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BOGDAN FEDELES: Hi, everyone.

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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.

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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

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00:13:43,089 --> 00:13:45,630
with the protons that are going
to be flowing through the ATP

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00:13:45,630 --> 00:13:47,640
synthase to generate ATP.

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00:13:47,640 --> 00:13:51,570
So in effect, we're
discharging this battery

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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.

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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,

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00:14:05,760 --> 00:14:08,510
the battery is going to heat
up because you're discharging

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00:14:08,510 --> 00:14:10,230
an electrochemical gradient.

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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.

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00:14:20,920 --> 00:14:23,850
Therefore, we can
answer Part C by saying

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00:14:23,850 --> 00:14:27,000
that the medium in which
these cells are growing

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00:14:27,000 --> 00:14:30,330
is going to heat up when
we add dinitrophenol to it.

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00:14:30,330 --> 00:14:33,840
The processes described in this
problem are fairly universal.

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00:14:33,840 --> 00:14:38,450
Now, in eukaryotes, like
more evolved organisms,

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00:14:38,450 --> 00:14:41,560
they would happen
in the mitochondria.

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00:14:41,560 --> 00:14:43,350
Now, if you look
back at this diagram,

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if this was the double
membrane of the mitochondria,

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00:14:46,320 --> 00:14:48,480
this would be the
inside of the cell that

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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,

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00:14:59,700 --> 00:15:02,490
who can dissipate the
electrochemical gradient

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00:15:02,490 --> 00:15:07,440
in the mitochondria and
cause the cell to heat up.

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00:15:07,440 --> 00:15:10,170
In fact, this process
is actually used

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00:15:10,170 --> 00:15:13,650
by a number of organisms
to generate heat instead

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00:15:13,650 --> 00:15:17,460
of chemical energy, or ATP.

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00:15:17,460 --> 00:15:23,250
For example, the brown fat
cells in newborns in mammals

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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

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00:15:27,010 --> 00:15:30,030
in the mitochondria
to generate heat.

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00:15:30,030 --> 00:15:34,890
Another good example is
the seeds of many plants.

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00:15:34,890 --> 00:15:36,930
When they germinate,
they actually

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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.

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00:15:41,940 --> 00:15:43,500
That's why some
of the plants can

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00:15:43,500 --> 00:15:45,730
start growing even
before the snow has

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00:15:45,730 --> 00:15:48,284
melt in the early spring.

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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

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00:15:58,830 --> 00:16:01,500
into an electrochemical
gradient, which

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00:16:01,500 --> 00:16:05,280
can then be used to generate
high energy compounds like ATP.

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00:16:05,280 --> 00:16:08,780
Or it can be dissipated
to generate heat.