MCDB 3101st Edition Lecture 7 Outline of Last Lecture I. Protein equilibrium constants (Kd and Ka)II. Myoglobin and Hemoglogin: what makes these proteins able to perform their functions?III. EnzymesOutline of Current Lecture I. Enzyme Kineticsa. Michaelis-Menton Plot and Propertiesb. LIneweaver-Burk plotsII. Enzyme Inhibitiona. Mechanisms of inhibtionIII. Enzyme regulation and modificationsCurrent LectureI. Enzyme Kineticsa. Remember: the only thing that enzymes that do is decrease the activation energybarrier and therefore the rate at which the reaction occurs. Enzymes do NOT affect the Gibbs Free Energy (spontaneity) or the Keq (equilibrium constant).b. Kinetics is the study of the rate at which compounds reactc. The rate at which an enzymatic reaction occurs is affected by:i. Enzyme concentration: increasing the enzyme concentration increases therate of the reaction. ii. Substrate: changing the substrate concentration changes the rate at whichthe reaction occurs. These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.iii. Effectors: Similar to modulators, they bind to enzymes and change the rate. iv. Temperatured. Why do we want to study enzyme kinetics? We want numbers:i. Quantify the catalysisii. Determine the order of binding substratesiii. Understand the catalytic mechanism (competitive, non-competitive, or mixed)iv. Find effective inhibitorsv. Understand how to regulate activitye. How do we perform kinetic measurements?i. Mix enzyme and substrateii. Record the rate of product formation as a function of time and plot it.iii. V0 is the velocity of the reaction at time=0 (slope of the graph at time=0) iv. Change the substrate concentration and determine the new V0. The plot that results is shown below. Each solid line represents a different initialsubstrate concentration. The dotted lines show the V0 for each of these reactions. 1. The reason we measure V0 instead of the velocity at any other point in time is because the V0 is the maximum velocity of a reaction (this is the point at which substrates are being converted into products the fastest)v. Then, we plot the INITIAL RATE (slope or V0) against SUBSTRATE CONCENTRATION to give us the Michaelis-Menten plot (shown below)1. We are looking for constants for chemical reactions (for the sake ofcomparison)2. We want to come up with a constant that reveals how efficient an enzyme really is3. The resulting graph will give us 2 important values: Km and Vmaxa. Km=Michaelis-Menten constant, easy to measureb. Vmax=not a constantf. The beginnings of enzymology:i. Adrian Brown (1902) discovered that low concentrations of sucrose, the reaction rate was directly dependent on the sucrose concentration1. He determined the reaction rate (Molarity/second) and the Rate constant K (1/second)ii. When the concentration of sucrose was very high, the reaction rate reached a maximum velocity (Vmax) and was no longer dependent on sucrose concentration.1. Conclusion: The reaction must be composed of 2 individual reactions, and one of these reactions is the rate limiting (slow) reaction and the other is the fast reactiona. The rate ultimately only depends on the slowest step of thereaction, NOT just concentrationiii. Michaelis-Menton Model1. Enzymes and substrate form an enzyme/substrate complex (ES)a. The k values for that reaction are another example of thoseused for binding affinity2. ES can either dissociate or substrate is converted into a product P 3. When substrate concentration is LOW —> the formation of ES is the rate-determining step 4. When substrate concentration is HIGH —> all the binding sites on the enzyme are bound to a substrate molecule —> adding more substrate does not change the rate —> the formation of the products is the rate determining step5. Shown below shows the dependence of the rate on the concentration of the substrateiv. What is the meaning of the Michaelis Constant (Km)?1. Km is the substrate concentration at which the reaction has reached half of its Vmax (usually taken from MM plot)a. From the equation above, when substrate concentration equals Km —> V0=Vmax/22. The units of Km is molarity3. Km tells us how much substrate we need to put in for the reaction to reach half of its maximum velocitya. Therefore, very good enzymes have LOW Km values (only need a little bit of substrate)4. Km is nothing but a combination of rate constants5. Km is important in determining the efficiency of an enzyme when only a small amount of substrate is presentv. Vmax tells us the maximum rate of the reaction when the enzyme is FULLY SATURATED1. Vmax=kcat*concentration of total enzyme2. kcat is a better constant than Vmax for an enzyme because Vmax varies with enzyme concentration3. kcat and Km are independent of enzyme concentration4. Every velocity calculation increases LINEARTLY with total enzyme concentrationa. However, Km will remain constantvi. kcat is the turn over number of the enzyme: it represents the total number of substrate molecules that are converted to product per second by one enzyme molecule under SATURATED conditions.1. There are enzymes whose kcat values are very large2. Example: carbonic anhydrase hydrates 400,000 molecules of CO2 per second. vii.kcat/Km=catalytic efficiency: how fast an enzyme catalyzes a reaction when it encounters its substrate1. If substrate concentration is very low, you can ignore the substrate concentration value in the denominator —> the equation looks like that of a second order reaction and the ratio of kcat/Km is the rate constant2. REMEMBER: there is an upper limit for this ratio. The rate can never be faster than the frequency with which E and S collide with each other in solutiona. Therefore the upper limit is the diffusion value (10^8 - 10^9 M^(-1)*s^(-1)b. When an enzyme’s ratio reaches this diffusion value, it has reached “catalytic perfection”i. This occurs when a reaction has either a high kcat or low Kmc. This means that the active site of the enzyme acts as a magnet, and the reaction is catalyzed almost every time the enzyme encounters a substrate molecule (the active site becomes the ONLY place that a favorable reaction can occur)d. Example: DNA contains negatively charged molecules and aromatic rings. The active site will therefore have positivelycharged amino acids (lys and arg) and aromatic residues (Phe, Trp, Tyr) so that the DNA is attracted to that