Voiceover: Today we're gonna talk about
Michaelis-Menten kinetics and the steady-state.
First, let's review the idea
that enzymes make reactions go faster
and that we can divide the enzymes catalysis
into two steps.
First the binding of enzyme to substrate
and second the formation of products.
Each of these reactions has its own rate.
Let's also review the idea
that if we keep the concentration of enzyme constant
then a really high substrate concentrations
will hit the maximum speed for a reaction
which we call Vmax.
First we'll talk about the Steady-State Assumption
and what that means.
Like I said before there are two steps
to an enzyme's catalysis.
Now when we use the term steady-state
what we mean is that we're at a point
where the concentration of ES
or enzyme substrate complex is constant
which means that the formation of ES
is equal to the loss or dissociation of ES.
Now notice that I've used equilibrium arrows
between these steps
and that was to show the idea
that these reactions like any reaction
can go forwards or backwards.
Our enzyme substrate complex
doesn't have to form products.
It could just as easily dissociate
back to an enzyme and a substrate molecule.
I'll call these reverse reactions minus one
and minus two.
If we look at that in terms of our rates
we can say that the rate of formation of ES
would be the sum of rate one and rate minus two
since both of these reactions lead to ES
and the rate of loss of ES
is equal to the sum of rates minus one and two
since both of these lead away from ES.
Now I also remember that products very rarely
go back to reactants
since these reactions are usually thermodynamically stable.
Rate minus two is going to be so small
in comparison to rate one
that we can really just cross it out.
Which means that we can swap out
that second double headed for a single headed arrow.
Using this information let's do some math.
Now I'm going to be deriving a new equation.
This can get a bit confusing
so don't worry if you have a little trouble with this.
Just rewind the video
and try watching it a couple more times if you need to.
I'll start up by drawing the same sequence I did before
with the three different reactions,
and I'll also write out
that steady-state equation I mentioned before
where we have rates forming ES equal to rates
taking away ES.
Now first thing I'll do is swap out those rate values
for their rate constants
times the reactants for those reactions.
Rate one will be equal to K one times E times S
and so on for the other two.
Next I'll introduce a new idea
and say that the total amount of enzyme available
which we'll call ET or E total
is equal to the free enzyme E
plus the enzyme bound to substrate or ES.
Using this equation I'm going to rewrite the E
on the left side of our equation
as the total E minus the ES
which would be equal to the E we had there before.
On the right side of the equation
I just factored out the common term ES.
Next I'm just going to expand
the left side of the equation
so take a moment to look at that.
Now what I'm going to do
is I'm going to divide both sides of the equation by K one.
K one will disappear on our left side
and on our right side I've put K one in
with all the other rate constants.
Now since all these rate constant are constant values
I'm going to combine them in this expression
of K minus one plus K two over K one
into a new term KM
which I'm going to talk a little bit more about later.
In this next line I've done two things.
First I've thrown in that KM value
that I just mentioned,
but I've also added ES times S
to both sides of the equations
and thus moved it from the left side to the right.
In the next line I've done two things.
First I switched the left sides
and right sides of the equation
just to keep things clear,
but I've also factored out the common term ES
on our new left side.
Then what I'm gonna do
is I'm gonna divide both sides of the equation
by KM plus S
so I can move that term to the right side.
I'll make some more room over here
and now what I'm gonna do is remind you
that the speed of our whole process
which I'll call Vo
is equal to the rate of formation of our product
which we called rate two before
which is also equal to K two times ES.
Now using our equation over here
I'm gonna multiply both sides of the equation
by K two.
Here's where it gets really tricky.
Remember that if we're if at our max speed
so our reaction speed Vo is equal to Vmax
which happens when our substrate concentration
is really high,
then our total enzyme concentration
is going to be equal to ES
since all of our enzyme is saturated by substrate
and there won't be any free enzyme left.
K two times ET instead of times ES
would be equal to Vmax instead of being equal to Vo
like you see at the top.
I'll make some room here
and then sub in K two ES for Vo
and K two E total for Vmax
and then we finally get to our end equation
which is called the Michaelis-Menten Equation
and is super important
when we talk about enzyme kinetics.
Let's take a few steps back
and talk about the Michaelis constant.
First I'll write out the Michaelis-Menten equation
and if you remember we created this new term
which I called KM,
but we never really talked about what it meant.
Let's get to that.
Now if you bear with me for a moment
and pretend that KM is equal
to our substrate concentration
then we can sub in that value
into our Michaelis-Menten equation
which would put two S on the bottom,
the sum of S plus S
and then the S will cancel out
and will be left with Vmax over two.
What this means that KM
which we call the Michaelis constant
is defined as the concentration of substrate
at which our reaction speed is half of the Vmax.
When Vo is equal to 1/2 of Vmax.
If we look at that on a graph from before
you'd see that KM is a substrate concentration
specific to our circumstances.
Where our rate is at half of its max
and the lower our KM,
the better our enzyme is at working
when substrate concentrations are small.
We can use this KM term to quantify
an enzyme's ability to catalyze reactions
which we call Catalytic Efficiency.
I'll rewrite the Michaelis-Menten Equation.
Remember we defined KM as a substrate concentration
where Vo is 1/2 Vmax.
Since it's a concentration it will be in units of molar
or moles per liter.
Now I'm going to throw a new term at you
called Kcat
which is equal to the maximum speed of a reaction
divided by the total enzyme available.
We call this the enzyme's turnover number.
All these term is, is how many substrates
and enzyme can turn into product
in one second at its maximum speed.
We measure it in units of seconds minus one
or per second as in reactions per second.
We can define an enzyme's catalytic efficiency
as a combination of KM and Kcat,
and we do this by saying it's equal to Kcat over KM.
A higher Kcat or a lower KM
would result in an increase
in an enzyme's catalytic efficiency.
Every different enzyme has a different
catalytic efficiency in certain conditions.
We can use this term to score enzymes
on how good they are.
We covered a lot of content in this video
but the really crucial points to remember
are first the idea of the Steady-State Assumption
that we make when looking at enzyme kinetics.
This is where we assume
that the ES concentration is constant.
We knew that the formation and loss of ES are equal.
Second, we derived the critically important
Michaelis-Menten Equation
which you should consider committing to memory.
Third, we talked about how you can score
how good an enzyme is at speeding up reactions
by looking at that enzyme's catalytic efficiency
which is a combination of two new terms
we learned about Kcat and KM.