Suggested languages for you:

Americas

Europe

|
|

Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken

Nie wieder prokastinieren mit unseren Lernerinnerungen.

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!

• 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.

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

1. Atmospheric Ozone Chemistry, Columbia.edu (July 26, 2022)

The steady-state approximation is used to derive the rate law of a multistep reaction. We do this by setting the concentration equal to zero. For example, the process that forms and consumes ozone (O3) has two intermediates: O3 and O (oxygen). We would set their concentrations equal to 0 to solve for the rate law of the total reaction.

You identify what species is the intermediate (both produced and consumed by the reaction) and set its change in concentration equal to zero.

In enzyme kinetics, the steady-state approximation is used to derive the Michaelis-Menten equation. This equation is used to find the rate of reaction for enzyme reactions

The steady-state is an approximation which helps use solve for the rate law of a multistep equation. Trying to solve for the rate law without this approximation is very difficult.

The steady-state approximation is used when the first step of a multistep reaction is very slow compared to the second. However, for the equilibrium approximation, the first step must be faster

60%

of the users don't pass the Steady State Approximation Example quiz! Will you pass the quiz?

Start Quiz

## Study Plan

Be perfectly prepared on time with an individual plan.

## Quizzes

Test your knowledge with gamified quizzes.

## Flashcards

Create and find flashcards in record time.

## Notes

Create beautiful notes faster than ever before.

## Study Sets

Have all your study materials in one place.

## Documents

Upload unlimited documents and save them online.

## Study Analytics

Identify your study strength and weaknesses.

## Weekly Goals

Set individual study goals and earn points reaching them.

## Smart Reminders

Stop procrastinating with our study reminders.

## Rewards

Earn points, unlock badges and level up while studying.

## Magic Marker

Create flashcards in notes completely automatically.

## Smart Formatting

Create the most beautiful study materials using our templates.