Question 1

How crucial is it to be able to derive the Michaelis equation for the MCAT?

Accepted Answer

you dont need to be able to derive it. just need to understand it, the variables, and the graphs

Question 2

at 5:07  1) 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?

Accepted Answer

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

Question 3

A 4:00 - Why does Vo=K2{ES]?  I would have assumed Vo would equal K1{E}{S} + K2{ES}.

Accepted Answer

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.

Question 4

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?

Accepted Answer

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.

Question 5

At 1:40 why 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?

Accepted Answer

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.

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.

Question 6

Is there any way to use the Michaelis Menten equation to determine the total amount of enzyme present? At 6:09 Kcat = 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?

Accepted Answer

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.

Question 7

At 1:41, 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

Accepted Answer

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.

Question 8

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].

Accepted Answer

(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 ^^

Question 9

at 3:05, 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?

Accepted Answer

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).

Course: MCAT > Unit 5

Steady states and the Michaelis Menten equation

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