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 --> 00:00:20,750 at ocw.mit.edu. 8 00:00:20,750 --> 00:00:23,870 BOGDAN FEDELES: Hi, I'm Dr. Bogdan Fedeles. 9 00:00:23,870 --> 00:00:25,940 I'm a research associate at MIT doing research 10 00:00:25,940 --> 00:00:28,040 with Professor John Essigman. 11 00:00:28,040 --> 00:00:29,990 I love biochemistry, and I'm going 12 00:00:29,990 --> 00:00:33,200 to help you solve some problems today. 13 00:00:33,200 --> 00:00:37,580 Let's work together through question 5 or problem set 5. 14 00:00:37,580 --> 00:00:41,420 This is an excellent question about sugar biochemistry. 15 00:00:41,420 --> 00:00:43,130 Specifically, we're going to find out 16 00:00:43,130 --> 00:00:46,370 how mannose an isomer of glucose gets 17 00:00:46,370 --> 00:00:49,640 to enter metabolism by being converted 18 00:00:49,640 --> 00:00:51,230 into fructose 6-phosphate. 19 00:00:54,040 --> 00:00:58,090 Now mannose is a common sugar that's 20 00:00:58,090 --> 00:01:02,200 found in polysaccharides and glycoproteins. 21 00:01:02,200 --> 00:01:06,550 Here is the transformation that converts mannose, shown here, 22 00:01:06,550 --> 00:01:09,430 to mannose 6-phosphate in first step. 23 00:01:09,430 --> 00:01:11,950 And then in the second step, to fructose 6-phosphate. 24 00:01:11,950 --> 00:01:16,420 So our goal here will be to figure out the mechanism 25 00:01:16,420 --> 00:01:19,000 for these two transformations. 26 00:01:19,000 --> 00:01:22,210 Let's take a look at the first transformation. 27 00:01:22,210 --> 00:01:25,710 We're converting mannose to mannose 6-phosphate. 28 00:01:25,710 --> 00:01:28,210 Now, as you guys have encountered in biochemistry, 29 00:01:28,210 --> 00:01:31,690 adding a phosphate group is a ubiquitous transformation 30 00:01:31,690 --> 00:01:35,440 and it's catalyzed by proteins called kinases. 31 00:01:35,440 --> 00:01:39,010 Now, kinases typically use ATP, the energy currency 32 00:01:39,010 --> 00:01:42,350 of the cell, as a phosphate donor, 33 00:01:42,350 --> 00:01:45,860 and transfer the furthest most phosphate, 34 00:01:45,860 --> 00:01:48,610 so-called gamma phosphate, onto the substrate, 35 00:01:48,610 --> 00:01:50,852 and leaving behind ADP. 36 00:01:50,852 --> 00:01:52,810 A similar transformation will be happening here 37 00:01:52,810 --> 00:01:54,430 in part one of the problem, where 38 00:01:54,430 --> 00:01:57,820 mannose is going to be converted to mannose 6-phosphate 39 00:01:57,820 --> 00:02:00,944 by a kinase. 40 00:02:00,944 --> 00:02:02,360 And the source of the phosphate is 41 00:02:02,360 --> 00:02:08,389 going to be ATP, which, in the process, will yield ADP. 42 00:02:08,389 --> 00:02:13,310 Now, in order for the kinase to work, to react with ATP, 43 00:02:13,310 --> 00:02:15,650 we also need magnesium. 44 00:02:15,650 --> 00:02:18,740 So a salt of magnesium with ATP is actually 45 00:02:18,740 --> 00:02:21,390 essential for this reaction to work. 46 00:02:21,390 --> 00:02:23,960 Now, let's take a look at the structure of ATP 47 00:02:23,960 --> 00:02:26,180 to gain a little bit more mechanistic insight. 48 00:02:35,220 --> 00:02:39,390 As you know, ATP, or adenosine triphosphate, 49 00:02:39,390 --> 00:02:42,270 has these three phosphate groups attached 50 00:02:42,270 --> 00:02:46,470 to a 5-carbon sugar called ribose 51 00:02:46,470 --> 00:02:52,470 and a base, adenine, which we're not going to draw out here. 52 00:02:52,470 --> 00:02:56,460 Now, the phosphates in ATP are typically 53 00:02:56,460 --> 00:03:03,250 labeled alpha, the one closest to the sugar, beta, and gamma. 54 00:03:03,250 --> 00:03:05,980 Now, in order for ATP to react, notice 55 00:03:05,980 --> 00:03:08,740 these four negative charges on the phosphates. 56 00:03:08,740 --> 00:03:12,160 They need to be neutralized in order for the molecule 57 00:03:12,160 --> 00:03:13,360 to become reactive. 58 00:03:13,360 --> 00:03:16,720 So this is what magnesium is doing. 59 00:03:16,720 --> 00:03:19,150 So at least early on in the reaction, 60 00:03:19,150 --> 00:03:23,230 magnesium is going to neutralize two of these four 61 00:03:23,230 --> 00:03:26,800 negative charges, while the other two negative charges will 62 00:03:26,800 --> 00:03:30,280 be neutralized by positively charged amino acids 63 00:03:30,280 --> 00:03:33,140 in the active site of the kinase. 64 00:03:33,140 --> 00:03:35,020 Once all the charges are neutralized, 65 00:03:35,020 --> 00:03:37,630 then the phosphorus at the gamma position 66 00:03:37,630 --> 00:03:40,990 will become available to be attacked by a nucleophile. 67 00:03:40,990 --> 00:03:46,600 In our case, the 6 hydroxyl of mannose. 68 00:03:46,600 --> 00:03:50,410 So therefore, the reaction proceeds as follows. 69 00:03:50,410 --> 00:03:55,930 A base in the active site will activate our hydroxyl, 70 00:03:55,930 --> 00:04:01,120 which then can attack the gamma position of the phosphate. 71 00:04:01,120 --> 00:04:05,020 And then the phosphate gets transferred to the mannose, 72 00:04:05,020 --> 00:04:07,940 and it's going to leave behind ADP. 73 00:04:07,940 --> 00:04:11,480 Now, notice during this reaction, magnesium, 74 00:04:11,480 --> 00:04:13,770 which was coordinating two of these negative charges, 75 00:04:13,770 --> 00:04:18,649 will probably move to coordinate these two charges or the newly 76 00:04:18,649 --> 00:04:21,320 formed charge here in ADP. 77 00:04:21,320 --> 00:04:26,350 And that would also help stabilize the ADP 78 00:04:26,350 --> 00:04:28,930 into the active site of the kinase. 79 00:04:28,930 --> 00:04:31,630 This mechanistic insight basically answers part 80 00:04:31,630 --> 00:04:32,440 one of the problem. 81 00:04:34,950 --> 00:04:37,050 To talk about part two of the problem, 82 00:04:37,050 --> 00:04:39,965 we need to remember something fundamental about sugar 83 00:04:39,965 --> 00:04:43,920 biochemistry, namely that 5- and 6-carbon sugars 84 00:04:43,920 --> 00:04:49,557 exist in equilibrium between a linear form and cyclic forms. 85 00:04:49,557 --> 00:04:51,390 Now, sugars that you've encountered already, 86 00:04:51,390 --> 00:04:53,730 like glucose and fructose, they are 87 00:04:53,730 --> 00:04:56,940 in equilibrium between linear and cyclic forms. 88 00:04:56,940 --> 00:05:00,340 So let's take a look at them first. 89 00:05:00,340 --> 00:05:03,010 This is glucose. 90 00:05:03,010 --> 00:05:05,860 As you know, it's an aldehyde that 91 00:05:05,860 --> 00:05:09,250 has hydroxyl group on all the other carbons 92 00:05:09,250 --> 00:05:10,800 in this stereochemistry. 93 00:05:10,800 --> 00:05:14,680 Now, this is a linear form of glucose, and as an aldehyde, 94 00:05:14,680 --> 00:05:16,720 it can react with good nucleophiles, 95 00:05:16,720 --> 00:05:21,550 like these hydroxyls, to form cyclic hemiacetals. 96 00:05:21,550 --> 00:05:25,960 If the reaction occurs with a hydroxyl position four, 97 00:05:25,960 --> 00:05:28,810 we're going to close a five-membered ring hemiacetal, 98 00:05:28,810 --> 00:05:30,350 which is shown here. 99 00:05:30,350 --> 00:05:33,010 These kind of structures are called furanoses. 100 00:05:33,010 --> 00:05:36,760 So this is the glucofuranose. 101 00:05:36,760 --> 00:05:41,520 Now if, instead, we react with a 5-carbon, 102 00:05:41,520 --> 00:05:44,270 then we're going to close a six-membered ring, which 103 00:05:44,270 --> 00:05:45,980 is called a pyranose. 104 00:05:45,980 --> 00:05:49,270 This is glucopyranose. 105 00:05:49,270 --> 00:05:51,940 Now, while the sugars are in equilibrium 106 00:05:51,940 --> 00:05:53,860 between the linear and the cyclic forms, 107 00:05:53,860 --> 00:05:56,950 the cyclic forms tend to be predominant. 108 00:05:56,950 --> 00:06:03,060 So in equilibrium, it will be primarily 99% cyclic form 109 00:06:03,060 --> 00:06:05,380 and only about 1% linear form. 110 00:06:05,380 --> 00:06:07,650 Nevertheless, the presence of the linear form 111 00:06:07,650 --> 00:06:11,010 allows the sugars to undergo some interesting 112 00:06:11,010 --> 00:06:15,930 transformations, one of which we are exploring in this question. 113 00:06:15,930 --> 00:06:18,930 Similarly, here is fructose. 114 00:06:18,930 --> 00:06:20,970 Fructose has the carbonyl at the 2 position. 115 00:06:20,970 --> 00:06:23,560 It's a keto group, and then hydroxyl groups 116 00:06:23,560 --> 00:06:26,070 on all the other carbons. 117 00:06:26,070 --> 00:06:32,110 And just like glucose, it can form cyclic hemiacetals. 118 00:06:32,110 --> 00:06:35,500 So if the reaction happens between the carbonyl 119 00:06:35,500 --> 00:06:37,980 and the hydroxyl at position 5, it's 120 00:06:37,980 --> 00:06:40,880 going to close a five-member ring shown here. 121 00:06:40,880 --> 00:06:43,330 This is too a furanose. 122 00:06:43,330 --> 00:06:46,060 This will be fructofuranose. 123 00:06:46,060 --> 00:06:50,260 If the reaction happens with the hydroxyl at the position 6, 124 00:06:50,260 --> 00:06:54,080 then we're going to be closing a six-member ring shown here. 125 00:06:54,080 --> 00:06:57,800 This is fructopyranose. 126 00:06:57,800 --> 00:07:01,040 Once again, just like for glucose, the equilibrium, 127 00:07:01,040 --> 00:07:04,510 it's strongly shifted towards the cyclic forms. 128 00:07:04,510 --> 00:07:07,490 Nevertheless, there is enough of the linear form 129 00:07:07,490 --> 00:07:10,310 to allow certain kind of chemistry to happen. 130 00:07:10,310 --> 00:07:16,310 Now, going back to solving part two of our problem, that 131 00:07:16,310 --> 00:07:18,800 is converting from mannose 6-phosphate, which 132 00:07:18,800 --> 00:07:23,570 is a sugar with a six-member ring cyclic structure, 133 00:07:23,570 --> 00:07:26,960 to fructose 6-phosphate, a sugar with a five-member 134 00:07:26,960 --> 00:07:28,478 ring cyclic structure. 135 00:07:31,306 --> 00:07:32,930 Keeping in mind what we just discussed, 136 00:07:32,930 --> 00:07:35,450 that sugars, they're in equilibrium 137 00:07:35,450 --> 00:07:38,730 between these cyclic structures and their linear structures, 138 00:07:38,730 --> 00:07:43,680 linear forms, one good place to start 139 00:07:43,680 --> 00:07:45,960 to figure out how this transformation will happen 140 00:07:45,960 --> 00:07:51,780 is to write the linear forms of the two molecules here. 141 00:07:51,780 --> 00:07:53,880 Now, for the mannose 6-phosphate, 142 00:07:53,880 --> 00:07:56,910 we notice we have a hemiacetal functionality at carbon 143 00:07:56,910 --> 00:07:59,700 1, which is attached to both the hydroxyl group 144 00:07:59,700 --> 00:08:03,420 and another oxygen. So this carbon 145 00:08:03,420 --> 00:08:09,010 will form an aldehyde in the linear form. 146 00:08:09,010 --> 00:08:12,503 So let's write the linear form of mannose 6-phosphate. 147 00:08:22,860 --> 00:08:25,700 All right, so notice at carbon 1, 148 00:08:25,700 --> 00:08:27,610 we're going to have an aldehyde. 149 00:08:27,610 --> 00:08:36,289 And then all the other carbons, we have a 6-carbon chain. 150 00:08:36,289 --> 00:08:38,400 Now, in terms of mechanism, this is just 151 00:08:38,400 --> 00:08:41,100 a reverse of the hemiacetal formation, 152 00:08:41,100 --> 00:08:44,740 and it's going to require a base to deprotonate 153 00:08:44,740 --> 00:08:47,430 the hydroxyl here in carbon 1. 154 00:08:47,430 --> 00:08:49,510 And then we're going to need to reprotonate 155 00:08:49,510 --> 00:08:55,200 at the hydroxyl in position 5 to form this hydroxyl here. 156 00:08:55,200 --> 00:08:57,210 Now, something to keep in mind, though, 157 00:08:57,210 --> 00:09:01,590 the way this enzyme works is we don't really 158 00:09:01,590 --> 00:09:04,770 know if the enzyme is catalyzing this transformation as it's 159 00:09:04,770 --> 00:09:07,980 written here, or the enzyme is just binding 160 00:09:07,980 --> 00:09:11,610 the linear form that will be in equilibrium in solution 161 00:09:11,610 --> 00:09:14,600 of mannose 6-phosphate. 162 00:09:14,600 --> 00:09:16,340 Similarly, for fructose 6-phosphate, 163 00:09:16,340 --> 00:09:20,010 we can write its linear form. 164 00:09:20,010 --> 00:09:22,970 Notice, fructose 6-phosphate has a carbon too, 165 00:09:22,970 --> 00:09:25,250 it's a hemiketal functionality. 166 00:09:25,250 --> 00:09:28,880 This carbon is attached to both a hydroxyl and an oxygen. 167 00:09:28,880 --> 00:09:30,950 So that will open up to form a ketone. 168 00:09:43,430 --> 00:09:49,190 All right, this is our fructose 6-phosphate in a linear form. 169 00:09:49,190 --> 00:09:53,220 And notice at 2 position, we have the ketone. 170 00:09:53,220 --> 00:09:56,370 Now, it would be a good idea at this point to number 171 00:09:56,370 --> 00:09:58,530 the carbons to see what do we need 172 00:09:58,530 --> 00:10:00,530 to do, going from this linear form 173 00:10:00,530 --> 00:10:03,610 of mannose 6-phosphate to this linear form of fructose 174 00:10:03,610 --> 00:10:05,170 6-phosphate. 175 00:10:05,170 --> 00:10:10,150 So here we have carbons 6, 5, 4, 3, 2, 1. 176 00:10:10,150 --> 00:10:16,630 And we have carbons 6, 5, 4, 3, 2, 1. 177 00:10:16,630 --> 00:10:19,360 And if we contrast the two structures, 178 00:10:19,360 --> 00:10:23,350 we notice that carbons from 3 to 6 in both cases 179 00:10:23,350 --> 00:10:25,150 are, in fact, the same. 180 00:10:25,150 --> 00:10:27,070 And we even have the hydroxyls in 181 00:10:27,070 --> 00:10:30,480 the right stereochemical orientation, 182 00:10:30,480 --> 00:10:33,250 in the right stereochemistry. 183 00:10:33,250 --> 00:10:38,140 And all we need to do is move the carbonyl group 184 00:10:38,140 --> 00:10:40,310 from the 1 position here in mannose 185 00:10:40,310 --> 00:10:45,850 6-phosphate to the 2 position in the fructose 6-phosphate. 186 00:10:45,850 --> 00:10:49,180 So this transformation can be accomplished 187 00:10:49,180 --> 00:10:52,870 by an intermediate, which you have seen already 188 00:10:52,870 --> 00:10:56,740 in glycolysis, the intermediate that allows to go from glucose 189 00:10:56,740 --> 00:10:58,900 6-phosphate into fructose 6-phosphate, which 190 00:10:58,900 --> 00:11:02,680 is a cis-enediol. 191 00:11:02,680 --> 00:11:09,520 So it's essentially enolization with the hydrogen on carbon 2. 192 00:11:09,520 --> 00:11:16,710 So if we were to form the enol here, 193 00:11:16,710 --> 00:11:19,980 which, as you know from the carbonyl video, 194 00:11:19,980 --> 00:11:23,850 so a base could deprotonate and allow the electrons 195 00:11:23,850 --> 00:11:29,110 to move and form another hydroxyl on the 1 position. 196 00:11:29,110 --> 00:11:33,225 Because this is a cis-enediol, the two hydroxyl 197 00:11:33,225 --> 00:11:36,990 are going to end up on the same side of the double bond. 198 00:11:48,970 --> 00:11:53,020 Once again, let's number our carbons. 199 00:11:53,020 --> 00:11:57,130 4, 3, 2, 1. 200 00:11:57,130 --> 00:11:59,876 And this is our cis-enediol. 201 00:12:02,860 --> 00:12:04,690 So we have two hydroxyls. 202 00:12:04,690 --> 00:12:07,180 They're both attached to a double bond. 203 00:12:07,180 --> 00:12:10,520 And cis means they're both on the same side of the molecule. 204 00:12:10,520 --> 00:12:15,130 And this allows the enzyme to basically 205 00:12:15,130 --> 00:12:18,010 switch back and forth between which one of these hydroxyl 206 00:12:18,010 --> 00:12:19,660 becomes the carbonyl. 207 00:12:19,660 --> 00:12:23,150 If we go backwards, we have the carbonyl at the 1 position. 208 00:12:23,150 --> 00:12:28,450 However, if we go forward, we can move the carbonyl 209 00:12:28,450 --> 00:12:30,020 on the 2 position. 210 00:12:30,020 --> 00:12:35,870 So that's just the reverse of the enolization reaction 211 00:12:35,870 --> 00:12:39,680 in which we will deprotonate this hydroxyl 212 00:12:39,680 --> 00:12:45,580 and reprotonate at carbon 1 to take us forward 213 00:12:45,580 --> 00:12:48,860 to generate the 2-keto group, which 214 00:12:48,860 --> 00:12:53,150 is fructose 6-phosphate in the linear form. 215 00:12:53,150 --> 00:12:55,340 Now from here, in the very last step, 216 00:12:55,340 --> 00:12:56,720 we just need to close the ring. 217 00:12:56,720 --> 00:12:59,930 So that's just a hemiketal formation. 218 00:12:59,930 --> 00:13:03,850 We have a base deprotonated hydroxyl at position 5. 219 00:13:03,850 --> 00:13:11,290 And this will attack the ketone to form the hemiketal group 220 00:13:11,290 --> 00:13:14,020 that we see in fructose 6-phosphate. 221 00:13:14,020 --> 00:13:18,022 So now this is basically the curved arrow mechanism 222 00:13:18,022 --> 00:13:19,480 from going from mannose 6-phosphate 223 00:13:19,480 --> 00:13:21,830 to fructose 6-phosphate. 224 00:13:21,830 --> 00:13:23,770 Now, some key points to notice here 225 00:13:23,770 --> 00:13:28,840 is that this enolization reaction where the base removes 226 00:13:28,840 --> 00:13:31,870 this alpha hydrogen would not have been possible 227 00:13:31,870 --> 00:13:33,970 in the cyclic structures. 228 00:13:33,970 --> 00:13:40,570 If you look here, this hydrogen, the PKA is about 30. 229 00:13:40,570 --> 00:13:44,020 Now, once we form the linear structure here, 230 00:13:44,020 --> 00:13:48,730 the PKA of this hydrogen, now it's about 18. 231 00:13:48,730 --> 00:13:51,620 So that's like 12 order of magnitude more acidic, 232 00:13:51,620 --> 00:13:55,450 and that allows the reaction to for this cis-enediol, 233 00:13:55,450 --> 00:13:59,380 which allows the formation of the fructose 6-phosphate. 234 00:13:59,380 --> 00:14:03,270 So the take home message here is that opening and closing 235 00:14:03,270 --> 00:14:07,440 the ring of the sugars allows a lot of interesting chemistry 236 00:14:07,440 --> 00:14:07,950 to happen. 237 00:14:13,370 --> 00:14:17,660 Part three of this question asks us to comment on 238 00:14:17,660 --> 00:14:20,630 why, when going from mannose to fructose, 239 00:14:20,630 --> 00:14:22,040 we have to go through the mannose 240 00:14:22,040 --> 00:14:24,750 6-phosphate intermediate. 241 00:14:24,750 --> 00:14:28,460 Now, this turns out to be a very common motif 242 00:14:28,460 --> 00:14:31,640 in carbohydrate chemistry, because adding 243 00:14:31,640 --> 00:14:36,950 a phosphate group is a way of regulation of which product 244 00:14:36,950 --> 00:14:38,660 is being formed. 245 00:14:38,660 --> 00:14:42,200 Taking a closer look at our transformation, 246 00:14:42,200 --> 00:14:44,830 we see that if we didn't have the phosphate group at the 6 247 00:14:44,830 --> 00:14:47,690 position, this hydroxyl would be available 248 00:14:47,690 --> 00:14:49,100 and could potentially be involved 249 00:14:49,100 --> 00:14:51,200 in forming cyclic forms. 250 00:14:51,200 --> 00:14:56,480 As we recall our discussion on fructose, whenever 251 00:14:56,480 --> 00:14:59,360 we have a 6-hydroxyl available, the fructose 252 00:14:59,360 --> 00:15:02,060 can form the fructopyranose form in addition 253 00:15:02,060 --> 00:15:04,010 to the fructofuranose form. 254 00:15:04,010 --> 00:15:07,060 So by blocking this position with a phosphate, 255 00:15:07,060 --> 00:15:10,340 we're limiting the reaction to produce only one product, 256 00:15:10,340 --> 00:15:12,950 namely the fructofuranose. 257 00:15:12,950 --> 00:15:15,590 Additionally, adding a phosphate group to a sugar 258 00:15:15,590 --> 00:15:18,260 imparts a negative charge, and that 259 00:15:18,260 --> 00:15:20,870 allows the sugar to be trapped inside the cell 260 00:15:20,870 --> 00:15:25,539 and allows metabolism to occur much more efficiently. 261 00:15:25,539 --> 00:15:27,080 This is something that you've already 262 00:15:27,080 --> 00:15:30,890 encountered in glycolysis, where glucose, in the first step, 263 00:15:30,890 --> 00:15:34,510 is first converted to glucose 6-phosphate.