What happens to Km when enzyme concentration is *very* high?

Question

Let Km be an empirical measurement of a certain enzyme with concentration [E]. Theoretically, this value is constant and shouldn't vary when [E] goes up or down. Now let [E']=10*Km. Under this concentration of enzyme, it's clear that if [S]=Km, V0 cannot be 1/2*Vmax (as there's only enough substrate to saturate 1/10-th of the enzyme molecules). What's happening here, and how does this relate to the derivation of the Michaelis-Menten equation?

Answer 1

Michaelis-Menten is often challenging for students because of the importance of the conditions that must exist in order for the model to hold. Walking through the derivation and the relevance of the assumptions can be helpful. Berg's Biochemistry is available via NCBI, and the section on Michaelis-Menten kinetics does a good job here, illustrating the following points:

Assumption 1: The rates of all the relevant reactions are linear.

This assumption is what allows us to write down the equations the derivation starts with. We look at each separate reaction, and say that the rate of that reaction is equal to some rate constant times the concentration of each thing on the left side. From the equation $E + S \rightleftharpoons ES \rightarrow E + P$, we get the following

1. $E + S \rightarrow ES$

   • $Rate = k_1[E][S]$

2. $ES \rightarrow E + S$

   • $Rate = k_{-1}[ES]$

3. $ES \rightarrow E + P$

   • $Rate = k_2[ES]$

This assumption (the rate is linear) only matches the experimental evidence early on in the reaction, while most of the substrate is free $[S]$ (rather than bound up in $ES$). Remember this. It will come into play when we evaluate assumption 3.

Assumption 2: The $ES \rightarrow E + P$ reaction is irreversible

In order for us to combine the above rate equations to derive the Michaelis-Menten equation, the only source of ES should be from $E + S \rightarrow ES$. We can't have any measurable $E + P \rightarrow ES$. Experimentally, for enzymes that follow Michaelis Menten kinetics, this is true when $[E]$ and $[P]$ are small relative to [ES]. So here we have our first requirement that $[E]$ be small. Biochemical reactions tend to not involve very high changes in free energy. The in situ free energy of ATP hydrolysis is the high end for a reaction in the cell, and that releases about $50kJ mol^{-1}$. We're not talking about combustion here. So in fact, these reactions are often reversible at appreciable levels of $[E]$ and $[P]$

Assumption 3: ES formation and breakdown is at steady state

This assumption is often described in a confusing way, i.e. the entire reaction ($E + S \rightarrow E + P$) is at steady state. Lets drill down to what we're actually talking about. In order to derive the MM equation. We need the rate of formation of $ES$ to equal the rate of breakdown of $ES$, that is, for $\frac{d[ES]}{dt} = 0$. That is steady state. Let's say that in words: the change in $[ES]$ needs to be zero. We only achieve steady state at an early point in the reaction (remember, we need most of the substrate to be free $[S]$ for assumption 1 to hold), if $[S]$ is substantially greater than $[E]$. $E$ needs to be saturated with $S$ while $S$ is still mostly in the free form. This is the key combination of assumptions that the situation you described violates.

Once we meet the three assumptions above, we can write down the following equality:

$k_1[E][S] = (k_{-1} + k_2)[ES]$

This equation says that the rate of formation of $ES$ (the rate from equation 1 at the top) is equal to the combined rate of the breakdown of $ES$, either into $E + S$ or $E + P$ (equations 2 and 3 at the top). Once we have this equality, we can derive the Michaelis-Menten equation: $V_o = V_{max} \frac{[S]}{[S] + K_m}$, where $K_m = \frac{k_{-1} + k_2}{k_1}$. Again, see Biochemistry to walk through this.

Answer 2

The derivation of the Michaelis–Menten equation makes a number of assumptions. The exact assumptions depend a bit on how you actually do the derivation, but most modern formulations depend on the steady state approximation. (For example, see here.)

In your situation, where the concentration of enzyme greatly exceeds the concentration of substrate, that assumption is violated. You'll never reach a steady-state. As such, applying the Michaelis–Menten equation to this particular situation gives inaccurate results. You certainly can model the kinetics of the reaction in these conditions, but you need to use other equations and approaches.

That's not to say that the K_m doesn't apply in those situations. The K_m of an enzyme is a property of the enzyme that applies even in the high-enzyme regime - it's just that the "conventional" interpretation of K_m ("the concentration of substrate at which reaction rate is half maximal") doesn't apply. That statement is only true in the particular conditions which the Michaelis–Menten derivation holds.

Answer 3

The effect of high concentrations of enzyme on the Michaelis-Menten equation has been considered by M. F. Chaplin, "The effect of enzyme concentration on the Michaelis-Menten equation" published in the January 1981 issue of Trends in Biochemical Sciences, page VI, under the heading 'Textbook Error' (but, as far as I can determine, this article is not indexed in Pubmed or the like).

The author draws two important conclusions:

1. The assumption that the total enzyme concentration, $E_o$, must be less than the initial substrate concentration ($S_o$) is unnecessarily restrictive. The Michaelis-Menten equation holds, to a very good approximation, when $E_o$ is greater than $S_o$ but is less than the Michaelis constant (!)
2. At high enzyme concentrations, a modified equation is proposed $$ v_i = \frac{V_{max}[S_o]}{K_m + [S_o] + [E_o]}$$

The derivation of this equation is given in the cited reference.

Edit

• A Dropbox link to the Chaplin reference. It does not seem to be available from anywhere, including the TIBS site (and it does not appear to be indexed in PubMed or the like)
• A user asked (question now deleted) if $K_m$ is in any way dependent on E${_o}$

Setting

$$ K^{'}_m = K_m + E_o$$

it is seen that the substrate concentration at half-V$_{max}$ is $K^{'}_m$

that is the 'true' Michaelis constant will indeed depend on $E_o$ 'just a little bit' as the User suggested under the less restrictive assumptions considered by M.F. Chaplin (where $K_m$ is defined algebraically above).

• Perhaps the User would consider un-deleting this question? IMO an interesting one.