Have you ever heard of a Rube Goldberg machine? In those machines, an initial action triggers a chain reaction of various tasks which all wind up performing one simple one, like turning on a light switch.
Reactions can work the same way! In a multistep reaction, one reaction kicks off another, and another, like dominoes to form a final product.
In this article, we will be continuing on from the Steady-State Approximation article. We will be looking at examples of the steady-state approximation, mostly through a biochemistry angle!
- This article is a partner article to Steady-State Approximation, and will be focusing on examples related to biochemistry
- We will look at how to use the steady-state approximation for the ozone reaction that takes place in our atmosphere
- We will also look at how the approximation is useful for the world of enzymes.
Steady-State Approximation for Ozone
Let's start by working on a steady-state approximation of ozone (O3) example. In this example, you'll see that we can also use the approximation to calculate the concentration of an intermediate.
Given the equations below, write the expression for the concentration of O3 :
\(O_2\xrightarrow{k_1}2O\)
\(O+O_2+M\xrightarrow{k_2}O_3+M\text{(M is any non-reactive species that can stabilize}O_3)\)
\(O_3\xrightarrow{k_3}O+O_2\)
\(O+O_3\xrightarrow{k_4}2O_2\)
Typically, "M" is another O2 (oxygen gas) or N2 (nitrogen gas). This mechanism is for the formation and depletion of ozone (called the Chapman mechanism)1. You'll notice that we have two intermediates: O (oxygen) and O3 .
In this case, we are just going to focus on O3 :
\(\frac{d[O_3]}{dt}=0\)
\(\frac{d[O_3]}{dt}=(k_3[O_3]+k_4[O][O_3])-k_2[O][O_2][M]\)
Now we can set up the equation for the concentrations of ozone (O3) :
\(k_3[O_3]+k_4[O][O_3]-k_2[O][O_2][M]=0\)
\(k_3[O_3]+k_4[O][O_3]=k_2[O][O_2][M]\)
\([O_3](k_3+k_4[O])=k_2[O][O_2][M]\)
\([O_3](k_3+k_4[O])=k_2[O][O_2][M]\)
\([O_3]=\frac{k_2[O][O_2][M]}{k_3+k_4[O]}\)
While it is okay to stop here, we are going to go ahead and simplify this expression some more.
\([O_3]=\frac{k_2[O][O_2][M]}{k_3+k_4[O]}*\frac{(\frac{1}{k_4[O]})}{(\frac{1}{k_4[O]})}\)
\([O_3]=\frac{\frac{(k_2[O_2][M])}{k_4}}{(\frac{k_3}{k_4[O]})+1}\)
So why did we do that? Well, we can simplify the denominator by assuming that the ratio \( \frac{k_3}{k_4[O]} \) is much greater than one, or that k3/k4[O] >>> 1.
The reason for this assumption is that this reaction mechanism is for the creation and destruction of the ozone in the atmosphere. If the destruction of the ozone layer was fast, it would deplete pretty quickly, which would be a big problem for us!
If we use this approximation, we get the final simplified equation:
\([O_3]\approx \frac{\frac{k_2[O_2][M]}{k_4}}{\frac{k_3}{k_4[O]}}\)
\([O_3]\approx \frac{k_2][O_2][M][O]}{k_3}\)
Using this series of calculations, we have derived an approximate expression for the concentration of ozone (O3).
The approximation is helpful here, since it can be difficult to monitor an intermediate. For this specific reaction, it is helpful for climate scientists, since the ozone layer is an important part of our atmosphere and for life on earth.
The Steady-State Approximation and Enzymes
There is a special type of multistep reaction used for enzyme mechanisms. An enzyme is a catalyst in living organisms.
A reactant called a substrate (S) binds to the enzyme and forms an enzyme-substrate (ES) complex, this complex then leads to the production of a product, P, and the unaltered enzyme. This process looks like this:
$$E+S\underset{k_{off}} {\stackrel{k_{on}}{\rightleftharpoons}}ES\xrightarrow{k_{cat}}P$$
Enzymes are like a cast or a mold. Let's say you have a mold that you put some clay in. The clay will warp to fit whatever shape the mold is. Once you open the mold, you now have a finished product, and the mold is unchanged. This mold is a "catalyst" since it is a lot faster to use a mold for clay instead of molding it yourself by hand (i.e. an enzyme speeds up the reaction).
We calculate the rate of reaction (here called velocity of reaction) by using the Michaelis-Menten equation. The formula is:
$$v=\frac{V_{max}[S]}{K_M+[S]}$$
Where
Vmax is the maximum reaction velocity.
[S] is the concentration of the substrate.
KM is the Michaelis constant. The Michaelis constant is the "rate constant" for enzymes.
Using this equation, we can graph the relationship between the concentration of the substrate and the reaction rate.
Fig. 1 - As the concentration increases, so does the rate until we hit the maximum reaction rate (Vmax).
We can derive this formula using the steady-state approximation.
\(\frac{d[ES]}{dt}=0\)
\(\frac{d[ES]}{dt}=(k_{cat}[ES]+k_{off}[ES])-k_{on}[E][S]\)
To get to our final equation, we need to convert [E] to [ET]. [E] is the concentration of the "free" enzyme, so basically the concentration of enzyme not bound to a substrate. However, [ET] is the concentration of all the enzymes, so it includes the bound enzymes.
\(\frac{d[ES]}{dt}=k_{cat}[ES]+k_{off}[ES]-k_{on}[E][S]\)
\([E]=[E_T]-[ES]\)
When we insert this into the equation we get:
\(k_{cat}[ES]+k_{off}[ES]-k_{on}([E_T]-[ES])[S]=0\)
Now we continue to simplify:
\(k_{cat}[ES]+k_{off}[ES]-k_{on}[E_T][S]+k_{on}[ES][S]=0\)
\(k_{cat}[ES]+k_{off}[ES]+k_{on}[ES][S]=k_{on}[E_T][S]\)
\([ES](k_{cat}+k_{off}+k_{on}[S])=k_{on}[E_T][S]\)
\([ES]=\frac{k_{on}[E_T][S]}{k_{cat}+k_{off}+k_{on}[S]}\)
\([ES]=\frac{k_{on}[E_T][S]}{k_{cat}+k_{off}+k_{on}[S]}*\frac{\frac{1}{k_{on}}}{\frac{1}{k_{on}}}\)
\([ES]=\frac{[E_T][S]}{(\frac{k_{cat}+k_{off}}{k_{on}})+[S]}\)
We multiply this expression by kcat:
\( ([ES]=\frac{[E_T][S]}{(\frac{k_{cat}+k_{off}}{k_{on}})+[S]})*k_{cat}\)
\([ES]k_{cat}=\frac{k_{cat}[E_T][S]}{(\frac{k_{cat}+k_{off}}{k_{on}})+[S]}\)
Since \(v=[ES]k_{cat}\),
\(v =\frac{k_{cat}[E_T][S]}{(\frac{k_{cat}+k_{off}}{k_{on}})+[S]}\)
Now it looks very similar to our original equation:
\( v=\frac{V_{max}[S]}{K_M+[S]} \)
So, by comparing the newly derived equation with our original equation, we get:
- \( V_{max}=[E_T]k_{cat}\)
- \( K_M=\frac{k_{cat}+k_{off}}{k_{on}} \)
The math here isn't super important, it's just to show you how important the steady-state approximation is to enzyme kinetics. By using this approximation, we now only have to focus on the concentration of the substrate, which is a lot easier to monitor.
Enzymes aren't the only kind of catalytic reactions. A catalyst is any species that can speed up a reaction, but isn't a product or a reactant. Unlike an intermediate, a catalyst is not produced during the reaction, and it is still present at the end of the reaction. However, it is not present in the net chemical equation.
We can use the steady-state approximation for these catalytic reactions in the same way as shown above, where we set the change in catalyst concentration equal to 0.
Steady-State Approximation: Biochemistry - Key takeaways
- When looking at multistep reactions, they may have multiple intermediates, such as in our ozone example
- The approximation is important for enzyme kinetics as it is used to derive the formula for the reaction velocity.
- The steady-state approximation can also be used for catalytic reactions (like in our enzyme example), since it is also not present in the net equations like intermediates.
References
- Atmospheric Ozone Chemistry, Columbia.edu (July 26, 2022)
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