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BOGDAN FEDELES: Hello, and welcome to 5.07 Bio Chemistry
Online.
I'm Dr. Bogdan Fedeles.
Let's metabolize some problems.
Today we're going to be talking about Problem 2
of Problem Set 4.
Here we're going to be discussing in detail
the mechanism of HMG-CoA synthase, a key enzyme
in central metabolism responsible for making the five
carbon building blocks from which
all sterols, such as cholesterol and steroid hormones are made.
HMG-CoA synthase catalyzes the following reaction.
It takes acetyl-CoA, which we're going
to encounter a lot in the central metabolism,
and combines it with acetoacetyl-CoA,
another thioester.
In this process, the one molecule of CoA is lost,
and we get the product hydroxymethylglutaryl
CoA or HMG-CoA, as you notice the initials H, M, and G are
part of this name.
To help us understand the mechanism,
a crystal structure of this HMG-CoA synthase
is provided in this problem, and it's shown here.
Question 1 of this problem is asking
us to provide a detailed curved arrow
mechanism of the reaction catalyzed
by the HMG-CoA synthase.
And for that we're going to use the information provided
in the crystal structure.
And we are also given an additional hint
and that is the acetyl CoA substrate
provides the CH2CO2 CO2 minus moiety in the HMG-CoA product.
Let's take a look.
So the hint is telling us that these two carbons, which
I'm going to label in blue from acetyl-CoA,
are, in fact, these two carbons, CH2 COO
minus in hydroxymethylglutaryl CoA.
That means that the other four carbons in the product
must be the carbons from the acetoacetyl-CoA.
Let's label these red.
So now since we have a thioester functionality in both cases,
these are probably the same, so the carbons
from a acetoacetyl-CoA are as follows.
So notice the HMG-CoA synthase, it
accomplishes the formation of a carbon-carbon bond,
and that is this bond right here.
It's between the CH3 carbon acetyl-CoA
and the carbonyl carbon of acetoacetyl-CoA.
And in the process, this carbonyl
will become a hydroxyl group.
Now, let's take a look at the crystal structure
and see what information we can gather there.
Now this is a picture of the active site
of the HMG-CoA synthase.
We notice the acetyl-CoA moiety is actually
now bound to this C111.
As you know C is a cysteine--
so it's actually covalently bound
to a cysteine-- in the active site.
Now, notice this is the other substrate,
which is the acetoacetyl-CoA and bound in the active site.
So as we've discussed before, we're
going to make this carbon-carbon bond,
and that's going to happen between this carbon
here and this carbon here.
So this bond needs to be formed.
However, let's take it one step at a time.
The reaction will start by forming
this thioester between the acetoacetyl-CoA carbons
and the cysteine in the active site of the enzyme.
In many reactions that use a cysteine in the active site,
which is used to form a covalent bond to the substrate,
we first need a base to deprotonate the cysteine
and make it a really good nucleophile, which
will subsequently attack the substrate
and form a covalent bond.
Now, in a lot of these cases, the base is a histidine.
Let's take a look if we see a histidine in our structure.
There is a histidine in the structure, H233.
But if you look at this histidine,
it's quite far from our cysteine here.
Well, obviously, because this is a projection,
this is a two-dimensional structure so that we only
see here a projection of a three-dimensional structure,
it's very hard for us to tell if this histidine is,
in fact, close enough to deprotonate this cysteine.
So even though we have this crystal structure,
from this one picture, we do not have enough information
to figure out, well, what is the base that
will deprotonate cysteine before it reacts with acetyl-CoA.
Remember this is often the case with crystal structures that
because of the resolution which we can collect them,
we cannot see the protons.
And if we cannot see the protons,
it's very hard to tell which amino acid residues are
protonated and can serve as a general acid or general bases.
A lot of the researcher's intuition
comes into play to draw these kind of structures.
And it's only with collecting many different kinds
of experimental evidence that we can put together
a more definitive mechanism.
Therefore, we might not know for sure
from this one picture, which is the general base that
deprotonates the cysteine at 111.
So we're not going to assign it for now.
Let's try to write the mechanism for that part.
Here is the acetyl-CoA substrate,
and here is the active site cysteine 111
with its thiol group.
And again, we're not going to assign it,
but there will be a base in the active site of the enzyme that
will need to deprotonate this cysteine before it can react.
Therefore, the reaction first involves
activation of the cysteine, is deprotonated, and then
cysteine can attack the thioester
and form a tetrahedral intermediate.
As we've seen it before, for a lot of these reactions
involving thioesters.
In the first step, we will form four bonds
to this carbonyl carbon, and we have
formed a new bond between the thiol of the cysteine 111
and the starting material, acetyl-CoA.
Of course, there is a negative charge here.
And our base in the active site of the enzyme
will now be protonated.
Now that the tetrahedral intermediate
is going to fall apart by kicking off the CoA moiety.
Of course, this will get protonated presumably
by the same base.
Now it's a general acid that will protonate a CoA.
So we obtain the thioester with the cysteine in the active site
and one molecule of CoA is going to be leaving.
And here is our base.
Notice the tetrahedral intermediate
that we're forming here.
We have an oxygen that develops a negative charge.
And this is very similar to the kinds of intermediate
you've seen in the serine protease mechanism.
Now in that case, such an intermediate
was stabilized with hydrogen bonds
from the backbone of the protein in a structure that
was called an oxyanion hole.
Now let's take a look at the crystal structure of HMG-CoA
to see how a tetrahedral intermediate might
be stabilized.
Here is the acetyl of the acetoacetyl-CoA substrate.
The acetyl now is bound to the cysteine
111 here in the active site.
Now, let's take a look if there are any other residues close
enough to this oxygen. One of them that's shown here
is this NH, which belongs to an amide bond, the backbone amide,
and that's part of the amino acid serine 307.
So this distance here, C3.06 Angstrom,
it's actually close enough for a fairly good hydrogen bond.
Presumably, this interaction will stabilize the binding
of acetyl-CoA in this region of the active site
and may also be involved in catalyzing
the reaction with the cysteine by stabilizing
the tetrahedral intermediate.
In that case, this distance will have
to become even smaller, that is to form a really good hydrogen
bond on the order of 2.6, 2.7 Angstrom.
Since this is only one snapshot of the reaction,
we don't have enough information to tell if this is
a key catalytical interaction.
Once acetyl-CoA has reacted with the enzyme,
hence, formed the thioester with the cysteine 111,
we're now ready to proceed with the reaction
and form a carbon-carbon bond.
Let's take a look.
So in the next step of the reaction,
we want to form a carbon-carbon bond between this methyl
group here and this carbon of the carbonyl
of the other substrate.
So first of all, we need to deprotonate this methyl group.
We're going to form an enolate.
Then the enolate is going to attack this carbonyl.
Once again, we look for a suitable base, and we see,
for example, this glutamate 79 may be, in fact,
serving as a general base to deprotonate the methyl group
here and form the enolate.
Once the enolate is formed, it's going to attack this carbonyl,
and this oxygen is going to start
developing a negative charge, which needs to be compensated
by a general acid.
And it looks like this histidine 233 is close enough
to donate a proton and generate a hydroxyl group here.
All right.
So let's try to write a mechanism based
on what we just said.
Here is the thioester between cysteine 111 and acetyl.
And let's show the general base.
It's going to be glutamate 79.
Because it's a base, I'm going to put a negative charge on it.
There you go.
And this will serve to deprotonate
the methyl from the acetyl, moiety here, and generate
an enolate.
So it's picking up a proton, electron move,
and it's going to generate, as before, a negative charge
on the oxygen, which may be stabilized
by that interaction with the [INAUDIBLE] hydrogen
that we just discussed.
Now, let's try this again.
And here is our enolate.
In the next step, the enolate now
is going to attack the carbonyl of the acetoacetyl-CoA
and form the carbon-carbon bond.
All right.
This is the acetoacetyl-CoA, and we
discussed that next to this oxygen
there is our histidine 233, which we're going
to show it as being protonated.
Therefore, the reaction proceeds as follows.
Here we're forming the carbon-carbon bond.
Then the oxygen is going to pick up
the proton from the protonated histidine
and it's going to form a hydroxyl group.
Therefore, we get cysteine 111 is still
attached to what is known the HMG-CoA product.
So this is our carbon.
This is the new carbon-carbon bond.
We have here a methyl group.
We have the new hydroxyl, which were formed here,
and the rest of the molecule.
And our histidine, it was protonated here,
now it's going to be deprotonated.
So we just saw how the carbon-carbon bond is formed
in the course of this reaction.
And now we're left with a six-carbon thioester
with a cysteine 111 in the active side of the enzyme.
Therefore, the last step of the reaction
would involve hydrolysis of this thioester to free the product
and regenerate the system.
So for the last step, we need to hydrolyze this thioester,
and for that, we're going to need
to activate a water molecule.
Once again, we don't know--
here is the water molecule--
we don't know what's the general base, the residue
in the active site, which removed this proton from water
to allow it to attack the carbonyl of the thioester.
So we're going to call it a general base attached
to the enzyme.
So this base is going to pick a proton from water,
and then the water is going to add to the carbonyl
and once again generate a tetrahedral intermediate.
Say, this the OH from water, and this
is the rest of the molecule.
Once again, this is a tetrahedral intermediate, very
similar to the one we saw in the first step
when we formed this thioester, and presumably, it's
going to be stabilized in a fairly similar manner.
Let's also show that this base attached to the enzyme
is now protonated, and it would probably
donate this proton to reprotonate the cysteine
and reform that thiol group.
So in the second step, thiol takes the electrons,
which picks up the proton from the general base
in the active site.
So at this point, we're just going
to release the thiol of the cysteine 111,
and the rest of the molecule is exactly our product,
which is the HMG-CoA.
The curved arrow mechanism we just wrote
answers part 1 of the problem.
Second question of this problem is
asking about the stabilization of
the tetrahedral intermediate, which actually we
have just discussed.
But let me reiterate as you guys saw
in the serine protease mechanisms,
whenever we're forming this tetrahedral intermediate,
they tend to be stabilized in an oxyanion hole, which
is basically a structure of the enzyme, which can form hydrogen
bonds with the partial negative charge or full negative charge
that develops in a tetrahedral intermediate.
Presumably, similar structure exists
for this HMG-CoA synthase, but because we're only given one
snapshot, one crystal structure, that is not
sufficient information to say for sure which
are the key interactions to stabilize
these tetrahedral intermediates.
A lot more work, a lot more experimental data
is necessary to figure out which are the key hydrogen bonds
and interactions that stabilize the tetrahedral intermediates.
Question 3 is asking us to review the mechanisms by which
enzymes can achieve their amazing rate
acceleration, which is on the order of 10 to the 6
to 10 to 15 times over the uncatalyzed reaction
and of course, to point out these mechanisms in the context
of the HMG-CoA synthase.
As you have seen over and over in this course,
the three general mechanisms by which
enzymes accelerate reactions are binding energy,
general acid/general base catalysis
and covalent catalysis.
Now let's take a look at our structure
and figure out how each one of these mechanisms
might be operating.
To start off, covalent catalysis,
it's pretty obvious here the acetyl-CoA substrate first
reacts and forms a covalent bond with the cysteine
of the HMG-CoA synthase.
So this covalent attachment to the enzyme
allows this residue to be positioned just right
so that it can react with the other substrate.
Now, general acid/general base catalysis,
we've seen all these residues that participate.
Obviously, in order for this to react,
we need a base to deprotonate the cysteine.
Then we need a base, presumably, this glutamate 79,
to deprotonate the methyl group here to form the enolate.
And we need an acid to stabilize and form
this hydroxyl group that will be developing on this oxygen.
So covalent catalysis and general acid/general base
catalysis, that's pretty obvious.
Now when it comes to binding energy,
this is not something that we can obviously
derive from the structure, but we
can postulate the number of ways in which binding energy
contributes to this reaction.
First of all, both of the substrates
need to bind to the enzyme.
In order to do that, they need to be desolvated,
that is to remove all the water molecules that surround them.
So that by itself requires energy.
The binding energy is also derived
by when we align the substrates in the active site
of the enzyme, we align them so closely the right geometry
and within a few tenths of an Angstrom
so that the right orbitals overlap and allow
the reaction to happen.
So also the ability to align this residue
so closely that also contributes to the binding energy.
And finally, the binding energy also
manifests when we're stabilizing, for example,
the transition state of the reaction
relative to the binding of the substrates.
So if, for example, for our tetrahedral intermediate
that will form here, if that transition state
or the tetrahedral intermediate is stabilized more
than the substrate, then the reaction is accelerated
and proceeds towards that pass.
Part 4 of the problem is asking us
to look up the structure of coenzyme A, or CoA
and then contrast the reactivity of say, acetyl-CoA,
the thioester with CoA, with the reactivity of a thioester
with a much smaller thiol group.
Let's see what the coenzyme A looks like.
Here we have the structure coenzyme A.
The business end of the molecule is
this thiol group, which is attached
to a substantially long arm.
And at the end here, we have, as you can recognize,
two phosphates, the ribose and the base adenine.
This is a nucleotide is ADP.
There's also another phosphate here.
But the business end of the molecule is this thiol group,
and we're asked to contrast whether a thioester
form with coenzyme A would behave similarly to a thioester
form with this right-hand portion of the molecule, which
I wrote here.
So it's a much shorter thiol.
Now, it turns out these thioesters will be very, very
similar because really it's only the thiol moiety that we
need to form thioester.
So those thioesters will behave very, very similarly.
Now, the advantage of having such a long arm
for the coenzyme A is that it provides a way
to insert the substrate, which will attach here,
to say, acetyl-CoA, to insert acetyl-CoA in very deep
into the active site of the enzyme.
Having a long arm to guide the thioester may be important.
As you will see later in the course,
fatty acid synthases, which are these mega dalton complexes,
have multiple active sites.
So the fatty acid attaches a thioester
to coenzyme A allows it to be moved
through different active sites and through a same kind
of chemistry in multiple steps over and over again.
The advantage of having a coenzyme A thioester
is that it may provide some additional binding energy.
That nucleotide portion of coenzyme A
may interact specifically with the enzyme
near the active site, whereas, a much smaller thiol
like the one we just saw will not
have that kind of interaction available.
Since we've been talking about thioesters
throughout this entire problem, the last question
is asking us to rationalize why we're
seeing thioesters in metabolism as opposed to oxygen esters.
Now, the key message you need to remember here
is that the resonance that we observe in oxygen esters
is almost completely absent in thioesters.
And this fact makes the thioesters less stable
and therefore, more reactive.
Not only their carbonyl group behaves more
like a ketone group, but their alpha hydrogens
are more acidic.
Let's take a look.
Here is an oxygen ester that shows
a proton in alpha position.
And as you know, the lone pairs on this oxygen
can conjugate with the carbonyl group
and form certain resonance structures, one of which
being this one.
So the electrons can move like this,
and then we're going to have a negative charge here
and a positive charge here.
And this is possible because the electrons on both oxygens
are found in orbitals of comparable energies.
By contrast, in the case of a thioester, we have a sulfur.
Now, the electrons on the sulfur are
in orbitals that are not in comparable energies
with the ones on the ketyl group and therefore,
this kind of conjugation does not, in fact, happen.
And because this doesn't happen, therefore, the electrons
on the carbonyl stay localized, and that
makes the carbonyl a better reactive site.
So it's a better electrophile to react as nucleophiles
behaves more like a ketone.
For the same reason, when we are to deprotonate
the alpha position on a thioester,
the density on the carbonyl is there
to stabilize the enolate much more so than it
would be for an oxygen ester.
Therefore, the pKa, the acidity constant
for this hydrogen on the thioesters
is close to 18, which is smaller, therefore
more acidic than the pKa of the alpha position in oxygen
esters, which is 22.
This sums up Problem 2 of Problem Set 4.
I hope you now have a much better feel of how
we can use crystal structured data
to propose a reasonable mechanism
for enzyme catalyzed reactions.
Keep in mind that writing mechanisms on paper
is relatively easy.
But to truly confirm that that mechanism is taking place
in real life inside the cells, it
takes a lot more experimental work and evidence.