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JOANNE STUBBE: So what we were doing last time is we were

still focused the first two lectures were trying

to understand the biosynthetic pathway for cholesterol bio--

it's good, thanks--

for cholesterol biosynthesis.

And we almost got to where we wanted to go,

but we didn't quite get there.

So what we've been focusing on is a new way

of forming carbon-carbon bonds using

C5 units, isopentenyl pyrophosphate

and dimethylallyl pyrophosphate.

And to do that, we had an initiation

process where these molecules were generated from acetyl CoA.

And then the last lecture we were

focused on how we did the elongation process where

we took a bunch of these IPP units,

strung them together to make farnesyl pyrophosphate, which

is C15, and I showed you that C15

could be linear or cyclized.

And we went through the general rules

of what you're going to see with all turpine chemistry, which

is quite diverse, given that there

are estimated to be 70,000 natural products

in the terpenome.

So we had gotten to production of farnesyl pryophosphate

and now the next step--

remember, cholesterol, if you look at its structure--

this is a precursor to its structure--

is a C30.

And so the next step is quite an interesting enzymatic reaction

which we're not going to talk about in any detail,

but those of you who are interested can go look it up.

But how do you take two C15s and form a C30

so you lose your pyrophosphates?

And you can see when you generate this,

now you have a linear c30, which,

of course, is a complete hydrocarbon and is insoluble.

So this now sort of defines that you

need to be in the membrane to be able to do

any further chemistry.

So those of you who are interested in mechanisms

of how things work, that's really

sort of a fascinating system it's really pretty well

worked out at this stage.

But today what I mean to do is focus on

the next step is, how do we get from C30, which

is this linear squalene hydrocarbon, into lanosterol,

which is then the precursor to steroids

but also the precursor to cholesterol,

which is what we're focusing on in this particular module.

So what we're going to be looking at

is how we went from two FPPs--

we're still doing chain elongation-- to a C30.

And then the question is, how do you

get from C30, which is linear, to a linear epoxide.

And I'm not going to draw the whole structure out,

but we're still linear.

And then the next step is the step I want to talk about.

So this is when lanosterol synthase.

So that's where we're going in the next few minutes

to get to our final product.

So if you look at this reaction, remember,

we're going to do a cyclization.

And what do you need to do to do cyclizations?

What was the general rule that I gave you last time?

Does anybody remember?

If you want to cyclize something, we talked about it.

We looked at a number of examples.

What did we do in those examples?

Anybody remember?

So here's a second example.

I gave you two rules.

If you go back and you look at your notes,

we protonated the olefin and that triggered off

the cyclization.

And here, perhaps you could have protonated the olefin

to trigger off the cyclization, but in the end,

cholesterol has a hydroxyl group in the C3 position.

So the next step in the pathway, which also will involve,

ultimately, protonation and ring cyclization,

so those are the two rules I gave you

during the last lecture, to get to this epoxide,

we have to do some chemistry.

Does anybody know what cofactors you

would use to do this reaction?

Anybody got any ideas from introductory biochemistry?

You have a vitamin bottle.

What vitamin would be involved in doing this transformation

or could be involved with doing this kind of a transformation?

It's an oxidation.

Requires oxygen gas.

So what are the possibilities?

AUDIENCE: NAD.

JOANNE STUBBE: So NAD.

Does NAD-- this is a good teaching point.

Does NAD react with oxygen?

Who suggested NAD?

Why doesn't it make you react with oxygen?

That's one of the things you learn

in any introductory course.

NAD does not react with oxygen. Why?

What is the chemistry of NAD/NADH?

Whoa.

Maybe I should be teaching 5.07.

So NAD/NADH, we just went through this

with conversion of acetyl CoA moiety of mevalonic acid

to the alcohol.

It involves hydride transfer.

And if you tried to do this chemistry instead of two

electrons at a time, one electron at a time,

and you looked at the reduction potentials,

it would be way uphill, thermodynamically.

So NAD/NADH never does one electron in chemistry.

So that's not going to be a possibility.

Yeah?

AUDIENCE: You could use something that's like iron?

JOANNE STUBBE: So that would be one thing.

And we're going to see iron--

heme irons play a key role in all of this process.

This turns out to be a flavoprotein.

That's the other redox active cofactor.

So this is a flavin monooxygenase.

You don't need to remember this.

We understand the details.

I'm not going to talk about the detailed mechanism,

but flavin cofactors are extremely well understood.

The chemistry of them is extremely well understood now.

So we've gotten to our oxidosqualene

and now we've finally gotten to this really cool step.

So how do we go from this step--

so this is this molecule here.

And what I'm emphasizing again is

we're going from a linear step into the cyclic product.

So remember, triggering off cyclization,

there were two rules-- protonation,

protonation of an olefin.

In this case, you have some kind of protonation of the epoxide.

Epoxides are not very good leaving groups.

You need to protonate it.

And that is then going to trigger off

this cascade of reactions to allow you to generate

a molecule with four rings.

And this all occurs with a single enzymatic step.

And so the way you can visualize this happening--

and again, you don't need to copy this down.

It's all-- if you look at your handouts ahead of time,

there's some things that are written down

that would take you 10 minutes to copy

and then you probably get it written down incorrectly

because you're looking like this is.

The hard things that are hard to write down

are all given to you in your handouts.

You can write it down if you want, that's fine.

So what we want to do is we want a ring open,

so we need to protonate the epoxide,

and that generates what?

A carbocation.

And then now what happens?

We generate another carbocation.

And now what happens?

We generate another carbocation.

And now what happens?

We generate another carbocation and we end up

with a carbocation at this position.

So I'm going to draw the structure of this.

So we have ring opened, and let me also

emphasize that the key to this process occurring

to give us lanosterol is the conformation

of the linear molecule.

So what do we see here?

What does this look like?

In terms of cyclohexanes, what does this look like?

If you have cyclohexyl rings, what kinds of conformations

do they have?

AUDIENCE: Chair?

JOANNE STUBBE: Chair and chair and boat.

So the key here is that you have a chair conformation here.

You have a chair conformation here,

but here you have a boat conformation.

And one of the general rules I told you

last time about terpene chemistry

in general was, what do the enzymes do in the active site

to transform something that's linear into something that's

cyclic?

They need to fold the molecule into the right conformation.

And that can, in part, be done, the fact

is the active site is very hydrophobic.

We talked about that.

And you can also have aromatics that could potentially--

I'm not drawing out all these intermediates,

but could potentially facilitate not only the conformation

but stabilization somewhat of the intermediates

that you observe along the reaction pathway.

So here's another example of the importance

of shape to defining the chemistry that's

actually going to happen.

And in contrast to the enzymes we

talked about last time, which were type I. You probably

don't remember that.

But this is, again, a different super family

involved that you observe, and it's observed quite frequently.

So these are type II.

So if you look up the structures, and in the article

you had to read by Christiansen, the second type of structure.

There are two general types of structure.

This is the second type of structure involved in making

interesting terpene molecules.

So what I'm doing now is showing you

how we've cyclized this to leave us with a carbocation.

And remember, if you have just a stick as opposed to a stick

with a hydrogen, that's a methyl group.

So here at the ring juncture, we have a hydrogen.

We have a trans ring juncture.

And again, if we have a stick with nothing on it,

it's a methyl group.

And we're into a chair conformation again,

and then we need to attach the last ring

so we have three six-membered rings and a five-membered ring.

And in the end, what have we generated?

We've generated a carbocation.

So I've written this as a single step.

Nobody has seen the intermediates.

You could write it is multiple steps.

I mean, the fact is it would be--

it's pretty hard to trap any of these carbocations,

and people have spent a lot of time trapping them.

So what you see, I think, is quite amazing,

but we aren't finished yet because we have a carbocation

and we need to get rid of that.

And what you need to do and this is-- you

will have one of these problems on the problem set

that will be due next week.

You'll be given something simple,

not as complicated as cholesterol.

But what you need to think about is

where do all these methyl groups end up in.

What's the stereochemistry of the reaction?

So then this geometry becomes critical

if you're thinking-- you need to think

about the stereo electronic control of hydride and methyl

anion equivalent migrations.

So what you have in this particular reaction

is you're going to have--

and I like this example because, again, I gave you