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In beer and wine production, enzymes in yeast
aid the conversion of sugar into ethanol.

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Enzymes are used in cheese-making to degrade
proteins in milk, changing their solubility,

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and causing the proteins to precipitate. Many
industrial processes ranging from fruit juice

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production to paper production to biofuel
production utilize enzymes. In this video

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you'll see that studying how enzymes catalyze
reactions can help us design better therapeutics.

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This video is part of the Differential Equations
video series. Laws that govern a system's

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properties can be modeled using differential
equations.

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Hi. My name is Krystyn Van Vliet and I am
a professor in the Materials Science and Engineering

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and Biological Engineering departments at
MIT.

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of reactions and after some practice, you
will be able to derive a rate law for a general

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enzyme-catalyzed reaction.

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In order to understand the topic of this video,
you should be familiar with determining rate

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laws from experimental data, predicting rate
laws from proposed chemical reaction mechanisms,

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and have some basic understanding of the effects
of catalysts on the kinetics of a reaction.

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The
study of enzyme catalysis has become important
for drug development. Many drugs work by inhibiting

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an enzyme.

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For example, enzymes thought to be important
for the survival and replication of the parasite

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that causes malaria are being explored as
therapeutic targets. The parasites enter the

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red blood cells of the host and use proteases
to catalyze the degradation of the hemoglobin

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inside of the red blood cells. Degrading hemoglobin,
a protein, yields amino acids, which can then

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be used by the parasite for its own protein
synthesis. Protease inhibitors could help

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slow the degradation of hemoglobin, which
in turn would slow the growth and reproduction

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of the malaria parasite.

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When a potential inhibitor is identified,
kinetic data is used to evaluate its efficacy.

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In order to understand how this can be done,
let's start by describing the kinetics of

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a simpler case -- that of an enzyme catalyzed
reaction without an inhibitor present. Then

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we'll think about what might happen if we
throw an inhibitor into the mix.

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This video will help you describe enzyme catalysis
mathematically by first reviewing some common

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biochemical terms, then by describing the
general characteristics of enzyme-catalyzed

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you will encounter throughout your undergraduate
experience.

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a reactant is called a substrate. This is
really just a matter of different fields using

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different terms, but it's important for you
to be aware of this. Next, catalysts found

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in biological organisms are called enzymes.
Textbooks frequently abbreviate "substrate"

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using the letter "S" and enzyme using the
letter "E."

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How do enzymes work? If we look at a reaction
coordinate, recall that the transition state

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represents a chemical species intermediate
between the reactant, or substrate in this

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case, and the product. The potential energy
difference between the transition state and

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the reactants is the activation energy for
the forward reaction. The potential energy

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difference between the transition state and
the products is the activation energy for

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the reverse reaction. Only reactants with
the energy to overcome the activation energy

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will form products.

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Like synthetic catalysts, enzymes accelerate
the rates of reactions by stabilizing the

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transition state, lowering its potential energy,
and providing a new pathway by which the reaction

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can occur. This new reaction pathway has a
lower activation energy than the uncatalyzed

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path making it more likely that a greater
number of reactant molecules will have the

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energy needed to overcome the activation energy
and proceed to product.

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Enzymes facilitate reactions, but, like other
catalysts, are not consumed in the reaction.

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Enzymes are typically proteins, but RNA has
also been shown to catalyze reactions.

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While there are different theories about how
enzymes work to catalyze reactions, there

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are a couple of things that are agreed upon.
One is that enzymes have a region called an

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active site. Substrates bind to this region.
The shape and chemistry of the active site

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determine the selectivity of the enzyme for
particular substrates.

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The second point is that when a substrate
binds to an enzyme, an enzyme-substrate complex

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is formed. This enzyme-substrate complex is
a reaction intermediate, meaning that it is

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formed and consumed in the reactions, but
does not appear in the overall chemical equation.

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For many enzyme-catalyzed reactions, if we
were to measure the rate of reaction at various

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substrate concentrations, we would see that
the rate of reaction appears to follow first

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order kinetics at low substrate concentrations
and then transitions to behavior that resembles

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zero-order kinetics at high substrate concentrations.
Please pause the video here, turn to the person

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beside you and discuss what it means for a
reaction to be first-order or zero-order.

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Remember, in a first-order reaction, the reaction
rate is directly proportional to the concentration

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of substrate. In a zero-order reaction, the
reaction rate is constant as the reaction

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progresses and is unaffected by substrate
concentration. So, if we look at this data,

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we see that, for a given enzyme concentration,
the addition of substrate above a certain

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value has no effect on the rate of reaction.

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So let's think about what reaction mechanism
might explain this data.

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Some scientists hypothesized that after an
enzyme and substrate combine to form the enzyme-substrate

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complex, this complex yields the product and
degrades to regenerate free enzyme. Does the

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rate law derived from this reaction mechanism
fit the experimental data?

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Let's derive the rate law and see. We can
derive this rate law just as we would for

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other chemical systems.

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Remember, each step of a reaction mechanism
is assumed to be an elementary reaction. Please

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pause the video here, turn to the person beside
you and discuss how you determine the rate

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law for an elementary reaction.

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Remember, for an elementary reaction, you
CAN predict the rate law from the chemical

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equation.

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With this in mind, can you write a differential
equation for the rate of product formation

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with time? Please pause the video, try to
write the equation, and continue playing the

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video to see if you are correct.

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From the 2nd step of the reaction mechanism,
we can write that the rate of formation of

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product is equal to the rate constant k2 multiplied
by the concentration of our enzyme-substrate

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complex.

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The enzyme-substrate complex is a reaction
intermediate and not something that is easily

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measured in experiments. How can we verify
if this expression is correct? It would be

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nice to restate the rate in terms of quantities
that are more easily measured such as the

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substrate concentration and the initial enzyme
concentration.

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To get an expression for the concentration
of the enzyme-substrate complex, let's write

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a differential equation for the net rate of
change of the enzyme-substrate complex concentration

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with time. Please pause the video here, try
to write the differential equation, and then

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continue playing the video to see if you are
correct.

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If we want to write an expression for d[ES]/dt,
we see that in the first step of our mechanism,

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the enzyme-substrate complex is produced in
the forward direction and consumed in the

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reverse direction. In the second step, the
enzyme-substrate complex is converted into

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product.

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This equation introduces two additional unknowns:
the free enzyme concentration and substrate

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concentration. So, we need two additional
equations in order to solve it.

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Can you write an equation for the rate of
change of substrate concentration with time?

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Please pause the video here, try to write
the differential equation, and then continue

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playing the video to see if you are correct.

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If we want to write an expression for d[S]/dt,
we see that the substrate will be consumed

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in the forward direction and produced in the
reverse direction of the first step of the

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reaction mechanism.

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We still need one more equation. This equation
is a simple relationship that says we know

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how much enzyme we added to our reaction mixture
at time zero, so at any time, we know that

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the amount of free enzyme plus the amount
of enzyme bound in the enzyme-substrate complex

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should equal the initial amount added. This
could also be written in differential form

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as d[E]/dt plus d[ES]/dt is equal to zero.

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So, we have a system of four ordinary differential
equations with four variables that all depend

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on time. Equation 1 and Equation 4 have analytic
solutions, but Equations 2 and 3 contain a

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non-linear term. When you learn about differential
equations in your future courses, you'll see

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why these equations are difficult to solve
and that you will need numerical methods to

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solve them.

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Here, we have integrated these equations numerically
using the boundary conditions that at time

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equals zero, the substrate concentration is
S0, the product concentration is zero, the

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enzyme concentration is E0 and the concentration
of enzyme bound to substrate is zero.

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We see that after a short start-up period,
the concentration of enzyme bound to substrate

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remains approximately constant. In this region,
where the substrate concentration is much

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greater than E0, as enzyme is released from
the enzyme-substrate complex, it quickly recombines

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with available substrate.

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As more substrate is converted to product
and the substrate concentration approaches

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E0, the concentration of enzyme bound to substrate
is no longer constant.

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In the regime where the substrate concentration
IS greater than E0, we can make a steady-state

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approximation and set d[ES]/dt equal to zero.
This approximation will allow us to obtain

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an analytic expression for the concentration
of enzyme bound to substrate.

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Setting d[ES]/dt equal to zero allows us to
solve for the concentration of enzyme-substrate

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complex, which is equal to the rate constant
k1 times the concentration of free enzyme

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times the concentration of substrate divided
by the sum of k-1 and k2.

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Let's lump the rate constants into a new term
called Km and substitute this into our expression

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for the concentration of enzyme-substrate
complex.

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Now we can use this expression... ... to rewrite
d[P]/dt.

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Let's try to eliminate the concentration of
free enzyme from this expression. We can use

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the enzyme balance that we wrote earlier,
which said that the concentration of free

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enzyme plus the concentration of enzyme-substrate
complex should equal the initial enzyme concentration.

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Substituting our expression for the concentration
of ES into the enzyme balance and doing a

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little bit of algebra, we can solve for the
concentration of free enzyme.

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And finally, we can substitute this into our
equation for d[P]/dt and obtain this expression.

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Let's check this rate law against our experimental
data. At low substrate concentration, the

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expression reduces to a first-order rate law
and at high substrate concentration; the expression

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reduces to a zero order rate law.

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It turns out that the rate law that we just
derived is called the Michaelis-Menten equation.

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To make our equation look more like the version
of the Michaelis-Menten equation that you

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will see in textbooks, there are a couple
of terms that we will use because they are

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the convention in enzyme kinetics. Instead
of saying d[P]/dt is the rate of the reaction,

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we will call it the velocity of the reaction
and use the notation of 'v.'

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Finally, we'll lump k2[E]o into a term called
vmax. At high substrate concentration, the

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reaction velocity approaches vmax.

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Remember that lump parameter Km? We can determine
Km from experimental data. If we look at the

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substrate concentration needed to reach half
of vmax, we see that the concentration of

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substrate required equals Km. Km values for
many enzyme-substrate pairs can be found in

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the literature allowing for comparison.

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You may be wondering why we went through all
of this. Let's go back to our drug development

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example.

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Let's say that you have designed a drug that
you think will inhibit one of the proteases

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used by the malaria parasite to degrade hemoglobin.
How will you test its effectiveness?

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First, you might add a fixed amount of protease
to solutions of varying hemoglobin concentration

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in order to measure reaction rates. Let's
say that this reaction follows Michaelis-Menten

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kinetics well and that you are able to estimate
vmax and Km.

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Next, you repeat the experiment, but this
time, you also add a fixed amount of the drug

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candidate to all of your solutions. You get
your rate data and see that the apparent vmax

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has decreased. But what does this mean? Can
you speculate how this inhibitor works?

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Is the drug interacting with the enzyme and
preventing it from binding substrate?

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Is the drug interacting with the enzyme-substrate
complex? Is it something else entirely?

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The thought process that you used today to
derive the Michaelis-Menten equation can be

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used to derive rate laws for other reaction
mechanisms. Then you can see which mechanism

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is best supported by the data.

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Today we learned a little bit about the importance
of enzymes and derived a rate law from a general

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reaction mechanism using differential equations
and a steady-state approximation. This rate

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law was compared against experimental rate
data for an enzyme catalyzing the conversion

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of a substrate to a product. The equation
that we derived is called the Michaelis-Menten

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equation. While all enzyme-catalyzed reactions
may not exhibit Michaelis-Menten kinetics,

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the same logic that you used to derive the
equation can be used to derive rate laws for

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other proposed reaction mechanisms.