1 00:00:03,770 --> 00:00:09,620 In beer and wine production, enzymes in yeast aid the conversion of sugar into ethanol. 2 00:00:09,620 --> 00:00:13,600 Enzymes are used in cheese-making to degrade proteins in milk, changing their solubility, 3 00:00:13,600 --> 00:00:18,789 and causing the proteins to precipitate. Many industrial processes ranging from fruit juice 4 00:00:18,789 --> 00:00:23,770 production to paper production to biofuel production utilize enzymes. In this video 5 00:00:23,770 --> 00:00:30,330 you'll see that studying how enzymes catalyze reactions can help us design better therapeutics. 6 00:00:30,330 --> 00:00:34,489 This video is part of the Differential Equations video series. Laws that govern a system's 7 00:00:34,489 --> 00:00:37,610 properties can be modeled using differential equations. 8 00:00:37,610 --> 00:00:43,070 Hi. My name is Krystyn Van Vliet and I am a professor in the Materials Science and Engineering 9 00:00:43,070 --> 00:00:45,380 and Biological Engineering departments at MIT. 10 00:00:45,380 --> 00:00:46,320 of reactions and after some practice, you will be able to derive a rate law for a general 11 00:00:46,320 --> 00:00:46,620 enzyme-catalyzed reaction. 12 00:00:46,620 --> 00:00:52,329 In order to understand the topic of this video, you should be familiar with determining rate 13 00:00:52,329 --> 00:00:58,450 laws from experimental data, predicting rate laws from proposed chemical reaction mechanisms, 14 00:00:58,450 --> 00:01:05,450 and have some basic understanding of the effects of catalysts on the kinetics of a reaction. 15 00:01:08,689 --> 00:01:15,689 The study of enzyme catalysis has become important for drug development. Many drugs work by inhibiting 16 00:01:23,039 --> 00:01:24,459 an enzyme. 17 00:01:24,459 --> 00:01:29,069 For example, enzymes thought to be important for the survival and replication of the parasite 18 00:01:29,069 --> 00:01:34,499 that causes malaria are being explored as therapeutic targets. The parasites enter the 19 00:01:34,499 --> 00:01:41,499 red blood cells of the host and use proteases to catalyze the degradation of the hemoglobin 20 00:01:42,240 --> 00:01:49,240 inside of the red blood cells. Degrading hemoglobin, a protein, yields amino acids, which can then 21 00:01:49,399 --> 00:01:54,709 be used by the parasite for its own protein synthesis. Protease inhibitors could help 22 00:01:54,709 --> 00:01:59,749 slow the degradation of hemoglobin, which in turn would slow the growth and reproduction 23 00:01:59,749 --> 00:02:02,369 of the malaria parasite. 24 00:02:02,369 --> 00:02:08,679 When a potential inhibitor is identified, kinetic data is used to evaluate its efficacy. 25 00:02:08,679 --> 00:02:12,770 In order to understand how this can be done, let's start by describing the kinetics of 26 00:02:12,770 --> 00:02:18,550 a simpler case -- that of an enzyme catalyzed reaction without an inhibitor present. Then 27 00:02:18,550 --> 00:02:21,720 we'll think about what might happen if we throw an inhibitor into the mix. 28 00:02:21,720 --> 00:02:23,200 This video will help you describe enzyme catalysis mathematically by first reviewing some common 29 00:02:23,200 --> 00:02:23,700 biochemical terms, then by describing the general characteristics of enzyme-catalyzed 30 00:02:23,700 --> 00:02:23,700 you will encounter throughout your undergraduate experience. 31 00:02:27,790 --> 00:02:33,040 a reactant is called a substrate. This is really just a matter of different fields using 32 00:02:33,040 --> 00:02:38,280 different terms, but it's important for you to be aware of this. Next, catalysts found 33 00:02:38,280 --> 00:02:43,829 in biological organisms are called enzymes. Textbooks frequently abbreviate "substrate" 34 00:02:43,829 --> 00:02:47,860 using the letter "S" and enzyme using the letter "E." 35 00:02:47,860 --> 00:02:53,700 How do enzymes work? If we look at a reaction coordinate, recall that the transition state 36 00:02:53,700 --> 00:02:59,190 represents a chemical species intermediate between the reactant, or substrate in this 37 00:02:59,190 --> 00:03:04,620 case, and the product. The potential energy difference between the transition state and 38 00:03:04,620 --> 00:03:10,340 the reactants is the activation energy for the forward reaction. The potential energy 39 00:03:10,340 --> 00:03:14,550 difference between the transition state and the products is the activation energy for 40 00:03:14,550 --> 00:03:20,480 the reverse reaction. Only reactants with the energy to overcome the activation energy 41 00:03:20,480 --> 00:03:22,760 will form products. 42 00:03:22,760 --> 00:03:28,000 Like synthetic catalysts, enzymes accelerate the rates of reactions by stabilizing the 43 00:03:28,000 --> 00:03:33,400 transition state, lowering its potential energy, and providing a new pathway by which the reaction 44 00:03:33,400 --> 00:03:39,500 can occur. This new reaction pathway has a lower activation energy than the uncatalyzed 45 00:03:39,500 --> 00:03:44,250 path making it more likely that a greater number of reactant molecules will have the 46 00:03:44,250 --> 00:03:50,620 energy needed to overcome the activation energy and proceed to product. 47 00:03:50,620 --> 00:03:57,040 Enzymes facilitate reactions, but, like other catalysts, are not consumed in the reaction. 48 00:03:57,040 --> 00:04:03,250 Enzymes are typically proteins, but RNA has also been shown to catalyze reactions. 49 00:04:03,250 --> 00:04:08,350 While there are different theories about how enzymes work to catalyze reactions, there 50 00:04:08,350 --> 00:04:12,620 are a couple of things that are agreed upon. One is that enzymes have a region called an 51 00:04:12,620 --> 00:04:18,959 active site. Substrates bind to this region. The shape and chemistry of the active site 52 00:04:18,959 --> 00:04:23,689 determine the selectivity of the enzyme for particular substrates. 53 00:04:23,689 --> 00:04:29,460 The second point is that when a substrate binds to an enzyme, an enzyme-substrate complex 54 00:04:29,460 --> 00:04:36,460 is formed. This enzyme-substrate complex is a reaction intermediate, meaning that it is 55 00:04:36,650 --> 00:04:43,650 formed and consumed in the reactions, but does not appear in the overall chemical equation. 56 00:04:43,889 --> 00:04:49,919 For many enzyme-catalyzed reactions, if we were to measure the rate of reaction at various 57 00:04:49,919 --> 00:04:56,919 substrate concentrations, we would see that the rate of reaction appears to follow first 58 00:04:57,080 --> 00:05:04,080 order kinetics at low substrate concentrations and then transitions to behavior that resembles 59 00:05:04,270 --> 00:05:11,270 zero-order kinetics at high substrate concentrations. Please pause the video here, turn to the person 60 00:05:11,830 --> 00:05:18,289 beside you and discuss what it means for a reaction to be first-order or zero-order. 61 00:05:18,289 --> 00:05:25,289 Remember, in a first-order reaction, the reaction rate is directly proportional to the concentration 62 00:05:28,360 --> 00:05:34,229 of substrate. In a zero-order reaction, the reaction rate is constant as the reaction 63 00:05:34,229 --> 00:05:40,449 progresses and is unaffected by substrate concentration. So, if we look at this data, 64 00:05:40,449 --> 00:05:45,330 we see that, for a given enzyme concentration, the addition of substrate above a certain 65 00:05:45,330 --> 00:05:50,139 value has no effect on the rate of reaction. 66 00:05:50,139 --> 00:05:55,180 So let's think about what reaction mechanism might explain this data. 67 00:05:55,180 --> 00:06:00,620 Some scientists hypothesized that after an enzyme and substrate combine to form the enzyme-substrate 68 00:06:00,620 --> 00:06:07,620 complex, this complex yields the product and degrades to regenerate free enzyme. Does the 69 00:06:08,089 --> 00:06:14,300 rate law derived from this reaction mechanism fit the experimental data? 70 00:06:14,300 --> 00:06:19,740 Let's derive the rate law and see. We can derive this rate law just as we would for 71 00:06:19,740 --> 00:06:21,889 other chemical systems. 72 00:06:21,889 --> 00:06:28,889 Remember, each step of a reaction mechanism is assumed to be an elementary reaction. Please 73 00:06:29,300 --> 00:06:35,389 pause the video here, turn to the person beside you and discuss how you determine the rate 74 00:06:35,389 --> 00:06:40,620 law for an elementary reaction. 75 00:06:40,620 --> 00:06:47,620 Remember, for an elementary reaction, you CAN predict the rate law from the chemical 76 00:06:48,439 --> 00:06:50,389 equation. 77 00:06:50,389 --> 00:06:55,490 With this in mind, can you write a differential equation for the rate of product formation 78 00:06:55,490 --> 00:07:01,849 with time? Please pause the video, try to write the equation, and continue playing the 79 00:07:01,849 --> 00:07:08,849 video to see if you are correct. 80 00:07:09,979 --> 00:07:16,979 From the 2nd step of the reaction mechanism, we can write that the rate of formation of 81 00:07:18,139 --> 00:07:24,979 product is equal to the rate constant k2 multiplied by the concentration of our enzyme-substrate 82 00:07:24,979 --> 00:07:27,659 complex. 83 00:07:27,659 --> 00:07:33,409 The enzyme-substrate complex is a reaction intermediate and not something that is easily 84 00:07:33,409 --> 00:07:40,409 measured in experiments. How can we verify if this expression is correct? It would be 85 00:07:41,080 --> 00:07:47,729 nice to restate the rate in terms of quantities that are more easily measured such as the 86 00:07:47,729 --> 00:07:52,830 substrate concentration and the initial enzyme concentration. 87 00:07:52,830 --> 00:07:58,499 To get an expression for the concentration of the enzyme-substrate complex, let's write 88 00:07:58,499 --> 00:08:05,499 a differential equation for the net rate of change of the enzyme-substrate complex concentration 89 00:08:05,759 --> 00:08:12,759 with time. Please pause the video here, try to write the differential equation, and then 90 00:08:14,270 --> 00:08:21,270 continue playing the video to see if you are correct. 91 00:08:23,889 --> 00:08:30,889 If we want to write an expression for d[ES]/dt, we see that in the first step of our mechanism, 92 00:08:30,899 --> 00:08:35,640 the enzyme-substrate complex is produced in the forward direction and consumed in the 93 00:08:35,640 --> 00:08:42,520 reverse direction. In the second step, the enzyme-substrate complex is converted into 94 00:08:42,520 --> 00:08:43,580 product. 95 00:08:43,580 --> 00:08:49,940 This equation introduces two additional unknowns: the free enzyme concentration and substrate 96 00:08:49,940 --> 00:08:56,150 concentration. So, we need two additional equations in order to solve it. 97 00:08:56,150 --> 00:09:01,190 Can you write an equation for the rate of change of substrate concentration with time? 98 00:09:01,190 --> 00:09:05,690 Please pause the video here, try to write the differential equation, and then continue 99 00:09:05,690 --> 00:09:12,690 playing the video to see if you are correct. 100 00:09:13,310 --> 00:09:19,150 If we want to write an expression for d[S]/dt, we see that the substrate will be consumed 101 00:09:19,150 --> 00:09:24,000 in the forward direction and produced in the reverse direction of the first step of the 102 00:09:24,000 --> 00:09:27,160 reaction mechanism. 103 00:09:27,160 --> 00:09:33,710 We still need one more equation. This equation is a simple relationship that says we know 104 00:09:33,710 --> 00:09:40,470 how much enzyme we added to our reaction mixture at time zero, so at any time, we know that 105 00:09:40,470 --> 00:09:46,710 the amount of free enzyme plus the amount of enzyme bound in the enzyme-substrate complex 106 00:09:46,710 --> 00:09:53,020 should equal the initial amount added. This could also be written in differential form 107 00:09:53,020 --> 00:09:58,000 as d[E]/dt plus d[ES]/dt is equal to zero. 108 00:09:58,000 --> 00:10:05,000 So, we have a system of four ordinary differential equations with four variables that all depend 109 00:10:05,580 --> 00:10:12,580 on time. Equation 1 and Equation 4 have analytic solutions, but Equations 2 and 3 contain a 110 00:10:16,800 --> 00:10:22,740 non-linear term. When you learn about differential equations in your future courses, you'll see 111 00:10:22,740 --> 00:10:27,930 why these equations are difficult to solve and that you will need numerical methods to 112 00:10:27,930 --> 00:10:30,600 solve them. 113 00:10:30,600 --> 00:10:37,600 Here, we have integrated these equations numerically using the boundary conditions that at time 114 00:10:38,010 --> 00:10:45,010 equals zero, the substrate concentration is S0, the product concentration is zero, the 115 00:10:47,280 --> 00:10:54,280 enzyme concentration is E0 and the concentration of enzyme bound to substrate is zero. 116 00:10:55,020 --> 00:11:01,350 We see that after a short start-up period, the concentration of enzyme bound to substrate 117 00:11:01,350 --> 00:11:07,810 remains approximately constant. In this region, where the substrate concentration is much 118 00:11:07,810 --> 00:11:14,810 greater than E0, as enzyme is released from the enzyme-substrate complex, it quickly recombines 119 00:11:15,510 --> 00:11:18,410 with available substrate. 120 00:11:18,410 --> 00:11:23,190 As more substrate is converted to product and the substrate concentration approaches 121 00:11:23,190 --> 00:11:30,190 E0, the concentration of enzyme bound to substrate is no longer constant. 122 00:11:30,540 --> 00:11:36,840 In the regime where the substrate concentration IS greater than E0, we can make a steady-state 123 00:11:36,840 --> 00:11:43,840 approximation and set d[ES]/dt equal to zero. This approximation will allow us to obtain 124 00:11:44,930 --> 00:11:51,600 an analytic expression for the concentration of enzyme bound to substrate. 125 00:11:51,600 --> 00:11:57,360 Setting d[ES]/dt equal to zero allows us to solve for the concentration of enzyme-substrate 126 00:11:57,360 --> 00:12:04,360 complex, which is equal to the rate constant k1 times the concentration of free enzyme 127 00:12:04,630 --> 00:12:11,630 times the concentration of substrate divided by the sum of k-1 and k2. 128 00:12:12,180 --> 00:12:18,950 Let's lump the rate constants into a new term called Km and substitute this into our expression 129 00:12:18,950 --> 00:12:24,100 for the concentration of enzyme-substrate complex. 130 00:12:24,100 --> 00:12:29,920 Now we can use this expression... ... to rewrite d[P]/dt. 131 00:12:29,920 --> 00:12:35,140 Let's try to eliminate the concentration of free enzyme from this expression. We can use 132 00:12:35,140 --> 00:12:39,260 the enzyme balance that we wrote earlier, which said that the concentration of free 133 00:12:39,260 --> 00:12:46,260 enzyme plus the concentration of enzyme-substrate complex should equal the initial enzyme concentration. 134 00:12:48,130 --> 00:12:53,680 Substituting our expression for the concentration of ES into the enzyme balance and doing a 135 00:12:53,680 --> 00:12:59,700 little bit of algebra, we can solve for the concentration of free enzyme. 136 00:12:59,700 --> 00:13:06,700 And finally, we can substitute this into our equation for d[P]/dt and obtain this expression. 137 00:13:08,070 --> 00:13:14,880 Let's check this rate law against our experimental data. At low substrate concentration, the 138 00:13:14,880 --> 00:13:20,400 expression reduces to a first-order rate law and at high substrate concentration; the expression 139 00:13:20,400 --> 00:13:24,320 reduces to a zero order rate law. 140 00:13:24,320 --> 00:13:28,310 It turns out that the rate law that we just derived is called the Michaelis-Menten equation. 141 00:13:28,310 --> 00:13:35,310 To make our equation look more like the version of the Michaelis-Menten equation that you 142 00:13:37,040 --> 00:13:42,530 will see in textbooks, there are a couple of terms that we will use because they are 143 00:13:42,530 --> 00:13:49,530 the convention in enzyme kinetics. Instead of saying d[P]/dt is the rate of the reaction, 144 00:13:51,960 --> 00:13:57,480 we will call it the velocity of the reaction and use the notation of 'v.' 145 00:13:57,480 --> 00:14:04,480 Finally, we'll lump k2[E]o into a term called vmax. At high substrate concentration, the 146 00:14:06,800 --> 00:14:10,370 reaction velocity approaches vmax. 147 00:14:10,370 --> 00:14:17,370 Remember that lump parameter Km? We can determine Km from experimental data. If we look at the 148 00:14:17,370 --> 00:14:23,620 substrate concentration needed to reach half of vmax, we see that the concentration of 149 00:14:23,620 --> 00:14:30,620 substrate required equals Km. Km values for many enzyme-substrate pairs can be found in 150 00:14:31,330 --> 00:14:36,910 the literature allowing for comparison. 151 00:14:36,910 --> 00:14:41,440 You may be wondering why we went through all of this. Let's go back to our drug development 152 00:14:41,440 --> 00:14:43,140 example. 153 00:14:43,140 --> 00:14:47,640 Let's say that you have designed a drug that you think will inhibit one of the proteases 154 00:14:47,640 --> 00:14:54,420 used by the malaria parasite to degrade hemoglobin. How will you test its effectiveness? 155 00:14:54,420 --> 00:15:01,420 First, you might add a fixed amount of protease to solutions of varying hemoglobin concentration 156 00:15:01,560 --> 00:15:07,240 in order to measure reaction rates. Let's say that this reaction follows Michaelis-Menten 157 00:15:07,240 --> 00:15:11,640 kinetics well and that you are able to estimate vmax and Km. 158 00:15:11,640 --> 00:15:17,470 Next, you repeat the experiment, but this time, you also add a fixed amount of the drug 159 00:15:17,470 --> 00:15:23,460 candidate to all of your solutions. You get your rate data and see that the apparent vmax 160 00:15:23,460 --> 00:15:30,460 has decreased. But what does this mean? Can you speculate how this inhibitor works? 161 00:15:31,690 --> 00:15:38,040 Is the drug interacting with the enzyme and preventing it from binding substrate? 162 00:15:38,040 --> 00:15:44,850 Is the drug interacting with the enzyme-substrate complex? Is it something else entirely? 163 00:15:44,850 --> 00:15:50,050 The thought process that you used today to derive the Michaelis-Menten equation can be 164 00:15:50,050 --> 00:15:56,750 used to derive rate laws for other reaction mechanisms. Then you can see which mechanism 165 00:15:56,750 --> 00:16:02,050 is best supported by the data. 166 00:16:02,050 --> 00:16:08,010 Today we learned a little bit about the importance of enzymes and derived a rate law from a general 167 00:16:08,010 --> 00:16:14,350 reaction mechanism using differential equations and a steady-state approximation. This rate 168 00:16:14,350 --> 00:16:19,590 law was compared against experimental rate data for an enzyme catalyzing the conversion 169 00:16:19,590 --> 00:16:25,960 of a substrate to a product. The equation that we derived is called the Michaelis-Menten 170 00:16:25,960 --> 00:16:32,290 equation. While all enzyme-catalyzed reactions may not exhibit Michaelis-Menten kinetics, 171 00:16:32,290 --> 00:16:37,240 the same logic that you used to derive the equation can be used to derive rate laws for 172 00:16:37,240 --> 00:16:43,279 other proposed reaction mechanisms.