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 00:00:28,572 --> 00:00:30,580 BOGDAN FEDELES: Greetings, and welcome to 5.07 9 00:00:30,580 --> 00:00:32,229 Biochemistry online. 10 00:00:32,229 --> 00:00:34,170 I'm Dr. Bogdan Fedeles. 11 00:00:34,170 --> 00:00:36,350 Let's metabolize some problems. 12 00:00:36,350 --> 00:00:39,500 Today we're going to be talking about problem one of problem 13 00:00:39,500 --> 00:00:40,670 set seven. 14 00:00:40,670 --> 00:00:43,160 Now this is a problem we are chasing labels 15 00:00:43,160 --> 00:00:45,240 through biochemical pathways. 16 00:00:45,240 --> 00:00:47,270 Although it sounds funny, it's actually 17 00:00:47,270 --> 00:00:49,370 one of the established ways through which we 18 00:00:49,370 --> 00:00:53,300 can test whether the mechanism we propose for these pathways 19 00:00:53,300 --> 00:00:56,810 is in fact consistent with what we observe inside the cells. 20 00:01:00,610 --> 00:01:02,690 In part a of this problem, we're going 21 00:01:02,690 --> 00:01:05,540 to be looking at glycogen, and try to figure out 22 00:01:05,540 --> 00:01:08,210 which carbon's in glycogen end up being 23 00:01:08,210 --> 00:01:12,200 lost as CO2 in the pyruvate dehydrogenase 24 00:01:12,200 --> 00:01:14,420 step of the metabolism. 25 00:01:14,420 --> 00:01:15,890 Here is a shorthand representation 26 00:01:15,890 --> 00:01:19,880 of glycogen. As you know glycogen is a polymer formed 27 00:01:19,880 --> 00:01:21,440 of glucose monomers. 28 00:01:21,440 --> 00:01:24,170 Here is a cyclic form of glucose, 29 00:01:24,170 --> 00:01:26,990 and in glycogen we have these one four linkages as 30 00:01:26,990 --> 00:01:27,950 shown here. 31 00:01:27,950 --> 00:01:30,890 Occasionally we'll have one six linkages, 32 00:01:30,890 --> 00:01:32,420 as in the case for branches. 33 00:01:32,420 --> 00:01:36,480 But for simplicity we're not going to represent them here. 34 00:01:36,480 --> 00:01:39,260 Now we want to figure out which one of these 35 00:01:39,260 --> 00:01:42,249 carbons in glycogen-- 36 00:01:42,249 --> 00:01:43,790 we can label them starting from here. 37 00:01:43,790 --> 00:01:48,800 One, two, three, four, five, and six. 38 00:01:48,800 --> 00:01:50,390 Which one of these carbons is going 39 00:01:50,390 --> 00:01:56,030 to be lost as CO2 in the pyruvate dehydrogenase step. 40 00:01:56,030 --> 00:01:58,280 Now let's take a look at the pyruvate dehydrogenase 41 00:01:58,280 --> 00:02:00,370 reaction. 42 00:02:00,370 --> 00:02:03,080 As you remember, one of the endings of glycolysis 43 00:02:03,080 --> 00:02:05,080 is the pyruvate dehydrogenase reaction, 44 00:02:05,080 --> 00:02:09,910 in which pyruvate loses a CO2 molecule and forms acetyl 45 00:02:09,910 --> 00:02:14,250 CoA which later can enter the TCA cycle. 46 00:02:14,250 --> 00:02:19,290 Now this carbon in the carboxyl or group of pyruvate-- 47 00:02:19,290 --> 00:02:21,155 I'm going to label with a red dot. 48 00:02:21,155 --> 00:02:25,700 This is the carbon that is being lost as CO2. 49 00:02:25,700 --> 00:02:30,620 So we want to find out which of the carbons in our glycogen 50 00:02:30,620 --> 00:02:34,460 molecule ends up being this red dotted carbon that's 51 00:02:34,460 --> 00:02:37,160 being lost as CO2. 52 00:02:37,160 --> 00:02:41,330 To figure this out we have to backtrack from pyruvate 53 00:02:41,330 --> 00:02:43,550 all the way to the beginning of glycolysis 54 00:02:43,550 --> 00:02:47,270 to figure out where this carbon is coming from. 55 00:02:47,270 --> 00:02:52,490 Shown here is a layout of the entire glycolysis pathways 56 00:02:52,490 --> 00:02:55,640 starting from glycogen. Let me walk you through it 57 00:02:55,640 --> 00:02:57,440 very quickly. 58 00:02:57,440 --> 00:03:01,220 Glycogen, shown here-- we've shown only a couple 59 00:03:01,220 --> 00:03:05,040 of monomers attached to the glycogen and protein. 60 00:03:05,040 --> 00:03:08,670 It's going to get hydrolyzed by glycogen-phosphorylase 61 00:03:08,670 --> 00:03:13,400 to form glucose-1-phosphate, which then mutates to glucose 62 00:03:13,400 --> 00:03:15,300 6-phosphate shown here. 63 00:03:15,300 --> 00:03:19,340 And then becomes fructose-6-phosphate, 64 00:03:19,340 --> 00:03:21,230 fructose 1,6-bisphosphate. 65 00:03:21,230 --> 00:03:22,730 Then the aldolase reaction splits it 66 00:03:22,730 --> 00:03:26,225 into dihydroxyacetone-phosphate and glycerol-dihy-3phosphate, 67 00:03:26,225 --> 00:03:27,827 or GAP. 68 00:03:27,827 --> 00:03:30,280 Then the gap dehydrogenase converts it 69 00:03:30,280 --> 00:03:32,540 to 1,3-bisphosphoglycerate. 70 00:03:32,540 --> 00:03:34,490 And then it would go down, 3-phosphoglycerate, 71 00:03:34,490 --> 00:03:37,880 2-phosphoglycerate, phosphopyruvate, and finally, 72 00:03:37,880 --> 00:03:39,080 pyruvate. 73 00:03:39,080 --> 00:03:41,920 And here, I've also written the pyruvate dehydrogenase 74 00:03:41,920 --> 00:03:42,770 reaction. 75 00:03:42,770 --> 00:03:46,460 Pyruvate becomes acetyl-CoA by losing the CO2. 76 00:03:46,460 --> 00:03:49,700 Once again, this carbon that's lost, the CO2, 77 00:03:49,700 --> 00:03:51,500 is the carbon that we want to track. 78 00:03:51,500 --> 00:03:53,690 So we're going to put a red dot on it. 79 00:03:53,690 --> 00:03:57,380 And as we just said, this carbon is the carboxyl group 80 00:03:57,380 --> 00:03:59,660 in the pyruvate. 81 00:03:59,660 --> 00:04:03,080 So the way to solve this problem is basically go backward 82 00:04:03,080 --> 00:04:04,760 through the pathway and figure out 83 00:04:04,760 --> 00:04:08,870 where this carbon is coming from in the original glycogen 84 00:04:08,870 --> 00:04:11,870 molecule. 85 00:04:11,870 --> 00:04:15,110 For these couple of steps, it's pretty clear. 86 00:04:15,110 --> 00:04:19,760 It's going to be the carboxyl in each one of these molecules all 87 00:04:19,760 --> 00:04:22,940 the way to 1,3-bisphosphoglycerate right 88 00:04:22,940 --> 00:04:24,290 there. 89 00:04:24,290 --> 00:04:26,240 So now this 1,3-bisphosphoglycerate is 90 00:04:26,240 --> 00:04:27,560 coming from GAP. 91 00:04:27,560 --> 00:04:32,690 So this carbon is, in fact, the aldehyde carbon in GAP. 92 00:04:32,690 --> 00:04:37,460 Now, as you guys know, trios isomerase interconverts 93 00:04:37,460 --> 00:04:39,410 between dihydroxyacetone phosphate and GAP. 94 00:04:39,410 --> 00:04:43,040 So this carbon in dihydroxyacetone phosphate 95 00:04:43,040 --> 00:04:45,710 is actually this carbon. 96 00:04:45,710 --> 00:04:49,490 Because the phosphate group is going to stay unchanged, 97 00:04:49,490 --> 00:04:52,820 and then the carbonyl group at C2 98 00:04:52,820 --> 00:04:57,270 can interconvert with C1 to form an aldehyde here. 99 00:04:57,270 --> 00:05:01,430 So any of these two carbons, if they were labeled, 100 00:05:01,430 --> 00:05:04,340 they would end up being lost as CO2 101 00:05:04,340 --> 00:05:07,650 in the pyruvate dehydrogenase step. 102 00:05:07,650 --> 00:05:12,290 Now, if we go backwards in the aldolase reaction, 103 00:05:12,290 --> 00:05:17,150 GAP and DHAP, when put together, these carbons 104 00:05:17,150 --> 00:05:20,920 are going to be carbons 3 and 4. 105 00:05:20,920 --> 00:05:23,290 So counting here, 1, 2, 3. 106 00:05:23,290 --> 00:05:27,026 This is a carbon that comes from dihyrdoxyacetone phosphate. 107 00:05:27,026 --> 00:05:28,400 And this is the carbon that comes 108 00:05:28,400 --> 00:05:31,650 from GAP, so carbons 3 and 4. 109 00:05:31,650 --> 00:05:34,250 And obviously, they're going to be staying carbons 3 110 00:05:34,250 --> 00:05:39,410 and 4 all the way back to glucose, 1, 2, 3, and 4 111 00:05:39,410 --> 00:05:40,550 right there. 112 00:05:40,550 --> 00:05:45,770 And also in glucose, 1 phosphate, 113 00:05:45,770 --> 00:05:51,410 and consequently in glycogen as well. 114 00:05:51,410 --> 00:05:56,330 So to answer part one, we can now 115 00:05:56,330 --> 00:06:02,150 write here that carbons 3 and 4 of glycogen 116 00:06:02,150 --> 00:06:04,700 are going to be lost at the pyruvate dehydrogenase 117 00:06:04,700 --> 00:06:06,510 step as carbon dioxide. 118 00:06:10,730 --> 00:06:14,810 Part B of the problem deals with the metabolism of glycerol. 119 00:06:14,810 --> 00:06:17,840 As you know, glycerol is formed by the hydrolysis 120 00:06:17,840 --> 00:06:19,990 of triacylglycerides. 121 00:06:19,990 --> 00:06:21,830 Now, we are asked to trace a label 122 00:06:21,830 --> 00:06:24,860 from the C2 carbon of glycerol all the way 123 00:06:24,860 --> 00:06:27,880 to the first step, in which this carbon is last 124 00:06:27,880 --> 00:06:29,400 as carbon dioxide. 125 00:06:29,400 --> 00:06:32,680 Let's first take a look at the metabolism of glycerol. 126 00:06:32,680 --> 00:06:34,820 As I've shown here, triacylglycerides 127 00:06:34,820 --> 00:06:37,310 can be hydrolyzed to form glycerol, 128 00:06:37,310 --> 00:06:41,460 which is 1, 2, 3 propane triol. 129 00:06:41,460 --> 00:06:45,480 Now, as you know, glycerol is metabolized in two steps. 130 00:06:45,480 --> 00:06:48,050 First, we have a glycerol kinase that's 131 00:06:48,050 --> 00:06:51,440 going to form a glycerol 3-phosphate, as shown here. 132 00:06:51,440 --> 00:06:54,260 And then, we're going to use a dehydrogenase that 133 00:06:54,260 --> 00:06:59,120 uses NAD to oxidize the second carbon of glycerol 134 00:06:59,120 --> 00:07:01,310 to dihydroxyacetone phosphate. 135 00:07:01,310 --> 00:07:06,050 Then it can enter glycolysis very conveniently right here. 136 00:07:06,050 --> 00:07:07,730 And then it's going to continue getting 137 00:07:07,730 --> 00:07:11,770 metabolized towards pyruvate and acetyl-CoA, 138 00:07:11,770 --> 00:07:13,890 as we've seen before. 139 00:07:13,890 --> 00:07:19,470 Now, the second carbon in glycerol is C2 right here. 140 00:07:19,470 --> 00:07:23,165 I'm going to mark it with a blue square. 141 00:07:25,670 --> 00:07:28,940 So this carbon is right here, and it's 142 00:07:28,940 --> 00:07:31,490 going to end up right there. 143 00:07:31,490 --> 00:07:37,230 Now, this second carbon in dihydroxyacetone phosphate 144 00:07:37,230 --> 00:07:40,480 is going to be the second carbon in GAP 145 00:07:40,480 --> 00:07:46,150 and then second carbon here, here, here. 146 00:07:46,150 --> 00:07:48,550 Isn't this fun? 147 00:07:48,550 --> 00:07:50,740 Then the second carbon in pyruvate. 148 00:07:50,740 --> 00:07:53,700 Now, once the pyruvate decarboxylates, 149 00:07:53,700 --> 00:07:58,560 then it's going to be the carbonyl of acetyl-CoA. 150 00:07:58,560 --> 00:08:00,320 It's this carbon right here. 151 00:08:00,320 --> 00:08:04,270 Now, so far, this carbon has followed the metabolic pathway, 152 00:08:04,270 --> 00:08:08,190 but it has not left yet as CO2. 153 00:08:08,190 --> 00:08:09,990 Now, what happens to acetyl-CoA? 154 00:08:09,990 --> 00:08:12,520 It's going to enter the TCA cycle. 155 00:08:12,520 --> 00:08:14,430 Now let's take a look at the TCA cycle. 156 00:08:14,430 --> 00:08:16,970 As we just said, we're looking now 157 00:08:16,970 --> 00:08:20,740 at the carbon, the first carbon in acetyl-CoA 158 00:08:20,740 --> 00:08:24,070 here, the carbon that has the carbonyl group on. 159 00:08:24,070 --> 00:08:27,820 So as I've shown here in the to TCA cycle, 160 00:08:27,820 --> 00:08:32,010 the two carbons in acetyl-CoA are marked with this red line. 161 00:08:32,010 --> 00:08:35,870 And they will enter and combine as oxaloacetate 162 00:08:35,870 --> 00:08:37,600 to form citrate. 163 00:08:37,600 --> 00:08:41,440 Then citrate isomerizes to isocitrate. 164 00:08:41,440 --> 00:08:43,780 Then we're going to lose a CO2 molecule, which 165 00:08:43,780 --> 00:08:47,140 is this middle guy, to form alpha-ketoglutarate. 166 00:08:47,140 --> 00:08:49,330 But notice, the two carbons from acetyl-CoA 167 00:08:49,330 --> 00:08:51,380 are still in the molecule. 168 00:08:51,380 --> 00:08:54,760 Then we're going to lose another CO2 with this bottom one 169 00:08:54,760 --> 00:08:56,470 to form succinyl-CoA. 170 00:08:56,470 --> 00:08:59,290 But once again, the two carbons that came from acetyl-CoA 171 00:08:59,290 --> 00:09:00,640 are still here. 172 00:09:00,640 --> 00:09:05,350 So in the first TCA cycle, none of these CO2 173 00:09:05,350 --> 00:09:10,000 will contain the label that came from the glycerol. 174 00:09:10,000 --> 00:09:12,880 Now, as we go through the TCA cycle, 175 00:09:12,880 --> 00:09:15,760 we reach this step where it's succinate. 176 00:09:15,760 --> 00:09:18,910 Now here, I stopped putting this red mark, 177 00:09:18,910 --> 00:09:21,730 because succinate is a symmetric molecule. 178 00:09:21,730 --> 00:09:25,120 So therefore, if these two carbons 179 00:09:25,120 --> 00:09:29,830 were coming from acetyl-CoA, at this point they will scramble. 180 00:09:29,830 --> 00:09:32,140 So we won't be able to tell whether it's these two 181 00:09:32,140 --> 00:09:35,200 carbons or these two carbons. 182 00:09:35,200 --> 00:09:38,370 Now, let me backtrack and put in the labels. 183 00:09:38,370 --> 00:09:41,680 So acetyl-CoA, we have this carbon 184 00:09:41,680 --> 00:09:43,760 came from the C2 of glycerol. 185 00:09:43,760 --> 00:09:49,725 So we'll find it in this carbon, this carbon, this carbon, 186 00:09:49,725 --> 00:09:51,670 this carbon. 187 00:09:51,670 --> 00:09:53,920 Now, we get to the succinate step, 188 00:09:53,920 --> 00:09:57,190 and we said, well, it was here. 189 00:09:57,190 --> 00:10:00,370 This would be the carbon that corresponds to succinyl-CoA. 190 00:10:00,370 --> 00:10:02,140 But because this molecule's symmetric, 191 00:10:02,140 --> 00:10:04,180 by the time we get to the malate step 192 00:10:04,180 --> 00:10:08,200 to add this hydroxyl group, it's going to be to the carbon 193 00:10:08,200 --> 00:10:12,100 next to the label or the carbon further from the label. 194 00:10:12,100 --> 00:10:17,320 Now, because the molecule is symmetric, we can't-- 195 00:10:17,320 --> 00:10:21,612 the fumarase enzyme cannot tell which carbon was labeled 196 00:10:21,612 --> 00:10:22,320 and which wasn't. 197 00:10:22,320 --> 00:10:24,377 Therefore, malate is going to be-- 198 00:10:24,377 --> 00:10:25,960 half of the molecules is going to have 199 00:10:25,960 --> 00:10:28,330 the label on this carbon, and half 200 00:10:28,330 --> 00:10:30,220 is going to have the label on this carbon. 201 00:10:30,220 --> 00:10:34,360 So I'm going to write like 1/2 square and 1/2 square. 202 00:10:34,360 --> 00:10:37,120 Similarly, when we get to oxaloacetate, 203 00:10:37,120 --> 00:10:41,770 the label is distributed 1/2 on one carboxyl group, 204 00:10:41,770 --> 00:10:45,840 and the other 1/2 is going to be on the other carboxyl group. 205 00:10:48,890 --> 00:10:51,940 So we've gone through the TCA cycle once, 206 00:10:51,940 --> 00:10:57,080 and we have not lost the carbon that came from the glycerol. 207 00:10:57,080 --> 00:11:00,280 But look what happens when we continue the TCA 208 00:11:00,280 --> 00:11:02,470 cycle a second time. 209 00:11:02,470 --> 00:11:04,930 So now let's say we combine with an acetyl-CoA that 210 00:11:04,930 --> 00:11:06,670 doesn't have any label at all. 211 00:11:06,670 --> 00:11:09,310 Now, these two carboxyl groups are 212 00:11:09,310 --> 00:11:12,970 going to be these two carboxyl groups in citrate. 213 00:11:12,970 --> 00:11:16,510 And as we discussed, both of these two groups 214 00:11:16,510 --> 00:11:19,310 are going to be lost as CO2 in these two steps. 215 00:11:19,310 --> 00:11:22,930 First is the middle carboxyl group that's being lost here. 216 00:11:22,930 --> 00:11:25,510 And this other carboxyl is going to be 217 00:11:25,510 --> 00:11:29,380 lost at the alpha-ketoglutarate dehydrogenase step. 218 00:11:29,380 --> 00:11:33,050 So the second time we go through the TCA cycle, 219 00:11:33,050 --> 00:11:37,660 we lose half of the label at the isocitrate dehydrogenase step 220 00:11:37,660 --> 00:11:40,040 and half of the label at the alpha-ketoglutarate 221 00:11:40,040 --> 00:11:42,910 dehydrogenase step. 222 00:11:42,910 --> 00:11:44,830 So I'm going to circle these. 223 00:11:44,830 --> 00:11:48,230 So to answer part B, the C2 carbon of glycerol 224 00:11:48,230 --> 00:11:51,550 is going to be lost as CO2 in the TCA cycle. 225 00:11:51,550 --> 00:11:56,650 But we have to go once through the cycle first, through which 226 00:11:56,650 --> 00:11:58,330 none of the label will be lost. 227 00:11:58,330 --> 00:12:00,580 And then the second time, first in 228 00:12:00,580 --> 00:12:02,410 the isocitrate dehydrogenase, then 229 00:12:02,410 --> 00:12:04,180 at the alpha-ketoglutarate dehydrogenase, 230 00:12:04,180 --> 00:12:06,820 we're going to be losing the CO2 that came 231 00:12:06,820 --> 00:12:08,905 from the C2 carbon of glycerol. 232 00:12:12,490 --> 00:12:14,590 As you might imagine, glycerol can also 233 00:12:14,590 --> 00:12:16,450 be used to produce energy. 234 00:12:16,450 --> 00:12:19,390 In fact, some bacteria can grow on glycerol 235 00:12:19,390 --> 00:12:21,970 using no other carbon source. 236 00:12:21,970 --> 00:12:23,680 Now, in part C of the problem, we're 237 00:12:23,680 --> 00:12:26,290 going to explore how much energy we can get 238 00:12:26,290 --> 00:12:28,290 from one molecule of glycerol. 239 00:12:28,290 --> 00:12:31,990 Let's first review the metabolism of glycerol. 240 00:12:31,990 --> 00:12:35,310 As we just discussed, glycerol enters metabolism 241 00:12:35,310 --> 00:12:38,830 with glycerol kinase, which then in two steps 242 00:12:38,830 --> 00:12:40,990 becomes dihydroxyacetone phosphate, which 243 00:12:40,990 --> 00:12:44,620 can enter glycolysis all the way to pyruvate, 244 00:12:44,620 --> 00:12:46,240 and then acetyl-CoA. 245 00:12:46,240 --> 00:12:51,340 Here, the pyruvate dehydrogenase allows us to lose one CO2, 246 00:12:51,340 --> 00:12:54,820 and then acetyl-CoA will enter the TCA 247 00:12:54,820 --> 00:12:58,540 cycle, where within one cycle, we're 248 00:12:58,540 --> 00:13:00,610 going to lose two more CO2s. 249 00:13:00,610 --> 00:13:03,760 So that's a total of three carbons that we lose. 250 00:13:03,760 --> 00:13:05,410 And that's exactly how many carbons 251 00:13:05,410 --> 00:13:07,210 we have in the glycerol. 252 00:13:07,210 --> 00:13:09,220 Now, what we need to keep track in order 253 00:13:09,220 --> 00:13:10,750 to evaluate how much energy we get 254 00:13:10,750 --> 00:13:15,070 from one molecule of glycerol is, whenever we need to, 255 00:13:15,070 --> 00:13:16,750 use ATP. 256 00:13:16,750 --> 00:13:19,090 For example, we need to put in energy, 257 00:13:19,090 --> 00:13:23,350 or whenever we generate NADH or FADH2 molecules, 258 00:13:23,350 --> 00:13:25,810 then we can then take to the electron transport chain 259 00:13:25,810 --> 00:13:28,610 and generate ATPs out of them. 260 00:13:28,610 --> 00:13:31,930 So I've put together a list of the steps in the pathway 261 00:13:31,930 --> 00:13:35,170 where the energy balance is affected, 262 00:13:35,170 --> 00:13:37,870 either we need to use energy or we are generating energy 263 00:13:37,870 --> 00:13:41,650 in the form or ATP or in the form of redox cofactors, 264 00:13:41,650 --> 00:13:45,490 such as NAD and FAD. 265 00:13:45,490 --> 00:13:47,860 So the first step, glycerol kinase, 266 00:13:47,860 --> 00:13:51,020 we're going to need to spend one molecule of ATP. 267 00:13:51,020 --> 00:13:55,040 So I'm going to put a minus 1 here for ATP equivalents. 268 00:13:55,040 --> 00:13:59,350 Now, in the glycerol 3-phosphate dehydrogenase step, 269 00:13:59,350 --> 00:14:02,950 which is shown right here, we are generating 270 00:14:02,950 --> 00:14:06,250 one molecule of NADH. 271 00:14:06,250 --> 00:14:09,130 So plus 1 NADH. 272 00:14:09,130 --> 00:14:12,130 Now, for the purpose of this problem, 273 00:14:12,130 --> 00:14:14,920 we're going to use the convention the 1 NADH is 274 00:14:14,920 --> 00:14:19,750 worth about 3 ATP equivalents. 275 00:14:19,750 --> 00:14:22,310 Now, later on in the pathway, we're 276 00:14:22,310 --> 00:14:26,790 going to get to the GAPDH step where we generate one more NADH 277 00:14:26,790 --> 00:14:27,900 molecule. 278 00:14:27,900 --> 00:14:32,910 So GAPDH, another NADH molecule. 279 00:14:32,910 --> 00:14:34,675 That's equivalent to 3 ATPs. 280 00:14:37,240 --> 00:14:42,000 So moving ahead, we have the phosphoglycerate kinase step, 281 00:14:42,000 --> 00:14:44,400 where we generate 1 ATP, and then we 282 00:14:44,400 --> 00:14:48,520 have the pyruvate kinase step where we generate 1 more ATP. 283 00:14:51,480 --> 00:14:52,920 I have these written here. 284 00:14:52,920 --> 00:14:58,500 So plus 1 ATP, and plus 1 ATP. 285 00:14:58,500 --> 00:15:02,430 Now, finally, we know that pyruvate, it's 286 00:15:02,430 --> 00:15:06,240 going to be decarboxylated by the pyruvate dehydrogenase, 287 00:15:06,240 --> 00:15:08,430 and here, too, we're generating 1 NADH. 288 00:15:11,400 --> 00:15:19,050 So plus 1 NADH, that's going to be equivalent to 3 ATPs. 289 00:15:19,050 --> 00:15:21,510 And finally, the TCA cycle. 290 00:15:21,510 --> 00:15:25,460 Now, we have one molecule of acetyl-CoA 291 00:15:25,460 --> 00:15:26,820 that enters the TCA cycle. 292 00:15:31,001 --> 00:15:33,250 As you guys know for every molecule of acetyl-CoA that 293 00:15:33,250 --> 00:15:34,790 enters the TCA cycle, we're going 294 00:15:34,790 --> 00:15:40,090 to be generating 1, 2, 3 NADHs. 295 00:15:40,090 --> 00:15:45,260 1 FADH2, and 1 GTP. 296 00:15:45,260 --> 00:15:54,470 So the tally is 3 NADH, 1 FADH2, and 1 GDP. 297 00:15:54,470 --> 00:15:57,680 Now, GDP is equivalent to an ATP. 298 00:15:57,680 --> 00:16:02,110 FADH2 counts as 2 ATPs, and NADH counts as 3 ATPs. 299 00:16:02,110 --> 00:16:07,370 So that's a grand total of 12 ATPs. 300 00:16:07,370 --> 00:16:09,890 So putting all of this together from one 301 00:16:09,890 --> 00:16:13,760 molecule of glycerol, when fully metabolized to CO2, 302 00:16:13,760 --> 00:16:17,972 we get 22 ATP equivalents. 303 00:16:17,972 --> 00:16:19,430 So that's the final answer for part 304 00:16:19,430 --> 00:16:21,540 C. From one molecule of glycerol, 305 00:16:21,540 --> 00:16:24,874 we get about 22 ATP equivalents. 306 00:16:27,840 --> 00:16:31,830 In part D of the problem, we're tracing the same labels 307 00:16:31,830 --> 00:16:35,970 we had in part A, but instead of tracing them to CO2, 308 00:16:35,970 --> 00:16:39,930 we're going to trace them to the amino acid alanine. 309 00:16:39,930 --> 00:16:42,630 As you know, one way to produce alanine 310 00:16:42,630 --> 00:16:45,750 is by transamination from pyruvate. 311 00:16:45,750 --> 00:16:48,120 Since we already tracked the label to pyrate, 312 00:16:48,120 --> 00:16:51,930 we need to know how do we convert pyruvate into alanine. 313 00:16:51,930 --> 00:16:54,740 Let's take a look at that reaction. 314 00:16:54,740 --> 00:16:57,810 As you know, alpha-keto acids, such as pyruvate, 315 00:16:57,810 --> 00:17:00,630 can be converted into amino acids 316 00:17:00,630 --> 00:17:02,910 by a transamination reaction. 317 00:17:02,910 --> 00:17:05,880 Here, we're going to use another amino acid 318 00:17:05,880 --> 00:17:09,810 to donate the amino group to the alpha-keto acid pyruvate 319 00:17:09,810 --> 00:17:12,906 to form alanine. 320 00:17:12,906 --> 00:17:14,530 Now, all these transamination reactions 321 00:17:14,530 --> 00:17:19,420 are catalyzed by PLP or pyridoxal 5-phosphate, which 322 00:17:19,420 --> 00:17:21,500 is a cofactor derived from vitamin B6. 323 00:17:24,099 --> 00:17:27,880 So when it's left to right, in this transformation 324 00:17:27,880 --> 00:17:30,400 is the other pair of amino acid-- 325 00:17:30,400 --> 00:17:31,270 alpha-keto acid. 326 00:17:31,270 --> 00:17:34,690 So basically, where is this amino group coming from? 327 00:17:34,690 --> 00:17:38,470 Typically for most transaminases the other pair 328 00:17:38,470 --> 00:17:44,020 is glutamate alpha-ketoglutarate. 329 00:17:44,020 --> 00:17:46,010 So I'm going to have glutamate is going 330 00:17:46,010 --> 00:17:48,710 to donate the amino group, and in the process is 331 00:17:48,710 --> 00:17:52,520 going to become an alpha-keto acid alpha-ketoglutarate. 332 00:17:52,520 --> 00:17:57,140 So in this way, pyruvate becomes alanine. 333 00:17:57,140 --> 00:18:04,120 Now, we were tracking the label from this carbon, so the carbon 334 00:18:04,120 --> 00:18:07,180 that will be lost as CO2 in the pyruvate dehydrogenase 335 00:18:07,180 --> 00:18:08,020 reaction. 336 00:18:08,020 --> 00:18:11,290 So that is this carbon right here in pyruvate. 337 00:18:11,290 --> 00:18:13,980 So in the transamination reaction, 338 00:18:13,980 --> 00:18:19,280 this carbon becomes the carboxyl carbon of alanine. 339 00:18:19,280 --> 00:18:22,730 So if you were to start with a glycogen that 340 00:18:22,730 --> 00:18:25,270 was labeled at the 3 or 4-- 341 00:18:25,270 --> 00:18:28,550 the carbons 3 and 4, that label would 342 00:18:28,550 --> 00:18:32,570 be lost as CO2 in the pyruvate dehydrogenase reaction 343 00:18:32,570 --> 00:18:35,420 that we saw in part A. But also, that label 344 00:18:35,420 --> 00:18:38,660 would be incorporated in alanine at the carboxyl group 345 00:18:38,660 --> 00:18:39,580 of this amino acid. 346 00:18:43,060 --> 00:18:44,950 Parts E and F of the problem deal 347 00:18:44,950 --> 00:18:48,930 with tracing labels to the amino acids glutamate and aspartate. 348 00:18:48,930 --> 00:18:50,650 Now, both of these amino acids have 349 00:18:50,650 --> 00:18:54,920 their corresponding alpha-keto acids as part of the TCA cycle. 350 00:18:54,920 --> 00:18:57,470 So let's first take a look at this TCA cycle. 351 00:18:57,470 --> 00:18:59,470 Here, we have the TCA cycle where 352 00:18:59,470 --> 00:19:03,400 I highlighted alpha-ketoglutarate going 353 00:19:03,400 --> 00:19:06,870 into glutamate through a transamination reaction. 354 00:19:06,870 --> 00:19:09,100 Alpha-ketoglutarate and alpha-keto acid 355 00:19:09,100 --> 00:19:12,310 can undergo a PLP-catalyzed transamination 356 00:19:12,310 --> 00:19:13,780 to form glutamate. 357 00:19:13,780 --> 00:19:17,140 And similarly, oxaloacetate, another alpha-keto acid 358 00:19:17,140 --> 00:19:21,290 can transaminate to form aspartate. 359 00:19:21,290 --> 00:19:23,090 Now, in part B of the problem, we 360 00:19:23,090 --> 00:19:25,100 were looking at the label present 361 00:19:25,100 --> 00:19:28,300 at carbon 1 in acetyl-CoA. 362 00:19:28,300 --> 00:19:31,010 And we said that this label will stay 363 00:19:31,010 --> 00:19:36,900 inside the intermediates of the TCA cycle for the whole round. 364 00:19:36,900 --> 00:19:40,670 Now, when this label gets the alpha-ketoglutarate, 365 00:19:40,670 --> 00:19:42,500 it's going to be on this carboxyl group. 366 00:19:42,500 --> 00:19:44,510 So in the transamination reaction, 367 00:19:44,510 --> 00:19:46,850 the label is going to end completely 368 00:19:46,850 --> 00:19:52,880 on the furthermost carboxyl in glutamate. 369 00:19:52,880 --> 00:19:55,790 So that takes care of part E. 370 00:19:55,790 --> 00:19:59,840 Now, if we continue chasing this label, 371 00:19:59,840 --> 00:20:02,480 once we get to the succinate, the label is going to split 372 00:20:02,480 --> 00:20:07,130 half and half between these two carboxyl groups, 373 00:20:07,130 --> 00:20:11,124 because we cannot tell which one-- 374 00:20:11,124 --> 00:20:13,040 because the molecule is going to be symmetric. 375 00:20:13,040 --> 00:20:19,880 Similarly, for fumarate and in malate as well. 376 00:20:23,710 --> 00:20:28,210 So in oxaloacetate, the label is going 377 00:20:28,210 --> 00:20:30,785 to be on both of the carboxyl group. 378 00:20:30,785 --> 00:20:32,410 Half of the molecules will have it one. 379 00:20:32,410 --> 00:20:34,540 Half of the molecules will have it on the other. 380 00:20:34,540 --> 00:20:36,250 So therefore, the aspartate is going 381 00:20:36,250 --> 00:20:40,720 to mirror that label distribution, 1/2 label on one 382 00:20:40,720 --> 00:20:44,590 carboxyl, 1/2 label on the other carboxyl. 383 00:20:44,590 --> 00:20:47,470 So that should answer the final part of this problem. 384 00:20:47,470 --> 00:20:49,000 Now, I hope this problem gave you 385 00:20:49,000 --> 00:20:52,330 a better understanding of what it means to chase labels 386 00:20:52,330 --> 00:20:56,170 through biochemical pathways, and also that chasing labels 387 00:20:56,170 --> 00:20:59,170 can help us better understand the mechanisms 388 00:20:59,170 --> 00:21:01,020 of biochemical transformations.