The following content is provided under a Creative
Commons license.
Your support will help MIT OpenCourseWare continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare
at ocw.mit.edu.
PROFESSOR: All right, so let's do our first clicker question.
All right, let's just do 10 more seconds.
Interesting.
So, in this problem you were given the maximum rate, Vmax,
which, by the way, you calculated in class last time.
And then said what will the substrate concentration be when
you have half of that rate.
And so, when you have half of the maximum rate, that's
the definition of k m.
So all you have to do is if a k m is given, you just
write down that value.
So that's the level of question you'll get on enzyme kinetics.
All right, so that should be a good tie breaker, too.
All right, so in this little review and we talk about the
material from the second half of the course, and why I think
this material is really important and cool, and that's
because it all represents the basic principles of
how enzymes work.
And as a biochemist, I recognize this material
is really important.
So that's what interests me particularly about it.
And the point that I want to make in general is that when
you're studying introductory material, sometimes it seems
like you're never going to really need that material
again, but that's because you haven't seen the connection
between it in other topics.
So, a lot of the things you've learned will be related to
things that you're going to see you later on.
So I'm going to give you examples and review examples of
how the material you've learned will allow you to understand
an enzyme, for example.
So, we talked last time that one reason why you care about
understanding how an enzyme works is because a lot of
people want to inhibit enzymes.
Well, why would you want to inhibit enzymes, and that's
because inhibiting enzymes is very useful to treat
a bunch of things.
A number of you will, in the next week, I'm pretty sure, be
interested in ways to inhibit enzymes to prevent headaches
or to decrease headaches.
And this is true of students and faculty during finals week.
And so, some of the treatments for headaches that a number of
those pharmaceuticals are geared toward enzymes called
prostaglandin synthesis, and they inhibit those enzyme from
working and that gets rid of your headache.
Arthritis, for example, is also treated by a lot of the same
medications that treat headaches.
Cancer, often chemotherapy or you're giving people inhibitors
of enzymes to decrease tumor growth.
And we talked last time about inhibiting HIV proteases as
part of a treatment for HIV.
So, understanding enzymes is very important for the
pharmaceutical industry, and so that's one example of why you
need to know about chemical equilibrium, why you need to
know about acid based oxidation reduction transition
metals, and also kinetics.
So, I'm going to give you a case study.
You've seen some of these examples before, but we're
going to put them all together now.
And I try in every lecture, if I can, to mention vitamin B12,
and here, we're going to tell you about a vitamin B12
dependent enzyme, and how knowing what you know now about
these topics would allow you to understand how this
enzyme works.
So, first, kinetics, this is easy, methionine
synthase, it's an enzyme.
So, what does this enzyme do?
So this particular enzyme converts an amino acid called
homocysteine to another amino acid, methionine.
It also converts another vitamin, a B vitamin, the
methyltetrahydrofolate form to tetrahydrofolate, so this
folate is one your B vitamins.
Why do you care?
Well, methionine is needed for proper formation of brain and
spinal column, and it's particularly important when
women are pregnant that they have enough folic acid so
that they can make enough methionine.
If they do not have enough folic acid, then you can
have babies born with neural tube defects.
Also, as I mentioned before, inhibiting methionine synthase
raises homocysteine levels, so you can't do this conversion,
and that increases risk of heart disease that homocysteine
is actually a really good indicator, high homocysteine
levels of whether someone's going to have heart
trouble or not.
This is also a target for chemotherapy, because you need
tetrahydrofolate to do DNA biosynthesis, and tumor cells
that are growing, tumors that are growing, need a lot
of DNA biosynthesis.
So if you inhibit this enzyme, you can inhibit
DNA biosynthesis, so it's a potential
chemotherapeutic target.
There is some use of this in Europe.
Actually there is a treatment that's given that's
specifically supposed to inhibit methione synthase, it's
not used so much in this country, but laughing gas will
inhibit methione synthase.
And so, in Europe some cancer patients are treated with
laughing gas as a way to decrease their tumors.
I'm not sure it's the most effective therapy, but they're
at least pretty happy with it.
So that's the counterpart.
So, this is an important enzyme.
And enzymes, of course, as we know are catalysts and they
catalyze reactions, so tell me what you know about catalysts.
Which is the following statements about
catalysts are true?
OK, let's take 10 more seconds.
Very good.
All right, so going back to the notes, statement 2 is true.
So, catalysts work by lowering the activation energy barrier
for both the forward and the reverse reaction.
So both rates would be changed.
And also, catalysts are not going to affect the
thermodynamics, they're not going to shift the
equilibrium toward products.
What's something that you could use to shift an equilibrium,
say, to products or reactants.
Temperature, yeah.
So, just a little review of kinetics.
Again, a catalyst works by lowering the activation energy
barrier, or another way to say that is by stabilizing a
transition state, so it's going to affect the activation energy
for the forward reaction.
And for the reverse reaction, it's going to decrease both
of them by the same amount.
So if you know how much one is decreased, you also know how
much the other is decreased.
And we also know that it doesn't affect
the thermodynamics.
So, does it change the equilibrium constant?
No, it doesn't change the equilibrium constant.
All right, so a little review of kinetics.
So now let's talk about transition metals.
So this particular enzyme needs 2 different kinds
of transition metals.
It needs cobalt in the form of vitamin B12 and
it also needs zinc.
So we talked about methylcobalamin, so
cobalamin is another name for vitamin B12.
And here we have a methyl ligand, c h 3 is a methyl
ligand coordinated to the cobalt in the center, and we
saw before that when you have a thing that forms attachments at
more than one point, it's called a chelate, so the corn
ring here is a chelating agent, and it's a tetradentate
ligand because there are 4 points of attachment.
So we can have monodentate, bidentate, tetradentate,
hexadentate, and tell you the number of points of attachment.
And again, this is a chelate, and what do you know
about a chelate effect?
I want someone to tell me what the chelate effect is, and
there's a special bonus prize of a teeshirt for
the best answer.
Who wants to give it a try and tell me what
the chelate effect is.
Any volunteers?
Bonus teeshirt.
Who wants to give it a try.
Not very brave.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, so chelate, it's something that's going to be
binding to a metal at multiple points of attachment.
And what do we know about how stable or how favorable that
kind of interaction is?
STUDENT: It makes it more stable because [INAUDIBLE].
PROFESSOR: Are you reading from your notes?
Well, you know what, I didn't say that was not allowed,
so congratulations.
Right, so chelates are unusually stable due
to entropic affects.
So it's an entropy affect.
So if you have a metal that's free in solution, it's going
to bind to a lot of water molecules, and often, metals
like to bind to thick things.
You see a lot of octahedral complexes.
So if you have a metal bound to 6 waters, and you bring in a
chelate that will blind with multiple points of attachment,
such as EDTA, which binds at 6 points of attachment, that
1 molecule of EDTA will displace 6 water molecules.
So that is going to be entropically favorable.
6 things floating around in solution have a lot more
entropy than 1 thing floating around in solution.
And this makes chelates very stable.
And so, if you want to get -- if someone is poisoned with the
metal and you want to get that metal out of your system, you
give them a chelate such as EDTA, hospitals have EDTA.
It's also good, as we learned, for cleaning bath tubs.
All right, so that's the chelate effect.
All right, so what about zinc.
This enzyme has a zinc and it has zinc in the plus
-- zinc plus 2 ion.
So, if you know that, you can calculate a d count.
And so again, we can go to our periodic table.
What group is zinc in?
12.
So what would be the d count?
10.
So 12 minus 2 equals 10.
So what would you think would be true about the color
of this d 10 system?
All right, so let's just take 10 more seconds.
All right, so it would be colorless.
And so, all of the d orbitals are going to be full, and so
there's no transition from one set of d orbitals to the other,
and so it would be a colorless compound.
And that is, in fact, true that it took years and years of
studying methionine synthase before people realized that it
had a zinc in it, no one knew that there was a zinc, because
they couldn't see it.
It didn't add a color, and so no one knew that a metal
was there for a long time.
All right, so what about oxidation reduction?
So, this enzyme does a lot of different oxidation
reduction reactions.
So, when it reacts with homocysteine to form
methionine, and the vitamin B12, which is shown in a
cartoon form down here, goes from a plus 3 state that has
the methyl associated, it gives the methyl group, the c h 3
group, up to homocysteine to form methionine, and
then you're in a plus 1 state of cobalt here.
But this state can lose an electron, and you can have
a what's called [? cobe ?]
2 form of the enzyme, which you need to put another electron in
and reduce it to go back, and you also need to
methylate it again.
And the methyl group comes from s adenosylmethionine.
So, we need to think about how we're going to reduce the
vitamin to get it back in its primary turnover state.
so, we know the redox potentials for this.
The standard reduction potential for vitamin B12 is
minus point 5 2 6 volts.
The potential reduction potential for flavodoxin, which
is the protein that gives it the 1 electron, and it's a
flavin protein, which is another B vitamin.
And that's minus point 2 3 0 volts.
So, what do we know about these?
Tell me which is the better reducing agent.
All right, let's just do 10 more seconds.
Click in your responses.
OK.
So, vitamin B12 is a better reducing agent.
If you have a large negative number, then the reduced
species is very reducing.
So it wants to reduce other things and get oxidized itself.
So, that's interesting.
It seems like vitamin B12 should be reducing flavodoxin
and not the other way around.
So we can look at how unfavorable this process is,
and we can calculate a standard potential for this, and we can
use this equation where we have reduction minus oxidation and
plug the values in, and so we have a minus point 5 2 6 minus
a minus of the floating potential, and that gives us a
negative value for delta e.
So is this reaction spontaneous?
No, and that's because if you have a negative standard
reduction potential, then you're going to have a positive
value for delta g, so it will not be spontaneous.
And again, you can calculate the value for delta g, so we
have minus n, number of moles of electrons, Faraday's
constant times the difference in potential, and it's a 1
electron process, we put in Faraday's constant and the
calculated value for delta e, and we get a positive
28 point 6.
And as I mentioned before in class, the way this reaction
goes is that at the same time an electron goes in, you also
cleave s adenosylmethionine, which gives the methyl group to
also return to the catalytic cycle, and that process is very
favorable minus 37 point 6 kilojoules per mole.
So it drives the unfavorable reaction.
So in the body you will have a favorable oxidation
reductions and you'll have unfavorable ones.
All right, so what do you a cell that requires
energy to catalyze a non-spontaneous reaction?
So, that's an electrolitic cell.
And a cell that catalyzes a spontaneous reaction?
Galvanic, right.
So again, with oxidation reduction, it doesn't matter if
you're talking about a battery, or if you're talking about a
cellular process, all the same equations apply.
All right, what about acid based equilibrium.
So in this particular -- in this catalytic cycle, we are
converting homocysteine to methionine, and that chemistry
involves acid base.
So what is true about the reaction is that the
protonation state of the substrate,
homocysteine, matters.
So, you can have a protonated homocysteine, and here's the
structure of homocysteine, a deprotonated homocysteine
where the sulfur does not have a proton anymore.
So we have an s minus here.
And this protonated form of homocysteine can be converted
to methionine where there's a methyl group attached
to the sulfur here.
So, we want to know at physiological p h 7 point 4,
how much of the homocysteine is deprotonated?
And here, if you're asking how much of something is in a
protonated state, how much of something is in a deprotonated
state, what you are asking is what is the p k a
of homocysteine.
So what is the p k a of homocysteine?
That's what you need to know to figure out how much would be in
the protonated state, how much would be in the
deprotonated state.
So here, the p k a is 10.
So now, without doing any calculations, I want you to
tell me what you expect about how much is protonated and how
much is deprotonated at physiological p h if
the p k a is 10.
All right, let's just take 10 more seconds.
I hope that I mentioned that acid base will
be on the final exam.
So good thing we're reviewing it right now.
All right, so now let's look at the math.
So if you're given a p k a and p h and asked about a ratio of
protonated to deprotonated it's OK to pull out your favorite
Henderson Hasselbalch equation, which gives you a sense of the
ratio, can predict the ratio, again it's approximation,
but predict the ratio of protonated to deprotonated.
H a, of course, being an abbreviation for protonated,
a minus for deprotonated.
And so we can calculate here that if you have a p h of 7
point 4, p k a of 10, you get a ratio of 400:1.
So that's the math, but you can also just think about it
in terms of what you know.
So let's just do a brief review.
So the answer here is that free homocysteine is protonated
and non-reactive at physiological p h.
But let's just think about this question a little more.
So this is not a figure in today's handout, but you've
seen this a few times, so let's just look at it for a minute.
So, we talked a lot about acid based titrations.
We talked about so here we have a weak acid tritrated
with a strong base.
In the beginning it's a weak acid problem, at the
equivalence point it's a weak base problem, because we've
added enough moles of our strong acid to convert all of
our weak acid to its conjugate base.
And in the middle when you've added half the number of the
moles, you've converted half of your weak acid to its
conjugate, then the p h is equal to the p k
a at this point.
So that would be right in the middle of this
buffering region.
Again, in a buffering region you have quite a bit -- you
have quite a bit of your weak acid and quite a bit of a
conjugate, so it can buffer.
If strong acid is added, then it can be used up, if strong
base is added, it can be used up without changing
the p h very much.
That's a buffer, so the p h curve is flat in here in
that buffering region.
And so, when you think about this, if you're at p h's that
are below the p k a, then you're going to be more
protonated, and p h is above the p k a, you'll be
more deprotonated.
So let's look at the figure now that's in today's handout
and think about this.
So when the p h equals the p k a, you will have equal number
of moles, of something that's protonated as deprotonated.
And if you're at p h's above the p k a, you'll
be more deprotonated.
P h's below the p k a, you'll be more protonated.
And in the particular example I gave, we have a p k a of 10, so
at 7 point 4 we're at a p h that is below the p k a, and so
here they'd be more protonated.
So if you do the math it's 400:1.
But even if you aren't doing any math, you should think
about the fact that that would have more of the
protonated form than the deprotonated form.
All right, but the enzyme has a problem, because only the
deprotonated form is active.
The p k a of the free homocysteine is 10, and you're
at physiological p h, so this reaction doesn't seem
like it should go.
But again, it's an enzyme, and enzymes catalyze reactions.
So they do things to make that reaction go faster.
And what this particular enzyme does is it lowers
the p k a of homocysteine.
So when homocysteine is bound to the enzyme, it's p k a
is no longer 10, but now its p k a is 6.
So, how does the enzyme do this?
Well, that's where the zinc comes in.
So, the zinc binds to the homocysteine and it acts as
a Lewis acid as it binds to the homocysteine.
So the homocysteine is your donor ligand, and you have
your metal as the Lewis acid, your acceptor.
And that alters the p k a, so it actually associates, zinc
associates with this sulfur, and that changes the p k a.
So, I just wanted to mention again, I was having dinner with
some of my faculty colleagues who teach organic
chemistry here.
We were interviewing another job candidate, and they looked
at me and said, they're teaching organic this semester
and they said we asked our class about p k a's, and they
all insisted that they had never heard about p k a's
in their freshman chemistry course.
And I, of course, said, well, they did not take 511-1 then.
So just remember, especially when it's Barbara Imperiali
asking, did you hear about p k a's, the answer is?
STUDENT: Yes.
PROFESSOR: Thank you.
OK, so just this week.
So it alters the p k a here.
So now what's our situation?
Well, now at physiological p h, we have a p k a of 6, and so
that gives us a very different ratio.
Instead of 400:1, we have 1:25.
So most of the homocysteine is deprotonated at physiological
p h, which means that it can react.
So, if we go back to this now, our p k a is now 6,
and so physiological p h is now above that p k a.
So when you're above the p k a, you should have more
deprotonated than protonated.
So that's how you can rationalize these things.
All right, now I have to ask, all right, we do
not need a tie breaker.
So at the end of the lecture, we will announce the winner.
But let's first do chemical equilibrium.
Chemical equilibrium.
So we didn't talk too much about this in chemical
equilibrium, but enzymes can have alternate conformation,
so the enzyme can change its shape during chemistry.
And that those conformation of the enzyme can
be in equilibrium.
So the enzyme itself can be an equilibrium with different
conformational states.
So here, there are a lot of states of the enzyme.
It needs to react with homocysteine, it needs to react
with methyltetrahydrofolate and with s adenosylmethionine.
And when the structure was determined to this enzyme, here
in green is the vitamin B12, and red is the methyl group.
You see that the methyl group is pretty buried, you can
barely see it, but yet it needs to interact with homocysteine
to give it a methyl group, it needs to take a methyl group
off methyltetrahydrofolate, it also needs to take a methyl
group off s adenosylmethionine.
But it doesn't seem like there's any room for any of
them to actually get in there.
So, this was a sign that there has to be some kind
of change in the structure.
So here's another picture of the structure.
Here's the vitamin B12 in green, the methyl group in red,
and there's this whole protein part up here that has to get
out of the way to do the chemistry.
And, in fact, we do know from experimental data that it
does move so you can do the chemistry.
So that means that the enzyme has to have a number of
different structures.
It's modular.
Here's the vitamin B12 binding domain, the vitamin B12,
there's this region that have to move called the methyl cap.
It needs to interact, the B12 here needs to interact with
a folate domain, a homocysteine domain, and an
activation domain.
So you need to have at least these 4 different structures,
and these 4 structures will be in equilibrium with each other.
So you need to have a structure with folate, a binding domain
on top of the red B12, you need to have homocysteine in yellow
on top of the B12, you need to have a resting state where
those helices are on top of the B12, and you need to have an
activation state where the [? adomet ?]
domain is on top of the B12, and all of these states are
going to be in equilibrium with each other, and they're going
to be moving, which means that enzymes are dynamic, which
means that chemistry is dynamic.
Chemistry in the body is not in the solid state.
Chemistry in the body is in solution.