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## MCAT

### Unit 5: Lesson 3

Enzyme kinetics# Steady states and the Michaelis Menten equation

Created by Ross Firestone.

## Want to join the conversation?

- How crucial is it to be able to derive the Michaelis equation for the MCAT?(76 votes)
- I do not believe you will need to derive the equation, although learning how the equation is derived and what all the terms mean is important to the understanding of the concepts and when you can apply the equation. It is also helpful for problem solving in case you need to do tweeks to the equation to get the information required.(8 votes)

- at5:071) Why can you just assume Km = [S]?

(2) Why would Km even have the same units as [S], i thought it was just a combination of all of the other constant values we had in the equation previously?(22 votes)- Because when [S]=Km, this is the optimum condition for the cell.

When S>>Km, V0=Vmax[S]/[S], this means that the reaction is always catalyzed at full speed and the enzyme cannot be fine tuned by the cell.

When S<<Km, V0= Vmax[S]/Km, this means that the enzyme can be fine tuned, but it will never reach its full potential(7 votes)

- A4:00- Why does Vo=K2{ES]? I would have assumed Vo would equal K1{E}{S} + K2{ES}.(12 votes)
- The speed of a reaction is usually defined by the rate of formation of products or the rate of consumption of reactants, which in this case would be equal to k2[ES].

Here's an interesting way to go about answering your question:

Imagine if you will that reaction 2 is the rate limiting step of the sequence. We always assume that the rate of an entire process is equal to the rate of the rate limiting step, which would make the rate of this entire process k2[ES].

Now imagine that the rate limited step is reaction 1. Since the only source of ES is from reaction 1, we could still define the rate of the entire process by k2[ES] (since the concentration of ES is dependent on its formation from reaction 1).

In both cases the rate of the whole process (E + S -> E + P) would be equal to k2[ES], but we can't say the same for k1[E][S] since reaction 2 could be the rate limited step of the sequence.(23 votes)

- Does anyone know if the Kcat and catalytic efficiency concepts will be on MCAT 2015? I don't see them mentioned in this content outline here: https://www.aamc.org/students/applying/mcat/map/mcat2015knowledgemapconcepts2/368562/cpfbfc5cc5e.html

I was a bit overwhelmed by this video frankly, but after looking at a practice test for the old MCAT, it appears that you really only need to understand the meaning of Vmax and Km and how enzymes and inhibitors affect them. Does anyone know if that's an accurate impression, and if it will likely hold true for the new MCAT 2015?(7 votes)- The official MCAT guide outline says "Michaelis-Menten" - I am assuming we'll have to understand what effects inhibition has on Vmax, Km, what the corresponding graphs look like. A sample question on the MCAT 2015 guide demands that we: "must recognize the

relationship between two variables in the context of an experiment. To answer the question, you must recognize KM, recall its significance in Michaelis-Menten enzyme kinetics, and relate it to another fundamental variable, Vmax. "

All in all, I'm not sure entirely, but from what the MCAT 2015 guide says, it SEEMS like we won't have to derive the equation, it's more about understanding it and particularly Vmax and Km.(5 votes)

- At1:41, it was mentioned that Rate(-2) is so small that we can cross it out, would appreciate if you could clarify that part? How do we know that rate-2 is negligible?

I understand that is more thermodynamically favoured to process from ES----> to E +P , but why? thanks ahead(4 votes)- If you have watched his earlier videos and have seen the reaction coordinate diagrams, it show you that the product is at a lower energy level. This implies that in order to convert the the product back into the reactants, you not only have to provide enough energy to overcome ∆G of the transition state, you also need to provide enough energy to overcome the free energy, ∆G0. In a biological system, this would be a significant amount of energy, likely so much so that you would not be able to catalyze the reverse reaction. That is why biological systems rely mainly on enzyme catalysis to perform chemical reactions.

Another thing to consider is that once the ES passes through the transition state, the product will begin to form, the enzymes association with the molecule in its active site will begin to weaken, and the product will eventually be released into the environment. The enzyme returns to its initial conformation. What you need to remember is that the active site of that enzyme likely only has a high degree of affinity to bind the substrate and not the product, as that is how the specificity of binding works, whether we look at it from the perspective of the induced fit or lock and key model. The enzyme may still be able to bind product and catalyze the reverse reaction, but the affinity for the product is likely such that a substrate molecule will always outcompete a product molecule for binding with the enzyme, or k-2 will be a very slow rate and therefore the enzyme will not bind to product molecule (substrate outcompetes) or it will take a much longer time for the enzyme to catalyze the reaction of product back into the ES transition state. Remember here that not only will the Enzyme need to overcome the ∆G energy barrier, but it also has to overcome ∆G0 to get back to ES, so it will be thermodynamically more difficult to catalyze the reverse reaction as well as all of the other factors that are mentioned. This is the reason that you can assume that ES <- EP is negligible and can be ignored.

In real examples, enzymes work incredibly quickly catalyzing substrate into products, sometimes catalyzing thousands or tens of thousands of reactions per second. Even if the reverse reaction took place 10 times a second, this is orders of magnitude slower than the forward reaction and therefore would be become a decreasing percentage of the total reactions as each second went by. E+S would so out pace <- E+P that nearly 100% of the formation of ES would come from E+S.(12 votes)

- At1:40why do you cross out the bottom arrow going from {E + P} to {ES) and not the arrow going from {ES} to {E + S}? I am asking this because you mentioned that products rarely go back to reactants and since {E + S} is a reactant, the first bottom arrow should be crossed out. Right?(5 votes)
- Remember the steady state assumption. Formation of ES = The Loss (Dissociation) of ES. It is not likely that product will go back to becoming the ES complex, but it is likely that the Enzyme can release the substrate before it is able to catalyze the reaction to E+P, so you cannot ignore Rate-1 as that will contribute an appreciable amount to the dissociation side of the steady state. This means that Rate 1 (formation of ES) = Rate-1 (Dissociation of ES to E+S) + Rate 2 (Dissociation of ES to E+P). As the active site of E opens up with the dissociations it is free to bind new substrate, so you remain in a steady state.

Even if Rate-1 is very rare, it will directly contribute to an Enzyme being freed up for ES formation, so in order to keep the equality intact we need to factor it in.(5 votes)

- Is there any way to use the Michaelis Menten equation to determine the total amount of enzyme present? At6:09Kcat = Vmax/[E]t so if the kcat was determined for the enzyme in a separate experiment, could that value then be used to determine [E]t in another experiment?(5 votes)
- Ideally yes, as long as your conditions meet the assumptions ( [substrate] > Km ) and the environment is the same. Although its more common and accurate to experimentally determine Km to calculate unknown enzyme concentration since you can rule out environmental differences between experiments.(3 votes)

- wait, why is the rate of E + P formation = k2[ES]?

I thought rate = k[products]/[reactants] therefore in this case the rate of E+P formation should = k2[E][P]/{ES] and NOT k2[ES]. Even if the reaction is 0th order with respect to E and P then the rate forward could only still be k2/[ES].(4 votes)- (Just a little note before i explain: i think the formula you are using is the correct formula for acid dissociation constant ka, its a bit confusing because they use the same letters k and a!)

If you look back to the last video when looking at enzyme kinetics define a reaction like:

A -> B

He defines rate = K (a)

(Rate = constant (k)×concentration of reactant (a))

So in this reaction

E + s <-> es -> e + p

[K1 (e) (s) = k-1 (es) + k2 (es)]

We specifically want to focus on the reaction that produces products

Es --> e +p

The product is p and the reactants are es

Hence if we sub this into

Rate = k (a) <--- a being reactants

rate = ks2 (es)

Hope this helps ^^(2 votes)

- At4:15why does he multiply everything by k2, even if we know that reaction rate is dependent on k2[ES] , I don't really understand why we have multiplied it. (Sorry I need to know why, so the whole derivative makes sense to me)(3 votes)
- 2:43mark, can you better explain what you did on left of equation? You said you were solving for [E] on left side, yet your right side doesn't (appear, at least to me) to equal [E]?

Thank you,(1 vote)

- at3:05, when the right side of the equation is divided by K1, why is it possible to just group it in with the rest of the rate constants as opposed to dividing the entire side of the equation by K1?(2 votes)
- It wouldn't matter whether K1 was grouped or divided through the entire equation, the outcome would be the same. Try plugging some numbers into the equation (for the right side only), and you will see that the answer is the same regardless of whether K1 is grouped or not. The reason it is grouped is to make easier to see how Km can be substituted for ((K-1 + K2)/K1).(2 votes)

## 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.