucose; some also work okay mannose and fructose • Muscle hexokinases have Km ~ 0.1mM so they work efficiently in blood, where [glucose] ~ 4 mM • Liver glucokinase has Km = 10 mM, which is around the liver [glucose] and can respond to fluctuations in liver [glucose] Biochem: Enzymes II

  45. Using kinetics to determine mechanisms • In a reaction involving substrates A and B, we hold [B] constant and vary [A]. • Then we move to a different [B] and again vary [A]. • Continue through several values of [B] • That gives us a family of Lineweaver-Burk plots of 1/v0 vs 1/[A] • How those curves appear on a single plot tells us which kind of mechanism we have. Biochem: Enzymes II

  46. L-B plots for ordered sequential reactions • http://www-biol.paisley.ac.uk/kinetics/Chapter_4/chapter4_3.html • Plot 1/v0 vs. 1/[A] for various [B] values;flatter slopes correspond to larger [B] • Lines intersect @ a pointin between X intercept and Y intercept Biochem: Enzymes II

  47. L-B plots for ping-pong reactions • Again we plot 1/v vs 1/[A] for various [B] • Parallel lines (same kcat/Km);lower lines correspond to larger [B] • http://www-biol.paisley.ac.uk/kinetics/Chapter_4/chapter4_3_2.html Biochem: Enzymes II

  48. Using exchange reactions to discern mechanisms • Example: sucrose phosphorylase and maltose phosphorylase both cleave disaccharides and add Pi to one product: • Sucrose + Pi glucose-1-P + fructose • Maltose + Pi glucose-1-P + glucose • Try 32P tracers with G-1-P:G-1-P + 32Pi Pi + G-1-32Pi • … so what happens with these two enzymes? Biochem: Enzymes II

  49. Sucrose & maltose phosphorylase • Sucrose phosphorylase doescatalyze the exchange;not maltose phosphorylase • This suggests that SucPase usesdouble-displacement reaction;MalPase uses a single-displacement • Sucrose + E  E-glucose + fructoseE-glucose + Pi E + glucose-1-P • Maltose + E + Pi Maltose:E:PiMaltose:E:Pi glucose-1P + glucose Sucrose phosphorylaseBifidobacterium113 kDa dimerPDB 1R7A, 1.77ÅEC 2.4.1.7 Biochem: Enzymes II

  50. Why study inhibition? • Let’s look at how enzymes get inhibited. • At least two reasons to do this: • We can use inhibition as a probe for understanding the kinetics and properties of enzymes in their uninhibited state; • Many—perhaps most—drugs are inhibitors of specific enzymes. • We'll see these two reasons for understanding inhibition as we work our way through this topic. Biochem: Enzymes II

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