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In Michaelis-Menten kinetics, it is assumed that the substrate concentration is much higher than the enzyme concentration. In which step of the derivation of the Michaelis-Menten rate law is this assumption used?

From the explanations given in different sources, I think this assumption is used in expressing the total enzyme concentration as the sum of free enzyme and enzyme-substrate complex. Is my understanding correct? What happens when the enzyme concentration is in the same range as that of substrate concentration? Why is it important to make this assumption?

Could someone explain?

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  • $\begingroup$ Well, that's certainly not the case near the end of the reaction... $\endgroup$ – Zhe May 25 '18 at 11:45
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Let us look at the derivation detail. I will try to note, all assumptions/approximations made. I hope I don't miss any. The basic model is as follows: First an enzyme substrate complex is formed. The substrate can either be released unchanged, or undergoes modification to form the product. This can be captured in the following two equations

$$\ce{E + S <=>[k_a][k'_{a}] ES } \tag{1}$$ $$\ce{ES ->[k_b] P } \tag{2}$$

First, assumption herein is that initially, so little product is formed that the reverse reaction (product combining with enzyme and re-forming substrate) can be ignored, and hence the unidirectional arrow.

Rate of formation of product is, $v = k_b \ce{[ES]}$

And we use the steady-state approximation to write, $$\frac{\mathrm{d}\ce{[ES]}}{\mathrm{d}t} = k_a\ce{[E][S]}-k'_a\ce{[ES]}-k_b\ce{[ES]} \approx 0$$

Re-arranging, $$ [ES] = \overbrace{\frac{k_a}{k'_a + k_b}}^{K_M}[E][S]$$

Now, let us re-cast this in terms of concentrations of enzymes, and substrate we started with.

From the explanations given in different sources, I think this assumption is used in expressing the total enzyme concentration as the sum of free enzyme and enzyme-substrate complex. Is my understanding correct?

$$\ce{[E]_0 = [E] + [ES]}$$

No. The equation given above holds regardless of the initial substrate concentration.

Substituting for $\ce{[ES]}$

$$\ce{[E]_0 = [E] + \frac{K_M [ES]}{[S]} = [ES]\left( 1+ \frac{K_M}{[S]}\right)}$$

In which step of the derivation of the Michaelis-Menten rate law is this assumption used?

Now, finally, the substrate concentration is so large, that $\ce{[S] \approx [S]_0}$

This, finally gives us

$$\ce{[ES] = \frac{[E]_0}{1+ K_M/[S]_0}} \tag{*}$$

Plugging this in the expression for $v$, we get the Michaelis-Menten equation

$$v = \frac{k_b\ce{[E]_0}}{1+ K_M/\ce{[S]_0}}$$

What happens when the enzyme concentration is in the same range as that of substrate concentration? Why is it important to make this assumption?

You don't get a nice expression for $\ce{[ES]}$ like equation $(*)$ which only involves constants on the right hand side.

$$\ce{[ES] = \frac{[E]_0}{1+ K_M/([S]_0-[ES])}} \tag{**}$$

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  • $\begingroup$ Many computational models that are used to study biological pathways have been built using Michaelis-Menten equation. In a living cell, the concentration of the substrate wouldn't be much higher than the substrate. What is the reasoning behind using Michaelis-Menten kinetics to model biological processes where these assumptions don't hold always?@getafix @Dr.J. $\endgroup$ – Natasha May 25 '18 at 15:37
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    $\begingroup$ @Natasha This can be a question in itself. Short answer, it has does fit experimental observations well. I’ll have to do more research because this domain is a bit foreign to me. But like I said, ask on the site, and perhaps other people can weigh in $\endgroup$ – getafix May 25 '18 at 15:40
  • $\begingroup$ @Natasha:I'm not sure I quite understand your question on this point. What are you saying about the living cell? Are you saying [S] is not >> [E]? I believe it is in most, if not all cases. $\endgroup$ – Dr. J. May 25 '18 at 15:57
  • $\begingroup$ Yes, I wish to know whether [S] >> [E] in biological systems? There are a few sources that state the enzyme concentration and substrate fall in the same range. Excuse me for the naive question. Since I am not able to find the enzyme concentration in the literature in which the substrate concentration is reported, I'm confused.@ Dr. J. $\endgroup$ – Natasha May 25 '18 at 16:14
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I think it is more about expressing[S]. If we assume the substrate concentration is much higher than the enzyme concentration, then we can ignore the substrate that is bound as ES in the derivation of the MM equation. If the enzyme concentration is in the same range as the substrate (or if the enzyme affinity is very high), then the amount of substrate bound as ES cannot be ignored, and the derivation is not simplified. In this latter case, [S] in the equation can't be considered as the total, but must be corrected by the amount bound.

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First we need to understand two things. (1) Since an Enzyme is a catalyst, we assume that the forward reaction is much, much, more favored than the reverse reaction. That is, k1 >> k-1. (2) It is impossible to measure the concentration of the Enzyme-Substrate Complex. We can only know how much Enzyme is present, how much Product is formed, and how much Substrate is consumed.

Having said that, we get to the heart of your question. The rate determining step is the dissociation of the Enzyme-Substrate Complex into the Product and the Free Enzyme. So essentially V = k2[ES]. But since we cannot know what [ES] is, we have to relate it in terms we know. So [ES] should be able to determined from [E] and [S]. In order to do this, [ES] must be at a steady state until most of the substrate is consumed. This is why it's important that [S] be greater than [E]. [S] will continue to decrease and [P] will continue to increase. This won't hold true if there is insufficient substrate to form [ES].

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    $\begingroup$ Catalysts don't change the equilibrium of a reaction, so I am not sure that your first point in the first paragraph is correct. $\endgroup$ – Curt F. May 25 '18 at 14:03

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