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# Catalytic efficiency of enzymes

1. 1. Introduction
3. 3. Problems

Enzymes exist in all biological systems in abundant numbers, but not all of their functions are fully understood. Enzymes are important for a variety of reasons, most significantly because they are involved in many chemical reactions that help us to maintain our daily lives. Increasing the reaction rate of a chemical reaction allows the reaction to become more efficient, and hence more products are generated at a faster rate. These products then become involved in some other biological pathway that initiates a certain function of the human body. This is known as catalytic efficiency of enzymes, which, by increasing the rates, results in a more efficient chemical reaction within a biological system.

### Introduction

An enzyme's active sites are usually composed of amino acid residues, depending on which amino acid residues are present, the specificity of the substrate can vary greatly. As we all know, depending on the pH level, the physical properties (mainly, the electric charge) of an enzyme can change. A change in the electric charge can alter the interaction between the active site amino acid residues and the incoming substrate. With that said, the substrate can bind to the active site via Hydrogen bonding or Van Der Waals Forces. Once the substrate binds to the active site it forms an enzyme-substrate complex that is then involved in further chemical reactions.

In order for an enzyme to be active and be energetically favorable to allow a chemical reaction to proceed forward, a substrate must bind to an enzyme's "active site". An active site can be thought of as a lock and the substrate as a key, known as the lock and key theory. A key (substrate) must be inserted and turned (chemical reactions), then the lock (enzyme) will open (production of products). Note that an enzyme might have more than one active site. Another theory on the active site-substrate relationship is the induced fit theory, which is quite opposite of the lock and key theory (where the active site is seemingly inflexible). In the induced fit theory, the active site of the enzyme is very flexible, and only changes its conformation when the substrate binds to it.

Enzymes work as a catalyst by lowering the Gibbs free energy of activation of the enzyme-substrate complex. Below are two figures showing a basic enzymatic reaction with and without a catalyst.

Notice the difference of the Gibbs Energy with the presence of a catalyst; it is higher in value when a catalyst is absent. We know that Gibbs Energy can tell us whether or not a reaction is spontaneous judging from its value. In this case  ∆G­,the Gibbs Free Energy, can be seen as the probability of whether or not a reaction would occur in nature, without any interference. With a lower Gibbs energy values, the probability of reaction is higher, and vice versa. Enzymes are here to increase the probability of the chemical reaction via many mechanisms, with one of the more prominent explanations being that it changes into a more favorable conformation.

Now that we have a basic understanding of enzyme kinetics, how does one determine the effiency of the enzyme? Lets start by looking at a simple enzymatic reaction:

Fig 3

German biochemist Leonor Michaelis and Canadian biochemist Maud Menten derived an equation later known as the "Michaelis-Menten Equation", shown below.

$v_0 = \dfrac{v_{max}[S]}{K_M + [S]}$

From a quick glance, this equation tells us the rate of the reaction at a given substrate concentration, given that we know Vmax , which is the maximum rate the reaction can proceed at, and also KM, the Michaelis constant. However, in a practical application of the Michaelis-Menten, V0 is often measured, and Vmax is observed when we see saturation or plateau in our data plot. Since we know the substrate concetration, KM is usually the calculated value we are after.

For KM we know that V0 = V­Max/2

Max/2 = (Vmax[S])/(KM + [S])

(KM + [S])(V­Max/2) =  (Vmax[S])

(KM + [S]) = (Vmax[S]) / (V­Max/2)

KM + [S] = 2[S]

KM = [S]

The Michaelis constant can be thought of as the rate at which the substrate becomes unbound to the enzyme, which can either be in the events of substrate-enzyme complex becoming the product, or the substrate becomes unbound to the enzyme. KM can be shown as an equation.

$K_M = \dfrac{k_{-1} + k_2}{k_1}$

Whereas k-1 is the rate constant at which the substrate becomes unbound to the enzyme, resulting in the dissociation of the enzyme-substrate complex, K2 is the rate constant where the substrate-enzyme complex dissappears and turns into product, and K1 is the rate constant for the formation of the the substrate-enzyme complex formation. Therefore, we can view KM as the rate of substrate-enzyme complex dissappearance divided by the rate of substrate-enzyme complex formation, which is the level at which half of the substrate is bound to the enzyme. KM is a useful indicator for the presence of an inhibitor because we can look for changes in KM and compare to our control (biological systems that we know have zero inhibitor presence). KM is a dependant variable, and its value can change due to many reasons, including the pH level of the system, temperature, or any other condition that might affect a chemical reaction. A small KM means that the substrate has a high affinity to the enzyme.

So in what ways can the Michaelis-Menten equation tell us about the efficiency of the enzyme? If we were to graph V0 vs [S], we would end up with something like this.

Vmax is the maximum rate the reaction can run at, regardless of [S], meaning that even if you add more substrate, the reaction would not go any faster. That is because at Vmax all of the active sites on the enzyme are occupied.  After all the explanations on various forms of enzyme kinetic equations, we arrive at our conclusion of catalytic efficiency. Referring back to Fig 3, we have:

Fig 3

Notice k2 is an irreversible reaction as oppose to an equilibrium expression, when compared to k-1 and k1. K2 here is also known as kcat the catalytic effiency of enzyme. From our previous discussion we learned that Vis the measured reaction rate, which is the product formation over time, so we can conclude that an equation would end up looking like the following:

­$v_0 = \dfrac{d[P]}{dt} = k_2[E]_0$

Where [E]is the total enzyme concentration.

We also know that Max is observed when all of the enzyme-substrate complex disappear and turn into products, so we can make the following assumption:

$V­_{Max} = k_2[E]_0$

and after rearrangment, we have this equation:

$k_{cat} = k_2 = \dfrac{V­_{Max} }{[E]_0}$

That is the equation for calculating catalytic efficiency, to be used after we obtain data from experiments and after using the Michaelis Menten Equation. With a larger kcat , the enzyme is efficient because less enzyme is needed.

### Problems

1. What is the rate of product formation if k2 is 4.3 min-1 and [ES] is 2.3 x 10-2 M?
2. Find the formation of ES if k1 = 3.3 x 103 min-1, k-1 = 1.1 x 10-10 min-1 and [E] and [S] are 1.2 x 10-6 M and 6.3 x 10-2 M respectively.
3. Given that k-1 = 8 x 10­­4 s-1, k2 = 9 x 10­­5 s-1, k1 = 7 x 10­­6 ­M-1s-1 find KM. What does the answer tell us about the affinity of the substrate?
4. Using the KM value from above, find V0 if we determine experimentally the Vmax value to be 1.5 x 10-4 M min-1 and the [S] is 5.1 x 10-4 M.
5. Judging from KM from problem 1, we hypothesize that [E]0 is half of [S], find k2 using the Vmax value given above.

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This module had zero development, I took the liberty to do some basic development by introducing the idea of an enzyme and a catalyst, but I feel the work I am trying to expand on are more Phase 2 related. So for Phase 1 I just created this module with some basic introduction, and as for Phase 2 this page will be fully developed with appropriate equations, figures, problems and references for further studies.
Posted 06:18, 7 Mar 2011
1. Synopsis: This module explains what enzymes are, how they work, and there efficiency as catalysts. It explains that they are necessary for making efficient chemical reactions, but only function if an active site is bound by its substrate. It also provides a mechanism on how the enzyme active site works with its substrate and the bonds associated between the two. Overall, the main point provided is how the enzyme makes a chemical reaction more efficient by reducing the Gibbs energy of activation.
2. Vet2 Level: No, I do not believe that this module is ready to move to the Vet2 level. There were several grammatical-type errors and spelling errors. Also, the module is clearly still in the draft stage and needs additional content.
3. Review Details: The format of the module is good in that it contains an introduction. However, it is incomplete and needs additional information to be added. The current information is relevant and useful and the author provides a preview to what other topics will be added. Since this is clearly a draft of an incomplete module, there is still a lot of missing information.
Figures: The figures are helpful in understanding the content of the module. However, they are not properly labeled (i.e. Figure 1, Figure 2, etc).
Table: No table provided.
Text: The wording of the module is not always clear nor does it always flow well. There are several English mistakes.
4. Mistakes: I corrected “effcient” to “efficient.”
5. Plagiarism: The below websites were used to search for plagiarism – none was detected. The CHE 107B textbook for Winter 2010 was also reviewed for plagiarism; none was found. Key words were googled and plagiarism was not found.
http://www.articlechecker.com/
http://www.dustball.com/cs/plagiarism.checker/
http://www.plagiarismchecker.com/url
6. Tineye.com was used to determine if any of the images were taken from commercial sites; all searches turned up zero results.
Posted 23:29, 7 Mar 2011
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