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

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

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

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

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

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

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

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

Graphical interpretation 10/27/2009 Biochem: Enzymes II

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

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

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

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

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 10/27/2009 Biochem: Enzymes II

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 10/27/2009 Biochem: Enzymes II

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. 10/27/2009 Biochem: Enzymes II

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! 10/27/2009 Biochem: Enzymes II

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 10/27/2009 Biochem: Enzymes II

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 10/27/2009 Biochem: Enzymes II

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 10/27/2009 Biochem: Enzymes II

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

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 10/27/2009 Biochem: Enzymes II

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

iClicker quiz question 2 kcat is a measure of (a) substrate binding (b) turnover (c) inhibition potential (d) none of the above 10/27/2009 Biochem: Enzymes II

Hexokinase conformational changes G&G Fig. 13.28 10/27/2009 Biochem: Enzymes II

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

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

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

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

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

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

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

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, 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

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

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

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 10/27/2009 Biochem: Enzymes II

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

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

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

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 point in between X intercept and Y intercept 10/27/2009 Biochem: Enzymes II

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 10/27/2009 Biochem: Enzymes II

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

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 10/27/2009 Biochem: Enzymes II

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

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 10/27/2009 Biochem: Enzymes II

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. 10/27/2009 Biochem: Enzymes II

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

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) 10/27/2009 Biochem: Enzymes II

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