1. Andy Howard Introductory Biochemistry 27 October 2008
Enzymes II
Andy Howard Introductory Biochemistry
27 October 2008
Biochem: Enzymes II
10/27/2009

2. What we’ll discuss Inhibition Enzymes Enzyme kinetics
Michaelis-Menten kinetics: overview
Kinetic Constants
Kinetic Mechanisms
Induced Fit
Bisubstrate reactions
Measurements and calculational tools
Inhibition
Why study it?
The concept
Types of inhibitors
10/27/2009
Biochem: Enzymes II

3. Michaelis-Menten kinetics
[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.
10/27/2009
Biochem: Enzymes II

4. 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!
10/27/2009
Biochem: Enzymes II

5. 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 vd = 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.
10/27/2009
Biochem: Enzymes II

6. Derivation, continued
Thus since vf = vd. 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])
10/27/2009
Biochem: Enzymes II

7. 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 Vmax = v0 = k2[ES] = k2[E]tot
10/27/2009
Biochem: Enzymes II

8. Using Vmax in M-M kinetics
Thus since Vmax = k2[E]tot,
v0 = Vmax [S] / (Km+[S])
That’s the famous Michaelis-Menten equation
10/27/2009
Biochem: Enzymes II

9. Graphical interpretation
10/27/2009
Biochem: Enzymes II

10. Physical meaning of Km
As we can see from the plot, the velocity is half-maximal when [S] = Km
Trivially derivable: if [S] = Km, then v0 = 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
10/27/2009
Biochem: Enzymes II

11. 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
10/27/2009
Biochem: Enzymes II

12. 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
10/27/2009
Biochem: Enzymes II

13. 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)
10/27/2009
Biochem: Enzymes II

14. 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
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Biochem: Enzymes II

15. 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
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Biochem: Enzymes II

16. 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 , 13.49, 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.
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Biochem: Enzymes II

17. Historical thought
Biochemists, 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!
10/27/2009
Biochem: Enzymes II

18. 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
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Biochem: Enzymes II

19. 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
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Biochem: Enzymes II

20. 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
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Biochem: Enzymes II

21. 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
10/27/2009
Biochem: Enzymes II

22. Kinase reactions unwanted reaction ATP + H-O-H ⇒ ADP + Pi
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
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Biochem: Enzymes II

23. 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
10/27/2009
Biochem: Enzymes II

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

25. Hexokinase conformational changes
G&G Fig
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Biochem: Enzymes II

26. 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.
10/27/2009
Biochem: Enzymes II

27. Lineweaver-Burk Simple linearization of Michaelis-Menten:
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
10/27/2009
Biochem: Enzymes II

28. 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!
10/27/2009
Biochem: Enzymes II

29. 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.
10/27/2009
Biochem: Enzymes II

30. 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
10/27/2009
Biochem: Enzymes II

31. 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.
10/27/2009
Biochem: Enzymes II

32. 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
10/27/2009
Biochem: Enzymes II

33. 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 M-1, then Km = -1/(X-intercept) =(-)(-1/5000 M-1) = 2*10-4M
Remember that the X intercept is negative, but Km is positive!
10/27/2009
Biochem: Enzymes II

34. 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
10/27/2009
Biochem: Enzymes II

35. 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.
10/27/2009
Biochem: Enzymes II

36. 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?
Answer
X-intercept
Y-intercept a
Moves toward origin
Unchanged b
Moves away from origin c d
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Biochem: Enzymes II

37. 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.
10/27/2009
Biochem: Enzymes II

38. 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.
10/27/2009
Biochem: Enzymes II

39. Example: hexokinase isozymes
Mutant human type I hexokinase
PDB 1DGK, 2.8Å 110 kDa monomer
Hexokinase catalyzes hexose + 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]
10/27/2009
Biochem: Enzymes II

40. L-B plots for ordered sequential reactions
kinetics/Chapter_4/chapter4_3.html
Plot 1/v0 vs. 1/[A] for various [B] values; flatter slopes correspond to larger [B]
Lines a point in between X intercept and Y intercept
10/27/2009
Biochem: Enzymes II

41. 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]
Chapter_4/chapter4_3_2.html
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Biochem: Enzymes II

42. 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?
10/27/2009
Biochem: Enzymes II

43. 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
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Biochem: Enzymes II

44. 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.
10/27/2009
Biochem: Enzymes II

45. 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
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Biochem: Enzymes II

46. 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.
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Biochem: Enzymes II

47. 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.
10/27/2009
Biochem: Enzymes II

48. Distinctions we can make
Inhibitors can be reversible or irreversible
Where do they bind?
At the enzyme’s active site
At a site distant from the active site.
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)
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Biochem: Enzymes II

49. Types of inhibitors Irreversible Reversible
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
10/27/2009
Biochem: Enzymes II