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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

  • a set of rules that you can see that are associated, typically,

  • with carbocation reactions in general,

  • and this one does all of them.

  • So one of the rules was that you have

  • hydrogen migrate with a pair of electrons, so that's a hydride.