Enzyme Kinetics - BIOC 2200 PDF

BIOC 2200 – Unit 3: Enzymes and Catalysis Enzyme Kinetics Enzyme kinetics is the study of factors that determine the speed of enzyme-catalysed reactions. It utilizes some mathematical equations that can be confusing to students when they first encounter them. However...

BIOC 2200 – Unit 3: Enzymes and Catalysis Enzyme Kinetics Enzyme kinetics is the study of factors that determine the speed of enzyme-catalysed reactions. It utilizes some mathematical equations that can be confusing to students when they first encounter them. However, the theory of kinetics is both logical and simple, and it is essential to develop an understanding of this subject to be able to appreciate the role of enzymes both in metabolism and in biotechnology. Recall Consult your notes from prior courses or the linked resources to re-acquaint yourself with the following topics BEFORE class. We will apply these concepts to biochemistry IN class. Chem Fundamentals of chemical equilibria (CHEM 1302 or OpenStax Chemistry 2e Ch13) Fundamentals of catalysis (CHEM 1302 or OpenStax Chemistry 2e Ch12.7) Biology Fundamentals of bioenergetics and enzymes. (BIOL 1103 or OpenStax Biology 2e Ch6) Read This document introduces the study of enzyme kinetics. Read the content below in order or use the following quick links below (Ctrl+ click) to navigate between sections of the document Enzyme Kinetics Enzymatic Reactions Studying Enzyme Kinetics What does the Michaelis Constant (Km) really mean? Determining the Michaelis Constant (Km) Turnover rate (Kcat) Perfect Enzymes? Enzyme Kinetics Recall our discussion of reaction equilibrium from the introduction to enzymes lesson. As we study biochemical reactions and reaction rates, it is important to remember that: 1) reactions do not generally start at equilibrium, 2) all reactions move in the direction of equilibrium; and 3) reactions in cells behave just like those in test tubes - they do not begin at equilibrium, but they move towards it. The reactions occurring in cells, are very dynamic and complex. In cells, the product of one reaction is often the substrate for another one. Reactions in cells are interconnected in this way, giving rise to what are called metabolic pathways. There are, in fact, thousands of different interconnected reactions going on continuously in cells. We will study these pathways in future units in the course. Attempts to study a single reaction in the chaos of a cell is daunting to say the least. In a test tube, they can be studied one at a time. For this reason, biochemists isolate enzymes from cells and study reactions individually. It is with this in mind that we begin our consideration of the phenomenon of enzymatic catalysis. Enzymatic Reactions We often describe an enzyme-catalysed reaction as proceeding through three stages as follows: The ES complex represents a position where the substrate (S) is bound to the enzyme (E) such that the reaction (whatever it might be) is made more favorable. As soon as the reaction has occurred, the product molecule (P) dissociates from the enzyme, which is then free to bind to another substrate molecule. As an enzymatic reaction proceeds toward equilibrium, the concentration of each of these components changes overtime (Figure 1). At the beginning of the reaction, substrate and free enzyme concentrations are high, while product and enzyme-substrate complex concentrations are low. In contrast by the end of the reaction, substrate and enzyme-substrate concentrations are low, while produce and free enzyme concentrations are high. Figure 1 – Concentration of product (P), substrate (S), enzyme (E), and enzyme-substrate complex (ES) versus time for an enzymatic reaction. Studying Enzyme Kinetics Assays (measurements) of enzyme activity can be performed in either a discontinuous or continuous fashion. Discontinuous methods involve mixing the substrate and enzyme together and measuring the product formed after a set period of time, so these methods are generally easy and quick to perform. In general, we would use such discontinuous assays when we know little about the system (and are making preliminary investigations), or alternatively when we know a great deal about the system and are certain that the time interval, we are choosing is appropriate. In continuous enzyme assays we would generally study the rate of an enzyme-catalysed reaction by mixing the enzyme with the substrate and continuously measuring the appearance of product over time. Of course, we could equally well measure the rate of the reaction by measuring the disappearance of substrate over time. Apart from the actual direction (one increasing and one decreasing), the two values would be identical. In enzyme kinetics experiments, for convenience we very often use an artificial substrate called a chromogen that yields a brightly coloured product, making the reaction easy to follow using a colorimeter or a spectrophotometer (see example below). However, we could in fact use any available analytical equipment that has the capacity to measure the concentration of either the product or the substrate. In almost all cases we would also add a buffer solution to the mixture. As we shall see, enzyme activity is strongly influenced by pH, so it is important to set the pH at a specific value and keep it constant throughout the experiment. Our first enzyme kinetics experiment may therefore involve mixing a substrate solution (chromogen) with a buffer solution and adding the enzyme. This mixture would then be placed in a spectrophotometer and the appearance of the coloured product would be measured. This would enable us to follow a rapid reaction which, after a few seconds or minutes, might start to slow down, as shown in Figure 2.  Figure 2 - Formation of product in an enzyme-catalyzed reaction, plotted against time. A common reason for this slowing down of the speed (rate) of the reaction is that the substrate within the mixture is being used up and thus becoming limiting. Alternatively, it may be that the enzyme is unstable and is denaturing over the course of the experiment, or it could be that the pH of the mixture is changing, as many reactions either consume or release protons. For these reasons, when we are asked to specify the rate of a reaction we do so early on, as soon as the enzyme has been added, and when none of the above-mentioned limitations apply. We refer to this initial rapid rate as the initial velocity (v0). Measurement of the reaction rate at this early stage is also quite straightforward, as the rate is effectively linear, so we can simply draw a straight line and measure the gradient (by dividing the concentration change by the time interval) to evaluate the reaction rate over this period. We may now perform a range of similar enzyme assays to evaluate how the initial velocity changes when the substrate or enzyme concentration is altered, or when the pH is changed. These studies will help us to characterize the properties of the enzyme under study. The relationship between enzyme concentration and the rate of the reaction is usually a simple one. If we repeat the experiment just described, but add 10% more enzyme, the reaction will be 10% faster, and if we double the enzyme concentration the reaction will proceed twice as fast. Thus, there is a simple linear relationship between the reaction rate and the amount of enzyme available to catalyze the reaction (Figure 3). This relationship applies both to enzymes in vivo and to those used in biotechnological applications, where regulation of the amount of enzyme present may control reaction rates.  Figure 3 - Relationship between enzyme concentration and the rate of an enzyme-catalyzed reaction. When we perform a series of enzyme assays using the same enzyme concentration, but with a range of different substrate concentrations, a slightly more complex relationship emerges, as shown in Figure 4. Initially, when the substrate concentration is increased, the rate of reaction increases considerably. However, as the substrate concentration is increased further the effects on the reaction rate start to decline, until a stage is reached where increasing the substrate concentration has little further effect on the reaction rate. At this point the enzyme is considered to be coming close to saturation with substrate and demonstrating its maximal velocity (Vmax). Note that this maximal velocity is, in fact, a theoretical limit that will not be truly achieved in any experiment, although we might come very close to it. Figure 4 - Relationship between substrate concentration and the rate of an enzyme-catalyzed reaction. The relationship described here is a common one, which a mathematician would immediately identify as a rectangular hyperbola. The equation that describes such a relationship is as follows: 𝑎×𝑥 𝑦= 𝑥+𝑏 The two constants a and b thus allow us to describe this hyperbolic relationship, just as with a linear relationship (y = mx + c), which can be expressed by the two constants m (the slope) and c (the intercept). We have in fact already defined the constant a — it is Vmax. The constant b is a little more complex, as it is the value on the x-axis that gives half of the maximal value of y. In enzymology we refer to this as the Michaelis constant (Km), which is defined as the substrate concentration that gives half-maximal velocity. Our final equation, usually called the Michaelis–Menten equation, therefore becomes: 𝑉𝑚𝑎𝑥 × [𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑡𝑒] 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 (𝑣0 ) = [𝑠𝑢𝑏𝑠𝑡𝑟𝑎𝑡𝑒] + 𝐾𝑚 In 1913, Leonor Michaelis and Maud Menten first showed that it was in fact possible to derive this equation mathematically from first principles, with some simple assumptions about the way in which an enzyme reacts with a substrate to form a product. Central to their derivation is the concept that the reaction takes place via the formation of an ES complex which, once formed, can either dissociate (productively) to release product, or else dissociate in the reverse direction without any formation of product. Thus, the reaction can be represented as follows, with k1, k−1 and k2 being the rate constants of the three individual reaction steps: The Michaelis–Menten derivation requires two important assumptions. The first assumption is that we are considering the initial velocity of the reaction (v0), when the product concentration will be negligibly small (i.e. [S] ≫ [P]), such that we can ignore the possibility of any product reverting to substrate. The second assumption is that the concentration of substrate greatly exceeds the concentration of enzyme (i.e. [S] ≫ [E]). The derivation begins with an equation for the expression of the initial rate, the rate of formation of product, as the rate at which the ES complex dissociates to form product. This is based upon the rate constant k2 and the concentration of the ES complex, as follows: 𝑑[𝑃] 𝑣0 = = 𝑘2 × [𝐸𝑆] 𝑑𝑡 Since ES is an intermediate, its concentration is unknown, but we can express it in terms of known values. In a steady-state approximation we can assume that although the concentration of substrate and product changes, the concentration of the ES complex itself remains constant (Figure 5, steady state).  Figure 5 – Steady state versus non-steady state conditions. The rate of formation of the ES complex and the rate of its breakdown must therefore balance, where: Rate of ES complex formation = Rate of ES complex breakdown where: 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐸𝑆 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 = 𝑘1 [𝐸][𝑆] and 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐸𝑆 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑏𝑟𝑒𝑎𝑘𝑑𝑜𝑤𝑛 = (𝑘−1 + 𝑘2 )[𝐸𝑆] Hence, at steady state: 𝑘1 [𝐸][𝑆] = 𝑘−1 + 𝑘2 [𝐸𝑆] The equation can be rearranged to yield [ES] as follows: 𝑘1 [𝐸][𝑆] [𝐸𝑆] = 𝑘−1 + 𝑘2 The Michaelis constant Km can be defined as follows: 𝑘−1 + 𝑘2 𝐾𝑚 = 𝑘1 The [ES] equation can then be simplified to: [𝐸][𝑆] [𝐸𝑆] = 𝐾𝑚 Since the concentration of substrate greatly exceeds the concentration of enzyme (i.e. [S] ≫ [E]), the concentration of uncombined substrate [S] is almost equal to the total concentration of substrate. The concentration of uncombined enzyme [E] is equal to the total enzyme concentration [E]T minus that combined with substrate [ES]. Introducing these terms to into the equation above and solving for ES gives us the following: [𝐸] 𝑇 [𝑆] [𝐸𝑆] = [𝑆] + 𝐾𝑚 We can then introduce this term for [ES] into the initial velocity equation above to yield: [𝑆] 𝑣0 = 𝑘2 [𝐸] 𝑇 [𝑆] + 𝐾𝑚 The term k2[E]T in fact represents Vmax, the maximal velocity. Thus, Michaelis and Menten were able to derive their final equation as: 𝑉𝑚𝑎𝑥 [𝑆] 𝑣0 = [𝑆] + 𝐾𝑚 Michaelis constants have been determined for many commonly used enzymes, and are typically in the lower millimolar range (Table 1). It should be noted that enzymes which catalyse the same reaction, but which are derived from different organisms, can have widely differing Km values. Furthermore, an enzyme with multiple substrates can have quite different Km values for each substrate. Table 1 – Typical Range of Values of the Michaelis Constant Enzyme Km (mmol l-1) Carbonic anhydrase 26 Chymotrypsin 15 Ribonuclease 8 Tyrosyl-tRNA synthase 0.9 Pepsin 0.3 A common mistake students make in describing Km is saying that Km = Vmax /2. This is, of course, not true. Km is a substrate concentration and is the amount of substrate it takes for an enzyme to reach V max /2. On the other hand, Vmax /2 is a velocity, and a velocity certainly cannot equal a concentration. What does the Michaelis Constant (Km) really mean? A low Km value indicates that the enzyme requires only a small amount of substrate to become saturated. Therefore, the maximum velocity is reached at relatively low substrate concentrations. A high Km value indicates the need for high substrate concentrations to achieve maximum reaction velocity. Thus, we generally refer to Km as a measure of the affinity of the enzyme for its substrate—in fact it is an inverse measure, where a high Km indicates a low affinity, and vice versa. The Km value tells us several important things about a particular enzyme: 1. An enzyme with a low Km value relative to the physiological concentration of substrate will probably always be saturated with substrate, and will therefore act at a constant rate, regardless of variations in the concentration of substrate within the physiological range.  2. An enzyme with a high Km value relative to the physiological concentration of substrate will not be saturated with substrate, and its activity will therefore vary according to the concentration of substrate, so the rate of formation of product will depend on the availability of substrate. 3. If an enzyme acts on several substrates, the substrate with the lowest Km value is frequently assumed to be that enzyme’s ‘natural’ substrate, although this may not be true in all cases. 4. If two enzymes (with similar Vmax) in different metabolic pathways compete for the same substrate, then if we know the Km values for the two enzymes we can predict the relative activity of the two pathways. Essentially the pathway that has the enzyme with the lower Km value is likely to be the ‘preferred pathway’, and more substrate will flow through that pathway under most conditions. For example, phosphofructokinase (PFK) is the enzyme that catalyses the first committed step in the glycolytic pathway, which generates energy in the form of ATP for the cell, whereas glucose-1-phosphate uridylyltransferase (GUT) is an enzyme early in the pathway leading to the synthesis of glycogen (an energy storage molecule). Both enzymes use hexose monophosphates as substrates, but the Km of PFK for its substrate is lower than that of GUT for its substrate. Thus, at lower cellular hexose phosphate concentrations, PFK will be active and GUT will be largely inactive. At higher hexose phosphate concentrations both pathways will be active. This means that the cells only store glycogen in times of plenty, and always give preference to the pathway of ATP production, which is the more essential function. Determining the Michaelis Constant (Km) Very often it is not possible to estimate Km values from a direct plot of velocity against substrate concentration (as shown in Figure 4 above) because we have not used high enough substrate concentrations to come even close to estimating maximal velocity, and therefore we cannot evaluate half-maximal velocity and thus Km. Fortunately, we can plot our experimental data in a slightly different way in order to obtain these values. The most used alternative is the Lineweaver–Burk plot (often called the double-reciprocal plot). This plot linearizes the hyperbolic curved relationship, and the line produced is easy to extrapolate, allowing evaluation of Vmax and Km. For example, if we obtained only the first seven data points in Figure 4, we would have difficulty estimating Vmax from a direct plot as shown in Figure 6a. However, as shown in Figure 6b, if these seven points are plotted on a graph of 1/velocity against 1/substrate concentration (i.e. a double-reciprocal plot), the data are linearized, and the line can be easily extrapolated to the left to provide intercepts on both the y-axis and the x-axis, from which Vmax and Km, respectively, can be evaluated.  Figure 6 - Michaelis Menton Kinetics. (a) Direct plot of data (b) Lineweaver-Burk plot of the same kinetic data. One significant practical drawback of using the Lineweaver–Burk plot is the excessive influence that it gives to measurements made at the lowest substrate concentrations. These concentrations might well be the most prone to error (due to difficulties in making multiple dilutions), and result in reaction rates that, because they are slow, might also be most prone to measurement error. Often, as shown in Figure 7, such points when transformed on the Lineweaver–Burk plot have a significant impact on the line of best fit estimated from the data, and therefore on the extrapolated values of both Vmax and Km. The two sets of points shown in Figure 7 are identical except for the single point at the top right, which reflects (because of the plot’s double-reciprocal nature) a single point derived from a very low substrate concentration and a low reaction rate. However, this single point can have an enormous impact on the line of best fit and the accompanying estimates of kinetic constants. Figure 7 - Lineweaver-Burk plot of similar kinetic data, which differ only in a single data point (the final data point (a) 1/v 0.03 at 1/S of 0.2 and (b) 1/v 0.031 at 1/S of 0.18). In fact, there are other kinetic plots that can be used, including the Eadie–Hofstee plot, the Hanes plot and the Eisenthal–Cornish-Bowden plot, which are less prone to such problems. We will not engage with these plot types in this course but you can anticipate encountering them in future classes. Turnover rate (kcat) It is desirable to have a measure of velocity that is independent of enzyme concentration. Remember that V max depended on the amount of enzyme used. For this, we use the kcat, also known as the turnover number. Kcat is a number that requires one to first determine Vmax for an enzyme and then divide the Vmax by the concentration of enzyme used to determine Vmax. Thus, Vmax kcat = [Enzyme] Since Vmax has units of concentration per time and [Enzyme] has units of concentration, the units on kcat are time-1. While that might seem unintuitive, it means that the value of kcat is the number of molecules of product made by each molecule of enzyme in the time given. So, a Kcat value of 1000/sec means each enzyme molecule in the reaction at Vmax is producing 1000 molecules of product per second. Note that since K cat is a calculated value, it cannot be read from a V vs [S] graph as Vmax and Km can. Perfect Enzymes? Measuring Catalytic Efficiency Now, if we think about what an ideal enzyme might be, it would be one that has a very high velocity and a very high affinity for its substrate. That is, it wouldn’t take much substrate to get to V max/2 and the Kcat would be very high. Such enzymes would have values of Kcat / Km (catalytic efficiency) that are maximum. Interestingly, there are several enzymes that have this property, and their maximal Kcat / Km values are all approximately the same (see table 2 below). Such enzymes are referred to as being “perfect” because they have reached the maximum possible value. Table 2 – kcat/Km values for perfect enzymes Enzyme kcat/Km (s-1 M-1) Acetylcholinesterase 1.6 x 108 Carbonic anhydrase 8.3 x 107 Catalase 4.0 x 107 Crotonase 2.8 x 108 Fumerase 1.6 x 108 Triose phosphate isomerase 2.4 x 108 Β-lactamase 1.0 x 108 Superoxide dismutase 7 x 109 Why should there be a maximum possible value of kcat / Km? The answer is that movement of substrate to the enzyme becomes the limiting factor for perfect enzymes. Movement of substrate by diffusion in water has a fixed rate at any temperature and that limitation ultimately determines the maximum speed an enzyme can catalyze at. In a macroscopic world analogy, factories can’t make products faster than suppliers can deliver materials. It is safe to say for a perfect enzyme that the only speed limit it has is the rate of substrate diffusion in water. References 1. Ahern K., Rajagopal I., & Tan T., Ch 4.1: Principles of Catalysis, Biochemistry Free for All, (CC BY-NC-SA 4.0) 2. Chemistry and Biology icons by Amethyst Studio from Noun Project (CC BY 3.0) 3. Flatt, P.M. (2019) Ch. 6: Enzyme Principals and biotechnological Applications, Biochemistry – Defining Life at the Molecular Level. Published by Western Oregon University, Monmouth, OR (CC BY-NC-SA). 4. Jakubowski H. and Flatt P., Ch 6: Enzyme Activity, Fundamentals of Biochemistry 5. Robinson, P. (2015) Enzymes: Principles and Biotechnological Applications. Essays in Biochemistry 59:1-41 (CC BY 3.0)