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# Steady states and the Michaelis Menten equation

## Video transcript

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.