1. 10/21/2008 Biochemistry: Enzyme Inhibition Enzyme Kinetics and Inhibition Andy Howard Introductory Biochemistry, Fall 2008 21 October 2008

2. 10/21/2008 Biochemistry: Enzyme Inhibition p. 2 of 51 Inhibition is important both conceptually and practically We study inhibition to clarify our understanding of enzyme mechanisms and because knowing how inhibition works helps us design pharmaceuticals.

3. 10/21/2008 Biochemistry: Enzyme Inhibition p. 3 of 51 Plans for Today Kinetics, concluded Induced fit Measurements and calculational tools Multisubstrate reactions Inhibition Why study it? The concept Types of inhibitors Kinetics of inhibition Pharmaceutical inhibitors What makes an inhibitor a useful drug?

4. 10/21/2008 Biochemistry: Enzyme Inhibition p. 4 of 51 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

5. 10/21/2008 Biochemistry: Enzyme Inhibition p. 5 of 51 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

6. 10/21/2008 Biochemistry: Enzyme Inhibition p. 6 of 51 Hexokinase conformational changes G&G Fig. 13.28

7. 10/21/2008 Biochemistry: Enzyme Inhibition p. 7 of 51 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.

8. 10/21/2008 Biochemistry: Enzyme Inhibition p. 8 of 51 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])) = (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

9. 10/21/2008 Biochemistry: Enzyme Inhibition p. 9 of 51 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!

10. 10/21/2008 Biochemistry: Enzyme Inhibition p. 10 of 51 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.

11. 10/21/2008 Biochemistry: Enzyme Inhibition p. 11 of 51 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

12. 10/21/2008 Biochemistry: Enzyme Inhibition p. 12 of 51 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.

13. 10/21/2008 Biochemistry: Enzyme Inhibition p. 13 of 51 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

14. 10/21/2008 Biochemistry: Enzyme Inhibition p. 14 of 51 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

15. 10/21/2008 Biochemistry: Enzyme Inhibition p. 15 of 51 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]=10 -7 M, V max = [E]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

16. 10/21/2008 Biochemistry: Enzyme Inhibition p. 16 of 51 iClicker quiz: question 1 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.

17. 10/21/2008 Biochemistry: Enzyme Inhibition p. 17 of 51 iClicker quiz #2 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

18. 10/21/2008 Biochemistry: Enzyme Inhibition p. 18 of 51 iClicker question 3 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.

19. 10/21/2008 Biochemistry: Enzyme Inhibition p. 19 of 51 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.

20. 10/21/2008 Biochemistry: Enzyme Inhibition p. 20 of 51 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

21. 10/21/2008 Biochemistry: Enzyme Inhibition p. 21 of 51 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

22. 10/21/2008 Biochemistry: Enzyme Inhibition p. 22 of 51 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

23. 10/21/2008 Biochemistry: Enzyme Inhibition p. 23 of 51 Using exchange reactions to discern mechanisms Example: sucrose phosphorylase and maltose phosphorylase both cleave disaccharides and add Pi 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?

24. 10/21/2008 Biochemistry: Enzyme Inhibition p. 24 of 51 Results with sucrose and maltose phosphorylase Sucrose phosphorylase does catalyze the exhange; not maltose phosphorylase This suggests that SucPase uses a double- displacement reaction; MalPase uses a single- displacement reaction Sucrose + E  E-glucose + fructose E-glucose + Pi  E + glucose-1-P Maltose + E + Pi  Maltose:E:Pi Maltose:E:Pi  glucose-1P + glucose

25. 10/21/2008 Biochemistry: Enzyme Inhibition p. 25 of 51 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.

26. 10/21/2008 Biochemistry: Enzyme Inhibition p. 26 of 51 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: sometimes at its active site sometimes at a site distant from the active site.

27. 10/21/2008 Biochemistry: Enzyme Inhibition p. 27 of 51 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.

28. 10/21/2008 Biochemistry: Enzyme Inhibition p. 28 of 51 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.

29. 10/21/2008 Biochemistry: Enzyme Inhibition p. 29 of 51 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

30. 10/21/2008 Biochemistry: Enzyme Inhibition p. 30 of 51 Types of reversible inhibition Competitive Inhibitor binds at active site Prevents binding of substrate Noncompetitive Inhibitor binds distant from active site Interferes with turnover Uncompetitive (rare?) Inhibitor binds to ES complex Removes ES, interferes with turnover Mixed (usually Competitive + Noncompetitive)

31. 10/21/2008 Biochemistry: Enzyme Inhibition p. 31 of 51 How to tell them apart Reversible vs irreversible dialyze an enzyme-inhibitor complex against a buffer free of inhibitor if turnover or binding still suffers, it’s irreversible Competitive vs. other reversible: Structural studies if feasible Kinetics

32. 10/21/2008 Biochemistry: Enzyme Inhibition p. 32 of 51 Competitive inhibition Put in a lot of substrate: ability of the inhibitor to get in the way of the binding is hindered: out-competed by sheer #s of substrate molecules. This kind of inhibition manifests itself as interference with binding, i.e. with an increase of K m

33. 10/21/2008 Biochemistry: Enzyme Inhibition p. 33 of 51 Competitive inhibitors don’t affect turnover 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.

34. 10/21/2008 Biochemistry: Enzyme Inhibition p. 34 of 51 Kinetics of competition Competitive inhibitor hinders binding of substrate but not reaction velocity: Affects the K m of the enzyme, not V max. Which way does it affect it? K m = amount of substrate that needs to be present to run the reaction velocity up to half its saturation velocity. Competitive inhibitor requires us to shove more substrate into the reaction in order to achieve that half-maximal velocity. So: competitive inhibitor increases K m

35. 10/21/2008 Biochemistry: Enzyme Inhibition p. 35 of 51 L-B: competitive inhibitor K m goes up so -1/ K m moves toward origin V max unchanged so Y intercept unchanged

36. 10/21/2008 Biochemistry: Enzyme Inhibition p. 36 of 51 Competitive inhibitor: Quantitation of K i Define inhibition constant K i to be the concentration of inhibitor that increases K m by a factor of two. K m,obs = K m (1+[ I c ]/K i ) So [ I c ] that moves K m halfway to the origin is K i. If K i = 100 nM and [ I c ] = 1 µM, then we’ll increase K m,obs elevenfold!

37. 10/21/2008 Biochemistry: Enzyme Inhibition p. 37 of 51 Noncompetitive inhibition Noncompetitive inhibitor has no influence on how available the binding site for substrate is, so it does not affect K m at all However, it has a profound inhibitory influence on the speed of the reaction, i.e. turnover. So it reduces V max and has no influence on K m. S I

38. 10/21/2008 Biochemistry: Enzyme Inhibition p. 38 of 51 L-B for non-competitives Decrease in V max  1/V max is larger X-intercept unaffected

39. 10/21/2008 Biochemistry: Enzyme Inhibition p. 39 of 51 K i for noncompetitives K i defined as concentration of inhibitor that cuts V max in half V max,obs = V max /(1 + [ I n ]/K i ) In previous figure the “high” concentration of inhibitor is K i If K i = K i ’, this is pure noncompetitive inhibition

40. 10/21/2008 Biochemistry: Enzyme Inhibition p. 40 of 51 Uncompetitive inhibition Inhibitor binds only if ES has already formed It creates a ternary ESI complex This removes ES, so by LeChatlier’s Principle it actually drives the original reaction (E + S  ES) to the right; so it decreases K m But it interferes with turnover so V max goes down If K m and V max decrease at the same rate, then it’s classical uncompetitive inhibition.

41. 10/21/2008 Biochemistry: Enzyme Inhibition p. 41 of 51 L-B for uncompetitives K m moves toward origin V max moves away from the origin Slope (  K m /V max ) is unchanged

42. 10/21/2008 Biochemistry: Enzyme Inhibition p. 42 of 51 K i for uncompetitives Defined as inhibitor concentration that cuts V max or K m in half Easiest to read from V max value I u labeled “high” is K i in this plot

43. 10/21/2008 Biochemistry: Enzyme Inhibition p. 43 of 51 iClicker quiz, question 4 4. Treatment of enzyme E with compound Y doubles K m and leaves V max unchanged. Compound Y is: (a) an accelerator of the reaction (b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

44. 10/21/2008 Biochemistry: Enzyme Inhibition p. 44 of 51 iClicker quiz, question 5 5. Treatment of enzyme E with compound X doubles V max and leaves K m unchanged. Compound X is: (a) an accelerator of the reaction (b) a competitive inhibitor (c) a non-competitive inhibitor (d) an uncompetitive inhibitor

45. 10/21/2008 Biochemistry: Enzyme Inhibition p. 45 of 51 Mixed inhibition Usually involves interference with both binding and catalysis K m goes up, V max goes down Easy to imagine the mechanism: Binding of inhibitor alters the active-site configuration to interfere with binding, but it also alters turnover Same picture as with pure noncompetitive inhibition, but with K i ≠ K i ’

46. 10/21/2008 Biochemistry: Enzyme Inhibition p. 46 of 51 Most pharmaceuticals are enzyme inhibitors Some are inhibitors of enzymes that are necessary for functioning of pathogens Others are inhibitors of some protein whose inappropriate expression in a human causes a disease. Others are targeted at enzymes that are produced more energetically by tumors than they are by normal tissues.

47. 10/21/2008 Biochemistry: Enzyme Inhibition p. 47 of 51 Characteristics of Pharmaceutical Inhibitors Usually competitive, i.e. they raise K m without affecting V max Some are mixed, i.e. K m up, V max down Iterative design work will decrease K i from millimolar down to nanomolar Sometimes design work is purely blind HTS; other times, it’s structure-based