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BOGDAN FEDELES: Hi, everyone.

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Welcome to 5.07 Bio
Chemistry Online.

00:00:32.810 --> 00:00:34.560
I'm Dr. Bogdan Fedeles.

00:00:34.560 --> 00:00:36.540
I'm going to help you
work through some more

00:00:36.540 --> 00:00:38.580
biochemistry problems today.

00:00:38.580 --> 00:00:42.405
I have here question
2 of Problem Set 8.

00:00:42.405 --> 00:00:44.760
Now, this is the
question I put together

00:00:44.760 --> 00:00:47.610
to get you thinking about
the electron transport chain.

00:00:47.610 --> 00:00:49.470
As you know, the
electron transport chain

00:00:49.470 --> 00:00:52.410
is a fundamental redox
process through which

00:00:52.410 --> 00:00:56.940
we convert the chemical
energy of the covalent bonds

00:00:56.940 --> 00:00:59.070
into an electrochemical
gradient.

00:00:59.070 --> 00:01:02.050
This electrochemical
gradient is like a battery,

00:01:02.050 --> 00:01:04.500
and it can be used inside
the cell to generate,

00:01:04.500 --> 00:01:09.000
for example, ATP, which is the
energy currency of the cell,

00:01:09.000 --> 00:01:12.420
or it can be dissipated
to generate heat.

00:01:12.420 --> 00:01:14.910
We're going to see both
of these modes in action

00:01:14.910 --> 00:01:15.960
in this problem.

00:01:15.960 --> 00:01:19.620
Now in most organisms, the
electron transport chain

00:01:19.620 --> 00:01:23.460
helps to transfer electrons all
the way to molecular oxygen.

00:01:23.460 --> 00:01:25.260
However, in this
problem, we're dealing

00:01:25.260 --> 00:01:30.000
with an organism that
lives deep inside the ocean

00:01:30.000 --> 00:01:33.510
where the atmospheric
oxygen is not available.

00:01:33.510 --> 00:01:36.990
And it turns out this organism
transfers its electrons

00:01:36.990 --> 00:01:38.190
to sulfate.

00:01:38.190 --> 00:01:40.020
Sulfate is the final
electron acceptor.

00:01:43.340 --> 00:01:45.460
Part A of this problem
asks us to write

00:01:45.460 --> 00:01:48.770
the order of the
electron carriers

00:01:48.770 --> 00:01:51.590
as they would function
in an electron transport

00:01:51.590 --> 00:01:54.230
chain for this organism.

00:01:54.230 --> 00:01:58.010
Now, for a number
of redox processes,

00:01:58.010 --> 00:02:01.880
the problem provides a table
with the electrochemical

00:02:01.880 --> 00:02:04.580
reducing potentials,
as you see here.

00:02:04.580 --> 00:02:07.400
Now, I've selected the ones that
are mentioned in the problem,

00:02:07.400 --> 00:02:11.300
and I put them into
a smaller table here.

00:02:11.300 --> 00:02:15.210
As you can see, we're dealing
with cytochrome A, B, C, C1.

00:02:15.210 --> 00:02:17.710
This is the flavin
mononucleotide.

00:02:17.710 --> 00:02:20.060
This is the sulfate, the
fine electron acceptor,

00:02:20.060 --> 00:02:22.130
and ubiquinol.

00:02:22.130 --> 00:02:25.430
Now, on this column here we
have the redox potential,

00:02:25.430 --> 00:02:29.180
which are the electrochemical
reduction potentials denoted

00:02:29.180 --> 00:02:32.120
by epsilon, or e0 prime.

00:02:32.120 --> 00:02:36.620
Now, e0, as you know from
physical chemistry or physics,

00:02:36.620 --> 00:02:38.870
denotes the
electrochemical potential

00:02:38.870 --> 00:02:40.440
in standard conditions.

00:02:40.440 --> 00:02:44.690
However, in biochemistry,
we use the e0 prime notation

00:02:44.690 --> 00:02:48.230
to denote that the pH
is taken into account,

00:02:48.230 --> 00:02:50.240
and it's not what
you would expect,

00:02:50.240 --> 00:02:53.030
like of hydrogen ion's
concentration equals 1 molar,

00:02:53.030 --> 00:02:54.990
but rather it's a pH of 7.

00:02:54.990 --> 00:02:59.510
The hydrogen ion's concentration
equals 10 to the minus 7.

00:02:59.510 --> 00:03:04.410
So therefore, these numbers are
adjusted to correspond to pH 7.

00:03:04.410 --> 00:03:05.810
The electrochemical
potentials we

00:03:05.810 --> 00:03:08.750
see in this table are
reduction potentials,

00:03:08.750 --> 00:03:13.340
and they tell us how easy it is
to reduce a particular species.

00:03:13.340 --> 00:03:16.070
Therefore, the higher
the number, the easier it

00:03:16.070 --> 00:03:18.950
is to reduce that particular
species and the more energy

00:03:18.950 --> 00:03:21.900
the reduction of that
species will generate.

00:03:21.900 --> 00:03:25.010
Therefore, the electron
transport chain

00:03:25.010 --> 00:03:27.050
will go from the
species that hardest

00:03:27.050 --> 00:03:32.540
to be reduce towards the species
that are easiest to be reduced.

00:03:32.540 --> 00:03:34.940
Therefore, the order of
the electron carriers

00:03:34.940 --> 00:03:37.730
will be from the ones that
have the lowest reductive

00:03:37.730 --> 00:03:40.010
potential to the ones
that have the highest

00:03:40.010 --> 00:03:41.490
reductive potential.

00:03:41.490 --> 00:03:44.090
So now if we're going to sort
all these electron carriers

00:03:44.090 --> 00:03:46.160
in order of their
potential, we're

00:03:46.160 --> 00:03:53.910
going to get the following
order as you see here.

00:03:53.910 --> 00:03:56.780
So the electrons are going
to flow from the flavin

00:03:56.780 --> 00:03:59.630
into the coenzyme Q,
and then the electrons

00:03:59.630 --> 00:04:03.320
are going to flow coenzyme
Q to cytochrome B, and then

00:04:03.320 --> 00:04:06.230
Cytochrome C1, C,
A, and sulfate.

00:04:06.230 --> 00:04:10.320
And as you can see, flavin has
a negative reduction potential.

00:04:10.320 --> 00:04:13.220
It's like the hardest
to be reduced.

00:04:13.220 --> 00:04:15.940
And the next one is ubiquinol.

00:04:15.940 --> 00:04:17.930
It's barely positive.

00:04:17.930 --> 00:04:22.340
And then the highest number
is sulfate 0.48 volts.

00:04:22.340 --> 00:04:24.500
Now, let's take a closer
look how the electrons

00:04:24.500 --> 00:04:27.740
are going to be transferred
through this proposed electron

00:04:27.740 --> 00:04:29.050
transport chain.

00:04:29.050 --> 00:04:31.940
In the first reaction,
here we have the flavin,

00:04:31.940 --> 00:04:35.390
I've written the flavin
adenine dinucleotide,

00:04:35.390 --> 00:04:38.060
FADH2, the reduced
version, is going

00:04:38.060 --> 00:04:43.760
to be converted to the
oxidized FAD version of it.

00:04:43.760 --> 00:04:45.860
And in this redox
reaction, we're

00:04:45.860 --> 00:04:48.380
going to use the coenzyme
Q, the oxidized version

00:04:48.380 --> 00:04:50.480
and reduce it in the process.

00:04:50.480 --> 00:04:54.930
So the electrons get transferred
from FADH2 to coenzyme Q.

00:04:54.930 --> 00:04:58.830
Now, in the next reaction, the
reduced version of coenzyme Q

00:04:58.830 --> 00:05:01.850
is going to get oxidized
back to coenzyme Q

00:05:01.850 --> 00:05:04.020
and in the process
cytochrome B is

00:05:04.020 --> 00:05:07.640
going to go from its oxidized
form to its reduced form.

00:05:07.640 --> 00:05:11.000
Now, this process continues
with every single step,

00:05:11.000 --> 00:05:15.650
every single electron carrier
up until we get to the sulfate

00:05:15.650 --> 00:05:19.100
where the reduced form
of the cytochrome A

00:05:19.100 --> 00:05:22.700
will donate its
electrons to the sulfate,

00:05:22.700 --> 00:05:25.770
and sulfate would get
reduced to its reduced form.

00:05:25.770 --> 00:05:28.560
It's called sulfite.

00:05:28.560 --> 00:05:31.610
So if we were to draw
how the electrons move

00:05:31.610 --> 00:05:33.980
through this chain,
the electrons

00:05:33.980 --> 00:05:37.460
are going to start at
FADH, and then they're

00:05:37.460 --> 00:05:42.020
going to be transferred to
coenzyme Q in the reduced form.

00:05:42.020 --> 00:05:45.620
And then coenzyme Q is going
to pass it to the cytochrome B.

00:05:45.620 --> 00:05:47.537
That's going to be
in its reduced form.

00:05:47.537 --> 00:05:49.120
And then cytochrome
B is going to pass

00:05:49.120 --> 00:05:53.870
it to cytochrome C1, and then
cytochrome C, cytochrome A,

00:05:53.870 --> 00:05:58.130
and finally, they're going
to end up in sulfite.

00:05:58.130 --> 00:06:00.680
Another thing to
notice here is that

00:06:00.680 --> 00:06:05.540
except for the initial
flavin and the final electron

00:06:05.540 --> 00:06:09.950
acceptor, sulfate, all the other
intermediates get regenerated.

00:06:09.950 --> 00:06:12.770
So we go from the oxidized
version to the reduced version

00:06:12.770 --> 00:06:14.480
and back to the
oxidized version.

00:06:14.480 --> 00:06:18.800
So all these electron carriers
are going to be sufficient only

00:06:18.800 --> 00:06:21.080
in catalytic amounts.

00:06:21.080 --> 00:06:22.910
So the only thing
that gets consumed

00:06:22.910 --> 00:06:26.060
is the FADH2 and the sulfate.

00:06:26.060 --> 00:06:29.030
These are two reactants.

00:06:29.030 --> 00:06:31.960
And we get in this
reaction FAD and sulfite.

00:06:35.970 --> 00:06:38.700
What we just said will
help us segue into the Part

00:06:38.700 --> 00:06:42.400
B of the problem, which asks us
to calculate how much energy do

00:06:42.400 --> 00:06:46.870
we get by converting
one molecule of FADH2

00:06:46.870 --> 00:06:53.360
and one molecule of sulfate into
FAD and sulfite, respectively.

00:06:53.360 --> 00:07:00.160
Now as we pointed out here,
only the FADH2 and sulfate

00:07:00.160 --> 00:07:02.890
are consumed in this reaction.

00:07:02.890 --> 00:07:07.290
All the other electron carriers
are recycled and regenerated

00:07:07.290 --> 00:07:10.300
in the course of the
electron transport chain.

00:07:10.300 --> 00:07:12.460
In order to
calculate the energy,

00:07:12.460 --> 00:07:15.640
it's useful first to write
the half reaction of the redox

00:07:15.640 --> 00:07:16.790
processes.

00:07:16.790 --> 00:07:20.840
Here are the two half reactions
of this redox process.

00:07:20.840 --> 00:07:27.400
FADH2 gets oxidized through FAD
and donates its two electrons.

00:07:27.400 --> 00:07:33.220
And the epsilon, or e0
prime is minus 0.22 volts.

00:07:33.220 --> 00:07:35.980
Now, this is the
potential from the table,

00:07:35.980 --> 00:07:38.330
and that's a
reduction potential.

00:07:38.330 --> 00:07:41.200
The equation as written
is an oxidation,

00:07:41.200 --> 00:07:42.820
and therefore, the
potential that we

00:07:42.820 --> 00:07:45.340
need to take into account
is the minus of this one.

00:07:49.000 --> 00:07:52.060
Sulfate is then going to
accept the two electrons

00:07:52.060 --> 00:07:55.330
and going to get reduced
to the sulfite and water.

00:07:55.330 --> 00:07:59.620
And the electrochemical
potential for this

00:07:59.620 --> 00:08:02.750
is 0.48 volts.

00:08:02.750 --> 00:08:05.150
So now when we add
these two together,

00:08:05.150 --> 00:08:13.270
we get the overall process where
FADH2 gets oxidized by sulfate

00:08:13.270 --> 00:08:15.730
to generate FAD and sulfite.

00:08:15.730 --> 00:08:20.110
And the electromotive force
is just the mathematical sum

00:08:20.110 --> 00:08:22.660
of these two keeping
in mind that this has

00:08:22.660 --> 00:08:24.010
to be taken as a negative sign.

00:08:27.190 --> 00:08:29.902
Because, again, as
written, this is

00:08:29.902 --> 00:08:32.110
an oxidation and this the
potential for the reduction

00:08:32.110 --> 00:08:32.990
reaction.

00:08:32.990 --> 00:08:37.120
So electromotive force
is actually 0.7 volts.

00:08:37.120 --> 00:08:41.740
Now, we can easily convert
from the electromotive force

00:08:41.740 --> 00:08:46.257
to a delta g0 prime value,
and the relationship

00:08:46.257 --> 00:08:47.590
is written here, delta g0 prime.

00:08:47.590 --> 00:08:53.800
It's minus nF delta e0 prime
and is the number of electrons

00:08:53.800 --> 00:08:55.280
in the process as we see here.

00:08:55.280 --> 00:08:58.960
Two, F is the Faraday's
constant and delta e0

00:08:58.960 --> 00:09:02.780
prime is going to be
the electromotive force.

00:09:02.780 --> 00:09:04.960
And if we go through
the number crunching,

00:09:04.960 --> 00:09:10.070
we get a delta g0 prime minus
135 kilojoules per mole.

00:09:10.070 --> 00:09:12.400
Notice because it's a
negative number that means

00:09:12.400 --> 00:09:15.880
there's a spontaneous
process as written.

00:09:15.880 --> 00:09:17.710
And as you know,
the negative delta g

00:09:17.710 --> 00:09:20.740
will correspond to a
positive electromotive force.

00:09:20.740 --> 00:09:23.050
Now, we're just one step
away from calculating

00:09:23.050 --> 00:09:26.620
how much ATP we can
produce with this energy.

00:09:26.620 --> 00:09:31.360
As you know, we generate ATP
out of ADP and phosphate,

00:09:31.360 --> 00:09:35.800
and this is the reaction that's
catalyzed by ATP synthase.

00:09:35.800 --> 00:09:38.950
And it takes about 30.5
kilojoules per mole

00:09:38.950 --> 00:09:42.560
to form ATP out of
ADP and phosphate.

00:09:42.560 --> 00:09:46.860
Therefore, the 135
kilojoules per mole

00:09:46.860 --> 00:09:50.380
that we generated
from 1 mole of FADH2,

00:09:50.380 --> 00:09:55.210
it's going to be enough for
about 4 molecules of ATPs.

00:09:55.210 --> 00:09:59.680
This is in contrast, which
was the normal processes that

00:09:59.680 --> 00:10:03.640
use oxygen as their
final electron acceptor

00:10:03.640 --> 00:10:06.695
where out of one FADH2
molecule, will generate

00:10:06.695 --> 00:10:09.800
at most 2 molecules of ATP.

00:10:09.800 --> 00:10:13.330
So in some ways, sulfate is
actually a better electron

00:10:13.330 --> 00:10:15.340
acceptor and can
give us more energy.

00:10:18.460 --> 00:10:21.430
Part C of these problem
deals with a culture

00:10:21.430 --> 00:10:24.710
of this microorganism
in the lab.

00:10:24.710 --> 00:10:27.420
And we're adding to this
culture dinitrophenol,

00:10:27.420 --> 00:10:31.430
a compound we're told
has a pKa of about 5.2.

00:10:31.430 --> 00:10:34.000
So let's explore what happens
to the electron transport

00:10:34.000 --> 00:10:37.180
chain of the organism
when we add dinitrophenol.

00:10:37.180 --> 00:10:39.430
Here I put together a
cartoon representation

00:10:39.430 --> 00:10:43.610
of the electron transport
chain of our organism.

00:10:43.610 --> 00:10:48.100
So as you can see here, this is
the extracellular environment.

00:10:48.100 --> 00:10:50.630
This is the outer membrane.

00:10:50.630 --> 00:10:54.000
This is the inner membrane where
we have all these complexes

00:10:54.000 --> 00:10:56.710
I denoted here with these
rectangles of the electron

00:10:56.710 --> 00:10:57.860
transport chain.

00:10:57.860 --> 00:11:01.310
And FADH2, for example, is
going to donate its electrons.

00:11:01.310 --> 00:11:04.810
They're going to be passed
along all the way to sulfate.

00:11:04.810 --> 00:11:07.870
And in the process,
protons are going

00:11:07.870 --> 00:11:11.230
to get pumped into this
intermembrane space.

00:11:11.230 --> 00:11:15.520
Now, these protons can be
used in the ATP synthase

00:11:15.520 --> 00:11:19.570
as they travel back into
the intercellular space.

00:11:19.570 --> 00:11:24.130
Their energy can be used to
convert ADP and organophosphate

00:11:24.130 --> 00:11:28.690
to ATP as we just
discussed in Part 2.

00:11:28.690 --> 00:11:32.230
Now, to this organism,
we said we're

00:11:32.230 --> 00:11:33.880
going to add dinitrophenol.

00:11:33.880 --> 00:11:36.095
Here is the structure
of dinitrophenol.

00:11:42.400 --> 00:11:45.850
And we're told the
pKa of this proton,

00:11:45.850 --> 00:11:49.629
right here, the
pKa is about 5.2.

00:11:49.629 --> 00:11:51.670
When this compound diffuses
through the membrane,

00:11:51.670 --> 00:11:54.280
it's going to go through this
intermembrane space, which

00:11:54.280 --> 00:12:01.330
has a very low pH and
also in the intercellular

00:12:01.330 --> 00:12:04.470
space in the cytosol,
which has a much higher pH.

00:12:04.470 --> 00:12:08.500
So because pKa 5.2, it's a
relatively low, much lower

00:12:08.500 --> 00:12:13.840
than 7, pKa, in the
intermembrane space where

00:12:13.840 --> 00:12:17.890
it's more acidic, it's
going to be protonated.

00:12:17.890 --> 00:12:24.400
So we can write, for example,
dinitrophenol OH in equilibrium

00:12:24.400 --> 00:12:27.915
with dinitrophenol O
minus plus a proton.

00:12:31.990 --> 00:12:34.730
Now, because here we
have a lot of protons,

00:12:34.730 --> 00:12:37.990
this equilibrium will
be shifted to the left.

00:12:37.990 --> 00:12:41.830
That is the protonated
form of dinitrophenol.

00:12:41.830 --> 00:12:48.320
However, here in the
cytosol, the NPOH,

00:12:48.320 --> 00:12:50.800
it's going to be in the
same equilibrium O minus

00:12:50.800 --> 00:12:52.540
plus H plus.

00:12:52.540 --> 00:12:55.180
But because the pH is
fairly high, that is

00:12:55.180 --> 00:12:58.000
there are not a lot of
protons, this equilibrium

00:12:58.000 --> 00:13:00.400
is going to be
shifted to the right.

00:13:00.400 --> 00:13:06.600
This equilibrium is going
to be shifted to the left.

00:13:06.600 --> 00:13:09.330
So now look what happens.

00:13:09.330 --> 00:13:12.360
So because this equilibrium
has shifted to the left,

00:13:12.360 --> 00:13:16.500
it's going to keep soaking
up a lot of these protons.

00:13:16.500 --> 00:13:18.630
Then the neutral
dinitrophenol molecule

00:13:18.630 --> 00:13:24.270
is going to diffuse through
the membrane as such

00:13:24.270 --> 00:13:28.440
and enter the intercellular
space to cytosol where

00:13:28.440 --> 00:13:30.930
it's going to be deprotonated.

00:13:30.930 --> 00:13:33.400
The equilibrium is
shifted to the right.

00:13:33.400 --> 00:13:35.340
So in effect,
dinitrophenol is going

00:13:35.340 --> 00:13:39.210
to carry the protons from
the intermembrane space

00:13:39.210 --> 00:13:41.470
inside the cell.

00:13:41.470 --> 00:13:43.089
Now it's going to
do that in parallel

00:13:43.089 --> 00:13:45.630
with the protons that are going
to be flowing through the ATP

00:13:45.630 --> 00:13:47.640
synthase to generate ATP.

00:13:47.640 --> 00:13:51.570
So in effect, we're
discharging this battery

00:13:51.570 --> 00:13:53.340
where the concentration
of protons

00:13:53.340 --> 00:13:57.770
is basically our
electrochemical gradient.

00:13:57.770 --> 00:14:02.280
It's going to be discharging the
battery without producing ATP.

00:14:02.280 --> 00:14:05.760
So as you know, if you
short circuit a battery,

00:14:05.760 --> 00:14:08.510
the battery is going to heat
up because you're discharging

00:14:08.510 --> 00:14:10.230
an electrochemical gradient.

00:14:10.230 --> 00:14:13.692
Similarly, dinitrophenol,
by taking these protons

00:14:13.692 --> 00:14:15.900
from the intermembrane space
and bringing them inside

00:14:15.900 --> 00:14:17.700
into the intercellular
space, it's

00:14:17.700 --> 00:14:20.920
going to be generating heat.

00:14:20.920 --> 00:14:23.850
Therefore, we can
answer Part C by saying

00:14:23.850 --> 00:14:27.000
that the medium in which
these cells are growing

00:14:27.000 --> 00:14:30.330
is going to heat up when
we add dinitrophenol to it.

00:14:30.330 --> 00:14:33.840
The processes described in this
problem are fairly universal.

00:14:33.840 --> 00:14:38.450
Now, in eukaryotes, like
more evolved organisms,

00:14:38.450 --> 00:14:41.560
they would happen
in the mitochondria.

00:14:41.560 --> 00:14:43.350
Now, if you look
back at this diagram,

00:14:43.350 --> 00:14:46.320
if this was the double
membrane of the mitochondria,

00:14:46.320 --> 00:14:48.480
this would be the
inside of the cell that

00:14:48.480 --> 00:14:50.100
contains the
mitochondria, this would

00:14:50.100 --> 00:14:51.780
be the intermembrane
space, and this

00:14:51.780 --> 00:14:53.640
will be the inside
of the mitochondria

00:14:53.640 --> 00:14:56.400
or the mitochondrial matrix.

00:14:56.400 --> 00:14:59.700
Similarly, by adding a
compound like dinitrophenol,

00:14:59.700 --> 00:15:02.490
who can dissipate the
electrochemical gradient

00:15:02.490 --> 00:15:07.440
in the mitochondria and
cause the cell to heat up.

00:15:07.440 --> 00:15:10.170
In fact, this process
is actually used

00:15:10.170 --> 00:15:13.650
by a number of organisms
to generate heat instead

00:15:13.650 --> 00:15:17.460
of chemical energy, or ATP.

00:15:17.460 --> 00:15:23.250
For example, the brown fat
cells in newborns in mammals

00:15:23.250 --> 00:15:25.170
have a special
protein that allows

00:15:25.170 --> 00:15:27.010
to dissipate this
electrochemical gradient

00:15:27.010 --> 00:15:30.030
in the mitochondria
to generate heat.

00:15:30.030 --> 00:15:34.890
Another good example is
the seeds of many plants.

00:15:34.890 --> 00:15:36.930
When they germinate,
they actually

00:15:36.930 --> 00:15:38.520
generate a lot of
heat that can be

00:15:38.520 --> 00:15:41.940
used to melt the ice or
the snow around them.

00:15:41.940 --> 00:15:43.500
That's why some
of the plants can

00:15:43.500 --> 00:15:45.730
start growing even
before the snow has

00:15:45.730 --> 00:15:48.284
melt in the early spring.

00:15:48.284 --> 00:15:49.950
I hope that working
through this problem

00:15:49.950 --> 00:15:52.530
will help you understand
better the inner workings

00:15:52.530 --> 00:15:54.660
of an electron
transport chain and how

00:15:54.660 --> 00:15:58.830
it can convert the chemical
energy of chemical bonds

00:15:58.830 --> 00:16:01.500
into an electrochemical
gradient, which

00:16:01.500 --> 00:16:05.280
can then be used to generate
high energy compounds like ATP.

00:16:05.280 --> 00:16:08.780
Or it can be dissipated
to generate heat.