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 00:00:31,706 --> 00:00:34,700 BOGDAN FEDELES: Hello, and welcome to 5.07 Bio Chemistry 9 00:00:34,700 --> 00:00:36,030 Online. 10 00:00:36,030 --> 00:00:37,360 I'm Dr. Bogdan Fedeles. 11 00:00:37,360 --> 00:00:38,660 Let's metabolize some problems. 12 00:00:42,300 --> 00:00:45,470 Today we're going to be talking about Problem 2 13 00:00:45,470 --> 00:00:47,300 of Problem Set 4. 14 00:00:47,300 --> 00:00:49,400 Here we're going to be discussing in detail 15 00:00:49,400 --> 00:00:54,080 the mechanism of HMG-CoA synthase, a key enzyme 16 00:00:54,080 --> 00:00:57,290 in central metabolism responsible for making the five 17 00:00:57,290 --> 00:00:59,570 carbon building blocks from which 18 00:00:59,570 --> 00:01:04,580 all sterols, such as cholesterol and steroid hormones are made. 19 00:01:04,580 --> 00:01:08,240 HMG-CoA synthase catalyzes the following reaction. 20 00:01:08,240 --> 00:01:11,870 It takes acetyl-CoA, which we're going 21 00:01:11,870 --> 00:01:14,450 to encounter a lot in the central metabolism, 22 00:01:14,450 --> 00:01:17,450 and combines it with acetoacetyl-CoA, 23 00:01:17,450 --> 00:01:19,420 another thioester. 24 00:01:19,420 --> 00:01:24,112 In this process, the one molecule of CoA is lost, 25 00:01:24,112 --> 00:01:25,945 and we get the product hydroxymethylglutaryl 26 00:01:25,945 --> 00:01:32,410 CoA or HMG-CoA, as you notice the initials H, M, and G are 27 00:01:32,410 --> 00:01:33,970 part of this name. 28 00:01:33,970 --> 00:01:35,850 To help us understand the mechanism, 29 00:01:35,850 --> 00:01:40,450 a crystal structure of this HMG-CoA synthase 30 00:01:40,450 --> 00:01:43,690 is provided in this problem, and it's shown here. 31 00:01:50,520 --> 00:01:53,130 Question 1 of this problem is asking 32 00:01:53,130 --> 00:01:55,830 us to provide a detailed curved arrow 33 00:01:55,830 --> 00:01:58,260 mechanism of the reaction catalyzed 34 00:01:58,260 --> 00:02:00,510 by the HMG-CoA synthase. 35 00:02:00,510 --> 00:02:03,750 And for that we're going to use the information provided 36 00:02:03,750 --> 00:02:05,560 in the crystal structure. 37 00:02:05,560 --> 00:02:07,950 And we are also given an additional hint 38 00:02:07,950 --> 00:02:10,889 and that is the acetyl CoA substrate 39 00:02:10,889 --> 00:02:19,350 provides the CH2CO2 CO2 minus moiety in the HMG-CoA product. 40 00:02:19,350 --> 00:02:20,730 Let's take a look. 41 00:02:20,730 --> 00:02:24,330 So the hint is telling us that these two carbons, which 42 00:02:24,330 --> 00:02:27,180 I'm going to label in blue from acetyl-CoA, 43 00:02:27,180 --> 00:02:30,990 are, in fact, these two carbons, CH2 COO 44 00:02:30,990 --> 00:02:35,070 minus in hydroxymethylglutaryl CoA. 45 00:02:35,070 --> 00:02:38,220 That means that the other four carbons in the product 46 00:02:38,220 --> 00:02:40,920 must be the carbons from the acetoacetyl-CoA. 47 00:02:43,710 --> 00:02:45,030 Let's label these red. 48 00:02:49,390 --> 00:02:54,470 So now since we have a thioester functionality in both cases, 49 00:02:54,470 --> 00:02:59,630 these are probably the same, so the carbons 50 00:02:59,630 --> 00:03:02,960 from a acetoacetyl-CoA are as follows. 51 00:03:02,960 --> 00:03:06,470 So notice the HMG-CoA synthase, it 52 00:03:06,470 --> 00:03:11,330 accomplishes the formation of a carbon-carbon bond, 53 00:03:11,330 --> 00:03:14,600 and that is this bond right here. 54 00:03:14,600 --> 00:03:19,220 It's between the CH3 carbon acetyl-CoA 55 00:03:19,220 --> 00:03:24,290 and the carbonyl carbon of acetoacetyl-CoA. 56 00:03:24,290 --> 00:03:26,300 And in the process, this carbonyl 57 00:03:26,300 --> 00:03:30,012 will become a hydroxyl group. 58 00:03:30,012 --> 00:03:31,970 Now, let's take a look at the crystal structure 59 00:03:31,970 --> 00:03:34,850 and see what information we can gather there. 60 00:03:34,850 --> 00:03:36,830 Now this is a picture of the active site 61 00:03:36,830 --> 00:03:38,890 of the HMG-CoA synthase. 62 00:03:38,890 --> 00:03:44,630 We notice the acetyl-CoA moiety is actually 63 00:03:44,630 --> 00:03:47,700 now bound to this C111. 64 00:03:47,700 --> 00:03:49,790 As you know C is a cysteine-- 65 00:03:49,790 --> 00:03:51,470 so it's actually covalently bound 66 00:03:51,470 --> 00:03:54,560 to a cysteine-- in the active site. 67 00:03:54,560 --> 00:03:57,080 Now, notice this is the other substrate, 68 00:03:57,080 --> 00:04:01,850 which is the acetoacetyl-CoA and bound in the active site. 69 00:04:01,850 --> 00:04:03,410 So as we've discussed before, we're 70 00:04:03,410 --> 00:04:05,630 going to make this carbon-carbon bond, 71 00:04:05,630 --> 00:04:08,030 and that's going to happen between this carbon 72 00:04:08,030 --> 00:04:10,700 here and this carbon here. 73 00:04:10,700 --> 00:04:14,780 So this bond needs to be formed. 74 00:04:14,780 --> 00:04:18,490 However, let's take it one step at a time. 75 00:04:18,490 --> 00:04:21,399 The reaction will start by forming 76 00:04:21,399 --> 00:04:25,480 this thioester between the acetoacetyl-CoA carbons 77 00:04:25,480 --> 00:04:29,800 and the cysteine in the active site of the enzyme. 78 00:04:29,800 --> 00:04:32,980 In many reactions that use a cysteine in the active site, 79 00:04:32,980 --> 00:04:36,250 which is used to form a covalent bond to the substrate, 80 00:04:36,250 --> 00:04:39,760 we first need a base to deprotonate the cysteine 81 00:04:39,760 --> 00:04:42,640 and make it a really good nucleophile, which 82 00:04:42,640 --> 00:04:44,680 will subsequently attack the substrate 83 00:04:44,680 --> 00:04:46,800 and form a covalent bond. 84 00:04:46,800 --> 00:04:50,920 Now, in a lot of these cases, the base is a histidine. 85 00:04:50,920 --> 00:04:54,960 Let's take a look if we see a histidine in our structure. 86 00:04:54,960 --> 00:04:59,760 There is a histidine in the structure, H233. 87 00:04:59,760 --> 00:05:01,350 But if you look at this histidine, 88 00:05:01,350 --> 00:05:04,530 it's quite far from our cysteine here. 89 00:05:04,530 --> 00:05:07,895 Well, obviously, because this is a projection, 90 00:05:07,895 --> 00:05:10,020 this is a two-dimensional structure so that we only 91 00:05:10,020 --> 00:05:12,311 see here a projection of a three-dimensional structure, 92 00:05:12,311 --> 00:05:14,460 it's very hard for us to tell if this histidine is, 93 00:05:14,460 --> 00:05:20,340 in fact, close enough to deprotonate this cysteine. 94 00:05:20,340 --> 00:05:25,080 So even though we have this crystal structure, 95 00:05:25,080 --> 00:05:28,640 from this one picture, we do not have enough information 96 00:05:28,640 --> 00:05:31,190 to figure out, well, what is the base that 97 00:05:31,190 --> 00:05:35,180 will deprotonate cysteine before it reacts with acetyl-CoA. 98 00:05:35,180 --> 00:05:38,690 Remember this is often the case with crystal structures that 99 00:05:38,690 --> 00:05:41,630 because of the resolution which we can collect them, 100 00:05:41,630 --> 00:05:43,535 we cannot see the protons. 101 00:05:43,535 --> 00:05:44,910 And if we cannot see the protons, 102 00:05:44,910 --> 00:05:47,900 it's very hard to tell which amino acid residues are 103 00:05:47,900 --> 00:05:52,810 protonated and can serve as a general acid or general bases. 104 00:05:52,810 --> 00:05:55,360 A lot of the researcher's intuition 105 00:05:55,360 --> 00:05:59,140 comes into play to draw these kind of structures. 106 00:05:59,140 --> 00:06:02,320 And it's only with collecting many different kinds 107 00:06:02,320 --> 00:06:05,410 of experimental evidence that we can put together 108 00:06:05,410 --> 00:06:09,250 a more definitive mechanism. 109 00:06:09,250 --> 00:06:11,770 Therefore, we might not know for sure 110 00:06:11,770 --> 00:06:15,910 from this one picture, which is the general base that 111 00:06:15,910 --> 00:06:19,510 deprotonates the cysteine at 111. 112 00:06:19,510 --> 00:06:22,840 So we're not going to assign it for now. 113 00:06:22,840 --> 00:06:25,900 Let's try to write the mechanism for that part. 114 00:06:25,900 --> 00:06:31,340 Here is the acetyl-CoA substrate, 115 00:06:31,340 --> 00:06:35,565 and here is the active site cysteine 111 116 00:06:35,565 --> 00:06:38,630 with its thiol group. 117 00:06:38,630 --> 00:06:41,270 And again, we're not going to assign it, 118 00:06:41,270 --> 00:06:47,190 but there will be a base in the active site of the enzyme that 119 00:06:47,190 --> 00:06:51,966 will need to deprotonate this cysteine before it can react. 120 00:06:51,966 --> 00:06:55,350 Therefore, the reaction first involves 121 00:06:55,350 --> 00:06:58,720 activation of the cysteine, is deprotonated, and then 122 00:06:58,720 --> 00:07:02,640 cysteine can attack the thioester 123 00:07:02,640 --> 00:07:05,181 and form a tetrahedral intermediate. 124 00:07:07,830 --> 00:07:10,890 As we've seen it before, for a lot of these reactions 125 00:07:10,890 --> 00:07:12,170 involving thioesters. 126 00:07:12,170 --> 00:07:15,780 In the first step, we will form four bonds 127 00:07:15,780 --> 00:07:24,050 to this carbonyl carbon, and we have 128 00:07:24,050 --> 00:07:29,060 formed a new bond between the thiol of the cysteine 111 129 00:07:29,060 --> 00:07:32,750 and the starting material, acetyl-CoA. 130 00:07:32,750 --> 00:07:36,070 Of course, there is a negative charge here. 131 00:07:36,070 --> 00:07:39,310 And our base in the active site of the enzyme 132 00:07:39,310 --> 00:07:41,051 will now be protonated. 133 00:07:44,149 --> 00:07:45,690 Now that the tetrahedral intermediate 134 00:07:45,690 --> 00:07:53,820 is going to fall apart by kicking off the CoA moiety. 135 00:07:53,820 --> 00:07:56,880 Of course, this will get protonated presumably 136 00:07:56,880 --> 00:07:58,840 by the same base. 137 00:07:58,840 --> 00:08:03,420 Now it's a general acid that will protonate a CoA. 138 00:08:03,420 --> 00:08:09,300 So we obtain the thioester with the cysteine in the active site 139 00:08:09,300 --> 00:08:15,980 and one molecule of CoA is going to be leaving. 140 00:08:15,980 --> 00:08:18,710 And here is our base. 141 00:08:18,710 --> 00:08:20,420 Notice the tetrahedral intermediate 142 00:08:20,420 --> 00:08:22,160 that we're forming here. 143 00:08:22,160 --> 00:08:25,232 We have an oxygen that develops a negative charge. 144 00:08:25,232 --> 00:08:27,440 And this is very similar to the kinds of intermediate 145 00:08:27,440 --> 00:08:31,730 you've seen in the serine protease mechanism. 146 00:08:31,730 --> 00:08:33,830 Now in that case, such an intermediate 147 00:08:33,830 --> 00:08:36,200 was stabilized with hydrogen bonds 148 00:08:36,200 --> 00:08:39,110 from the backbone of the protein in a structure that 149 00:08:39,110 --> 00:08:42,110 was called an oxyanion hole. 150 00:08:42,110 --> 00:08:45,260 Now let's take a look at the crystal structure of HMG-CoA 151 00:08:45,260 --> 00:08:48,050 to see how a tetrahedral intermediate might 152 00:08:48,050 --> 00:08:49,700 be stabilized. 153 00:08:49,700 --> 00:08:55,330 Here is the acetyl of the acetoacetyl-CoA substrate. 154 00:08:55,330 --> 00:08:58,070 The acetyl now is bound to the cysteine 155 00:08:58,070 --> 00:09:00,590 111 here in the active site. 156 00:09:00,590 --> 00:09:04,640 Now, let's take a look if there are any other residues close 157 00:09:04,640 --> 00:09:08,450 enough to this oxygen. One of them that's shown here 158 00:09:08,450 --> 00:09:13,580 is this NH, which belongs to an amide bond, the backbone amide, 159 00:09:13,580 --> 00:09:16,985 and that's part of the amino acid serine 307. 160 00:09:16,985 --> 00:09:20,870 So this distance here, C3.06 Angstrom, 161 00:09:20,870 --> 00:09:26,860 it's actually close enough for a fairly good hydrogen bond. 162 00:09:26,860 --> 00:09:29,980 Presumably, this interaction will stabilize the binding 163 00:09:29,980 --> 00:09:35,080 of acetyl-CoA in this region of the active site 164 00:09:35,080 --> 00:09:39,150 and may also be involved in catalyzing 165 00:09:39,150 --> 00:09:42,242 the reaction with the cysteine by stabilizing 166 00:09:42,242 --> 00:09:43,450 the tetrahedral intermediate. 167 00:09:43,450 --> 00:09:45,220 In that case, this distance will have 168 00:09:45,220 --> 00:09:51,190 to become even smaller, that is to form a really good hydrogen 169 00:09:51,190 --> 00:09:54,550 bond on the order of 2.6, 2.7 Angstrom. 170 00:09:54,550 --> 00:09:59,740 Since this is only one snapshot of the reaction, 171 00:09:59,740 --> 00:10:02,560 we don't have enough information to tell if this is 172 00:10:02,560 --> 00:10:05,500 a key catalytical interaction. 173 00:10:05,500 --> 00:10:07,930 Once acetyl-CoA has reacted with the enzyme, 174 00:10:07,930 --> 00:10:11,260 hence, formed the thioester with the cysteine 111, 175 00:10:11,260 --> 00:10:13,550 we're now ready to proceed with the reaction 176 00:10:13,550 --> 00:10:15,940 and form a carbon-carbon bond. 177 00:10:15,940 --> 00:10:16,990 Let's take a look. 178 00:10:16,990 --> 00:10:19,510 So in the next step of the reaction, 179 00:10:19,510 --> 00:10:23,590 we want to form a carbon-carbon bond between this methyl 180 00:10:23,590 --> 00:10:27,630 group here and this carbon of the carbonyl 181 00:10:27,630 --> 00:10:30,070 of the other substrate. 182 00:10:30,070 --> 00:10:35,530 So first of all, we need to deprotonate this methyl group. 183 00:10:35,530 --> 00:10:37,780 We're going to form an enolate. 184 00:10:37,780 --> 00:10:40,090 Then the enolate is going to attack this carbonyl. 185 00:10:42,670 --> 00:10:45,580 Once again, we look for a suitable base, and we see, 186 00:10:45,580 --> 00:10:50,770 for example, this glutamate 79 may be, in fact, 187 00:10:50,770 --> 00:10:54,850 serving as a general base to deprotonate the methyl group 188 00:10:54,850 --> 00:10:56,800 here and form the enolate. 189 00:10:56,800 --> 00:11:02,050 Once the enolate is formed, it's going to attack this carbonyl, 190 00:11:02,050 --> 00:11:03,580 and this oxygen is going to start 191 00:11:03,580 --> 00:11:08,050 developing a negative charge, which needs to be compensated 192 00:11:08,050 --> 00:11:11,050 by a general acid. 193 00:11:11,050 --> 00:11:14,620 And it looks like this histidine 233 is close enough 194 00:11:14,620 --> 00:11:19,010 to donate a proton and generate a hydroxyl group here. 195 00:11:19,010 --> 00:11:19,510 All right. 196 00:11:19,510 --> 00:11:22,360 So let's try to write a mechanism based 197 00:11:22,360 --> 00:11:24,070 on what we just said. 198 00:11:24,070 --> 00:11:31,720 Here is the thioester between cysteine 111 and acetyl. 199 00:11:31,720 --> 00:11:35,620 And let's show the general base. 200 00:11:35,620 --> 00:11:39,240 It's going to be glutamate 79. 201 00:11:39,240 --> 00:11:44,300 Because it's a base, I'm going to put a negative charge on it. 202 00:11:44,300 --> 00:11:45,960 There you go. 203 00:11:45,960 --> 00:11:50,720 And this will serve to deprotonate 204 00:11:50,720 --> 00:11:57,770 the methyl from the acetyl, moiety here, and generate 205 00:11:57,770 --> 00:11:58,865 an enolate. 206 00:12:01,595 --> 00:12:05,540 So it's picking up a proton, electron move, 207 00:12:05,540 --> 00:12:11,050 and it's going to generate, as before, a negative charge 208 00:12:11,050 --> 00:12:13,060 on the oxygen, which may be stabilized 209 00:12:13,060 --> 00:12:18,430 by that interaction with the [INAUDIBLE] hydrogen 210 00:12:18,430 --> 00:12:20,773 that we just discussed. 211 00:12:20,773 --> 00:12:27,160 Now, let's try this again. 212 00:12:30,980 --> 00:12:33,692 And here is our enolate. 213 00:12:33,692 --> 00:12:35,630 In the next step, the enolate now 214 00:12:35,630 --> 00:12:39,920 is going to attack the carbonyl of the acetoacetyl-CoA 215 00:12:39,920 --> 00:12:42,680 and form the carbon-carbon bond. 216 00:12:47,220 --> 00:12:47,720 All right. 217 00:12:47,720 --> 00:12:52,030 This is the acetoacetyl-CoA, and we 218 00:12:52,030 --> 00:12:54,630 discussed that next to this oxygen 219 00:12:54,630 --> 00:13:04,980 there is our histidine 233, which we're going 220 00:13:04,980 --> 00:13:07,420 to show it as being protonated. 221 00:13:07,420 --> 00:13:12,360 Therefore, the reaction proceeds as follows. 222 00:13:12,360 --> 00:13:14,960 Here we're forming the carbon-carbon bond. 223 00:13:14,960 --> 00:13:19,350 Then the oxygen is going to pick up 224 00:13:19,350 --> 00:13:22,260 the proton from the protonated histidine 225 00:13:22,260 --> 00:13:24,430 and it's going to form a hydroxyl group. 226 00:13:27,920 --> 00:13:32,140 Therefore, we get cysteine 111 is still 227 00:13:32,140 --> 00:13:40,870 attached to what is known the HMG-CoA product. 228 00:13:40,870 --> 00:13:43,520 So this is our carbon. 229 00:13:43,520 --> 00:13:45,700 This is the new carbon-carbon bond. 230 00:13:45,700 --> 00:13:48,080 We have here a methyl group. 231 00:13:48,080 --> 00:13:52,440 We have the new hydroxyl, which were formed here, 232 00:13:52,440 --> 00:13:55,742 and the rest of the molecule. 233 00:13:59,120 --> 00:14:02,210 And our histidine, it was protonated here, 234 00:14:02,210 --> 00:14:06,110 now it's going to be deprotonated. 235 00:14:06,110 --> 00:14:10,590 So we just saw how the carbon-carbon bond is formed 236 00:14:10,590 --> 00:14:12,450 in the course of this reaction. 237 00:14:12,450 --> 00:14:16,240 And now we're left with a six-carbon thioester 238 00:14:16,240 --> 00:14:20,764 with a cysteine 111 in the active side of the enzyme. 239 00:14:20,764 --> 00:14:23,170 Therefore, the last step of the reaction 240 00:14:23,170 --> 00:14:27,790 would involve hydrolysis of this thioester to free the product 241 00:14:27,790 --> 00:14:31,000 and regenerate the system. 242 00:14:31,000 --> 00:14:34,845 So for the last step, we need to hydrolyze this thioester, 243 00:14:34,845 --> 00:14:36,220 and for that, we're going to need 244 00:14:36,220 --> 00:14:39,190 to activate a water molecule. 245 00:14:39,190 --> 00:14:41,740 Once again, we don't know-- 246 00:14:44,400 --> 00:14:46,170 here is the water molecule-- 247 00:14:46,170 --> 00:14:49,080 we don't know what's the general base, the residue 248 00:14:49,080 --> 00:14:52,350 in the active site, which removed this proton from water 249 00:14:52,350 --> 00:14:57,450 to allow it to attack the carbonyl of the thioester. 250 00:14:57,450 --> 00:15:03,360 So we're going to call it a general base attached 251 00:15:03,360 --> 00:15:04,010 to the enzyme. 252 00:15:07,380 --> 00:15:10,250 So this base is going to pick a proton from water, 253 00:15:10,250 --> 00:15:14,680 and then the water is going to add to the carbonyl 254 00:15:14,680 --> 00:15:18,630 and once again generate a tetrahedral intermediate. 255 00:15:24,630 --> 00:15:31,350 Say, this the OH from water, and this 256 00:15:31,350 --> 00:15:33,091 is the rest of the molecule. 257 00:15:39,480 --> 00:15:41,790 Once again, this is a tetrahedral intermediate, very 258 00:15:41,790 --> 00:15:43,950 similar to the one we saw in the first step 259 00:15:43,950 --> 00:15:46,770 when we formed this thioester, and presumably, it's 260 00:15:46,770 --> 00:15:49,800 going to be stabilized in a fairly similar manner. 261 00:15:52,340 --> 00:15:55,430 Let's also show that this base attached to the enzyme 262 00:15:55,430 --> 00:16:02,270 is now protonated, and it would probably 263 00:16:02,270 --> 00:16:04,920 donate this proton to reprotonate the cysteine 264 00:16:04,920 --> 00:16:07,240 and reform that thiol group. 265 00:16:07,240 --> 00:16:15,130 So in the second step, thiol takes the electrons, 266 00:16:15,130 --> 00:16:18,889 which picks up the proton from the general base 267 00:16:18,889 --> 00:16:19,680 in the active site. 268 00:16:23,075 --> 00:16:24,910 So at this point, we're just going 269 00:16:24,910 --> 00:16:30,120 to release the thiol of the cysteine 111, 270 00:16:30,120 --> 00:16:38,490 and the rest of the molecule is exactly our product, 271 00:16:38,490 --> 00:16:40,200 which is the HMG-CoA. 272 00:16:44,090 --> 00:16:47,240 The curved arrow mechanism we just wrote 273 00:16:47,240 --> 00:16:48,960 answers part 1 of the problem. 274 00:16:52,240 --> 00:16:53,920 Second question of this problem is 275 00:16:53,920 --> 00:16:55,570 asking about the stabilization of 276 00:16:55,570 --> 00:16:57,940 the tetrahedral intermediate, which actually we 277 00:16:57,940 --> 00:16:59,710 have just discussed. 278 00:16:59,710 --> 00:17:02,500 But let me reiterate as you guys saw 279 00:17:02,500 --> 00:17:05,121 in the serine protease mechanisms, 280 00:17:05,121 --> 00:17:07,329 whenever we're forming this tetrahedral intermediate, 281 00:17:07,329 --> 00:17:12,190 they tend to be stabilized in an oxyanion hole, which 282 00:17:12,190 --> 00:17:17,500 is basically a structure of the enzyme, which can form hydrogen 283 00:17:17,500 --> 00:17:20,829 bonds with the partial negative charge or full negative charge 284 00:17:20,829 --> 00:17:24,130 that develops in a tetrahedral intermediate. 285 00:17:24,130 --> 00:17:26,470 Presumably, similar structure exists 286 00:17:26,470 --> 00:17:31,960 for this HMG-CoA synthase, but because we're only given one 287 00:17:31,960 --> 00:17:34,720 snapshot, one crystal structure, that is not 288 00:17:34,720 --> 00:17:37,630 sufficient information to say for sure which 289 00:17:37,630 --> 00:17:41,080 are the key interactions to stabilize 290 00:17:41,080 --> 00:17:43,660 these tetrahedral intermediates. 291 00:17:43,660 --> 00:17:46,050 A lot more work, a lot more experimental data 292 00:17:46,050 --> 00:17:51,820 is necessary to figure out which are the key hydrogen bonds 293 00:17:51,820 --> 00:17:55,231 and interactions that stabilize the tetrahedral intermediates. 294 00:17:58,060 --> 00:18:02,200 Question 3 is asking us to review the mechanisms by which 295 00:18:02,200 --> 00:18:05,200 enzymes can achieve their amazing rate 296 00:18:05,200 --> 00:18:08,170 acceleration, which is on the order of 10 to the 6 297 00:18:08,170 --> 00:18:13,470 to 10 to 15 times over the uncatalyzed reaction 298 00:18:13,470 --> 00:18:16,650 and of course, to point out these mechanisms in the context 299 00:18:16,650 --> 00:18:19,260 of the HMG-CoA synthase. 300 00:18:19,260 --> 00:18:22,040 As you have seen over and over in this course, 301 00:18:22,040 --> 00:18:24,360 the three general mechanisms by which 302 00:18:24,360 --> 00:18:29,230 enzymes accelerate reactions are binding energy, 303 00:18:29,230 --> 00:18:31,540 general acid/general base catalysis 304 00:18:31,540 --> 00:18:34,330 and covalent catalysis. 305 00:18:34,330 --> 00:18:36,730 Now let's take a look at our structure 306 00:18:36,730 --> 00:18:39,370 and figure out how each one of these mechanisms 307 00:18:39,370 --> 00:18:41,670 might be operating. 308 00:18:41,670 --> 00:18:44,520 To start off, covalent catalysis, 309 00:18:44,520 --> 00:18:49,900 it's pretty obvious here the acetyl-CoA substrate first 310 00:18:49,900 --> 00:18:54,070 reacts and forms a covalent bond with the cysteine 311 00:18:54,070 --> 00:18:58,080 of the HMG-CoA synthase. 312 00:18:58,080 --> 00:19:01,260 So this covalent attachment to the enzyme 313 00:19:01,260 --> 00:19:05,760 allows this residue to be positioned just right 314 00:19:05,760 --> 00:19:10,870 so that it can react with the other substrate. 315 00:19:10,870 --> 00:19:13,230 Now, general acid/general base catalysis, 316 00:19:13,230 --> 00:19:16,110 we've seen all these residues that participate. 317 00:19:16,110 --> 00:19:18,300 Obviously, in order for this to react, 318 00:19:18,300 --> 00:19:20,920 we need a base to deprotonate the cysteine. 319 00:19:20,920 --> 00:19:24,540 Then we need a base, presumably, this glutamate 79, 320 00:19:24,540 --> 00:19:28,560 to deprotonate the methyl group here to form the enolate. 321 00:19:28,560 --> 00:19:32,100 And we need an acid to stabilize and form 322 00:19:32,100 --> 00:19:38,550 this hydroxyl group that will be developing on this oxygen. 323 00:19:38,550 --> 00:19:42,080 So covalent catalysis and general acid/general base 324 00:19:42,080 --> 00:19:43,980 catalysis, that's pretty obvious. 325 00:19:43,980 --> 00:19:46,360 Now when it comes to binding energy, 326 00:19:46,360 --> 00:19:49,740 this is not something that we can obviously 327 00:19:49,740 --> 00:19:51,330 derive from the structure, but we 328 00:19:51,330 --> 00:19:54,660 can postulate the number of ways in which binding energy 329 00:19:54,660 --> 00:19:56,610 contributes to this reaction. 330 00:19:56,610 --> 00:19:58,380 First of all, both of the substrates 331 00:19:58,380 --> 00:20:00,030 need to bind to the enzyme. 332 00:20:00,030 --> 00:20:03,110 In order to do that, they need to be desolvated, 333 00:20:03,110 --> 00:20:06,720 that is to remove all the water molecules that surround them. 334 00:20:06,720 --> 00:20:11,640 So that by itself requires energy. 335 00:20:11,640 --> 00:20:14,620 The binding energy is also derived 336 00:20:14,620 --> 00:20:18,970 by when we align the substrates in the active site 337 00:20:18,970 --> 00:20:23,530 of the enzyme, we align them so closely the right geometry 338 00:20:23,530 --> 00:20:26,470 and within a few tenths of an Angstrom 339 00:20:26,470 --> 00:20:30,120 so that the right orbitals overlap and allow 340 00:20:30,120 --> 00:20:31,810 the reaction to happen. 341 00:20:31,810 --> 00:20:35,140 So also the ability to align this residue 342 00:20:35,140 --> 00:20:38,250 so closely that also contributes to the binding energy. 343 00:20:40,910 --> 00:20:43,520 And finally, the binding energy also 344 00:20:43,520 --> 00:20:47,210 manifests when we're stabilizing, for example, 345 00:20:47,210 --> 00:20:52,760 the transition state of the reaction 346 00:20:52,760 --> 00:20:55,560 relative to the binding of the substrates. 347 00:20:55,560 --> 00:20:58,090 So if, for example, for our tetrahedral intermediate 348 00:20:58,090 --> 00:21:01,830 that will form here, if that transition state 349 00:21:01,830 --> 00:21:04,860 or the tetrahedral intermediate is stabilized more 350 00:21:04,860 --> 00:21:09,270 than the substrate, then the reaction is accelerated 351 00:21:09,270 --> 00:21:11,420 and proceeds towards that pass. 352 00:21:14,760 --> 00:21:16,890 Part 4 of the problem is asking us 353 00:21:16,890 --> 00:21:20,580 to look up the structure of coenzyme A, or CoA 354 00:21:20,580 --> 00:21:24,380 and then contrast the reactivity of say, acetyl-CoA, 355 00:21:24,380 --> 00:21:27,810 the thioester with CoA, with the reactivity of a thioester 356 00:21:27,810 --> 00:21:31,220 with a much smaller thiol group. 357 00:21:31,220 --> 00:21:34,050 Let's see what the coenzyme A looks like. 358 00:21:34,050 --> 00:21:37,070 Here we have the structure coenzyme A. 359 00:21:37,070 --> 00:21:38,930 The business end of the molecule is 360 00:21:38,930 --> 00:21:41,270 this thiol group, which is attached 361 00:21:41,270 --> 00:21:44,660 to a substantially long arm. 362 00:21:44,660 --> 00:21:47,640 And at the end here, we have, as you can recognize, 363 00:21:47,640 --> 00:21:50,810 two phosphates, the ribose and the base adenine. 364 00:21:50,810 --> 00:21:53,390 This is a nucleotide is ADP. 365 00:21:53,390 --> 00:21:56,330 There's also another phosphate here. 366 00:21:56,330 --> 00:21:59,480 But the business end of the molecule is this thiol group, 367 00:21:59,480 --> 00:22:02,120 and we're asked to contrast whether a thioester 368 00:22:02,120 --> 00:22:06,200 form with coenzyme A would behave similarly to a thioester 369 00:22:06,200 --> 00:22:10,100 form with this right-hand portion of the molecule, which 370 00:22:10,100 --> 00:22:11,490 I wrote here. 371 00:22:11,490 --> 00:22:14,270 So it's a much shorter thiol. 372 00:22:14,270 --> 00:22:18,620 Now, it turns out these thioesters will be very, very 373 00:22:18,620 --> 00:22:21,980 similar because really it's only the thiol moiety that we 374 00:22:21,980 --> 00:22:26,800 need to form thioester. 375 00:22:26,800 --> 00:22:31,020 So those thioesters will behave very, very similarly. 376 00:22:31,020 --> 00:22:33,380 Now, the advantage of having such a long arm 377 00:22:33,380 --> 00:22:37,700 for the coenzyme A is that it provides a way 378 00:22:37,700 --> 00:22:41,360 to insert the substrate, which will attach here, 379 00:22:41,360 --> 00:22:46,220 to say, acetyl-CoA, to insert acetyl-CoA in very deep 380 00:22:46,220 --> 00:22:49,940 into the active site of the enzyme. 381 00:22:49,940 --> 00:22:53,240 Having a long arm to guide the thioester may be important. 382 00:22:53,240 --> 00:22:55,640 As you will see later in the course, 383 00:22:55,640 --> 00:22:59,720 fatty acid synthases, which are these mega dalton complexes, 384 00:22:59,720 --> 00:23:01,730 have multiple active sites. 385 00:23:01,730 --> 00:23:06,640 So the fatty acid attaches a thioester 386 00:23:06,640 --> 00:23:09,920 to coenzyme A allows it to be moved 387 00:23:09,920 --> 00:23:13,670 through different active sites and through a same kind 388 00:23:13,670 --> 00:23:17,450 of chemistry in multiple steps over and over again. 389 00:23:17,450 --> 00:23:20,900 The advantage of having a coenzyme A thioester 390 00:23:20,900 --> 00:23:23,900 is that it may provide some additional binding energy. 391 00:23:23,900 --> 00:23:25,960 That nucleotide portion of coenzyme A 392 00:23:25,960 --> 00:23:30,590 may interact specifically with the enzyme 393 00:23:30,590 --> 00:23:34,640 near the active site, whereas, a much smaller thiol 394 00:23:34,640 --> 00:23:37,190 like the one we just saw will not 395 00:23:37,190 --> 00:23:39,374 have that kind of interaction available. 396 00:23:42,160 --> 00:23:44,590 Since we've been talking about thioesters 397 00:23:44,590 --> 00:23:47,170 throughout this entire problem, the last question 398 00:23:47,170 --> 00:23:48,940 is asking us to rationalize why we're 399 00:23:48,940 --> 00:23:55,630 seeing thioesters in metabolism as opposed to oxygen esters. 400 00:23:55,630 --> 00:23:59,140 Now, the key message you need to remember here 401 00:23:59,140 --> 00:24:03,070 is that the resonance that we observe in oxygen esters 402 00:24:03,070 --> 00:24:06,470 is almost completely absent in thioesters. 403 00:24:06,470 --> 00:24:09,520 And this fact makes the thioesters less stable 404 00:24:09,520 --> 00:24:11,560 and therefore, more reactive. 405 00:24:11,560 --> 00:24:14,110 Not only their carbonyl group behaves more 406 00:24:14,110 --> 00:24:17,290 like a ketone group, but their alpha hydrogens 407 00:24:17,290 --> 00:24:19,060 are more acidic. 408 00:24:19,060 --> 00:24:20,620 Let's take a look. 409 00:24:20,620 --> 00:24:28,430 Here is an oxygen ester that shows 410 00:24:28,430 --> 00:24:32,060 a proton in alpha position. 411 00:24:32,060 --> 00:24:36,320 And as you know, the lone pairs on this oxygen 412 00:24:36,320 --> 00:24:41,260 can conjugate with the carbonyl group 413 00:24:41,260 --> 00:24:45,710 and form certain resonance structures, one of which 414 00:24:45,710 --> 00:24:47,930 being this one. 415 00:24:55,540 --> 00:24:57,216 So the electrons can move like this, 416 00:24:57,216 --> 00:24:59,340 and then we're going to have a negative charge here 417 00:24:59,340 --> 00:25:02,770 and a positive charge here. 418 00:25:02,770 --> 00:25:07,780 And this is possible because the electrons on both oxygens 419 00:25:07,780 --> 00:25:12,010 are found in orbitals of comparable energies. 420 00:25:12,010 --> 00:25:18,640 By contrast, in the case of a thioester, we have a sulfur. 421 00:25:18,640 --> 00:25:23,835 Now, the electrons on the sulfur are 422 00:25:23,835 --> 00:25:27,430 in orbitals that are not in comparable energies 423 00:25:27,430 --> 00:25:31,180 with the ones on the ketyl group and therefore, 424 00:25:31,180 --> 00:25:41,310 this kind of conjugation does not, in fact, happen. 425 00:25:41,310 --> 00:25:44,040 And because this doesn't happen, therefore, the electrons 426 00:25:44,040 --> 00:25:46,740 on the carbonyl stay localized, and that 427 00:25:46,740 --> 00:25:49,170 makes the carbonyl a better reactive site. 428 00:25:49,170 --> 00:25:52,620 So it's a better electrophile to react as nucleophiles 429 00:25:52,620 --> 00:25:55,320 behaves more like a ketone. 430 00:25:55,320 --> 00:25:58,380 For the same reason, when we are to deprotonate 431 00:25:58,380 --> 00:26:02,010 the alpha position on a thioester, 432 00:26:02,010 --> 00:26:04,230 the density on the carbonyl is there 433 00:26:04,230 --> 00:26:07,980 to stabilize the enolate much more so than it 434 00:26:07,980 --> 00:26:10,560 would be for an oxygen ester. 435 00:26:10,560 --> 00:26:13,590 Therefore, the pKa, the acidity constant 436 00:26:13,590 --> 00:26:16,050 for this hydrogen on the thioesters 437 00:26:16,050 --> 00:26:18,960 is close to 18, which is smaller, therefore 438 00:26:18,960 --> 00:26:23,400 more acidic than the pKa of the alpha position in oxygen 439 00:26:23,400 --> 00:26:26,350 esters, which is 22. 440 00:26:26,350 --> 00:26:30,210 This sums up Problem 2 of Problem Set 4. 441 00:26:30,210 --> 00:26:32,280 I hope you now have a much better feel of how 442 00:26:32,280 --> 00:26:35,430 we can use crystal structured data 443 00:26:35,430 --> 00:26:39,020 to propose a reasonable mechanism 444 00:26:39,020 --> 00:26:41,190 for enzyme catalyzed reactions. 445 00:26:41,190 --> 00:26:43,440 Keep in mind that writing mechanisms on paper 446 00:26:43,440 --> 00:26:44,930 is relatively easy. 447 00:26:44,930 --> 00:26:49,740 But to truly confirm that that mechanism is taking place 448 00:26:49,740 --> 00:26:53,340 in real life inside the cells, it 449 00:26:53,340 --> 00:26:58,020 takes a lot more experimental work and evidence.