1 00:00:00,500 --> 00:00:02,840 The following content is provided under a Creative 2 00:00:02,840 --> 00:00:04,380 Commons license. 3 00:00:04,380 --> 00:00:06,680 Your support will help MIT OpenCourseWare 4 00:00:06,680 --> 00:00:11,070 continue to offer high quality educational resources for free. 5 00:00:11,070 --> 00:00:13,670 To make a donation or view additional materials 6 00:00:13,670 --> 00:00:17,630 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,630 --> 00:00:18,800 at ocw.mit.edu. 8 00:00:36,060 --> 00:00:39,930 BOGDAN FEDELES: Hi, and welcome to 5.07 biochemistry online. 9 00:00:39,930 --> 00:00:42,260 I'm Dr. Bogdan Fedeles. 10 00:00:42,260 --> 00:00:45,810 Let's metabolize some problems. 11 00:00:45,810 --> 00:00:47,570 Now, I have here problem two of problem 12 00:00:47,570 --> 00:00:50,370 set three, which is an excellent exercise 13 00:00:50,370 --> 00:00:53,070 about the mechanism of inhibition of enzymes, 14 00:00:53,070 --> 00:00:55,380 specifically proteases. 15 00:00:55,380 --> 00:00:57,820 Now, this deals with the same protease 16 00:00:57,820 --> 00:01:02,220 from problem one of this problem set, which is interleukin 17 00:01:02,220 --> 00:01:05,430 converting enzyme, or ICE. 18 00:01:05,430 --> 00:01:07,560 Therefore, it's best that you familiarize yourself 19 00:01:07,560 --> 00:01:10,560 with the mechanism of action of this enzyme, ICE, 20 00:01:10,560 --> 00:01:13,300 by solving problem 1 and then continuing with this video, 21 00:01:13,300 --> 00:01:13,800 here. 22 00:01:19,270 --> 00:01:21,620 As you found out by solving problem one, 23 00:01:21,620 --> 00:01:24,110 ICE is a cysteine protease. 24 00:01:24,110 --> 00:01:28,520 It features cysteine and a histidine in its active site. 25 00:01:28,520 --> 00:01:30,740 The mechanism starts by the histidine 26 00:01:30,740 --> 00:01:34,460 acting as a general base and deprotonates the cysteine SH 27 00:01:34,460 --> 00:01:35,780 group. 28 00:01:35,780 --> 00:01:38,180 The thiolate anion then can attack 29 00:01:38,180 --> 00:01:42,020 the substrate, forming first, a tetrahedral intermediate. 30 00:01:42,020 --> 00:01:44,210 When this tetrahedral intermediate collapses, 31 00:01:44,210 --> 00:01:47,270 it cleaves the peptide bond and accomplishes 32 00:01:47,270 --> 00:01:50,330 the chemical reaction of the protease. 33 00:01:50,330 --> 00:01:52,910 Then, the newly formed thioester is 34 00:01:52,910 --> 00:01:57,200 hydrolyzed to release the other half of the product. 35 00:01:57,200 --> 00:01:59,060 Let's take a look at the mechanism. 36 00:01:59,060 --> 00:02:01,060 Here is our peptide substrate. 37 00:02:06,090 --> 00:02:08,100 And this is the peptide bond that's 38 00:02:08,100 --> 00:02:10,860 going to be cleaved by protease. 39 00:02:10,860 --> 00:02:14,820 As we mentioned, we have a cysteine in the active site 40 00:02:14,820 --> 00:02:17,130 and we have a histidine that's just going 41 00:02:17,130 --> 00:02:19,795 to function as a general base. 42 00:02:19,795 --> 00:02:23,590 I'm going to denote it as B. 43 00:02:23,590 --> 00:02:25,650 In the first step, the histidine is going 44 00:02:25,650 --> 00:02:28,970 to deprotonate the cysteine. 45 00:02:28,970 --> 00:02:32,540 The cysteine is then going to attack 46 00:02:32,540 --> 00:02:35,500 the carbonyl of our peptide bond, 47 00:02:35,500 --> 00:02:38,380 and the electrons are going to go to the oxygen, 48 00:02:38,380 --> 00:02:40,415 and form a tetrahedral intermediate. 49 00:02:49,820 --> 00:02:58,090 Here, notice the histidine is going to be protonated, 50 00:02:58,090 --> 00:03:01,900 and the oxygen is going to have a negative charge. 51 00:03:01,900 --> 00:03:04,330 In the next step, this tetrahedral intermediate 52 00:03:04,330 --> 00:03:07,630 is going to fall apart by breaking 53 00:03:07,630 --> 00:03:10,690 the peptide bond that the protease is supposed to cleave. 54 00:03:21,460 --> 00:03:25,270 There is the thioester with the cysteine in the active site 55 00:03:25,270 --> 00:03:34,450 and one half of our product is going to be released, here. 56 00:03:34,450 --> 00:03:36,280 Once again, the histidine is going 57 00:03:36,280 --> 00:03:38,670 to be deprotonated at this step. 58 00:03:38,670 --> 00:03:40,470 In the second step, the thioester 59 00:03:40,470 --> 00:03:42,340 we just formed in the active site 60 00:03:42,340 --> 00:03:45,310 will be hydrolyzed by a water molecule, which 61 00:03:45,310 --> 00:03:50,470 will be activated by the same histidine in the active site. 62 00:03:50,470 --> 00:03:52,843 Here is our water molecule. 63 00:03:57,190 --> 00:04:01,600 So, the histidine is going to deprotonate the water, 64 00:04:01,600 --> 00:04:06,340 activating it for attack on the carbonyl, forming once again, 65 00:04:06,340 --> 00:04:08,065 a tetrahedral intermediate. 66 00:04:18,970 --> 00:04:21,870 I have a negative charge here and a positive charge 67 00:04:21,870 --> 00:04:24,690 on the histidine. 68 00:04:24,690 --> 00:04:28,950 Finally, this tetrahedral intermediate 69 00:04:28,950 --> 00:04:37,380 will collapse, restoring the cysteine in the active site, 70 00:04:37,380 --> 00:04:40,350 which can be reprotonated by the protonated histidine, 71 00:04:40,350 --> 00:04:46,505 and releasing the second half of our substrate, here. 72 00:05:02,426 --> 00:05:03,300 So there you have it. 73 00:05:03,300 --> 00:05:06,250 Now we restored the active site with the cysteine 74 00:05:06,250 --> 00:05:07,580 and the histidine. 75 00:05:07,580 --> 00:05:13,330 And this is the second half of our peptide that we cleaved. 76 00:05:13,330 --> 00:05:16,610 So notice the carboxyl side is right here 77 00:05:16,610 --> 00:05:22,040 and the amine side was released a little bit earlier. 78 00:05:22,040 --> 00:05:23,720 The mechanism that you just saw is 79 00:05:23,720 --> 00:05:26,600 very similar to the serine protease 80 00:05:26,600 --> 00:05:29,570 mechanism that is described in the book 81 00:05:29,570 --> 00:05:31,980 and in the lecture notes. 82 00:05:31,980 --> 00:05:34,340 Notice, every time we form a tetrahedral intermediate, 83 00:05:34,340 --> 00:05:36,840 this is probably stabilized in an oxyanion 84 00:05:36,840 --> 00:05:40,400 hole formed by some of the residues 85 00:05:40,400 --> 00:05:42,080 on the backbone of the enzyme. 86 00:05:45,790 --> 00:05:47,470 Now let's take a look at a couple 87 00:05:47,470 --> 00:05:53,020 of strategies for inhibiting a cysteine protease like ICE. 88 00:05:53,020 --> 00:05:56,150 This problem is proposing two strategies. 89 00:05:56,150 --> 00:05:58,810 One involves an aldehyde inhibitor, 90 00:05:58,810 --> 00:06:03,040 the other involving an acyl methyl ketone inhibitor. 91 00:06:03,040 --> 00:06:04,130 Let's take a look. 92 00:06:04,130 --> 00:06:08,800 Here is the structure of a proposed aldehyde inhibitor. 93 00:06:08,800 --> 00:06:13,780 Notice here, this is a aspartate residue, or aspartate looking 94 00:06:13,780 --> 00:06:18,220 residue, which together with the other couple of amino acids, 95 00:06:18,220 --> 00:06:21,610 forms the recognition portion of the inhibitor 96 00:06:21,610 --> 00:06:25,180 that we're allowing to bind the protease. 97 00:06:25,180 --> 00:06:27,610 The R group is going to be an aldehyde, which 98 00:06:27,610 --> 00:06:32,080 will be crucial for actually inhibiting the enzyme. 99 00:06:32,080 --> 00:06:34,210 Question one of this problem is asking 100 00:06:34,210 --> 00:06:35,920 us to propose a mechanism by which 101 00:06:35,920 --> 00:06:38,890 the aldehyde inhibitor works. 102 00:06:38,890 --> 00:06:40,630 We're given an important clue that this 103 00:06:40,630 --> 00:06:43,550 is a mechanism based inhibitor. 104 00:06:43,550 --> 00:06:45,610 A mechanism based inhibitor means 105 00:06:45,610 --> 00:06:49,030 that the inhibitor binds in the same fashion 106 00:06:49,030 --> 00:06:52,770 as the normal substrate of the enzyme. 107 00:06:52,770 --> 00:06:55,120 Therefore, after we have reviewed 108 00:06:55,120 --> 00:06:57,550 the mechanism of the cysteine protease 109 00:06:57,550 --> 00:07:00,400 and remembering some of our carbonyl chemistry, 110 00:07:00,400 --> 00:07:03,880 we should be able to propose the following chemical reaction. 111 00:07:03,880 --> 00:07:07,260 Here is the active site of our protease. 112 00:07:07,260 --> 00:07:13,330 This is the cysteine and this is the histidine, 113 00:07:13,330 --> 00:07:19,410 which we're denoting as a general base, B. 114 00:07:19,410 --> 00:07:23,090 And here's our aldehyde inhibitor, 115 00:07:23,090 --> 00:07:26,750 which I'm going to just show the aldehydic group, right here. 116 00:07:31,400 --> 00:07:36,950 Since the R group next to this aldehyde 117 00:07:36,950 --> 00:07:39,290 resembles very closely the natural substrate 118 00:07:39,290 --> 00:07:40,970 of the enzyme, this aldehyde group 119 00:07:40,970 --> 00:07:44,510 will be positioned in the same place where we would normally 120 00:07:44,510 --> 00:07:48,770 find the peptide bond that will be cleaved by the enzyme. 121 00:07:48,770 --> 00:07:52,960 Therefore, this thiolate group, once it forms, 122 00:07:52,960 --> 00:07:57,250 will be in great position to react with the aldehyde 123 00:07:57,250 --> 00:07:59,557 and form a tetrahedral intermediate. 124 00:08:02,360 --> 00:08:04,710 Therefore, the histidine deprotonates the cysteine, 125 00:08:04,710 --> 00:08:08,480 and the cysteine can then attack the carbonyl 126 00:08:08,480 --> 00:08:12,710 to form a tetrahedral intermediate. 127 00:08:12,710 --> 00:08:22,060 Therefore, we get this tetrahedral intermediate 128 00:08:22,060 --> 00:08:27,600 and a protonated histidine base. 129 00:08:27,600 --> 00:08:35,940 Now normally, the reaction would proceed from here 130 00:08:35,940 --> 00:08:39,220 to form a thioester, but because this is an aldehyde, 131 00:08:39,220 --> 00:08:42,010 the reaction stops here, and that's 132 00:08:42,010 --> 00:08:45,290 how the enzyme will be inhibited because we have now 133 00:08:45,290 --> 00:08:47,270 bound this inhibitor. 134 00:08:47,270 --> 00:08:50,110 We'll have it covalently bound in the active site 135 00:08:50,110 --> 00:08:53,230 to this cysteine. 136 00:08:53,230 --> 00:08:55,390 An interesting observation, which you can't really 137 00:08:55,390 --> 00:08:57,940 tell from the problem, when people looked 138 00:08:57,940 --> 00:09:00,950 at the x-ray structure of this tetrahedral intermediate, 139 00:09:00,950 --> 00:09:04,960 they noticed that the negative charge on the oxygen is not, 140 00:09:04,960 --> 00:09:08,230 in fact, stabilized in the oxyanion hole 141 00:09:08,230 --> 00:09:10,630 that would stabilize such tetrahedral intermediates 142 00:09:10,630 --> 00:09:13,540 for the normal reaction. 143 00:09:13,540 --> 00:09:15,400 The ability of the aldehyde group 144 00:09:15,400 --> 00:09:17,380 to react with the cysteine in the active site 145 00:09:17,380 --> 00:09:21,040 and form a covalent bond can readily 146 00:09:21,040 --> 00:09:24,560 explain why the molecule would function as an inhibitor. 147 00:09:24,560 --> 00:09:27,820 Nevertheless, the reaction between the aldehyde 148 00:09:27,820 --> 00:09:30,220 and the nucleophile, the thiolate, 149 00:09:30,220 --> 00:09:32,740 is readily reversible. 150 00:09:32,740 --> 00:09:38,200 So whenever the inhibitor is in its carbonyl form, 151 00:09:38,200 --> 00:09:43,150 it can potentially fall off from the active site of the enzyme. 152 00:09:43,150 --> 00:09:45,280 Therefore, how good of an inhibitor 153 00:09:45,280 --> 00:09:48,460 this molecule is will depend on how tight 154 00:09:48,460 --> 00:09:52,000 it binds to the enzyme, and not necessarily on the fact 155 00:09:52,000 --> 00:09:55,190 that it forms a covalent bond in the active site. 156 00:09:55,190 --> 00:09:57,490 This is an example of a reversible inhibitor, 157 00:09:57,490 --> 00:10:01,400 even though it forms a covalent bond with the enzyme. 158 00:10:01,400 --> 00:10:04,090 Therefore, its ability to inhibit an enzyme 159 00:10:04,090 --> 00:10:06,340 will depend on the relative concentration 160 00:10:06,340 --> 00:10:09,040 between the inhibitor and the natural substrate 161 00:10:09,040 --> 00:10:10,060 of the enzyme. 162 00:10:10,060 --> 00:10:13,600 Let's remember the Michaelis-Menten equation 163 00:10:13,600 --> 00:10:20,110 written for a reversible inhibitor. 164 00:10:20,110 --> 00:10:23,710 As you recall, we have an enzyme reacting with a substrate. 165 00:10:23,710 --> 00:10:27,010 We have k1 and k minus 1 the rate 166 00:10:27,010 --> 00:10:31,030 constant to form the enzyme substrate complex, which 167 00:10:31,030 --> 00:10:35,710 then with k2, is going to form the product 168 00:10:35,710 --> 00:10:38,710 and reform the enzyme. 169 00:10:38,710 --> 00:10:42,430 But the inhibitor will react with the enzyme 170 00:10:42,430 --> 00:10:49,300 in the absence of the substrate in an equilibrium, 171 00:10:49,300 --> 00:10:52,540 forming an enzyme inhibitor complex which 172 00:10:52,540 --> 00:10:54,880 does not lead to any product. 173 00:10:54,880 --> 00:10:57,050 The constant of this equilibrium, 174 00:10:57,050 --> 00:10:58,840 I'm going to call it Ki, and this 175 00:10:58,840 --> 00:11:04,090 is the dissociation constant of the enzyme inhibitor complex. 176 00:11:04,090 --> 00:11:06,990 As you have seen in the notes, the rate, 177 00:11:06,990 --> 00:11:10,450 taking into account the inhibition constant here, 178 00:11:10,450 --> 00:11:14,315 the rate v is going to be Vmax times the concentration 179 00:11:14,315 --> 00:11:17,710 of the substrate over Km. 180 00:11:17,710 --> 00:11:22,870 This is Michaelis constant for the enzyme. 181 00:11:22,870 --> 00:11:27,670 Times 1 plus concentration of inhibitor 182 00:11:27,670 --> 00:11:34,510 over Ki plus concentration of substrate, s. 183 00:11:34,510 --> 00:11:41,170 Now, this equation tells us exactly how 184 00:11:41,170 --> 00:11:44,320 the rate is going to change as we increase 185 00:11:44,320 --> 00:11:46,720 or decrease the concentration of the inhibitor. 186 00:11:46,720 --> 00:11:50,800 Notice here, that this term is always greater than 1 187 00:11:50,800 --> 00:11:54,800 because concentration and Ki are going to be positive numbers. 188 00:11:54,800 --> 00:11:56,660 So this is 1 plus something positive. 189 00:11:56,660 --> 00:11:58,540 It's always going to be greater than 1, 190 00:11:58,540 --> 00:12:03,190 therefore, the denominator is going to be bigger 191 00:12:03,190 --> 00:12:07,900 than if we had Km times 1 plus s. 192 00:12:07,900 --> 00:12:10,300 So in the absence of the inhibitor, 193 00:12:10,300 --> 00:12:14,320 the denominator is going to be Km plus s. 194 00:12:14,320 --> 00:12:16,270 Therefore, when we add the inhibitor, 195 00:12:16,270 --> 00:12:19,210 this denominator gets bigger, and therefore 196 00:12:19,210 --> 00:12:21,490 the whole fraction gets smaller. 197 00:12:21,490 --> 00:12:23,530 We get a smaller rate. 198 00:12:23,530 --> 00:12:27,370 This is the basis for why the inhibitor will inhibit 199 00:12:27,370 --> 00:12:30,190 the enzyme, and therefore the rate of the reaction 200 00:12:30,190 --> 00:12:32,110 is going to be smaller. 201 00:12:32,110 --> 00:12:33,970 To see this graphically, we can write 202 00:12:33,970 --> 00:12:37,540 the reciprocal of the equation and look at the Lineweaver-Burk 203 00:12:37,540 --> 00:12:38,560 plot. 204 00:12:38,560 --> 00:12:43,540 The reciprocal of the equation is going to be 1/v equals-- 205 00:12:43,540 --> 00:12:46,350 and if we crunch the numbers, going 206 00:12:46,350 --> 00:12:59,950 to come up with Km/Vmax times 1 plus I/Ki times 1/s 207 00:12:59,950 --> 00:13:01,330 plus 1/Vmax. 208 00:13:06,440 --> 00:13:07,510 Let's plot this. 209 00:13:15,470 --> 00:13:21,260 I'm going to have 1/v and here I'm going to have 1/s. 210 00:13:21,260 --> 00:13:23,450 Now, if the concentration of substrate 211 00:13:23,450 --> 00:13:28,790 is really, really high, 1/s is going to be almost zero. 212 00:13:28,790 --> 00:13:33,940 So at the limit, when 1/s is zero, then we should get v 213 00:13:33,940 --> 00:13:35,030 equals Vmax. 214 00:13:35,030 --> 00:13:39,300 So therefore, let's say here it's 1/Vmax, 215 00:13:39,300 --> 00:13:42,110 and therefore without any inhibitor, 216 00:13:42,110 --> 00:13:47,340 we're going to get a line that looks like this. 217 00:13:51,680 --> 00:13:57,640 Now, as we're adding an inhibitor, this slope-- 218 00:13:57,640 --> 00:13:59,710 that is the coefficient or 1/s-- 219 00:13:59,710 --> 00:14:01,390 is going to be increasing. 220 00:14:01,390 --> 00:14:05,950 Therefore, we should get higher slopes. 221 00:14:05,950 --> 00:14:09,540 The higher the concentration of I, the bigger the slope. 222 00:14:09,540 --> 00:14:13,480 So it's going to look like this. 223 00:14:13,480 --> 00:14:18,430 So as concentration of I increases, 224 00:14:18,430 --> 00:14:22,090 the slope of this graph will increase. 225 00:14:22,090 --> 00:14:25,210 Notice however, that all these lines, 226 00:14:25,210 --> 00:14:28,750 even though correspond to slower rates, 227 00:14:28,750 --> 00:14:30,790 as the concentration of substrate increases-- 228 00:14:30,790 --> 00:14:34,030 that is 1/s gets closer to 0-- 229 00:14:34,030 --> 00:14:38,110 they will all converge to the same Vmax. 230 00:14:38,110 --> 00:14:43,960 This is the key feature of a competitive inhibition 231 00:14:43,960 --> 00:14:45,870 because the substrate in high quantities 232 00:14:45,870 --> 00:14:49,500 can out compete the inhibitor. 233 00:14:49,500 --> 00:14:51,220 Nevertheless, this mechanism only 234 00:14:51,220 --> 00:14:54,700 applies when the binding of an inhibitor to the enzyme 235 00:14:54,700 --> 00:14:57,430 is fast and reversible. 236 00:14:57,430 --> 00:15:00,070 If the binding is not reversible, 237 00:15:00,070 --> 00:15:04,330 obviously the enzyme will be inactivated forever, 238 00:15:04,330 --> 00:15:07,600 and then we will see a time dependent inactivation 239 00:15:07,600 --> 00:15:09,250 of the enzyme. 240 00:15:09,250 --> 00:15:13,750 The same phenomenon will happen if the reverse reaction, that 241 00:15:13,750 --> 00:15:16,150 is the dissociation of the inhibitor from the enzyme, 242 00:15:16,150 --> 00:15:18,310 is a slow process as well. 243 00:15:18,310 --> 00:15:21,310 All these considerations form a comprehensive answer 244 00:15:21,310 --> 00:15:23,110 for part one of the problem. 245 00:15:26,640 --> 00:15:31,290 Question two asked us to provide several reasons for which 246 00:15:31,290 --> 00:15:33,240 aldehyde inhibitors are not actually 247 00:15:33,240 --> 00:15:36,450 desirable as therapeutics. 248 00:15:36,450 --> 00:15:40,740 Once again, we have to think about the carbonyl chemistry. 249 00:15:40,740 --> 00:15:43,260 We saw here that aldehydes can react very well 250 00:15:43,260 --> 00:15:46,470 with nucleophiles such as thiols and thiolates. 251 00:15:46,470 --> 00:15:53,370 But aldehydes in solution can react with water and form 252 00:15:53,370 --> 00:15:55,020 what we call geminal diols. 253 00:15:59,400 --> 00:16:00,940 That look like this. 254 00:16:00,940 --> 00:16:05,470 Therefore, the effective concentration 255 00:16:05,470 --> 00:16:08,410 of the aldehyde in solution will be diminished because 256 00:16:08,410 --> 00:16:11,740 of this equilibrium, and that inhibitor 257 00:16:11,740 --> 00:16:16,000 might not be efficient at that lower concentration. 258 00:16:16,000 --> 00:16:21,340 Another consideration involves the oxidation of aldehydes. 259 00:16:21,340 --> 00:16:23,680 These could be enzymatically or even 260 00:16:23,680 --> 00:16:28,570 non-enzymatically oxidized to form carboxylic acids 261 00:16:28,570 --> 00:16:30,010 or carboxylates. 262 00:16:30,010 --> 00:16:35,590 These would obviously not be very reactive 263 00:16:35,590 --> 00:16:39,430 and any inhibitor that gets oxidized 264 00:16:39,430 --> 00:16:42,130 will stop being an inhibitor. 265 00:16:42,130 --> 00:16:44,020 Additionally, owing to their reactivity, 266 00:16:44,020 --> 00:16:47,860 aldehyde group could react with many other biomolecules. 267 00:16:47,860 --> 00:16:50,380 Think about the amino acid side chains. 268 00:16:50,380 --> 00:16:52,720 Many of them are actually, in fact, 269 00:16:52,720 --> 00:16:55,030 capable of reacting with aldehydes. 270 00:16:55,030 --> 00:16:57,820 This will also diminish the effective concentration 271 00:16:57,820 --> 00:17:02,470 of the inhibitor and it may even cause side effects. 272 00:17:02,470 --> 00:17:04,849 Therefore taking all these into considerations, 273 00:17:04,849 --> 00:17:08,681 aldehydes may not be a great solution for therapeutics. 274 00:17:11,839 --> 00:17:15,319 Question 3 is asking us to propose a mechanism by which 275 00:17:15,319 --> 00:17:19,190 the second kind of inhibitor, the acyloxymethyl ketone 276 00:17:19,190 --> 00:17:22,670 is inhibiting the protease. 277 00:17:22,670 --> 00:17:26,270 As you see here, the acyloxymethyl ketone 278 00:17:26,270 --> 00:17:30,730 is actually very similar to the aldehyde inhibitor. 279 00:17:30,730 --> 00:17:33,890 Notice this group is exactly the same 280 00:17:33,890 --> 00:17:36,560 as the carbonyl group of the aldehyde, 281 00:17:36,560 --> 00:17:40,040 but instead of having just the hydrogen, 282 00:17:40,040 --> 00:17:48,380 we have a methylene next to a aryloxy or acyloxy group. 283 00:17:48,380 --> 00:17:51,530 This, as you know, is a very good leaving group 284 00:17:51,530 --> 00:17:55,940 and provides a second reactive site, this methyl group here, 285 00:17:55,940 --> 00:17:58,280 that can react with the enzyme. 286 00:17:58,280 --> 00:18:00,260 The second kind of inhibitor is in fact 287 00:18:00,260 --> 00:18:03,890 very similar to the tosyl, phenyl, chloro ketone inhibitor 288 00:18:03,890 --> 00:18:06,560 of serine proteases, which is discussed 289 00:18:06,560 --> 00:18:10,240 at length in the lecture notes and in the book. 290 00:18:10,240 --> 00:18:12,260 These inhibitors fall in a general class 291 00:18:12,260 --> 00:18:15,080 of alpha substituted ketones and they 292 00:18:15,080 --> 00:18:16,910 feature two reactive sites. 293 00:18:16,910 --> 00:18:18,770 One is the carbonyl and the other one 294 00:18:18,770 --> 00:18:21,860 is the methylene group, which has attached a good leaving 295 00:18:21,860 --> 00:18:22,480 group. 296 00:18:22,480 --> 00:18:25,910 Now, here is a general form of a alpha substituted ketone. 297 00:18:28,870 --> 00:18:32,210 Here is the ketone, here is the methylene group attached 298 00:18:32,210 --> 00:18:35,720 to a good leaving group, which I'm going to denote x. 299 00:18:35,720 --> 00:18:38,570 Now, here are the residues in the active site. 300 00:18:38,570 --> 00:18:42,320 Here is the cysteine with the thiol group, 301 00:18:42,320 --> 00:18:46,980 and here is the histidine, which I'm going to draw out. 302 00:18:51,180 --> 00:18:53,820 Histidine, all right. 303 00:18:53,820 --> 00:18:56,010 So as we saw before, the reaction 304 00:18:56,010 --> 00:19:00,900 will start by the histidine acting as a general base. 305 00:19:00,900 --> 00:19:04,530 The histidine deprotonates the cysteine, which then 306 00:19:04,530 --> 00:19:09,960 can attack the ketone carbonyl. 307 00:19:09,960 --> 00:19:12,868 This leads to the formation of a tetrahedral intermediate. 308 00:19:22,440 --> 00:19:25,500 And we have a positive charge, here on the histidine, 309 00:19:25,500 --> 00:19:30,060 and a negative charge on the O minus, here. 310 00:19:30,060 --> 00:19:32,670 Now, this tetrahedral intermediate presumably 311 00:19:32,670 --> 00:19:35,250 will be stabilized in an oxyanion hole, 312 00:19:35,250 --> 00:19:38,640 as you guys have seen before with some kind of hydrogen 313 00:19:38,640 --> 00:19:43,030 bonds to the backbone of the protein. 314 00:19:43,030 --> 00:19:44,730 Now, in the next step, this is something 315 00:19:44,730 --> 00:19:49,139 that you can't really anticipate or know 316 00:19:49,139 --> 00:19:50,430 without doing some experiments. 317 00:19:50,430 --> 00:19:55,200 But it turns out this O minus is a good SN2 nucleophile 318 00:19:55,200 --> 00:19:58,470 to displace the good leaving group, x. 319 00:19:58,470 --> 00:20:02,770 So we're going to have an S2 reaction, 320 00:20:02,770 --> 00:20:05,070 This O minus attacks the carbon, and then 321 00:20:05,070 --> 00:20:09,390 the x takes the electrons and leaves. 322 00:20:09,390 --> 00:20:11,770 What we're going to form here is an epoxide. 323 00:20:18,860 --> 00:20:24,380 All right, so this is the epoxide 324 00:20:24,380 --> 00:20:29,840 and we still have our protonated histidine here. 325 00:20:36,220 --> 00:20:37,780 And there's a positive charge here, 326 00:20:37,780 --> 00:20:42,080 and of course, the x group is taking its electrons and leaves 327 00:20:42,080 --> 00:20:45,140 as an anion. 328 00:20:45,140 --> 00:20:49,430 Now, in the next step, because this is sitting actually 329 00:20:49,430 --> 00:20:51,770 close enough to the epoxide, the epoxide 330 00:20:51,770 --> 00:20:53,631 is going to get protonated. 331 00:20:56,580 --> 00:21:01,482 This takes the proton from the histidine. 332 00:21:09,210 --> 00:21:16,300 So now we have a protonated epoxide, 333 00:21:16,300 --> 00:21:25,860 and the histidine now is in its deeper native form, 334 00:21:25,860 --> 00:21:31,570 and the epoxide has a positive charge. 335 00:21:31,570 --> 00:21:35,070 Now, because we have a protonated epoxide, 336 00:21:35,070 --> 00:21:38,520 this acts as a very good leaving group, 337 00:21:38,520 --> 00:21:43,020 and this carbon becomes very susceptible for an SN2 338 00:21:43,020 --> 00:21:44,250 reaction. 339 00:21:44,250 --> 00:21:46,440 And it turns out, the histidine, this nitrogen, 340 00:21:46,440 --> 00:21:51,460 is a good enough nucleophile to react in an SN2 type reaction. 341 00:21:51,460 --> 00:22:03,569 And in fact, the reaction is helped by this other nitrogen. 342 00:22:03,569 --> 00:22:05,110 I should have used a different color. 343 00:22:10,520 --> 00:22:12,410 So what we're getting out here is 344 00:22:12,410 --> 00:22:15,270 a covalent bond between the histidine 345 00:22:15,270 --> 00:22:21,200 and the alpha position of the original alpha substituted 346 00:22:21,200 --> 00:22:22,470 ketone. 347 00:22:22,470 --> 00:22:27,230 Now, this reaction will not be, in fact, reversible. 348 00:22:27,230 --> 00:22:29,750 So we should just say one arrow only. 349 00:22:50,270 --> 00:22:53,550 So there we have our histidine. 350 00:22:53,550 --> 00:22:56,240 Now it's covalently attached to our inhibitor 351 00:22:56,240 --> 00:22:59,790 and this step is, in fact, irreversible. 352 00:22:59,790 --> 00:23:01,480 So this prevents the inhibitor from ever 353 00:23:01,480 --> 00:23:04,440 dissociating once this reaction has taken place. 354 00:23:04,440 --> 00:23:07,520 Now, even though this tetrahedral intermediate 355 00:23:07,520 --> 00:23:12,350 around the carbonyl carbon, it can fall apart 356 00:23:12,350 --> 00:23:18,260 as we saw before, reforming the carbonyl 357 00:23:18,260 --> 00:23:23,295 and the cysteine thiol, the covalent bond to the histidine 358 00:23:23,295 --> 00:23:23,795 remains. 359 00:23:32,860 --> 00:23:37,890 And this is, in fact, the reason for which these inhibitors 360 00:23:37,890 --> 00:23:42,210 are irreversible and will show a time dependent inhibition. 361 00:23:42,210 --> 00:23:44,880 Regardless of the more complicated mechanistic detail 362 00:23:44,880 --> 00:23:47,940 I just showed you, the take home message 363 00:23:47,940 --> 00:23:51,420 here is that when using an alpha substituted ketone, 364 00:23:51,420 --> 00:23:54,870 the end result will be a covalent bond 365 00:23:54,870 --> 00:23:56,730 between the enzyme and the inhibitor 366 00:23:56,730 --> 00:23:58,600 that is not reversible. 367 00:23:58,600 --> 00:24:02,160 Therefore, we expect to see a time dependent inactivation 368 00:24:02,160 --> 00:24:03,480 of the enzyme. 369 00:24:03,480 --> 00:24:06,300 The more enzyme is being taken out of the reaction 370 00:24:06,300 --> 00:24:09,480 by the inhibitor, the slower the overall reaction will 371 00:24:09,480 --> 00:24:14,100 be until there is no more enzyme left to catalyze the reaction. 372 00:24:14,100 --> 00:24:17,220 The problem also provides some kinetic data, 373 00:24:17,220 --> 00:24:21,720 which, in fact, supports the time dependent inhibition 374 00:24:21,720 --> 00:24:25,720 features of the second kind of inhibitor. 375 00:24:25,720 --> 00:24:31,200 In this figure, we see on the y-axis, the concentration 376 00:24:31,200 --> 00:24:34,500 of the product that is formed, and on the x-axis 377 00:24:34,500 --> 00:24:36,030 we see the time. 378 00:24:36,030 --> 00:24:39,330 Now, if we look at the dark circles, which 379 00:24:39,330 --> 00:24:40,890 is the control reaction in which we 380 00:24:40,890 --> 00:24:43,110 have some substrate, but no inhibitor, 381 00:24:43,110 --> 00:24:47,670 we see that the product is produced and increases linearly 382 00:24:47,670 --> 00:24:49,050 with time. 383 00:24:49,050 --> 00:24:51,330 However, once we add the inhibitor 384 00:24:51,330 --> 00:24:56,730 at a certain concentration, and this would be the dark squares, 385 00:24:56,730 --> 00:25:00,750 then we see that the amount of product increases for a while, 386 00:25:00,750 --> 00:25:04,800 but then it grows slower and slower until it eventually 387 00:25:04,800 --> 00:25:06,750 stops. 388 00:25:06,750 --> 00:25:11,280 Now at this point, if we add an excess of substrate, 389 00:25:11,280 --> 00:25:15,060 we see that the reaction does not restart, 390 00:25:15,060 --> 00:25:18,570 meaning the entire amount of enzyme has been inactivated, 391 00:25:18,570 --> 00:25:22,860 and this inactivation is irreversible. 392 00:25:22,860 --> 00:25:26,550 However, we're also provided this additional piece 393 00:25:26,550 --> 00:25:27,360 of the data. 394 00:25:27,360 --> 00:25:31,500 If we run the reaction in an excess of substrate, 395 00:25:31,500 --> 00:25:36,180 we see here the open circles, the amount of product 396 00:25:36,180 --> 00:25:40,200 produced increases linearly with time. 397 00:25:40,200 --> 00:25:43,440 But if we add the same amount of inhibitor, 398 00:25:43,440 --> 00:25:47,130 these open squares actually are on top or right 399 00:25:47,130 --> 00:25:51,690 next to the open circles on this line, which 400 00:25:51,690 --> 00:25:54,930 says that the inhibitor has virtually no effect 401 00:25:54,930 --> 00:25:57,990 at this level of concentration. 402 00:25:57,990 --> 00:25:59,520 Which tells us that the inhibitor 403 00:25:59,520 --> 00:26:01,500 is in fact competing with the substrate 404 00:26:01,500 --> 00:26:03,770 and we will have an excess of the substrate. 405 00:26:03,770 --> 00:26:07,470 The inhibitor does not get a chance to bind 406 00:26:07,470 --> 00:26:12,180 and inhibit the enzyme, at least within this interval of time. 407 00:26:12,180 --> 00:26:18,300 Now this way of plotting kinetic data is perhaps 408 00:26:18,300 --> 00:26:22,440 a little misleading because the amount of product 409 00:26:22,440 --> 00:26:24,480 that we obtained from the reaction 410 00:26:24,480 --> 00:26:28,940 does not reflect how much of the enzyme gets inhibited. 411 00:26:28,940 --> 00:26:34,560 We want, in fact, to look at the percentage of enzyme activity 412 00:26:34,560 --> 00:26:38,970 that is remaining after a given amount of time. 413 00:26:38,970 --> 00:26:48,480 Therefore, if we were to plot percentage enzyme 414 00:26:48,480 --> 00:26:58,520 activity versus time, for a controlled reaction 415 00:26:58,520 --> 00:27:00,710 we expect the reaction to proceed, 416 00:27:00,710 --> 00:27:04,430 and the enzyme to stay just as active at any given 417 00:27:04,430 --> 00:27:08,640 point in time, so we will see a straight line. 418 00:27:08,640 --> 00:27:11,450 However, when we add an inhibitor, which 419 00:27:11,450 --> 00:27:14,120 shows a time dependent inhibition, 420 00:27:14,120 --> 00:27:17,600 then the percentage enzyme activity 421 00:27:17,600 --> 00:27:22,770 that remains at every point in time will be decreasing. 422 00:27:22,770 --> 00:27:29,108 And in fact, will be decreasing to the point 423 00:27:29,108 --> 00:27:31,450 that it reaches the maximum of slope 424 00:27:31,450 --> 00:27:36,730 for the maximum amount of inhibitor present 425 00:27:36,730 --> 00:27:38,980 in the reaction mixture. 426 00:27:38,980 --> 00:27:41,650 So this line represents the fastest 427 00:27:41,650 --> 00:27:44,740 that the enzyme can be completely inactivated 428 00:27:44,740 --> 00:27:46,180 by an inhibitor. 429 00:27:46,180 --> 00:27:49,540 It will be governed by the binding affinity 430 00:27:49,540 --> 00:27:52,000 of the inhibitor to the enzyme. 431 00:27:52,000 --> 00:27:56,620 This answers the third and last question of this problem. 432 00:27:56,620 --> 00:27:58,630 I hope you enjoyed this exploration 433 00:27:58,630 --> 00:28:02,230 of the various strategies by which inhibitors 434 00:28:02,230 --> 00:28:04,520 can inhibit proteases. 435 00:28:04,520 --> 00:28:06,670 This problem highlights, in fact, 436 00:28:06,670 --> 00:28:09,010 the importance of understanding the mechanism of action 437 00:28:09,010 --> 00:28:10,420 of enzymes. 438 00:28:10,420 --> 00:28:15,130 Only then we can begin to design therapeutically useful 439 00:28:15,130 --> 00:28:17,280 inhibitors.