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 00:00:28,214 --> 00:00:29,770 BOGDAN FEDELES: Hi, everyone. 9 00:00:29,770 --> 00:00:32,810 Welcome to 5.07 Bio Chemistry Online. 10 00:00:32,810 --> 00:00:34,560 I'm Dr. Bogdan Fedeles. 11 00:00:34,560 --> 00:00:36,540 I'm going to help you work through some more 12 00:00:36,540 --> 00:00:38,580 biochemistry problems today. 13 00:00:38,580 --> 00:00:42,405 I have here question 2 of Problem Set 8. 14 00:00:42,405 --> 00:00:44,760 Now, this is the question I put together 15 00:00:44,760 --> 00:00:47,610 to get you thinking about the electron transport chain. 16 00:00:47,610 --> 00:00:49,470 As you know, the electron transport chain 17 00:00:49,470 --> 00:00:52,410 is a fundamental redox process through which 18 00:00:52,410 --> 00:00:56,940 we convert the chemical energy of the covalent bonds 19 00:00:56,940 --> 00:00:59,070 into an electrochemical gradient. 20 00:00:59,070 --> 00:01:02,050 This electrochemical gradient is like a battery, 21 00:01:02,050 --> 00:01:04,500 and it can be used inside the cell to generate, 22 00:01:04,500 --> 00:01:09,000 for example, ATP, which is the energy currency of the cell, 23 00:01:09,000 --> 00:01:12,420 or it can be dissipated to generate heat. 24 00:01:12,420 --> 00:01:14,910 We're going to see both of these modes in action 25 00:01:14,910 --> 00:01:15,960 in this problem. 26 00:01:15,960 --> 00:01:19,620 Now in most organisms, the electron transport chain 27 00:01:19,620 --> 00:01:23,460 helps to transfer electrons all the way to molecular oxygen. 28 00:01:23,460 --> 00:01:25,260 However, in this problem, we're dealing 29 00:01:25,260 --> 00:01:30,000 with an organism that lives deep inside the ocean 30 00:01:30,000 --> 00:01:33,510 where the atmospheric oxygen is not available. 31 00:01:33,510 --> 00:01:36,990 And it turns out this organism transfers its electrons 32 00:01:36,990 --> 00:01:38,190 to sulfate. 33 00:01:38,190 --> 00:01:40,020 Sulfate is the final electron acceptor. 34 00:01:43,340 --> 00:01:45,460 Part A of this problem asks us to write 35 00:01:45,460 --> 00:01:48,770 the order of the electron carriers 36 00:01:48,770 --> 00:01:51,590 as they would function in an electron transport 37 00:01:51,590 --> 00:01:54,230 chain for this organism. 38 00:01:54,230 --> 00:01:58,010 Now, for a number of redox processes, 39 00:01:58,010 --> 00:02:01,880 the problem provides a table with the electrochemical 40 00:02:01,880 --> 00:02:04,580 reducing potentials, as you see here. 41 00:02:04,580 --> 00:02:07,400 Now, I've selected the ones that are mentioned in the problem, 42 00:02:07,400 --> 00:02:11,300 and I put them into a smaller table here. 43 00:02:11,300 --> 00:02:15,210 As you can see, we're dealing with cytochrome A, B, C, C1. 44 00:02:15,210 --> 00:02:17,710 This is the flavin mononucleotide. 45 00:02:17,710 --> 00:02:20,060 This is the sulfate, the fine electron acceptor, 46 00:02:20,060 --> 00:02:22,130 and ubiquinol. 47 00:02:22,130 --> 00:02:25,430 Now, on this column here we have the redox potential, 48 00:02:25,430 --> 00:02:29,180 which are the electrochemical reduction potentials denoted 49 00:02:29,180 --> 00:02:32,120 by epsilon, or e0 prime. 50 00:02:32,120 --> 00:02:36,620 Now, e0, as you know from physical chemistry or physics, 51 00:02:36,620 --> 00:02:38,870 denotes the electrochemical potential 52 00:02:38,870 --> 00:02:40,440 in standard conditions. 53 00:02:40,440 --> 00:02:44,690 However, in biochemistry, we use the e0 prime notation 54 00:02:44,690 --> 00:02:48,230 to denote that the pH is taken into account, 55 00:02:48,230 --> 00:02:50,240 and it's not what you would expect, 56 00:02:50,240 --> 00:02:53,030 like of hydrogen ion's concentration equals 1 molar, 57 00:02:53,030 --> 00:02:54,990 but rather it's a pH of 7. 58 00:02:54,990 --> 00:02:59,510 The hydrogen ion's concentration equals 10 to the minus 7. 59 00:02:59,510 --> 00:03:04,410 So therefore, these numbers are adjusted to correspond to pH 7. 60 00:03:04,410 --> 00:03:05,810 The electrochemical potentials we 61 00:03:05,810 --> 00:03:08,750 see in this table are reduction potentials, 62 00:03:08,750 --> 00:03:13,340 and they tell us how easy it is to reduce a particular species. 63 00:03:13,340 --> 00:03:16,070 Therefore, the higher the number, the easier it 64 00:03:16,070 --> 00:03:18,950 is to reduce that particular species and the more energy 65 00:03:18,950 --> 00:03:21,900 the reduction of that species will generate. 66 00:03:21,900 --> 00:03:25,010 Therefore, the electron transport chain 67 00:03:25,010 --> 00:03:27,050 will go from the species that hardest 68 00:03:27,050 --> 00:03:32,540 to be reduce towards the species that are easiest to be reduced. 69 00:03:32,540 --> 00:03:34,940 Therefore, the order of the electron carriers 70 00:03:34,940 --> 00:03:37,730 will be from the ones that have the lowest reductive 71 00:03:37,730 --> 00:03:40,010 potential to the ones that have the highest 72 00:03:40,010 --> 00:03:41,490 reductive potential. 73 00:03:41,490 --> 00:03:44,090 So now if we're going to sort all these electron carriers 74 00:03:44,090 --> 00:03:46,160 in order of their potential, we're 75 00:03:46,160 --> 00:03:53,910 going to get the following order as you see here. 76 00:03:53,910 --> 00:03:56,780 So the electrons are going to flow from the flavin 77 00:03:56,780 --> 00:03:59,630 into the coenzyme Q, and then the electrons 78 00:03:59,630 --> 00:04:03,320 are going to flow coenzyme Q to cytochrome B, and then 79 00:04:03,320 --> 00:04:06,230 Cytochrome C1, C, A, and sulfate. 80 00:04:06,230 --> 00:04:10,320 And as you can see, flavin has a negative reduction potential. 81 00:04:10,320 --> 00:04:13,220 It's like the hardest to be reduced. 82 00:04:13,220 --> 00:04:15,940 And the next one is ubiquinol. 83 00:04:15,940 --> 00:04:17,930 It's barely positive. 84 00:04:17,930 --> 00:04:22,340 And then the highest number is sulfate 0.48 volts. 85 00:04:22,340 --> 00:04:24,500 Now, let's take a closer look how the electrons 86 00:04:24,500 --> 00:04:27,740 are going to be transferred through this proposed electron 87 00:04:27,740 --> 00:04:29,050 transport chain. 88 00:04:29,050 --> 00:04:31,940 In the first reaction, here we have the flavin, 89 00:04:31,940 --> 00:04:35,390 I've written the flavin adenine dinucleotide, 90 00:04:35,390 --> 00:04:38,060 FADH2, the reduced version, is going 91 00:04:38,060 --> 00:04:43,760 to be converted to the oxidized FAD version of it. 92 00:04:43,760 --> 00:04:45,860 And in this redox reaction, we're 93 00:04:45,860 --> 00:04:48,380 going to use the coenzyme Q, the oxidized version 94 00:04:48,380 --> 00:04:50,480 and reduce it in the process. 95 00:04:50,480 --> 00:04:54,930 So the electrons get transferred from FADH2 to coenzyme Q. 96 00:04:54,930 --> 00:04:58,830 Now, in the next reaction, the reduced version of coenzyme Q 97 00:04:58,830 --> 00:05:01,850 is going to get oxidized back to coenzyme Q 98 00:05:01,850 --> 00:05:04,020 and in the process cytochrome B is 99 00:05:04,020 --> 00:05:07,640 going to go from its oxidized form to its reduced form. 100 00:05:07,640 --> 00:05:11,000 Now, this process continues with every single step, 101 00:05:11,000 --> 00:05:15,650 every single electron carrier up until we get to the sulfate 102 00:05:15,650 --> 00:05:19,100 where the reduced form of the cytochrome A 103 00:05:19,100 --> 00:05:22,700 will donate its electrons to the sulfate, 104 00:05:22,700 --> 00:05:25,770 and sulfate would get reduced to its reduced form. 105 00:05:25,770 --> 00:05:28,560 It's called sulfite. 106 00:05:28,560 --> 00:05:31,610 So if we were to draw how the electrons move 107 00:05:31,610 --> 00:05:33,980 through this chain, the electrons 108 00:05:33,980 --> 00:05:37,460 are going to start at FADH, and then they're 109 00:05:37,460 --> 00:05:42,020 going to be transferred to coenzyme Q in the reduced form. 110 00:05:42,020 --> 00:05:45,620 And then coenzyme Q is going to pass it to the cytochrome B. 111 00:05:45,620 --> 00:05:47,537 That's going to be in its reduced form. 112 00:05:47,537 --> 00:05:49,120 And then cytochrome B is going to pass 113 00:05:49,120 --> 00:05:53,870 it to cytochrome C1, and then cytochrome C, cytochrome A, 114 00:05:53,870 --> 00:05:58,130 and finally, they're going to end up in sulfite. 115 00:05:58,130 --> 00:06:00,680 Another thing to notice here is that 116 00:06:00,680 --> 00:06:05,540 except for the initial flavin and the final electron 117 00:06:05,540 --> 00:06:09,950 acceptor, sulfate, all the other intermediates get regenerated. 118 00:06:09,950 --> 00:06:12,770 So we go from the oxidized version to the reduced version 119 00:06:12,770 --> 00:06:14,480 and back to the oxidized version. 120 00:06:14,480 --> 00:06:18,800 So all these electron carriers are going to be sufficient only 121 00:06:18,800 --> 00:06:21,080 in catalytic amounts. 122 00:06:21,080 --> 00:06:22,910 So the only thing that gets consumed 123 00:06:22,910 --> 00:06:26,060 is the FADH2 and the sulfate. 124 00:06:26,060 --> 00:06:29,030 These are two reactants. 125 00:06:29,030 --> 00:06:31,960 And we get in this reaction FAD and sulfite. 126 00:06:35,970 --> 00:06:38,700 What we just said will help us segue into the Part 127 00:06:38,700 --> 00:06:42,400 B of the problem, which asks us to calculate how much energy do 128 00:06:42,400 --> 00:06:46,870 we get by converting one molecule of FADH2 129 00:06:46,870 --> 00:06:53,360 and one molecule of sulfate into FAD and sulfite, respectively. 130 00:06:53,360 --> 00:07:00,160 Now as we pointed out here, only the FADH2 and sulfate 131 00:07:00,160 --> 00:07:02,890 are consumed in this reaction. 132 00:07:02,890 --> 00:07:07,290 All the other electron carriers are recycled and regenerated 133 00:07:07,290 --> 00:07:10,300 in the course of the electron transport chain. 134 00:07:10,300 --> 00:07:12,460 In order to calculate the energy, 135 00:07:12,460 --> 00:07:15,640 it's useful first to write the half reaction of the redox 136 00:07:15,640 --> 00:07:16,790 processes. 137 00:07:16,790 --> 00:07:20,840 Here are the two half reactions of this redox process. 138 00:07:20,840 --> 00:07:27,400 FADH2 gets oxidized through FAD and donates its two electrons. 139 00:07:27,400 --> 00:07:33,220 And the epsilon, or e0 prime is minus 0.22 volts. 140 00:07:33,220 --> 00:07:35,980 Now, this is the potential from the table, 141 00:07:35,980 --> 00:07:38,330 and that's a reduction potential. 142 00:07:38,330 --> 00:07:41,200 The equation as written is an oxidation, 143 00:07:41,200 --> 00:07:42,820 and therefore, the potential that we 144 00:07:42,820 --> 00:07:45,340 need to take into account is the minus of this one. 145 00:07:49,000 --> 00:07:52,060 Sulfate is then going to accept the two electrons 146 00:07:52,060 --> 00:07:55,330 and going to get reduced to the sulfite and water. 147 00:07:55,330 --> 00:07:59,620 And the electrochemical potential for this 148 00:07:59,620 --> 00:08:02,750 is 0.48 volts. 149 00:08:02,750 --> 00:08:05,150 So now when we add these two together, 150 00:08:05,150 --> 00:08:13,270 we get the overall process where FADH2 gets oxidized by sulfate 151 00:08:13,270 --> 00:08:15,730 to generate FAD and sulfite. 152 00:08:15,730 --> 00:08:20,110 And the electromotive force is just the mathematical sum 153 00:08:20,110 --> 00:08:22,660 of these two keeping in mind that this has 154 00:08:22,660 --> 00:08:24,010 to be taken as a negative sign. 155 00:08:27,190 --> 00:08:29,902 Because, again, as written, this is 156 00:08:29,902 --> 00:08:32,110 an oxidation and this the potential for the reduction 157 00:08:32,110 --> 00:08:32,990 reaction. 158 00:08:32,990 --> 00:08:37,120 So electromotive force is actually 0.7 volts. 159 00:08:37,120 --> 00:08:41,740 Now, we can easily convert from the electromotive force 160 00:08:41,740 --> 00:08:46,257 to a delta g0 prime value, and the relationship 161 00:08:46,257 --> 00:08:47,590 is written here, delta g0 prime. 162 00:08:47,590 --> 00:08:53,800 It's minus nF delta e0 prime and is the number of electrons 163 00:08:53,800 --> 00:08:55,280 in the process as we see here. 164 00:08:55,280 --> 00:08:58,960 Two, F is the Faraday's constant and delta e0 165 00:08:58,960 --> 00:09:02,780 prime is going to be the electromotive force. 166 00:09:02,780 --> 00:09:04,960 And if we go through the number crunching, 167 00:09:04,960 --> 00:09:10,070 we get a delta g0 prime minus 135 kilojoules per mole. 168 00:09:10,070 --> 00:09:12,400 Notice because it's a negative number that means 169 00:09:12,400 --> 00:09:15,880 there's a spontaneous process as written. 170 00:09:15,880 --> 00:09:17,710 And as you know, the negative delta g 171 00:09:17,710 --> 00:09:20,740 will correspond to a positive electromotive force. 172 00:09:20,740 --> 00:09:23,050 Now, we're just one step away from calculating 173 00:09:23,050 --> 00:09:26,620 how much ATP we can produce with this energy. 174 00:09:26,620 --> 00:09:31,360 As you know, we generate ATP out of ADP and phosphate, 175 00:09:31,360 --> 00:09:35,800 and this is the reaction that's catalyzed by ATP synthase. 176 00:09:35,800 --> 00:09:38,950 And it takes about 30.5 kilojoules per mole 177 00:09:38,950 --> 00:09:42,560 to form ATP out of ADP and phosphate. 178 00:09:42,560 --> 00:09:46,860 Therefore, the 135 kilojoules per mole 179 00:09:46,860 --> 00:09:50,380 that we generated from 1 mole of FADH2, 180 00:09:50,380 --> 00:09:55,210 it's going to be enough for about 4 molecules of ATPs. 181 00:09:55,210 --> 00:09:59,680 This is in contrast, which was the normal processes that 182 00:09:59,680 --> 00:10:03,640 use oxygen as their final electron acceptor 183 00:10:03,640 --> 00:10:06,695 where out of one FADH2 molecule, will generate 184 00:10:06,695 --> 00:10:09,800 at most 2 molecules of ATP. 185 00:10:09,800 --> 00:10:13,330 So in some ways, sulfate is actually a better electron 186 00:10:13,330 --> 00:10:15,340 acceptor and can give us more energy. 187 00:10:18,460 --> 00:10:21,430 Part C of these problem deals with a culture 188 00:10:21,430 --> 00:10:24,710 of this microorganism in the lab. 189 00:10:24,710 --> 00:10:27,420 And we're adding to this culture dinitrophenol, 190 00:10:27,420 --> 00:10:31,430 a compound we're told has a pKa of about 5.2. 191 00:10:31,430 --> 00:10:34,000 So let's explore what happens to the electron transport 192 00:10:34,000 --> 00:10:37,180 chain of the organism when we add dinitrophenol. 193 00:10:37,180 --> 00:10:39,430 Here I put together a cartoon representation 194 00:10:39,430 --> 00:10:43,610 of the electron transport chain of our organism. 195 00:10:43,610 --> 00:10:48,100 So as you can see here, this is the extracellular environment. 196 00:10:48,100 --> 00:10:50,630 This is the outer membrane. 197 00:10:50,630 --> 00:10:54,000 This is the inner membrane where we have all these complexes 198 00:10:54,000 --> 00:10:56,710 I denoted here with these rectangles of the electron 199 00:10:56,710 --> 00:10:57,860 transport chain. 200 00:10:57,860 --> 00:11:01,310 And FADH2, for example, is going to donate its electrons. 201 00:11:01,310 --> 00:11:04,810 They're going to be passed along all the way to sulfate. 202 00:11:04,810 --> 00:11:07,870 And in the process, protons are going 203 00:11:07,870 --> 00:11:11,230 to get pumped into this intermembrane space. 204 00:11:11,230 --> 00:11:15,520 Now, these protons can be used in the ATP synthase 205 00:11:15,520 --> 00:11:19,570 as they travel back into the intercellular space. 206 00:11:19,570 --> 00:11:24,130 Their energy can be used to convert ADP and organophosphate 207 00:11:24,130 --> 00:11:28,690 to ATP as we just discussed in Part 2. 208 00:11:28,690 --> 00:11:32,230 Now, to this organism, we said we're 209 00:11:32,230 --> 00:11:33,880 going to add dinitrophenol. 210 00:11:33,880 --> 00:11:36,095 Here is the structure of dinitrophenol. 211 00:11:42,400 --> 00:11:45,850 And we're told the pKa of this proton, 212 00:11:45,850 --> 00:11:49,629 right here, the pKa is about 5.2. 213 00:11:49,629 --> 00:11:51,670 When this compound diffuses through the membrane, 214 00:11:51,670 --> 00:11:54,280 it's going to go through this intermembrane space, which 215 00:11:54,280 --> 00:12:01,330 has a very low pH and also in the intercellular 216 00:12:01,330 --> 00:12:04,470 space in the cytosol, which has a much higher pH. 217 00:12:04,470 --> 00:12:08,500 So because pKa 5.2, it's a relatively low, much lower 218 00:12:08,500 --> 00:12:13,840 than 7, pKa, in the intermembrane space where 219 00:12:13,840 --> 00:12:17,890 it's more acidic, it's going to be protonated. 220 00:12:17,890 --> 00:12:24,400 So we can write, for example, dinitrophenol OH in equilibrium 221 00:12:24,400 --> 00:12:27,915 with dinitrophenol O minus plus a proton. 222 00:12:31,990 --> 00:12:34,730 Now, because here we have a lot of protons, 223 00:12:34,730 --> 00:12:37,990 this equilibrium will be shifted to the left. 224 00:12:37,990 --> 00:12:41,830 That is the protonated form of dinitrophenol. 225 00:12:41,830 --> 00:12:48,320 However, here in the cytosol, the NPOH, 226 00:12:48,320 --> 00:12:50,800 it's going to be in the same equilibrium O minus 227 00:12:50,800 --> 00:12:52,540 plus H plus. 228 00:12:52,540 --> 00:12:55,180 But because the pH is fairly high, that is 229 00:12:55,180 --> 00:12:58,000 there are not a lot of protons, this equilibrium 230 00:12:58,000 --> 00:13:00,400 is going to be shifted to the right. 231 00:13:00,400 --> 00:13:06,600 This equilibrium is going to be shifted to the left. 232 00:13:06,600 --> 00:13:09,330 So now look what happens. 233 00:13:09,330 --> 00:13:12,360 So because this equilibrium has shifted to the left, 234 00:13:12,360 --> 00:13:16,500 it's going to keep soaking up a lot of these protons. 235 00:13:16,500 --> 00:13:18,630 Then the neutral dinitrophenol molecule 236 00:13:18,630 --> 00:13:24,270 is going to diffuse through the membrane as such 237 00:13:24,270 --> 00:13:28,440 and enter the intercellular space to cytosol where 238 00:13:28,440 --> 00:13:30,930 it's going to be deprotonated. 239 00:13:30,930 --> 00:13:33,400 The equilibrium is shifted to the right. 240 00:13:33,400 --> 00:13:35,340 So in effect, dinitrophenol is going 241 00:13:35,340 --> 00:13:39,210 to carry the protons from the intermembrane space 242 00:13:39,210 --> 00:13:41,470 inside the cell. 243 00:13:41,470 --> 00:13:43,089 Now it's going to do that in parallel 244 00:13:43,089 --> 00:13:45,630 with the protons that are going to be flowing through the ATP 245 00:13:45,630 --> 00:13:47,640 synthase to generate ATP. 246 00:13:47,640 --> 00:13:51,570 So in effect, we're discharging this battery 247 00:13:51,570 --> 00:13:53,340 where the concentration of protons 248 00:13:53,340 --> 00:13:57,770 is basically our electrochemical gradient. 249 00:13:57,770 --> 00:14:02,280 It's going to be discharging the battery without producing ATP. 250 00:14:02,280 --> 00:14:05,760 So as you know, if you short circuit a battery, 251 00:14:05,760 --> 00:14:08,510 the battery is going to heat up because you're discharging 252 00:14:08,510 --> 00:14:10,230 an electrochemical gradient. 253 00:14:10,230 --> 00:14:13,692 Similarly, dinitrophenol, by taking these protons 254 00:14:13,692 --> 00:14:15,900 from the intermembrane space and bringing them inside 255 00:14:15,900 --> 00:14:17,700 into the intercellular space, it's 256 00:14:17,700 --> 00:14:20,920 going to be generating heat. 257 00:14:20,920 --> 00:14:23,850 Therefore, we can answer Part C by saying 258 00:14:23,850 --> 00:14:27,000 that the medium in which these cells are growing 259 00:14:27,000 --> 00:14:30,330 is going to heat up when we add dinitrophenol to it. 260 00:14:30,330 --> 00:14:33,840 The processes described in this problem are fairly universal. 261 00:14:33,840 --> 00:14:38,450 Now, in eukaryotes, like more evolved organisms, 262 00:14:38,450 --> 00:14:41,560 they would happen in the mitochondria. 263 00:14:41,560 --> 00:14:43,350 Now, if you look back at this diagram, 264 00:14:43,350 --> 00:14:46,320 if this was the double membrane of the mitochondria, 265 00:14:46,320 --> 00:14:48,480 this would be the inside of the cell that 266 00:14:48,480 --> 00:14:50,100 contains the mitochondria, this would 267 00:14:50,100 --> 00:14:51,780 be the intermembrane space, and this 268 00:14:51,780 --> 00:14:53,640 will be the inside of the mitochondria 269 00:14:53,640 --> 00:14:56,400 or the mitochondrial matrix. 270 00:14:56,400 --> 00:14:59,700 Similarly, by adding a compound like dinitrophenol, 271 00:14:59,700 --> 00:15:02,490 who can dissipate the electrochemical gradient 272 00:15:02,490 --> 00:15:07,440 in the mitochondria and cause the cell to heat up. 273 00:15:07,440 --> 00:15:10,170 In fact, this process is actually used 274 00:15:10,170 --> 00:15:13,650 by a number of organisms to generate heat instead 275 00:15:13,650 --> 00:15:17,460 of chemical energy, or ATP. 276 00:15:17,460 --> 00:15:23,250 For example, the brown fat cells in newborns in mammals 277 00:15:23,250 --> 00:15:25,170 have a special protein that allows 278 00:15:25,170 --> 00:15:27,010 to dissipate this electrochemical gradient 279 00:15:27,010 --> 00:15:30,030 in the mitochondria to generate heat. 280 00:15:30,030 --> 00:15:34,890 Another good example is the seeds of many plants. 281 00:15:34,890 --> 00:15:36,930 When they germinate, they actually 282 00:15:36,930 --> 00:15:38,520 generate a lot of heat that can be 283 00:15:38,520 --> 00:15:41,940 used to melt the ice or the snow around them. 284 00:15:41,940 --> 00:15:43,500 That's why some of the plants can 285 00:15:43,500 --> 00:15:45,730 start growing even before the snow has 286 00:15:45,730 --> 00:15:48,284 melt in the early spring. 287 00:15:48,284 --> 00:15:49,950 I hope that working through this problem 288 00:15:49,950 --> 00:15:52,530 will help you understand better the inner workings 289 00:15:52,530 --> 00:15:54,660 of an electron transport chain and how 290 00:15:54,660 --> 00:15:58,830 it can convert the chemical energy of chemical bonds 291 00:15:58,830 --> 00:16:01,500 into an electrochemical gradient, which 292 00:16:01,500 --> 00:16:05,280 can then be used to generate high energy compounds like ATP. 293 00:16:05,280 --> 00:16:08,780 Or it can be dissipated to generate heat.