biokim

THE EFFECT OF INHIBITION TOWARD ENZYME ACTIVITY

By: Ni Luh Ayu Sriningsih (0913031055),Chemistry Education Department, Ganesha University of Education

ABSTRACTEnzymes are biocatalyst that can increase the velocity of reaction. Enzyme is

influenced by many factors when it does its task, one of them is inhibitor. Inhibitor is a compound that can block and decrease reaction rate that catalyzed by enzymes. This inhibitor will bind with enzyme to produce enzyme-inhibitor complex. Inhibition complex consist of reversible and irreversible process. Reversible inhibition was determined quantitatively using Michaelis-Menten kinetics equation, that consist of competitive, non competitive, and uncompetitive.

In this experiment was done testing of the influence of competitive inhibitor to the enzyme activity, by looking Vmaks and KM value of reaction without inhibitor and with inhibitor. Sodium succinate that used in this experiment as substrate, while the piruvate as inhibitor. Enzymes that would be inhibited were got from extract (homogenate) chickens liver. The aim of this experiment is to know the effect of inhibitor to enzyme activity. According to the discussion, it can be conclude that increasing in concentration causing increase in KM value and Vmak value not change. KM value of without inhibitor reaction is 6,754x 10-2 M, while reaction with inhibitor is 0,133 M.Key words: inhibitor, enzyme, enzyme activity.

INTRODUCTION

Enzyme is a biocatalyst that can increase the speed of reactions in biological systems and the enzyme itself does not change during the reaction (Redhana, 2004). As with any other catalysts, enzymes affect reaction rates when equilibrium is reached, but did not affect the total equilibrium of the reaction. Enzymes assist the reaction by providing a reaction pathway that has a lower activation energy for the transition than the substrate into a product that is not catalyzed process (Tika, 2007).

Based on its structure, the enzyme consists of a component called apoenzim of protein and other component called a nonprotein prosthetic group. Prosthetic group is divided into coenzymes and cofactors. Coenzyme form of organic groups in general are vitamins, like vitamin B1, B2, NAD+ (Nicotinamide Adenine Dinucleotide). Cofactor in the form of inorganic clusters that are usually in the form of metal ions such as Cu2+, Mg2+, dan Fe2+. The enzyme is completely made up of active protein called apoenzim and coenzyme, which was then called the united holoenzyme (Isnan, 2007).

Here the enzyme has a catalytic role in lowering the activation energy of a reaction. Activation can be interpreted as a number of energy or calories are derived

by a mole of substance at a given temperature to bring the molecule into its active (Wirahadikusuma, 1989).

Protein that acts as an enzyme is formed with slits that act as a catalytic. Many compounds can enter the gap and then out again, but once a compound according to infiltrate, the enzyme will close the gap around the compound and put a certain chemical groups in a certain order of geometric conformation. The basic function of the enzyme, the catalytic effectiveness of an amino group, carboxyl group, and other binding groups increased several hundred-fold by placing it in a certain spatial arrangement so as to lock the affected compounds. Place the catalytic (catalytic site) of an enzyme is part of the binding site (binding site). With technical terms can be stated that an enzyme can indicate a high specificity towards their substrate.Substrate is a compound whose reaction is catalyzed by the enzyme. This meant that only one enzyme can catalyze the reaction from several different substrates or only a few compounds that can act as a substrate for an enzyme. The existence of the powerful forces that this is a very important factor to facilitate the transformation of the substrate so that the bonds in the substrate molecule can be disconnected and then forming new bonds (McGilvery, 1996).

The enzyme works in two ways, namely according to the theory of Lock and Key Theory and the suitability of Induction Theory (Induced Fit Theory). According to a key-lock theory, the reaction between the substrate with the enzyme because of the suitability of the substrate to form a space between the active site (active site) of the enzyme so that the enzyme active tend to be rigid. The substrate acts as a key entry into the active site, which acts as a lock, resulting in enzyme-substrate complex. At the time of the complex enzyme-substrate bonding is lost, the reaction product is removed and the enzyme will return to its original configuration. Meanwhile, according to the theory of induced fit, the reaction between the enzyme to the substrate takes place due to the induction of the enzyme substrate to the active side in such a way that they are structures that complement or complement each other. According to this theory is not the active site is rigid, but more flexible. The models of both theories can be described as follows.

Figure 1. Model of Lock and Key Theory and Induced Fit Theory

Action of the enzyme can be influenced by several factors. The factors are as follows, (Poedjiati, 1994).a. Concentration of enzyme

At a given substrate concentration, reaction rate increases with increasing concentration of the enzyme.

b. Substrate concentrationIncreased the substrate concentration can increase the speed of reaction when the amount of enzyme fixed. Yet at the active side of all the enzymes bind to the substrate, the addition of the substrate cannot further increase the speed of enzyme reactions.

c. Temperatured. The enzyme is composed of protein, so it is very sensitive to temperature.

Increase in temperature causes the kinetic energy of the substrate and enzyme molecules increases, so the reaction rate also increased. But the temperature is too high can cause damage to the enzyme, called denaturation, while the temperature is too low to inhibit the action of the enzyme. In general, the enzyme will work well at the optimum temperature, between 30o-40oC.

e. Effect of pHf. Changes in pH can affect key amino acid changes in the active enzyme, thereby

blocking the active side join the substrate. Each enzyme can work well on the pH optimum, each enzyme has a pH optimum different. Structure of the ion depends on the pH environment of enzymes, enzyme can form a positive or negative ions or charged double (zwitter ion). pH can cause the denaturation process can result in decreased enzyme activity.

g. Effect of activators and inhibitorsActivators are molecules that facilitate the bond between the enzyme to the substrate. While the inhibitor is a molecule that inhibits the enzyme with substrate binding.

As described above, activators and inhibitors can affect the action of the enzyme. Where inhibitors are molecules that can decrease the activity of the enzyme. In other words, inhibitors can inhibit or decrease the rate of enzyme catalyzed reactions (Tika, 2007).

There are two main types of inhibitors, the inhibitor is not working behind the (irreversible) and inhibitors that work in reverse (reversible). E + I EI.Irreversible inhibitors, work by binding to the active enzyme through irreversible reaction resulting reaction Therefore, after the inhibitor binds to the enzyme, inhibitor cannot be separated from the inactive enzyme. This condition causes the enzyme cannot bind substrate or inhibitor damaging some components (functional groups) on the side of the catalytic enzyme molecule.

Reversible or irreversible inhibitors work by binding to the side of enzyme active through a reversible reaction and inhibitor can be separated or released back of the bond, for example by dialysis. Reversible inhibitors consists of three types, namely a competitive inhibitor that works, non-competitive, and un-competitive.

substrateEnzyme

Competitiveinhibition

Product

Figure 2. Competitive inhibition at the work of enzyme

A. Inhibitor That Works Competitive Competitive inhibitors in general has a chemical structure of inhibitor (I) is

similar to the substrate (S). Therefore, in a reaction mixture, inhibitors compete with substrate for the enzyme bound to the active side. If the substrate concentration is high, then the inhibitor will be forced out of the active enzyme by the substrate molecules. Binding inhibitors of enzymes that have been unable to react with the substrate to produce a product, while the enzyme that has bound substrate can produce a product, it cannot bind to the inhibitor. The model of the reaction mechanism of competitive inhibition and the picture is as follows.

E + S ES E + PE + I EI

In the reversible competitive inhibition does not change Vmax, but the affinity of the enzyme to the substrate is reduced. Thus the KM will increase (Girindra, 1993). One example is the competitive inhibitor malonate is inhibite reaction catalyzed by succinate dehydrogenase. Where the enzyme succinate dehydrogenase catalyzes release of two hydrogen atoms of succinate, which is one of each of the two groups metilennya (-CH2-). Dehydrogenation of succinate was inhibited by malonate of succinic resemble because both have a negatively charged carboxyl groups is appropriate so as to occupy the enzyme active. However, the hydrogenation of succinic dehydrogenase by the enzyme so that the malonate only occupy the enzyme active side and lock it so that enzymes cannot work on the substrate. Malonate inhibited structure that is different than the structure of the succinic methylene group is as follows.

- I + I

FADH2E EFAD+

COO-

CH

CH

COO-

COO-

CH2

CH2

COO-

suksinat dehidrogenase

+

suksinat fumarat

FAD+

COO-

CH2

COO-

tak bereaksi

suksinat dehidrogenase

Succinate dehydrogenase enzyme catalyzes the release of two hydrogen atoms of succinate, which is one of each of the two groups metilennya (-CH2-). Enzyme succinate dehydrogenase was inhibited by malonate of succinic resemble because both have a negative charge of carboxyl group to occupy the enzyme active. However, malonate is not hydrogenated by the enzyme succinate dehydrogenase. Malonate only occupy the enzyme active side and lock it so that enzymes cannot work with the substrate (Tika, 2010).

Effect of competitive inhibitors in the reaction depends inhibitor concentration, substrate concentration, as well as the relative affinity of the substrate, and inhibitor of the enzyme. At a certain concentration of the inhibitor and the enzyme, when substrate concentration is low, the inhibitor has a greater capacity than competing with the substrate in the enzyme binding to a high level or degree its inhibition. However, the concentration of inhibitor and the same enzyme, when substrate concentration is high, then the inhibitor has the ability to compete is smaller than the substrate in the enzyme that binds to a low degree of its inhibition. If the substrate concentration is very high, then the number of substrate molecules far exceeds the number of inhibitor molecules, the influence of inhibitors of this state becomes very small or negligible.

B. Un-competitive InhibitionUn-competitive inhibitor can only bind the enzyme-substrate complex, and

cannot bind to free enzyme molecules. Complex enzyme-inhibitor-substrate complex is inactive because it cannot produce a product.

E + S ES E + P

No reaction

+ I- I- I + I

+ S

Enzym Substrate

Product

Figure 3. Non competitive Inhibition at the work of enzyme

C. Non-competitive InhibitionIn the non-competitive inhibition is competition between S and I in getting the

active site on the substrate. Inhibitors that work in a non-competitive with molecules able to bind both free enzyme and the complex. The enzyme-substrate complex generates an inactive inhibitor (not produce). Structure of the inhibitor is usually little or no resemblance to the enzyme and can be considered to bind to the substrate at different places. With the bound inhibitor, the catalytic activity of the enzyme to be defective. The model of noncompetitive inhibition of the reaction mechanism is as follows.

E + S ES E + P

Non-competitive inhibitor binds to the enzyme at different sides of the substrate binding. With the bound inhibitor, the enzyme activity becomes damaged. This may be caused by inhibition bound to the enzyme catalytic atu inhibitor bound to the other side (not the catalytic side). However, its binding causes conformational changes of enzymes that affect the state of the catalytic side while not affecting the binding of the substrate. Effect of competitive inhibitors in the reaction depends on menginhibisi inhibitor concentration, substrate concentration, the relative affinity of the substrate, and inhibitor of the substrate.

Figure 4. The relatioship between Competitive Inhibition and Michaelis-Manten Plot

At a certain concentration of inhibitor and enzyme, if the substrate concentration is low, the inhibitor has a greater ability to compete in binding to the enzyme than the substrate so that the high its inhibition. However, if the same concentration of enzyme inhibitors and, if the substrate concentration is high then the inhibitor has the ability to compete is smaller than the substrate binds to the enzyme so that the low-power its inhibition. If the substrate concentration is very high, then the number of substrate molecules far exceeds the number of inhibitor molecules. In this situation, the effect of inhibitors can be very small (negligible).

Figure 5. Plot Line Weaver-Burk 1/V toward 1/ [S] between reaction that used inhibition and without inhibition

Michaelis-Menten plot above shows that at high substrate concentrations, the effect of inhibitors can be eliminated, Vmax=Vmax reaction without inhibitor. Competitive inhibition provides a set of lines with 1/Vmax the same cut point on the axis but with different angles. Because the intersection of the axis equal to the 1/V = 1/Vmax =Vmax unchanged in the presence of competitive inhibitors. While the non-competitive inhibition, providing a set of lines that have the same cut point on the axis 1/[S], suggesting that the KM for the substrate does not change but the Vmax

decreased.

MATERIALS AND METHODS

This experiment was conducted on Thursday, 27st of April 2012 at Organic Laboratory of Ganesha University of Education start on 07.30 am until 12.30 pm. The equipment that used are test tube, spatula, beaker glass, volumetric pipette, filler. The chemical that used are metilene blue solution, sodium suksinate powder, piruvate solution, liver homogenate, NaOH crystal, and buffer phosphate.

Procedure1. Test tubes as many as 5 pieces was set with shelves.2. Test tube is filled with substances that exist in the table, and then stirred until

homogeneous (before added liver homogenate).3. Liver homogenates was added to 2 mL in each tube, then stirred quickly (slowly)

and added a few drops of paraffin oil to cover the surface of the solution. All times are recorded adding paraffin oil (t=to).

4. Change the color of each tube were observed and recorded. Time (t) is recorded when the color change is complete (compared to the tube 5 as the blank).

Table 1. Composition of chemical for each test tubeTest tube

Metilen blue 0,02%(mL)

H2O(mL)

Buffer fosfat pH

7,4(mL)

Na-suksinat(mL)

Pyruvate(mL)

LiverHomogenate

(mL)

1 1 3 2 2 - 22 1 2,8 2 2 0,2 23 1 2,6 2 2 0,4 24 1 0,6 2 4 0,4 25 1 6 2 - - 2

RESULT AND DISCUSSION

In conducting its enzyme is influenced by several factors, one of which is the inhibition. Inhibition is a compound that can inhibit and reduce the rate of reaction catalyzed by the enzyme. (Tika, 2007). Where this inhibitor binds to the enzyme to form enzyme-inhibitor complex. Inhibitors can be reversible and irreversible

resistance. Reversible inhibition is quantitatively determined using the kinetic equation of Michaelis - Menten, which consists of competitive, non-competitive, and un-competitive.

In this experiment was carried out analysis of the influence of a competitive inhibitor of the enzyme activity, judging from the price of Vmax and KM reaction without inhibitor and reaction with inhibitors. In practice this solution is used as an inhibitor of pyruvate and sodium succinate as substrate. The enzyme is derived from extracts that inhibited liver homogenates. The extract contains the enzyme dehydrogenase. This enzyme catalyzed the following:

Dehydrogenase enzyme activity from liver homogenates, can diinhibisi by using the solution as an inhibitor of pyruvate. Pyruvate in the reaction of the above acts as a competitive inhibitor. Therefore, when succinate was added to pyruvate and liver homogenates was added, then the species of pyruvate (as inhibitor) and succinate (as substrate) will compete to be able to bind to the active enzyme. When pyruvate binds to the enzyme succinate hydrogenases, then the product will not be generated.

In this practice, the first thing done is prepare five test tubes. In each tube is filled with a solution of methylene blue amount in accordance with the procedures recommended in the volume of work. This solution was used as a redox indicator. Then add distilled water, which is used as a solvent, followed by addition of 0.1 M phosphate buffer pH 7.4. The purpose of the addition of phosphate buffer pH 7.4 is to stabilize the pH of the solution. The enzyme has a pH optimum is typical, as well as the enzyme succinate dehydrogenase. This enzyme has maximum activity at pH 7.4 so that the pH of the enzyme have not denature or undergo a change of electric charge due to changes in the activity.

Then fed a solution of sodium succinate and pyruvate served as substrate which acts as an inhibitor. Concentrations of substrates and inhibitors were added to each tube varies. The goal is for the future can be made between the plot of 1/V and 1/[S] to determine the effect of inhibitors on enzyme activity. In the fifth addition of the reagent is no significant color change, where the initial color is blue (from methylene blue), which after the fifth reagent is added to the resulting solution remains blue but the color is more pale than before, and any added substance. The reaction between succinate with methylene blue can be written as follows.

FADH2E EFAD+

COO-

CH

CH

COO-

COO-

CH2

CH2

COO-

suksinat dehidrogenase

+

suksinat fumarat

Furthermore the addition of liver homogenates and the addition of paraffin oil. The addition of liver homogenate causes the color of the solution in a test tube into whitish blue. This is due to a combination of methylene blue color in the state was reduced by liver homogenates color (brown).

The next step is to add paraffin oil to last the five test tubes. Time (t0) are recorded at the start of adding paraffin oil. Paraffin oil serves to protect the solution from ambient air (oxygen) so that the solution does not occur by the oxidation of oxygen that can interfere with enzymatic reactions (catalysis process carried out by enzymes).

At certain intervals, it changes color in each tube, wherein the resulting solution and precipitate a light brown, and formed a ring at the top of the blue and very thin. This color change indicates that some methylene inidikator in reduced form. The color combination is the case because of the color brown liver homogenates. Brown color is due to the formation of brown complexes formed by the reaction between fumarate formed with Mb2+ ion of methylene blue. The reaction is as follows.

Metylene bluesuksinate Blue complex

Figure 6. Addition of liver homogenate to the mixture

Figure 6. After addition of parafin

Blue complexMetylene blue

The reduction of the redox indicator methylene blue caused by the substrate reaction catalyzed by the enzyme succinate dehydrogenase generate the reduced form of fumarate. The reduction of the substrate followed by reduction of methylene blue indicator so that the indicator is in the form of tereduksinya, which is colorless. The process can be drawn as follows.

Potential changes resulting in the indicator color change due to methylene tereduksinya substrate produces a product, wherein when the substrate has been depleted in certain parts is reduced, then the next is to be reduced methylene blue. In this experiment the determination of the influence of inhibitors can be seen on the tube 2, 3, and 4, while the tubes 1 shows the situation without the inhibitor and the tube 5 is used as a blank, ie without substrate and without inhibitor. Based on the experiments that have been conducted, the results obtained in the form of time (t) required by the solution in each test tube to become discolored.

Test tube 1: M1 substrate = 0,1 MV1 substrate = 2 mLV2 substraet = 5 mL(volume suksinate, piruvate, and water)

M2 =

2 mL x 0,1 M5 mL

= 0,04 M

1[ S ]

= 10,04 M

= 25 M−1

M1 inhibitor = 0,1 MV1 inhibitor = 0 mL

[ inhibitor ] = 0 mL x 0,1 M5 mL

= 0 M

Changing potensial

Oxidated methylne blue(blue)

Oxidated methylne blue(colorless)

Test tube 2:M1 substrate = 0,1 MV1 substrate = 2 mLV2 substrate = 5 mL(volume suksinate, piruvate, and water)

M2 =

2 mL x 0,1 M5 mL

= 0,04 M

1[ S ]

= 10,04 M

= 25 M−1

M1 inhibitor = 0,1 MV1 inhibitor = 0,2 mL

[ inhibitor ] = 0,2 mL x 0,1 M5 mL

= 0,004 M

Test tube 3:M1 substrate = 0,1 MV1 substratee = 2 mLV2 substrat = 5 mL (volume suksinate, piruvate, and water)

M2 =

2 mL x 0,1 M5 mL

= 0,04 M

1[ S ]

= 10,04 M

= 25 M−1

M1 inhibitor = 0,1 MV1 inhibitor = 0,4 mL

[ inhibitor ] = 0,4 mL x 0,1 M5 mL

= 0,008 M

While, for the test tune number 5 did not done the addition of sodium suksinate and sodium malonat so the substrate concentration and inhibitor are 0 M. the amount of rate of enzyme is represent by V, as calculaation bellow.

V =

Ct , where C = concentration; t = time

So, V

1t or

1V t (time), and can be asummed that the value of

1V that used in

making graph is propotional with t (time). To make easy, the amount of time used is in minute. Test tube 1 : t = 16 minutes 30 second = 16.50 minutesTest tube 2 : t = 26 minutes 32 second = 26.53 minutesTest tube 3 : t = 28 minutes 15 second = 28.25 minutesTest tube 4 : t = 16 minutes 20 second = 16.33 minutes

Plot of value

1[ S ] and

1V (identical with t) is represent by the following graph.

Test tube 4:M1 substrate = 0,1 MV1 substrate = 4 mLV2 substrat = 5 mL (volume suksinat, piruvat, dan air)

M2 =

4 mL x 0,1 M5 mL

= 0,08 M

1[ S ]

= 10,08 M

= 12,5 M−1

M1 inhibitor = 0,1 MV1 inhibitor = 0,4 mL

[ inhibitor ] = 0,4 mL x 0,1 M5 mL

= 0,008 M

-20 -15 -10 -5 0 5 10 15 20 25 300

10

20

30

f(x) = 0.4148 x + 6.13R² = 1

f(x) = 0.816 x + 6.13R² = 1

PLOT LINE WEAVER-BURK 1/V TER-HADAP 1/ [S] ANTARA REAKSI MEN-GUNAKAN INHIBITOR DAN TANPA

INHIBITOR

dengan inhibitor

1/[S]

1/V

Based on the graph above, can be take the data in test tube 2 and 4 because of both of thata test tubes the changing of concentration was occured. Both of two line is relate with the line and extrapolated until perpendicular with the axis of 1/V and 1/[S]. the point of 1/V is 6.13 that as the value of 1/Vmax for inhibition reaction. Harga ini didapat dari persamaan garis baik persamaan dengan inhibitor (y = 0.816x + 6.13) maupun tanpa inhibitor (y = 0.414x + 6.13) dengan harga x = 0, sehingga harga Vmaks-nya adalah

Vmaks=

16,13

= 0 ,163 menit − 1

Sedangkan titik potong pada sumbu

1[ S ] adalah merupakan -

1K M untuk reaksi yang

diinhibisi diperoleh -7,512 M-1. Harga tersebut didapat dari persamaan garis dengan inhibitor pada y = 0. Dengan demikian harga KM-nya adalah

-

1K M = -

1-7,512 = 0,133 M

Untuk menentukan harga

1K M untuk reaksi tanpa inhibisi, dapat diketahui dengan

menarik garis antara

1V maks reaksi dengan inhibitor dengan hasil titik potong pada

tabung 1. Garis yang diperoleh diteruskan mancapai sumbu

1[ S ] , dimana Vmaks tanpa

inhibitor sama dengan harga Vmaks dengan inhibitor yaitu 0,163 menit-1. -

1K M tanpa

inhibitor diperoleh pada -14,807. Harga tersebut didapat dari persamaan garis tanpa inhibitor dengan y = 0, maka harga KM adalah

-

1K M = -

1-14,807 = 6,754x 10-2 M

Based on the above chart can be seen that the price of Vmax without inhibitor with inhibitor reaction is the same, namely 0.163 minute-1, the price of KM in the reaction without inhibitor is 6.754 x 10-2 M, whereas the reaction with the inhibitor 0.133 M. This is because the inhibitor competes with substrate for pyruvate can bind to the active enzyme succinate dehydrogenase. Based on experimental results obtained that the addition of inhibitors apparently affect the price of KM. In this experiment we found that KM increases. KM is a measure of price stability complex ES, which is when the speed of decomposition of the complex ES with ES formation rates. The increasing price of KM shows the weak binding of the enzyme to the substrate. This is certainly caused by the competition between the substrate and inhibitor binding to the enzyme active. The higher the concentration of substrate (tubes 2 and 4) the enzyme activity increases, this can be seen from the increase in the price of 1/V on the tube 2 and 4, because at high concentrations, inhibitors will be forced out of the active enzyme. Similarly, when the fixed substrate concentration and inhibitor concentration increased prices (tubes 2 and 3), the enzyme activity will decrease, so the time required completing the enzymatic reaction will grow old.

SIMPULANBerdasarkan pembahasan di atas, dapat disimpulkan bahwa penambahan

konsentrasi inhibitor menyebabkan peningkatan harga KM dan harga Vmaks tidak berubah. Harga KM pada reaksi tanpa inhibitor adalah 6,754x 10-2 M, sedangkan reaksi dengan menggunakan inhibitor adalah 0,133 M.

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