1. 10/21/2010Biochem: Enzymes II Enzymes II Andy Howard Introductory Biochemistry 21 October 2010

2. 10/21/2010 Biochem: Enzymes II P. 2 of 55 What we’ll discuss 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

3. 10/21/2010 Biochem: Enzymes II P. 3 of 55 Kinetics, continued In most situations more product will be produced per unit time if A 0 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. [B] t

4. 10/21/2010 Biochem: Enzymes II P. 4 of 55 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!

5. 10/21/2010 Biochem: Enzymes II P. 5 of 55 Bimolecular reaction If in the reaction A + D  B the 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

6. 10/21/2010 Biochem: Enzymes II P. 6 of 55 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 k 1, the backward rate typically designated as k -1.

7. 10/21/2010 Biochem: Enzymes II P. 7 of 55 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  C we can write rate constants k 1, k -1, k 2, and k -2 as the rate constants associated with converting A to B, converting B to A, converting B to C, and converting C to B.

8. 10/21/2010 Biochem: Enzymes II P. 8 of 55 Michaelis-Menten kinetics 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. [ES] t

9. 10/21/2010 Biochem: Enzymes II P. 9 of 55 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 = v f = k 1 ([E] tot - [ES])[S] because the enzyme that is already substrate-bound is unavailable!

10. 10/21/2010 Biochem: Enzymes II P. 10 of 55 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 is v d = k -1 [ES] + k 2 [ES] = (k -1 + k 2 )[ES] This rate of disappearance should be equal to the rate of appearance Under these conditions v f = v d.

11. 10/21/2010 Biochem: Enzymes II P. 11 of 55 Derivation, continued Thus since v f = v d by assumption, k 1 ([E] tot - [ES])[S] = (k -1 + k 2 )[ES] K m  (k -1 + k 2 )/k 1 = ([E] tot - [ES])[S] / [ES] [ES] = [E] tot [S] / (K m + [S]) But the rate-limiting reaction is the formation of product: v 0 = k 2 [ES] Thus v 0 = k 2 [E] tot [S] / (K m + [S])

12. 10/21/2010 Biochem: Enzymes II P. 12 of 55 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 get V max = v 0 = k 2 [ES] = k 2 [E] tot

13. 10/21/2010 Biochem: Enzymes II P. 13 of 55 Using V max in M-M kinetics Thus since V max = k 2 [E] tot, v 0 = V max [S] / (K m +[S]) That’s the famous Michaelis-Menten equation

14. 10/21/2010 Biochem: Enzymes II P. 14 of 55 Graphical interpretation

15. 10/21/2010 Biochem: Enzymes II P. 15 of 55 Physical meaning of K m As we can see from the plot, the velocity is half-maximal when [S] = K m Trivially derivable: if [S] = K m, then v 0 = V max [S] / ([S]+[S]) = V max /2 We can turn that around and say that the K m is defined as the concentration resulting in half-maximal velocity K m is a property associated with binding of S to E, not a property of turnover Michaelis Constant: British Christian hiphop band

16. 10/21/2010 Biochem: Enzymes II P. 16 of 55 k cat We’ve already discussed what V max 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 k cat = V max / [E] tot Oh wait: that’s just the rate of our rate- limiting step, i.e. k cat = k 2

17. 10/21/2010 Biochem: Enzymes II P. 17 of 55 Physical meaning of k cat Describes turnover of substrate to product: Number of product molecules produced per sec per molecule of enzyme More complex reactions may not have k cat = k 2, but we can often approximate them that way anyway Some enzymes very efficient: k cat > 10 6 s -1

18. 10/21/2010 Biochem: Enzymes II P. 18 of 55 Specificity constant, k cat /K m k cat /K m 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)

19. 10/21/2010 Biochem: Enzymes II P. 19 of 55 Dimensions K m must have dimensions of concentration (remember it corresponds to the concentration of substrate that produces half-maximal velocity) V max must have dimensions of concentration over time (d[A]/dt) k cat must have dimensions of inverse time k cat / K m must have dimensions of inverse time divided by concentration, i.e. inverse time * inverse concentration

20. 10/21/2010 Biochem: Enzymes II P. 20 of 55 Typical units for kinetic parameters Remember the distinction between dimensions and units! K m typically measured in mM or µM V max typically measured in mMs -1 or µMs -1 k cat typically measured in s -1 k cat / K m typically measured in s -1 M -1

21. 10/21/2010 Biochem: Enzymes II P. 21 of 55 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.

22. 10/21/2010 Biochem: Enzymes II P. 22 of 55 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!

23. 10/21/2010 Biochem: Enzymes II P. 23 of 55 Sequential, ordered reactions Substrates, products must bind in specific order for reaction to complete A B P Q _____________________________ E EA (EAB) (EPQ) EQ E W.W.Cleland

24. 10/21/2010 Biochem: Enzymes II P. 24 of 55 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

25. 10/21/2010 Biochem: Enzymes II P. 25 of 55 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 Q E EA FA F FB FQ E

26. 10/21/2010 Biochem: Enzymes II P. 26 of 55 Induced fit 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. Daniel Koshland Cartoon from textbookofbacteriology.net

27. 10/21/2010 Biochem: Enzymes II P. 27 of 55 Kinase reactions unwanted reaction ATP + H-O-H ⇒ ADP + P i will compete with the desired reaction ATP + 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

28. 10/21/2010 Biochem: Enzymes II P. 28 of 55 iClicker quiz, question 1 The Michaelis constant K m has dimensions of (a) concentration per unit time (b) inverse concentration per unit time (c) concentration (d) inverse concentration (e) none of the above

29. 10/21/2010 Biochem: Enzymes II P. 29 of 55 iClicker quiz question 2 k cat is a measure of (a) substrate binding (b) turnover (c) inhibition potential (d) none of the above

30. 10/21/2010 Biochem: Enzymes II P. 30 of 55 Hexokinase conformational changes G&G Fig. 13.28

31. 10/21/2010 Biochem: Enzymes II P. 31 of 55 Measurements and calculations The standard Michaelis-Menten formulation is v 0 =f([S]), but it’s not linear in [S]. We seek linearizations of the equation so that we can find K m and k cat, and so that we can understand how various changes affect the reaction.

32. 10/21/2010 Biochem: Enzymes II P. 32 of 55 Lineweaver-Burk Simple linearization of Michaelis-Menten: v 0 = V max [S]/(K m +[S]). Take reciprocals: 1/v 0 = (K m +[S])/(V max [S]) = K m /(V max [S]) + [S]/(V max [S]) 1/v 0 = (K m /V max )*1/[S] + 1/V max Thus a plot of 1/[S] as the independent variable vs. 1/v 0 as the dependent variable will be linear with Y-intercept = 1/V max and slope K m /V max Dean Burk Hans Lineweaver

33. 10/21/2010 Biochem: Enzymes II P. 33 of 55 How to use this Y-intercept is useful directly: computeV max = 1/(Y-intercept) We can get K m /V max from slope and then use our knowledge of V max to get K m ; or X intercept = -1/ K m … that gets it for us directly!

34. 10/21/2010 Biochem: Enzymes II P. 34 of 55 Demonstration that the X-intercept is at -1/K m X-intercept means Y = 0 In Lineweaver-Burk plot, 0 = (K m /V max )*1/[S] + 1/V max For nonzero 1/V max we divide through: 0 = K m /[S] + 1, -1 = K m /[S], [S] = -K m. But the axis is for 1/[S], so the intercept is at 1/[S] = -1/ K m.

35. 10/21/2010 Biochem: Enzymes II P. 35 of 55 Graphical form of L-B 1/[S], M -1 1/v 0, s L mol -1 1/V max, s L mol -1 -1/K m, L mol -1 Slope=K m /V max

36. 10/21/2010 Biochem: Enzymes II P. 36 of 55 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.

37. 10/21/2010 Biochem: Enzymes II P. 37 of 55 Advantages and disadvantages of L-B plots Easy conceptual reading of K m and V max (but remember to take the reciprocals!) Suboptimal error analysis [S] and v 0 values have errors Error propagation can lead to significant uncertainty in K m (and V max ) Other linearizations available (see homework) Better ways of getting K m and V max available

38. 10/21/2010 Biochem: Enzymes II P. 38 of 55 Don’t fall into the trap! When you’re calculating K m and V max from Lineweaver-Burk plots, remember that you need the reciprocal of the values at the intercepts If the X-intercept is -5000 M -1, then K m = -1/(X-intercept) =(-)(-1/5000 M -1 ) = 2*10 -4 M Remember that the X intercept is negative, but K m is positive!

39. 10/21/2010 Biochem: Enzymes II P. 39 of 55 Sanity checks Sanity check #1: typically 10 -7 M < K m < 10 -2 M (table 13.3) Typically k cat ~ 0.5 to 10 7 s -1 (table 13.4), so for typical [E] tot =10 -7 M, V max = [E] tot k cat = 10 -6 Ms -1 to 1 Ms -1 If you get V max or K m values outside of these ranges, you’ve probably done something wrong

40. 10/21/2010 Biochem: Enzymes II P. 40 of 55 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.

41. 10/21/2010 Biochem: Enzymes II P. 41 of 55 iClicker quiz #4 If we alter the kinetics of a reaction by increasing K m but leaving V max alone, how will the L-B plot change? AnswerX-interceptY-intercept aMoves toward originUnchanged bMoves away from origin Unchanged c Moves away from origin dUnchangedMoves toward origin

42. 10/21/2010 Biochem: Enzymes II P. 42 of 55 iClicker question 5 Enzyme E has a tenfold stronger affinity for substrate A than for substrate B. Which of the following is true? (a) K m (A) = 10 * K m (B) (b) K m (A) = 0.1 * K m (B) (c) V max (A) = 10 * V max (B) (d) V max (A) = 0.1 * V max (B) (e) None of the above.

43. 10/21/2010 Biochem: Enzymes II P. 43 of 55 Another physical significance of K m Years of experience have led biochemists to a general conclusion: For its preferred substrate, the K m value of an enzyme is usually within a factor of 50 of the steady-state concentration of that substrate. So if we find that K m = 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.

44. 10/21/2010 Biochem: Enzymes II P. 44 of 55 Example: hexokinase isozymes Hexokinase catalyzes hexose + ATP  hexose-6-P + ADP Most isozymes of hexokinase prefer glucose; some also work okay mannose and fructose Muscle hexokinases have K m ~ 0.1mM so they work efficiently in blood, where [glucose] ~ 4 mM Liver glucokinase has K m = 10 mM, which is around the liver [glucose] and can respond to fluctuations in liver [glucose] Mutant human type I hexokinase PDB 1DGK, 2.8Å 110 kDa monomer

45. 10/21/2010 Biochem: Enzymes II P. 45 of 55 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.

46. 10/21/2010 Biochem: Enzymes II P. 46 of 55 L-B plots for ordered sequential reactions http://www-biol.paisley.ac.uk/ kinetics/Chapter_4/chapter4_3.html http://www-biol.paisley.ac.uk/ kinetics/Chapter_4/chapter4_3.html Plot 1/v 0 vs. 1/[A] for various [B] values; flatter slopes correspond to larger [B] Lines intersect @ a point in between X intercept and Y intercept

47. 10/21/2010 Biochem: Enzymes II P. 47 of 55 L-B plots for ping- pong reactions Again we plot 1/v vs 1/[A] for various [B] Parallel lines (same k cat /K m ); lower lines correspond to larger [B] http://www-biol.paisley.ac.uk/kinetics/ Chapter_4/chapter4_3_2.html http://www-biol.paisley.ac.uk/kinetics/ Chapter_4/chapter4_3_2.html

48. 10/21/2010 Biochem: Enzymes II P. 48 of 55 Using exchange reactions to discern mechanisms Example: sucrose phosphorylase and maltose phosphorylase both cleave disaccharides and add P i to one product: Sucrose + P i  glucose-1-P + fructose Maltose + P i  glucose-1-P + glucose Try 32 P tracers with G-1-P: G-1-P + 32 P i  P i + G-1- 32 P i … so what happens with these two enzymes?

49. 10/21/2010 Biochem: Enzymes II P. 49 of 55 Sucrose & maltose phosphorylase Sucrose phosphorylase does catalyze the exchange; not maltose phosphorylase This suggests that SucPase uses double-displacement reaction; MalPase uses a single-displacement Sucrose + E  E-glucose + fructose E-glucose + P i  E + glucose-1-P Maltose + E + P i  Maltose:E:P i Maltose:E:P i  glucose-1P + glucose Sucrose phosphorylase Bifidobacterium 113 kDa dimer PDB 1R7A, 1.77Å EC 2.4.1.7

50. 10/21/2010 Biochem: Enzymes II P. 50 of 55 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.

51. 10/21/2010 Biochem: Enzymes II P. 51 of 55 The concept of inhibition An enzyme is a biological catalyst, i.e. a substance that alters the rate of a reaction without itself becoming permanently altered by its participation in the reaction. The ability of an enzyme (particularly a proteinaceous enzyme) to catalyze a reaction can be altered by binding small molecules to it

52. 10/21/2010 Biochem: Enzymes II P. 52 of 55 Inhibitors and accelerators Usually these alterations involve a reduction in the enzyme's ability to accelerate the reaction; less commonly, they give rise to an increase in the enzyme's ability to accelerate a reaction.

53. 10/21/2010 Biochem: Enzymes II P. 53 of 55 Why more inhibitors than accelerators? Natural selection: if there were small molecules that can facilitate the enzyme's propensity to speed up a reaction, nature probably would have found a way to incorporate those facilitators into the enzyme over the billions of years that the enzyme has been available. Most enzymes are already fairly close to optimal in their properties; we can readily mess them up with effectors, but it's more of a challenge to find ways to make enzymes better at their jobs.

54. 10/21/2010 Biochem: Enzymes II P. 54 of 55 Distinctions we can make Inhibitors can be reversible or irreversible If they’re reversible, where do they bind? At the enzyme’s active site At a site distant from the active site. If they’re reversible, to what do they bind? To the unliganded enzyme E To the enzyme-intermediate complex or the enzyme-substrate complex (ES) To both (E or ES)

55. 10/21/2010 Biochem: Enzymes II P. 55 of 55 Types of inhibitors Irreversible Inhibitor binds without possibility of release Usually covalent Each inhibition event effectively removes a molecule of enzyme from availability Reversible Usually noncovalent (ionic or van der Waals) Several kinds Classifications somewhat superseded by detailed structure-based knowledge of mechanisms, but not entirely