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PROFESSOR: Today is our final clicker competition.
The first question is up.
And I would just like to go over the current standing for
the clicker competition as we're getting started here.
So, recitation 6 has two wins, and there are other
recitations and have one win, so recitation 2, 3, 5, 7, and
10 have one win.
If any of those recitation wins today, we're going to go
to a tie breaker to decide the overall recitation champion.
There are also recitations 1, 4, 9, and 11, who are looking
for their first win.
This is their last chance today.
So, if they win today, they can not be overall champions,
but they will have snacks, and they will assure that
recitations 2, 3, 5, 7, and 10 do not win the overall
competition.
So, you have a chance to foil those recitations.
And if we need to go to a tie breaker, it will not be from
current material, it's going to be past material or past
things that were mentioned.
So, just keep that in mind.
All right, and for the overall winner, I just would like to
show you what you will win, so members of the recitation will
each get their own chemistry t-shirt, and at some schools,
of course, you would look at this and say oh, they're on
the varsity team, but at MIT, it means that you are
associated with course 5.
So, Department of Chemistry at Massachusetts Institute of
Technology on the back, and we have different colors and
different sizes.
All right, so time to get started with the clicker
competition.
Go ahead and click in your first responses.
OK, let's just take 10 more seconds.
All right.
Big improvement from the first time we asked this question
last class where it's about 33%, now we're up to 89.
And no, it is not temperature dependent.
And we're going to be talking more about
activation energy today.
All right, so today we're going to finish up kinetics,
and we're going to start with the material that we didn't
quite get to at the end of last class.
And I love this part, because I love it when we can think
about what we learned before in a slightly different way.
So, when I first started lecturing, I was talking about
LeChatelier principle and, of course, encouraged students to
sort of -- you know it's not too intuitive to MIT students,
but LeChatelier's principle says that if you apply stress
to a system, it tends to respond in such a way to
minimize that stress.
So it's all about minimizing stress, which is something
that's hard to kind to imagine this time of year, a little
easier to imagine activation energy barriers, a little hard
to imagine minimizing stress.
But we're back to talking about minimizing stress.
And we talked about adding reagents and removing
products, and we talked about how reactions will shift to
respond to stresses.
And one of the things we talked about is rationalizing
the shift of reactions at equilibrium when temperature
has changed.
So, if you increase the temperature of a particular
reaction, it tends to shift in the endothermic direction to
remove that added heat from the reaction.
So, that's what we learned before, but now we're going to
think about this in a slightly different way.
So we're going to think about this in terms of these
reaction coordinate diagrams. So, on one axis we have p e
plotted -- what does that stand for?
Potential energy, right, so we have potential energy going up
on both sides.
And here we have a view of an endothermic reaction, and over
here is an exothermic reaction.
So, let's just take a look at this.
For the endothermic reaction, we have reactants and
products, we have a delta e between those, and we have a
very large activation energy barrier
for the forward direction.
And, of course, for anything to react, there always has to
be some barrier, you need some amount of critical energy,
because as the molecules come close to each other, and if
they're going to react and form some kind of product, you
have to sort of distort bonds and sort of rearrange things,
and that takes a certain amount of energy.
So if they come together and they have that energy, they'll
react and go on to products, if they don't they'll break
apart and go back and be reactants.
And how much temperature the system has, is more likely
that they'll be able to react.
So it's more likely, if there's high temperature, they
could overcome this activation energy barrier.
All right, so in an endothermic reaction, you have
a big one in the forward direction, and a much smaller
activation energy in the reverse direction.
For exothermic reaction it's the reverse -- there's a
smaller barrier for the forward direction, and a much
larger barrier for the reverse direction.
So, we can think about the sign of delta e.
So delta e, this difference here which is related to delta
h for liquids and solids, it's pretty much the same for gases
is 1% or 2% different.
So you have a positive value here, and you can get a
positive delta e here if you have a big number for the
forward activation energy barrier minus a small number
for the reverse, so that's going to give
you a positive number.
For an exothermic reaction, delta e's going to be
negative, and you get a negative number if you take a
small number here for the forward activation energy, and
subtract it from a larger number here for the reverse
activation energy.
And remember, this is one of the equations that we're not
going to give you, you need to have that memorized because
it's part of understanding what's going on in these
diagrams.
So now let's think about what increasing the temperature is
going to do.
So, it will help the molecules if the is temperatures
increase, they'll have enough energy to overcome barriers.
Now if the barriers are pretty small, then it's not very hard
for the molecules to get over them, regardless of the
temperature.
But when the barrier is big, then temperature is going to
make a big difference.
And you can sort of think about this in your own life as
you have a small assignment to do, you just jump in and do it
and get it done and it's fine.
But if it's a very, very large thing that you have to do, you
often need to procrastinate longer before
you start doing it.
So, a small barrier's not so much of a problem, but if it's
a big barrier, you need to increase that temperature to
help the molecules get over the bigger barriers.
So, let's think about what this means in terms of the
direction that a reaction might shift.
So, if you increase a temperature over here with an
endothermic reaction, it makes it easier to overcome the
bigger barrier.
The forward reaction barrier is the bigger one, so
increasing the temperature will help molecules get over
that barrier.
They're already doing fine with the reverse barrier, it's
the forward barrier that we're having trouble with.
So, if you increase the temperature, that's going to
shift the equilibrium toward products, or it will shift the
reaction in the endothermic direction.
That's what we had learned before with LeChatelier's
principle, but now there's a new way of thinking about why
that is true.
So, for an exothermic reaction, if you increase the
temperature, that helps with the bigger barrier.
Again, here the bigger barrier is for the reverse reaction.
So, if you increase the temperature of an exothermic
reaction, it's going to shift it toward reactants.
More molecules can overcome that reverse barrier, and so
you'll form more reactants, it'll shift toward reactants
or it'll shift in the endothermic direction.
Again, these are the same things that we already knew
from LeChatelier, but it's a different way of thinking
about why those things are true.
So, a large activation energy barrier means that rate
constants are very sensitive to temperature.
And so, if you have a big activation energy barrier,
increasing the temperature makes a big difference -- if
it's a big barrier, then there's going to have a big
difference if you have higher temperature.
If it's a small barrier, then increasing the temperature
doesn't make much of a difference.
And so you can think about that in terms of these
diagrams as well.
So, now we're done with temperature and we're going to
talk about catalysis and the use of catalysts.
And so, if you remember there were factors affecting the
reaction, and we've talked about everything on this list
except for catalysts.
So today we're going to talk about catalysts.
So, a catalyst is a substance that speeds up a reaction, but
it doesn't get consumed in the reaction, it doesn't undergo
any permanent change itself.
It's just added to speed up the reaction.
So catalysts do not appear in the overall balanced equation.
So let's look at what a catalyst is going to do, and
we're going to look at in terms of these potential
energy diagrams as well.
So here we here we have a reaction, we're going from
reactants down here to products.
We have the delta e for the reaction, we have a forward
activation energy barrier and a reverse one.
So, up here, at the top, that's our barrier without a
catalyst. This is the transition state or the
activated complex up here, so you have to overcome the
barrier, you form some kind of activated complex, which then,
if there's enough energy to overcome that barrier, goes on
to products.
So when you add a catalyst what happens is
it lowers this barrier.
So, in the blue line here is the new
activation energy barrier.
This is the barrier with a catalyst, and so this would
then be the new transition state with a catalyst.
So when it lowers the barrier, it's going to lower the
barrier for the forward reaction, so we have a new
activation energy for the forward reaction, and we're
going to have a new activation energy
for the reverse direction.
So catalysts work by reducing both the forward and the
reverse activation energy barrier.
And another way that you can say this is that they
stabilize the activated complex or
the transition state.
So here you have an activated complex with a much higher
potential energy without the catalysts, and by stabilizing,
you lower in energy that transition state.
So that's another way of expressing
what a catalyst does.
So, catalysts have no effect on the thermodynamics of the
reaction, they affect the kinetics of the reaction, but
they don't affect the
thermodynamics of the reaction.
And so, of course, when you think of thermodynamics, you
often think of delta g.
So, why don't you tell me what you think, what would happen
about, so delta g is a state function, it's independent of
path, and so therefore, what can you tell me about the
equilibrium constant in the presence of a catalyst?
OK, let's just take 10 more seconds.
Yup, very good.
So, the equilibrium constant is not changed.
The thermodynamics, which includes delta g and the
equilibrium constant are not affected, the rates of the
reaction are affected.
All right, so opposite of a catalyst is an inhabitor, and
we'll talk more about this at the end of class, so it would
-- inhibitor is slow or sometimes stop the rate of the
reaction, and one way that they do this is by increasing
the activation energy.
So let's consider types of catalysts now.
You can have a homogeneous catalyst, which is in the same
phase as the reaction that it's
catalyzing as the reactants.
An example of this is depletion of the ozone layer
by chlorofluorocarbons.
And so, this was one of the big environmental challenges
that the U.S. has faced, whether it should ban these
chlorofluorocarbons, and there was a lot of debate on what
the data really was about what how much they affected the
ozone layer.
And I guess that debate is somewhat still going on,
although I think most people now recognize that this is a
serious problem and that legislation is really needed
to help correct it.
So that's an example.
In this case they're all gases, so that's homogeneous
catalyst, not a happy one.
You could also have heterogeneous catalysts, which
are a different phase.
And here's another example that has to do with the
environment.
So a catalytic converter is an example of a heterogeneous
catalyst. So here you have a solid metal surface, you can
have palladium or platinum that will catalyze reactions
with gases, and so they catalyze oxidation of
hydrocarbons, carbon monoxide, also
reduction of nitrogen oxide.
So this is all to reduce pollution.
So that's an example of a heterogeneous catalyst. And
let me just show you a little movie of how that might work.
So in this movie, in grey here we have a metal surface, and
this metal surface can break the hydrogen bond of h 2.
And so here, it is already broken, the h 2 bond, so
there's a little hydrogen there and hydrogen there, and
so that activates the hydrogen to react.
And so, then we have ethene molecule come in, and oh,
there goes the hydrogen, and oh, there goes the hydrogen,
oh, there goes the other one and you can form ethane.
So it speeds up the reaction by breaking the h 2 bond so
it's more ready to react.
That would be an example of a heterogeneous catalyst there.
All right, and, of course, you may all have guessed that my
favorite kind of catalysts are enzymes.
They are the catalyst of life.
And so, enzymes are made up of protein, or mostly protein
molecules -- you can have an enzyme that's actually made of
RNA, but most are protein molecules.
And they're typically about 20,000 grams per mole or more,
and they're capable of carrying
out specific reactions.
And so, they're made up of amino acids, and just to take
a quick look at that, an amino acid has an amino group and it
has a carboxyl group, and it has what's called an alpha
carbon that has a side chain on it, which is abbreviated r,
and there are 20 different types of r.
The simplest is just a hydrogen, you could also have
a hydroxide, etc.
And so, this makes up the alphabet of proteins, there
are 20 different ones of these that get connected via a
peptide bond.
So at the end of this amino acid you'd stick on the amino
group of another amino acid and form this peptide bond.
And then you put together hundreds and hundreds of these
to form an enzyme complex.
And so, the long chain of amino acids folds up and forms
a compact structure.
So, in this particular picture, there are about 200
amino acids in each of these colored units, so this is a
tetramer, so there are four -- there's green,
red, yellow and blue.
And so, this is what's called a ribbon diagram.
So, these ribbons, this is an alpha helix -- the beta
strands are long.
It traces out the position of the alpha carbons.
And overall, this structure is about 90 angstroms by 70
angstroms by 50 angstroms. Of course, 1 angstroms is 1 times
10 the minus 10 meters.
So these are fairly small.
These protein molecules are, of course, in your body.
And this particular one is an enzyme that makes an
antibiotic, and that antibiotic is fosfomycin,
shown here.
And so, fosfomycin is used in antibiotic combination
therapies to treat staph infections and other kinds of
very difficult infections to treat.
And so, I always talk about the things that I'm most
concerned about.
There are a lot of things that are big threats that I don't
worry too much about.
Antibiotic resistance is something I
actually worry a lot about.
I guess I go to too many meetings were they talk about
problems. But the rate at which different kinds of
bacteria are becoming resistant to antibiotics seems
to be increasing.
So it used to take a lot longer before a resistance
would appear than it does now.
And there really haven't been, I think since about 1980s or
so, really new antibiotics.
So, we're still using the same antibiotics that we have been,
which causes more things to become immune or resistant to
those antibiotics, and I think this is really dangerous.
And there's not enough money in it for the pharmaceutical
industry to really go after this.
And it's not just a problem for biological warfare,
although that is a possibility, but also in
hospitals, you go in for some kind of surgery, you have to
worry about the fact that even if the surgery goes great, you
might get an infection that could really compromise your
health in the hospital.
And so, whether you have a heart problem, if you have
cancer, you become immune compromised, there's a lot, a
lot of cases where people end up not dying of cancer
directly, they end up dying of the bacterial infection that
can't be treated.
So this is really a huge problem.
So I always like to let you know all the possible problems
that you can solve in your future.
This is one that I think is particularly important, and I
hope some of you will focus on this, because we really need
different kinds of antibiotics, or we need to
change the way we do medicine in this country so that
resistance doesn't become as much of a
problem as it has been.
All right, so that enzyme made an antibiotic, so we like it,
and let's talk about how that works.
How that particular series of amino acids folds up to do
that reaction.
So now we have some new nomenclature, and sometimes
when people get into the biochemistry world they get
scared because there are all these words that they don't
know what they mean.
So most of them are not -- they are things that you can
relate back to something you already know.
So, we've been talking about reactants, and when you have a
reactant with an enzyme, it's usually called a substrate.
So this is just another name for a reactant molecule.
And then substrates tend to bind in what we call the
active site of the enzyme, which is the part of the
enzyme that's going to do the chemistry.
So those are two terms that you'll hear about in
biochemistry.
All right, so we have an enzyme, we'll call it e, it
binds a substrate, we'll call that s, and then it forms an
enzyme substrate complex, which we'll call an e s
complex, for enzyme substrate.
And then that enzyme substrate complex will go to enzyme and
product, which we'll call p.
And if you look in the state-of-the-art biochemistry
books, these are the abbreviations you'll see in
there as well.
So, this pains me greatly to show this cartoon, because I
spend my career determining 3-dimensional structures of
proteins, and so to describe it as a little squiggly is
painful for me.
But nonetheless, there you go.
Also, the substrate is almost about as big as the enzyme.
That is also usually not true.
But nonetheless, here's a little cartoon.
Here's our substrate binding to our enzyme, it's forming an
e s complex.
And then the enzyme is going to move the ears of the
substrate around to form product, and
now product is released.
So, this is actually pretty simple in terms
of writing a mechanism.
And so, what we can do is write a mechanism the way that
we have been writing a mechanism
so far in this class.
And so we're going to go through this, we're going to
derive expression mostly to show you that all the things
you've been learning are related to chemistry, they're
also related to biochemistry.
So, you already know a lot of biochemistry from just
studying freshman chemistry in 511-1.
All right, so we have 2 steps in our mechanism.
We have enzyme binding substrate to form an
intermediate enzyme substrate complex, which then goes on to
form enzyme plus product in step 2.
So the first step is reversible, the second step is
not as drawn.
All right, so now we can come up with the rates for each of
the individual steps in this overall mechanism, and since
they are elementary reactions or steps in an overall
mechanism, we can write the rate laws directly from what
we observe.
So for the forward direction, we're going to
have what rate constant?
K 1 times the concentration of what?
And?
Yeah.
So there we go.
We have k 1 times the concentration of e and the
concentration of s.
See, you already knew how to do biochemistry.
All right, so now at the rate of the reverse direction is --
what's our rate constant?
K minus 1 times e s.
And then for step 2, we have what rate constant?
K 2 times e s.
All right, so now we can talk about the rate
formation of product.
And so you'll see this expressed as d concentration
of p d t, so the change in product that's being formed
and that's going to be equal to the second step
here, k 2 times e s.
But, as has been the case before, e s is an
intermediate, and we need to solve our rate laws, our rate
expressions, in terms of reactants or
products and rate constants.
So we need to solve for the intermediate.
All right, so let's think about solving for the
intermediate.
And why don't you tell me how to do that.
OK, let's just take 10 more seconds.
Very good.
All right, so let's just take a look at
why that is the case.
So, when we solve for this, it's going to be equal to the
formation of the intermediate, which happens in the first
step with this rate that you told me, k 1 times the
concentration of e s.
And then we have the decomposition in the reverse
of step one, so that's minus k minus 1 e s.
And then we also have the consumption of the
intermediate, which is -- so we have minus this step, which
is k 2 times e s.
So again, formation minus decomposition minus
consumption.
And now, we can use the steady state approximation.
So the steady state approximation applies in
biochemistry just as well as in all of your chemistry
problems that you've done in this course.
And so, the steady state approximation allows us to set
that whole term equal to zero.
So it says that the net rate of intermediate formation and
decomposition and consumption is 0, or you can express it as
the rate of formation of the intermediate equals the rates
of decomposition and consumption -- those are
equivalent.
So we can set all of this equal to zero.
So again, the same as what we've done before.
All right, so now we're going to have a slight change of
what we've done before, and this has to do with the
practical consideration.
So you can solve for these in terms of reactants and
products and rate constants, but it's actually easier to
solve for e s in terms of your total enzyme rather than your
free enzyme.
So what we've had before is our free enzyme, but we often
don't know how much enzyme is free, and by that we mean not
bound to the substrate.
And so, we know how much total enzyme we put in, but we might
not know how much of that is unbound.
So it's actually easier for the experimentalist to solve
for things in terms of total enzyme.
And so what we're going to do then is we're going to replace
our free enzyme term with the following, total enzyme minus
bound, and total enzyme minus the bound
enzyme is the free enzyme.
So we're going to do this change that makes it easier
for the experimentalist.
So here's the term we had before, but now we want to
replace this term of e with e to 0 minus e s.
So we're going to put that, here's the term e up here, so
we're going to replace that with e 0 minus e s.
And then we're going to multiply that out, so we get k
1, e 0, s minus k 1, e s, and then the concentration of
substrate here, and the rest of it is the same.
So we got rid of our term e.
So now we can solve for e s.
So this is what we just had, and so we're going to
rearrange so the e s terms are on one side, so we have an e s
term here, here, and here.
So all that is going to be on one side, and then this other
term will be on the other side of the equation.
So now we have moved all the e s terms over here, so we've
got rid of the negative numbers, we've added them all
to the other side, and then on this side we just have this k
1 term left.
Now we're going to pull out the e s terms. So,
we'll solve for e s.
We'll pull that out, that leaves us with k 1 times the
concentration of s, k minus 1, and then k 2.
And now we're going to take this and divide by that term
in parentheses, and so now we've solved for e s in terms
of total enzyme substrate and our rate constants.
All right, now we're going to do another thing
that's a bit different.
We're going to introduce a term in biochemistry, which is
k m, also known as the Michaelis-Menten constant, and
so k m is equal to the rate constant for the
forward/reverse direction, k minus 1 plus k 2 over k 1.
So we're going to now try to use this in our expression.
So we're going to put this term in there, and so to have
this term of k m in there, we need to have this
divided by k 1.
So we can do that, it's OK, as long as we divide everything
by k 1, so we're going to divide the top by k 1, and
this by k 1, and that by k 1.
So now, we divide all by k 1, and so we have this top with k
1, this term with k 1, and this with k 1, and
that's our k m value.
But now we can simplify this, we've got a lot of k 1's here,
and so if we simplify this, we can cross out
those k 1's over there.
And we can also get rid of these k 1's over here.
And that's going to leave us with a much simpler
expression.
So that's going to leave us with the total enzyme times
substrate over substrate plus k m.
And that looks a whole lot better than that other term
that we had before.
OK, so we can now take this and substitute it back into
the expression we had earlier.
So here we had the rate of product formation, k 2 times
this intermediate, we solved for the intermediate using
total enzyme and using this k m term.
And now we just put all of this times k 2.
And if you put all of that times k 2, you have this
expression, which is the Michaelis-Menten equation.
So the only difference of what you did before was that you
solved in terms of total enzyme, and also, we have this
new k m term.
And so now we have an equation that's used in a lot of
biochemistry research to think about enzyme kinetics.
So let's think about enzyme kinetics for a minute.
This is what a plot often looks like -- we are looking
at the product formed by an enzyme versus the
concentration of substrate.
And it often goes up.
It's pretty steep, and then it starts to level off.
So let's think about what's happening here.
So if you have low substrate, so that's way down here, not
much substrate is available, adding more substrate will
increase this rate significantly, you'll form a
lot more product, it increases the rate of product formation.
Because a lot of the enzyme is free and
can react with substrate.
So it goes up, but then the rate starts to level off.
So when you got to higher substrate concentrations,
adding more substrate does not affect the rate much, the rate
is leveling off.
And we say that the enzyme is saturated.
All the enzyme that you have in there already has substrate
going on, it's already doing the chemistry.
So if you're throwing more at it, the enzyme can't bind that
substrate, it's busy with a different substrate molecule.
So all the active sites in the enzyme are full, and so not
much happens.
So this kind of graph is what you often
see in enzyme kinetics.
Now let's think about that in terms of our
Michaelis-Menten equation.
So let's think about conditions when there's a lot
of substrate, when substrate is much greater than k m.
So here we have substrate and k m on the bottom of our
equation, and what's going to happen if this term is much,
much bigger than this k m term?
What happens to the k m term?
It goes away.
It's really small compared to this, it's kind of
insignificant.
So, if that happens, then you can cancel substrate as well.
And so, that means that under conditions where substrate is
much greater than k m, the rate is a much simpler term,
it's just equal to k 2 times your total enzyme.
And so this has a name, this is called Vmax or the maximum
velocity of the reaction.
So, under these conditions, when the substrate is much
greater than that k m value, this expression simplifies and
you get just k 2, your rate constant, times your total
enzyme, and that'll give you the maximum
velocity of the reaction.
And so, that's just written again here.
Vmax equals k 2 times total enzyme, and that equation will
be given to you on an equation sheet.
So what's happening?
What's happening is that you're up here.
So when substrate concentration is much greater
than k m, then you're in these conditions, and you're at some
maximum velocity.
So the velocity will only go, the rate of the reaction will
only go so fast, it will saturate, it'll level off up
here, and so you can calculate that maximum
rate of that reaction.
Now let's consider what happens when your substrate
concentration equals your k m.
So down here, if k m is equal to substrate, then we're going
to have two substrates down here.
And so, we can write this, if they're equal, we can write
this as substrate plus substrate, and then we can do
some canceling.
So this will be equal to 2 substrates and then we can
cancel that out, and so we get the rate of product formation
under conditions where substrate equals k m of 1/2 k
2 times the total enzyme concentration.
And this is referred to the half maximal rate.
Remember, the maximal rate was k 2 times this e knot, and so
half of that is half the maximal rate here.
So, the definition of k m is the concentration of substrate
for which the rate is half maximal.
So if your rate is half of the maximum rate, that substrate
concentration that gives you half the maximum rate, is the
value for k m.
Substrate concentration equals k m at the half maximal rate.
So let's just look at what that meant.
So if this is Vmax up here, half of Vmax is here, half of
that value, and the substrate concentration at half Vmax is
equal to k m.
So you can write those in on your diagram.
Vmax is this velocity up here, half of that, when you're at
half of the maximal rate, k m equals the substrate
concentration.
And there will be problems that you will work in your
book very shortly, and often, the problems just have to do,
they'll give you the information in words and the
calculations can often be pretty simple.
Sometimes it'll say, figure out what the k m for this
reaction is.
At this substrate concentration, the rate is
half maximal.
Well, at that substrate concentration, that's the k m.
So, a lot of the problems are sort of word problems, they
give you the information and you need to know what the
definitions are.
But let's try an example.
So here in this example, we've talked about buffering in the
blood, so the conversion of your ingredients that make
your buffering agent in the blood, that can be catalyzed
by an enzyme called carbonicanhydrase, and the
following Michaelis-Menten constants are known, a k m is
known, and a k 2 is known, and if you're doing experiment and
you have a total concentration of your enzyme of 5 times 10
to the minus 6 molar, then you should be able to figure out
what the maximum rate of the reaction would be.
So, why don't you go ahead and tell me what the
maximum rate would be.
OK, let's just take 10 more seconds.
Very good.
Everyone's doing very good today.
All right, so all you have to do, you have remember Vmax
equals your total enzyme concentration times your k 2.
And pretty much no one, hardly anyone was fooled
by k m value there.
All right, very good.
So there are extra problems on enzyme kinetics posted on the
extra problems posted on the Web.
So there's a few there, so you can see the type of problems,
again, they're like that, they're pretty simple.
If you remember the definition of k m, and you'll be given
the equation for Vmax, you should be able to handle the
enzyme/kinetics problems pretty well.
Often, also I'll ask questions like explain why at high
substrate concentrations, the rate does not increase very
much, or why the graph levels off, or I might ask you to
draw the plot and tell me about low substrate
concentrations, what's true there, and high substrate
concentrations, what's true.
So those are the types of questions you're going to get
on enzyme kinetics, they're actually pretty simple.
All right, so let's just talk briefly about enzyme
inhibition.
So the opposite of catalysis -- instead of binding a
substrate, you'll bind an inhibitor.
So often, these are actually pretty simple.
An inhibitor sometimes will look a lot like a substrate or
like a transition complex, and it'll blind to the enzyme.
And when you have that inhibitor stuck in the active
site, substrate physically can't bind, so it occupies the
place that substrate goes, substrate will come in but it
can't bind anywhere.
So this is actually the mechanism by which a number of
pharmaceuticals work.
So, one thing that people who were designing enzyme
inhibitors think about is the fact that enzymes do tend to
stabilize a transition state in the reaction.
So if you make an inhibitor that resembles a transition
state, it should bind to the enzyme more tightly than
either the reactants, the substrates, or the products.
So, a lot of the pharmaceutical industry likes
to try to figure out what the transition state might look
like, and then try to make a molecule that looks like that
transition state that hopefully will blind to the
enzyme active site and prevent the enzyme from doing what
it's supposed to do.
So, just to kind of look back at this figure for a minute
that you had earlier in your notes, so you have the
transition state is often stabilized by a catalyst, and
so an enzyme will also stabilize a transition state.
So if you make your drug look like a transition state, it
will hopefully bind very tightly.
And so this is one of the sort of principles that's behind a
lot of the pharmaceuticals that have made to treat HIV
infections.
So I mentioned last week we had world AIDS day, that
understanding kinetics was actually very
important in HIV research.
And so, many of the pharmaceuticals that are given
to HIV patients are what are called protease inhibitors.
So they inhibit enzymes that are called proteases, and if
you have ase at the end of the name that means it's an
enzyme, and so protease means that it's an enzyme that
cleaves proteins.
It's a protein ase.
And so, how do these enzymes work.
Well, they cleave other proteins, they cleave peptide
bonds, and so you often have some kind of either activated
water or other hydroxide molecule that will attack here
the carbonyl of a peptide bond, and it forms a
tetrahedral intermediate, which then collapses and it
cleaves that peptide bond.
So the peptide bond is then broken, so that's what a
protease does, and a protease inhibitor
prevents that cleavage.
So often, protease inhibitors look like tetrahedral
intermediates, and so they'll bind to the protease active
site and prevent the chemistry from occurring.
So if it's a tetrahedral intermediate, what kind of
angles should some pharmaceutical company be
looking for in its compounds?
109 .
5, yes.
So, molecules with this stable tetrahedral intermediate
somewhere on the molecule could bind to the active site
and prevent a catalysis.
So, let me just show you one example of an improved HIV
drug, and there it is.
This is the tetrahedral site that binds at the active site
of that enzyme, and the enzyme can't cleave this tetrahedral
intermediate, and so this just binds, but the enzyme can't
work on it, so it just sits there and prevents substrate
from binding.
So that's how a lot of these compounds work.
And so, this is just a little picture of the enzyme active
site and there's an inhibitor bound to the enzyme active
site, and so a number of companies are working on
trying to come up with better and better inhibitors that
will sit in the HIV protease active site.
So, knowledgeable of a reaction mechanism can lead to
new therapeutic treatments.
So, one question with this, with protease, there are a lot
of proteases, not just the ones involved with HIV, and so
one real problem is specificity and toxicity, so
you want to have a drug that inhibits one enzyme and not
all the enzymes in that category.
So that's a big problem in the pharmaceutical industry.
So, apparently, clicker results, we actually need to
break a tie.
But we can't do it now.
OK, so Justin's section, woah!
So, apparently we're having, we have the questions ready,
but apparently they're not ready to be used
right at the moment.
So, we're going to use low tech here and indicate the
results for today.
And then on Wednesday, last day of class -- oh, my
goodness, so close.
Recitation 6, you were second.
All right, so on Wednesday then, we're going to have the
tie breaker, and review and evaluations.