AC(Enzyme Kinetic) Rev

7/28/2019 AC(Enzyme Kinetic) Rev

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ENZYME KINETIC

By: Luh Lian Pertiwi (0913031030)

Chemistry Education Department, Faculty of Mathematics and Natural Sciences

Ganesha Education University

Abstract

The enzyme is a biocatalyst that can increase the speed of reactions in biological systems and

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 low activation energy for the substrate into a product transition. Reaction rate of the

reaction catalyzed by the enzyme increases with the addition of substrate concentration, until

reaching a state with the addition of the substrate concentration does not affect the initial

reaction rate. This situation is referred to as the maximum reaction rate (Vmax). The aim of this

experiment is to determine Vmax and KM for enzymetic rate reaction and substrat concentration.

According to graph of relationship 1/Vo to 1/[S], it can be determined Vmax and KM value.According to the experiment, KM value and VMax value of enzymetic rate recation of tripsin with

kasein is 8,84 mg/mL and 1,938 mol/menit.

Keyword : enzyme, Vmax and KM.

Introduction

In the reaction, enzyme can react with the reactants and also can be recovered at the time

of formation of the product. The phenomenon is what causes the enzyme can be regarded as a

catalyst. Enzymes are the catalysts of biochemical reactions that play a role in the specific

enzyme is often referred to as a biocatalyst. As a biocatalyst, the enzyme can increase the speed

of reactions in biological systems and the enzyme itself does not change during the reaction. Inaddition, the enzyme also affects the reaction rate 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 low activation energy for the transition into a product compared to the substrate without

the catalysis process. The catalytic activity of the enzyme is very sensitive to conditions such as

temperature, pH, and the strength of the product. Enzyme reaction kinetics assumes that a

complex between the enzyme and the substrate is formed immediately and reversibly. This

complex then breaks down slowly and produces enzymes.

The speed of biochemical reactions that are catalyzed by an enzyme commonly used as

the value at time zero (Vo symbol, within minutes mol-1), because the highest velocity occurs in a

state where the reaction product has not been established. In these circumstances, the greatest

concentration of substrate and enzyme is not inhibited by its product feedback. Plot of the

reaction products formed with time on the enzymatic reaction showed that the initial stage occurs

a linear graph, and then followed by a decrease in reaction rate because most of the substrate has

been used in the reaction or the enzyme activity decreased. Initial reaction velocity (Vo) is

obtained by drawing a straight line through the linear curve at zero time point and the slope of a

straight line equal to Vo.


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Reaction rate depends on the concentration of enzyme that acts as a catalyst in a reaction.

At low substrate concentration, if the substrate concentration is increased two times, then the

initial reaction velocity (Vo) has doubled. This means that at low substrate concentration of

enzymatic reaction rates of order one. However, at high concentrations, increasing the substrate

concentration will lead to changes in Vo is very small and at very high substrate concentrations,

increasing the substrate concentration will not affect the price of V o (Vo constant prices). This

situation is caused because the enzyme was saturated by the substrate, or in other words the

entire active enzyme has a substrate binding.

The reaction rate increases as substrate concentration increased up to a point where said

enzyme is saturated with substrate. The measured initial velocity reaches a maximum speed and

is not affected by the subsequent increase in substrate concentration, because the substrate is in

molar excess. Overall reaction rate depends on the speed of product dissociation from the

enzyme, and the addition of the substrate will not affect Vo. Plot Vo against [S]-shaped

hyperbolic.

(a) (b)

(b)Figure 1 (a). The relation concentration of substrate and Vo plot

(b). The relation of 1/Vo and 1/[S] plot Lineweaver Burk

Michaelis and his friends claimed that the enzymatic reaction at various concentrations of

substrates having two phases, namely: (1) when a low substrate concentration ([S]) is low, then

not all of the active enzyme bound to the substrate and (2) if molecule of the substrate increases,

the active side was entirely bound to the substrate. At this time the enzyme has been working

complete. From the capacity Michaelis and friends statement can also be proven by metematics.

The curve in Figure 1 (a) Michaelis-Menten equation satisfies is follows:

[S]K

[S]VV

M

maks

o+

=

Where : Vo = the velocity of enzyme by the concentration of substrate [S]

KM = Michaelis Menten constant (mol perliter)

Vmaks = Maximum velocity

From the equation above its can be:

Kinetic Zero-order (Fase II)

Mixture kinetic zero and

first-order

KineticFirst-order(Fase I)

Maximum Velocity (V)

V0

Vmaks

Vmaks

KM

[S]


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E + Sk1

k2

k3k4

ES E + P

The specificity of an enzyme catalyzed by the enzyme appears to the formation of the

enzyme substrate complex (ES), which then decomposes into enzyme and products (E and P). In

this case k1, k2, k3, and K4 is the reaction rate constants. Initially be achieved disintegration of thestate of the enzyme rate equal to the speed of formation of ES.

k1[S][E] + k4 [E][P] = k2 [ES] + k3 [ES]

32

4

32

1

kk

[P]k

kk

[S]k

[E]

[ES]

+

+

+

=

In the condition of concentration is P (product) is very small so the constant of k1, k2, k3

and k4 can be written is as follows:

KM =1

32

k

kk +.

The equation is:

[S]

K

[ES]

[E] M=

If the amount of total concentration of enzyme [E]t and [ES] so the concentration [E] =

[E]t [ES], as follows:

[S]

K1

[ES]

[E]

[ES]

[ES][E]

[ES]

[E] Mtt==

=

1[S]

K

[ES]

[E] Mt+=

Maximum celocity (V if maks) if the enzyme the shape is complete if all of enzyme in

complex form with the substrate. But Vo directly proportional with [ES]. Its can be written is as

follows:

1[S]

K

V

VsO

[ES]

[E]

V

V Mmakstmaks+==

In the figure 1 (a) can be seen that KM as mole perliter. And KM is higher so can be written

is as follows:

M

maks

K

[S]VV=

In this case V depends on the substrate concentration (filtrate reaction order) and in

Figure 1 (a) can also be seen kinetic and first-order zero. Thus, the Michaelis-Menten equation iseligible for a simple reaction that is catalyzed by the enzyme. KM is often defined as the

dissociation constant of the reaction catalyzed by the enzyme. In the simple reaction can be seen:

E + Sk1

k2ES maka K s=

[E][S]

[ES]=

k2k1


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Because KM is1

32

k

kk +so Ks =

1

32

k

kk +

So, KM will always same or bigger than Ks. can be seen from the equation:

maksmaks

M

M

maks

V

1

[S]

1

V

K

V

1So

[S]K

[S]VV +

=

+

=

The equation above is identical with line equation is y = ax + b, where:

y =maksmaks

M

V

1bdan

V

Ka;

[S]

1x;

V

1===

The equation can be formulated by Lineweaver-Burk and like figure 1 (b). Based in the

figure 1 (b), line equation with the gradient inoV

1is

maksV

1, in

[S]

1is

MK

1 . The plot of

line ismaks

M

V

K. Plot Lineweaver-Burk can be used to determine inhibition that occurs in the

enzymatic reaction.

Spectrophotometer

Spectrophotometer is the method that used to qualitative and quantitative analysis in

solution. Spectroscopy imply the measurements how far the light energy absorption as a function

of the wavelength of radiation. Instruments that used in this method its called by

spectrophotometer. In analyzing the uses of UV-Vis spectrophotometer should note the

following:

a. Formation of colored compounds

This step is conducted if the analyzed compounds not absorbing visible region. In thiscase the compound must be converted into other compounds that can do the absorption or

reacted with a reagent that form color so can absorb visible light. Reagents that cause

color, must meet several requirements such as: reaction with a substance that is analyzed

to be selective and sensitive, do not form colored compounds with other substances that

exist in solution, they react with other substances which must be analyzed quickly and

quantitatively (perfect), color or compounds that are formed must be sufficiently stable

for a specified period, and not too fast changes with changes in pH.

b. Selection of wavelength

Wavelengths required in a quantitative analysis by spectrophotometry is the wavelengthcorresponding to maximum absorbance (peak absorption). This is caused by changes in

absorbance for each unit of concentration is greatest at the maximum wavelength, it will

obtain the maximum sensitivity as well. Wavelengths required in a quantitative analysis

by spectrophotometry.

Description:

Violet : 400 - 420 nm


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Indigo : 420 - 440 nm

Blue : 440 - 490 nm

Green : 490 - 570 nm

Yellow : 570 - 585 nm

Orange : 585 - 620 nm

Red : 680 780 nm

c. Preparation of calibration curve

For calibration curves, created a standard solution with various concentrations of the

unknown. Absorbance of standard solution is measured, and then plots the absorbance

(A) against concentration (C), the curve formed is called the calibration curve.

In the analysis using UV-Vis spectroscopy is used the basic law of Lambert-Beer.

Lambert-Beer law states the relationship between the intensity of light absorbed by the

concentration and the thick solution through which the beam. If a beam of light with a

certain length is passed in a solution containing absorbent material, some light will be

absorbed and some light transmitted. Simply, Lambert-Beer law can be shown in thefollowing scheme.

Light or radiation with the intensity Po of that passes through a medium-thick b

containing a solution with concentration C, will result in reduced intensity of P so that P


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Information obtained from ultraviolet and visible spectra:

1. The maximum wave length ( maks )

Quantitative analyse, for all the measure based on analisis kuatitatif, maks , but forthe compound that already know frequently used approach Woodward-Fieser.

Qualitative analyses : less informatif.

2. Absorptivity maximum of molar( maks ); qualitative analyses is the information transisi

electronic.

Lambert-Beer Laws

If a beam of light with specific wavelength is passed in a solution containing absorbent

material, then some light will be absorbed and some will be forwarded. The relationship between

the intensity of light absorbed by the concentration and the thick solution through which thebeam expressed by Lambert-Beer equation as follows.

A = -log I / Io = bC

(Muderawan,2009)

Where, A = absorbance

I = Intensity of light emitted by the solution in the cell

Io = intensity of light emitted by the solvent in the same cell at I

= coefficient of extinct species absorbent or constant comparison.

b = Length of solution through which the light (typically 1 cm)

C = Concentration of absorbing species in units of mol L-1 (M)

Slope = b

The relation between concentration

and absorbance

The relation between

concentrations with transmittance

C

(mg/

L)

T

Lambert-Beer diagram


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(Ismono,1978)

The greater the intensity of light is absorbed then the value of A will become bigger and

the intensity of the transmitted beam will be smaller. Often times in real terms in the

measurement result, polt. Beer law is not linear throughout the entire concentration observed.

Curvature states that is not a constant, which depends on the concentration. value is expected

to depend on the nature absorbance agent species in solution and at a wavelength of radiation. In

practice the deviation is usually caused by the deviation on the basis of chemical and diversion of

the instruments used.

Materials and Method

In this experiment about the determine the protein content of the egg whites in the

sample solution through the use of methods of Lowry at Thursday 13 April, 2012. This

experiment conducted in Organic Laboratory Undiksha.

Materials and equipment

The materials that used in this experiment are casein solution, Folin-Ciocalteu reagent,

aquades, trypsine solution, NaOH solution, and TCA solution and the equipment that used are

test tube, volumetric pipette, volumetric flask, drop pipette, graduated cylinder glass and beaker

glass.

Procedure

Prepare 10 test tubes. In each of test tube added by casein, buffer and trypsin in different

concentration like is as follows:

No Test tube Casein (mL) Buffer phosphate

(mL)

Trypsin (mL)

It= 0 minute 0,1 5,9 1

t = 20 minute 0,1 5,9 1

IIt= 0 minute 0,5 5,5 1

t = 20 minute 0,5 5,5 1

IIIt= 0 minute 1,0 5,0 1

t = 20 minute 1,0 5,0 1

IVt= 0 minute 3,0 3,0 1

t = 20 minute 3,0 3,0 1

V t= 0 minute 5,0 1,0 1


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t = 20 minute 5,0 1,0 1

At the incubation 20 minutes

1. In each test tube that contain casein incube 5 minute in water incubator 35oC. stirrer and

add buffer phosphate solution pH 8,0 and trypsin solution (based on the table).

2. The solution 20 minutes in water incubator35oC after added trypsin.

3. After incubation 20 minutes add 3 mL of TCA 20% solution into each of test tube. Than

cooled in 30 minute in ice batch to get the precipitate perfect.

4. After that filter the solution by Anson method. 2 mL of filtrate add by 4 mL NaOH 0,5

M, then add the reagent Folin-Ciocalteu wait 10 minute. Measure the absorbance by

spektronik 20+.

At the incubation 0 minute

1. In each test tube add buffer fosfat solution and trypsin. Than add the TCA solution 20%

incube 5 minute in water incubator 35oC.

2. In each test tube added by casein solution ditambahkan larutan kasein. Than cooled in 30

minute in ice batch to get the precipitate perfect.

3. After that filter the solution by Anson method. 2 mL of filtrate add by 4 mL NaOH 0,5

M, then add the reagent folin ciocalteu wait 10 minute. Measure the absorbance by

spektronik 20+.

Result and Discussion

In this experiment tests performed is the enzyme kinetics reaction, where the value of

these tests can be determined Vmax and KM (Michaelis-Menten constant) over the curve of therelationship between the reaction rate and substrate concentration.

In this experiment used as the substrate is a casein solution, while being used as the

enzyme is trypsin solution. Trypsin is a proteolytic enzyme, this enzyme catalyzes the hydrolysis

of the peptides bond, where in the peptide bonds were broken in the substrate casein is located on

the carboxyl side of lysine or arginine residues are charged.

N C

H

H

C

R1

O

N

H

C

H

R2

C

O

+ H2O N

H

C C

R1

O

O-

H

++H3N C C

H

R2

O

Peptides Carboxcyl Amine

There are two treatments (time) is done in this experiment the incubation time of 0

minutes and 20 minutes. In this experiment provided 10 tubes, for an incubation time of 0

minutes, while the tubes for 20 min incubation. In this case also do a variation of the

concentration of substrate in each tube. For the incubation of 20 minute, casein solution is

incubation first


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Incubation of casein in 5 minute

This treatment of substrate conditions at maximum temperature is 35oC. Into the casein

solution that incube add the buffer fosfat solution (pH 8,0) so the volume become 6 mL in each

test tube. The objective of dilution is to make the substrate condition in each test tube are

different so the plot of reaction rate is different. Then the solution is added by trypsine solution

1 mL in each test tube. The addition of trypsin solution aims to catalyze the hydrolysis reaction

under on the peptide bonds in casein. Peptide bonds were broken by trypsin is the bond that lies

at the carboxyl side of lysine and arginine residues.

Incubation at 20 minute

After the incubation the solution is added by TCA ( tri chloro acetat). The aim of this

solution is to stop the trypsine enzyme so can be analysis the effect of substrate concentration

in enzyme of reaction rate. The addition of TCA 20% solution that caused the precipitate form

in the test tube after cooled 30 minute.

Brown precipitate which formed a precipitate of casein protein and the enzyme trypsin. The

addition of a solution of TCA was able to stop the activity of the enzyme and substrate solution

of casein given TCA solution is acidic, so the addition of either TCA can cause denaturation of

the enzyme and the casein. Structures that cause clotting TCA are as follows.

C CCl

Cl

Cl

O

OH

Structure of TCA

With the addition of acid to the solution, causing denaturation protein both the substrate and the

enzyme trypsin. Also, the addition of TCA solution causes the pH of the acidic solution to be

down (acid) so that the substrate complex formed is unstable and the activity of the enzyme to


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catalyze the hydrolysis of casein in a standstill due to changes in pH. This is because the enzyme

trypsin to work at optimum pH is pH 8.0.

After filtering the precipitate, the resulting filtrate treated with the method of Anson,

where the filtrate to each tube to 2 mL after adding 4 ml of 0.5 M NaOH was added Folin-

Ciocalteu reagent. Folin-Ciocalteu reagent is a mixture between fosfotungstat and fosfomolibat

in yellow. These reagents can be reduced by the phenolic group on the amino acid tyrosine in

the filtrate contained tungstate and molybdenum producing blue-green. Based on the results of

the experiment, when the filtrate was added Folin-Ciocalteu reagent green in solution. The

resulting solution is then measured by using UV vis 20+.

For the 0-minute incubation time, the order is different from the incubation of 20

minutes, which at time 0 minute incubation in a row was added phosphate buffer, trypsin

solution, a solution of TCA to each tube. After incubation for 30 minutes then added a solution

of casein at varying volume for each test tube. Calculation of substrate concentration for each

tube as follows.

The filtrate obtained is done by Anson, where the filtrate to each tube to 2 mL after

adding 4 ml of 0.5 M NaOH was added Folin-Ciocalteu reagent. Folin-Ciocalteu reagent is a

mixture between fosfotungstat and fosfomolibat in yellow. These reagents can be reduced by the

phenolic group on the amino acid tyrosine in the filtrate contained tungstate and molybdenum

producing blue-green. Based on the results of the experiment, when the filtrate was added Folin-

Ciocalteu reagent the color is green. The resulting solution is then measured using a

spectrophotometer at a wavelength of 700 nm. Based on the experiments conducted, the

following is a comparison of observations absorbance for tube t = 0 and t = 20.

Table. Observation result of absorbance

No Test tube Absorbance Transmitant (%)

I t = 0 minute 0.096 81t = 20 minute 0.139 85

IIt = 0 minute 0.184 92

t = 20 minute 0.187 83

IIIt = 0 minute 0.211 72

t = 20 minute 0.229 89

IVt = 0 minute 0.259 75

t = 20 minute 0.387 88


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To calculate and determine the value of KM and VMax, so be must determine the value of

[S] and [Vo]. The value [S] concentration of casein in the test tube is different value. The

calculation is as follows:

Massa of casein that added is 5 gram in 250 mL aquades.

[casein] = mL500

gram5

= 10 mg/mL

Test tube 1

V1 = 0,1 mL; V2 = 6 mL; M1 = 10

mg/mL

M2=

mg/mL6

1

mL6

mg/mL10xmL0,1=

[S] = mg/mL6

1

6.mL/mg[S]

1=

Test tube 4

V1 = 3,0 mL; V2 = 6 mL; M1 = 10

mg/mL

M2 =

mg/mL5mL6

mg/mL10xmL3,0=

[S] = mg/mL5

mL/mg2.0[S]

1=

Test tube 2

V1 = 0,5 mL; V2 = 6 mL; M1 = 10

mg/mL

M2 =

mg/mL6

5

mL6

mg/mL10xmL0,5=

[S] = mg/mL6

5

mL/mg1.2[S]1 =

Test tube 3

V1 = 1,0 mL; V2 = 6 mL; M1 = 10

mg/mL

M2 =

mg/mL3

5

mL6

mg/mL10xmL1,0=

[S] = mg/mL3

5

mL/mg0.6[S]1 =

For the calculation of Vo can be determine the value of absorbance by spectrophotometer

dapat ditentukan dengan besaran absorbansi ketika diukur dengan spektronik 20+. The absorbance

is as follows:

A = .b.C


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Because .b is constant, so can assume that .b is constants. So, A = k x C, where A =

absorbance C = concentration k = constants. Based on the equation so AC

So the equation of velocity iso

ot

[C]V =

ot

AV =

oV

1

A

1=

The calculation of absorbance can be calculating the absorbance in t=20 and t=0 so theabsorbance:

Incubation time o minute Incubation time 20 minute

Test tube I

%T = 81; T = 0.81

A = -log T = -log 0.81 = 0.096

1/A = 1/0.096 = 10.41

1/V0 = 10.41

Test tube II

%T = 92; T = 0.92

A = -log T = -log 0.92 = 0.184

1/A = 1/0.184= 5.43

1/V0 = 5.43

Test tube III

%T = 72; T = 0.72

A = -log T = -log 0.72= 0.211

1/A = 1/0.211= 4.73

1/V0 = 4.73

Test tube IV

%T = 75; T = 0.75

A = -log T = -log 0.75=0.259

1/A = 1/0.259= 3.86

1/V0 = 3.86

Test tube I

%T = 85; T = 0.85

A = -log T = -log 0.85 = 0.139

1/A = 1/0.139 = 7.19

1/V0 = 7.19

Test tube II

%T = 83; T = 0.83

A = -log T = -log 0.83 = 0.187

1/A = 1/0.187= 5.34

1/V0 = 5.34

Test tube III

%T = 89; T = 0.89

A = -log T = -log 0.89= 0.229

1/A = 1/0.229= 4.36

1/V0 = 4.36

Test tube IV

%T = 88; T = 0.88

A = -log T = -log 0.88= 0.387

1/A = 1/0.387 = 2.58

1/V0 = = 2.58


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Based the calculation the table is as follows:

Tabel 2. Calculation table of 1/A

No Test tube %T T A 1/A

It = 20 minute 81 0.81 0.096 10.41

t = 0 minute 85 0.85 0.139 7.19

IIt = 20 minute 92 0.92 0.184 5.43

t = 0 minute 83 0.83 0.187 5.34

IIIt = 20 minute 72 0.72 0.211 4.73

t = 0 minute 89 0.89 0.229 4.36

IVt = 20 minute 75 0.75 0.259 3.86

t = 0 minute 88 0.88 0.387 2.58

Based on the table can make curve of the relation between 1/V 0 versus 1/[S] at the time of

incubation 0 minute and 20 minute.

The curve of the relation between 1/V0 versus 1/[S] at the incubation time o minute is as

follows:

Test Tube 1/V0 1/[S] (mL/mg )

1 10.41 6

2 5.43 1.2

3 4.73 0.6

4 3.86 0.2


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From the curve we y get the line equation is 1.086x + 3.934The relation with the Lineweaver-Burk can be calculate by 1/Vmax, 1/[S] and -1/KM. Lineweaver-

Burk equation:

maksmaks

M

oV

1

[S]

1

V

K

V

1+

=

From the equation, the value ofmaksV

1when the x = 0, so

y = 1.086x + 3.934

y = 1.086 0 + 3.934

y = 3.934so that,

maksV

1= 3.934

Vmaks = utemol min/25.03.934

1=

So when the[S]

1is -

MK

1, its same with y = 0, so

y = 1.086x + 3.934

0 = 1.086x + 3.934

1.086x = -3.934

x = -3.62

its means the -MK

1= -3.62

KM =3.62

1= 0.276 mg/mL


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The curve of the relation between 1/V0 versus 1/[S] at the incubation time 20 minute is as

follows:

Test Tube 1/V0 1/[S] (mL/mg )

1 7.19 6

2 5.34 1.2

3 4.36 0.6

4 2.58 0.2

From the curve we get the line equation is 0.629x+ 3.609

The relation with the Lineweaver-Burk can be calculate by 1/Vmax, 1/[S] and -1/KM. Lineweaver-

Burk equation:

maksmaks

M

oV

1

[S]

1

V

K

V

1+

=

From the equation, the value ofmaksV

1when the x = 0, so

y = 0.629x+ 3.609y = 0.629 0+ 3.609

y = 3.609

so that,

maksV

1= 3.609


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Vmaks = utemol min/277.03.609

1=

So when the[S]

1is -

MK

1, its same with y = 0, so

y = 0.629x+ 3.609

0 = 0.629x+ 3.609

0.629x = -3.609

x = -5.73

its means the -MK

1= -5.73

KM =5.73

1= 0.174mg/mL

The value of KM at the incubation 20 minutes is 0.174 mg/mL and the incubation time at 0

minute is 0.276 mg/mL. From the experiment data the value of KM at the incubation 20 minutelower than the incubation in 0 minute. The value of KM show the stability complex of enzyme

and the substrate where the value of KM is higer its mean the bond of enzyme and substrate is

less. From the value of KM can be concluded that if the time is increase so the complex enzyme

and substrate will stable because the bond of the enzyme and substrate is strong.

Conclusion

Based on the data result and data analyze that we have been done, so can be concluded

the value of KM from the experiment that have been done is 0.174 mg/mL in 20 minute and at 0

minute the value of KM is 0.276 mg/mL.

Acknowledgment

Thanks to the god almighty who has bless me to finishing this article with the title Enzyme

Kinetic. This article is made for me assignment in Biochemistry Practice lesson. In process to

create this paper, the author found some problem. It because of my knowledge still need to

improvement. But because of the guidance and big support from many peoples, so I can finish it.

Therefore, the authors want to say thanks to the:

1. Mr. Dr. I Nyoman Tika, M.Si. as my lecturer who has guide me,

2. Mrs. Dra. Siti Maryam, M.Kes. as my lecturer who has guide me,

3. My best friend in RKBI Class who has supported me, and


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4. Many other peoples who has supported me.

This article is still need improvement and still contains many of mistakes, because of that

authors need advice and critics to make this article will better and can useful for many people.

Thank you.

References

Ismono. 1978. Cara-cara Optik Dalam Analisis Kimia. Diktat. Bandung: Jurusan Kimia ITB.

Poedjadi, Anna dan Titin Supriyanti. 1994. Dasar-Dasar Biokimia. Jakarta: Universitas

Indonesia

Redhana, I Wayan dan Siti Maryam. 2003.Penuntun Praktikum Biokimia. Singaraja : IKIP

Negeri Singaraja

Tika, I Nyoman. 2007.Penuntun Praktikum Biokimia. Singaraja: Universitas Pendidikan

Ganesha

Wirahadikusuma, Muhamad. 2001.Biokimia :Protein, Enzim, dan Asam Nukleat. Bandung :

Penerbit ITB

Muderawan, I Wayan. 2009.Analisis Instrumen. Singaraja : Undiksha Press

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