NUR ISTIANAH,ST,MT,M.Eng

Kinetika Reaksi

Enzimatis

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Outline

Dasar Kinetika Reaksi Enzimatis

Penentuan aktivitas enzim

Faktor yang mempengaruhi reaksi enzimatis, inhibisi

Kinetika Reaksi Homogen: Penentuan Km

Reaksi penggantian tunggal

Reaksi Penggantian ganda

Mekanisme Ping-pong Bi Bi

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ENZYME REACTION KINETICS

Most enzymes catalyse reactions and follow Michaelis–Menten kinetics. For a single enzyme and single substrate, the rate equation is:

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For batch reaction, there is no inlet or outlet stream where V is the volume of batch reactor which is constant volume:

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S

S S = KM

S

Sm

Ss

CK

Cr

dt

dCr

max

Change of Cs with time, t

Batch operation with stirring

•A foam breaker may be installed to disperse foam

Batch Operation

Case study

• Contoh soal 1:

• Sebuah bioreaktor digunakan untuk memproduksi enzim dengan konsentrasi substrat awal sebesar 12x10-3 M dan konversi bioreaktor 90%. Berapakah waktu yang diperlukan untuk produksi enzim jika data kinetik reaktor diberikan sebagai berikut:

• rmax = 2.5 mmol/m3.det; Km = 8.9x10-3 M

Solution

• Konsentrasi substrat akhir,

• Cs = (1-XA)Cs0 = (1-0.9)( 12x10-3 M) = 1.2x10-3 M

• rmax = 2.5 mmol/m3det= 2.5 x10-6 mol/L.det

jamt

t

CCC

CK

rt ss

s

s

m

5.3det12517

x102.1x1012x102.1

x1012ln8.9x10

x105.2

1

ln1

3- 3-

3-

3- 3-

6-

0

0

max

Case study

• Contoh soal 2:

• Hitunglah laju fermentasi bioetanol dari nira sorgum jika konsentrasi gula awal sebesar 0.1 M dan di akhir fermentasi diperoleh konsentrasi gula sebesar 0.025 M. Fermentasi berlangsung selama 4 jam dengan konstanta Monod 5x10-3 M.

Solution

• Laju fermentasi bioetanol tersebut adalah 0.017 M.jam-

1

1

max

3-

max

0

0

max

.02.0

025.01.0025.0

1.0ln5x10

4

1

ln1

jamMr

r

CCC

CK

tr ss

s

s

m

1

3-

max

.017.0

025.05x10

)025.0(02.0

jamMr

r

CK

Crr

Sm

S

dt

dXVVrXXF x )( 0

F, Cs0

F, Cs

V

onAccumulati GOutput -Input eneration

the ratio of biomass rate of generation to biomass concentration, rx/X, that is the specific growth rate; μ

X

rx

D

0dt

dX sSteady state:

DV

F

1

00 XNo cell in inlet:

D

XDX

rXXV

Fx

.

)( 0

-3. Continuous stirred-tank – cont.

Monod rate: ss

s

CK

C

max

ss

s

CK

CD

max

D

DKC s

s

max

maxmax

111

s

s

C

K

D

At steady state, substrate utilisation is balanced with a rate equation:

VCK

CCCF

ss

s

ss

max

0 )(

XCK

CCCD

ss

s

ss

max

0 )(

CsCs

XXY

0

0

D

DKCYXX s

s

max

00

Case study

• Contoh soal 4:

• Kinetika Monod (Ks = 3 g/L) telah dipertimbangkan dalam operasi CSTR untuk memproduksi yeast dengan konsentrasi substrat awal 50 g/L. Jika laju pertumbuhan yeast spesifik maksimal sebesar 0.5 jam-1, berapakah laju pertumbuhan yeast fermentasi dan dilution rate untuk konversi reaksi sebesar 90% ? asumsikan tidak ada yeast di awal fermentasi.

Solution

Cs = (1-XA)Cs0 = (1-0.9)( 50 g/L) = 5 g/L

1

1

max

3125.0

g/L5g/L3

g/L55.0

jamD

xjamD

CsK

CsD

s

Case study

• Contoh soal 5:

• Hitunglah laju dilusi dari contoh soal 4 jika terdapat konsentrasi yeast di awal dan akhir fermentasi sebesar 0.1 g/L dan 35 g/L!

Solution

• Pada kondisi ini berlaku persamaan berikut:

X

rx

xrXXD )( 0SK

S

s max

)(

)(

0

0

XX

XD

XX

rD x

1

0

max

3134.01.035

)35(3125.0

)(

jamD

XX

XSK

S

D s

Case study

• Contoh soal 6:

• Berapakah yield fermentor pada contoh soal 5?

– Solution:

%56.777756.0550

1.035

0

0

Y

SS

XXY

Solution

• Pada kondisi ini berlaku persamaan berikut:

X

rx

xrXXD )( 0SK

S

s max

)(

)(

0

0

XX

XD

XX

rD x

1

0

max

3134.01.035

)35(3125.0

)(

jamD

XX

XSK

S

D s

Case study 7 • A chemostat study was performed with yeast. The med

ium flow rate was varied and the steady state concentration of cells and glucose in ther fermenter were measured and recorded. The inlet concentration of glucose was set at 100 g/L. The volume of the fermenter contents was 500 mL. the inlet stream was sterile.

• Find the rate equation of cell growth

Flow rate F, mL/ hr

Cell concentration Cx, g/ L

Substrate concentration Cs, g/ L

31 5.97 0.5

50 5.94 1.0

71 5.88 2.0

91 5.76 4.0

200 0 100

Solution

• Pada kondisi ini berlaku persamaan berikut:

maxmax

111

s

s

C

K

DChemostat:

1/D = V/F 1/ Cs

16.13 2.00 10.00 1.00 7.04 0.50 5.49 0.25 2.50 0.01

y = 3.1764x + 1.2959

R² = 0.9407

0,00

5,00

10,00

15,00

20,00

0,01 0,25 0,50 1,00 2,00

78.0

13.1

max

max

46.2

2.3max

Ks

Ks

Persamaan laju reaksi

s

s

ss

s

C

XCr

CK

XCr

Xr

46.2

78.0

max

NUR ISTIANAH,ST,MT,M.Eng

DUAL SUBTRATE

REACTION

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Mechanisms of Single Enzyme with Dual Substrates

The kinetics of double substrates with defined dissociation constants are given as:

Similarly, for a second substrate, the reaction is carried out and the second product is formed

K =equilibrium or dissociation constant.

The total enzyme concentration

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The intermediates, complexes of ES1 and ES2, are defined based on equilibrium constants The initial and total enzyme concentrations are defined based on measurable components given below: The free enzyme can also be defined based on the following equation:

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1. A carbohydrate (S) decomposes in the presence of an enzyme (E). The Michaelis-Menten kinetic parameters were found to be as follows;

Km = 200 mol/m3

rmax = 100 mol/m3 min

Chemostat (continuosly stirred-tank reactor) runs with various flow rates were carried out. If the inlet substrate cocentration is 300 mol/m3 and the flow rate is 100 cm3/min, what is the steady-state substrate concentration of the outlet? The reactor volume is 300 cm3. Assume that the enzyme concentration in the reactor is constant so that the same kinetic parameters can be used.

Contoh Soal

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CSTR equation

…….1

…….2

…….3

…….4

Subtitusi pers 2 dan 3 :

SOLUTION

3. The enzym, cathepsin, hydrozes L-glutamyl – L – tyrosine to carbobenzoxy – L – glutamic acid and L-tyrosine. It has been found (Frantz and Stephenson, J. Biol. Chem., 169, 359, 1947) that the glutamic acid formed in the hydrolysis, inhibits (competitively) the progress of the reaction by forming a complex with cathepsin. The course of the reaction is followed by adding tyrosine decarboxylase which envolvesCO2.

Substrat Inhibitor Laju pembentuka

n CO2

mmol/ml mmol/ml mmol/ml.menit

4,7 0 0,0434

4,7 7,57 0,0285

4,7 30,3 0,0133

10,8 0 0,0713

10,8 7,58 0,0512

10,8 30,3 0,0266

30,3 0 0,1111

30,3 7,58 0,0909

30,3 30,3 0,0581

Calculate (a) the value of Michaelis – Menten constant of the enzyme, Ks and (b) the dissosiation constant of enzyme – inhibitor complex, KI. (contributed by Professor Gary F. Bennett, The University of Toledo, Toledo, OH)

Solution

Plot grafik antara Cs dan Cs/r sehingga didapatkan grafik dengan slope = 1/rmax

y = 0,0022x + 0,0216

0

0,02

0,04

0,06

0,08

0,1

0,12

0 5 10 15 20 25 30 35

Cs/

r

Cs

Plot CI dan dan CI/r sehingga didapatkan KI= 265,9084 dan Ks = 254733,8 dari persamaan

y = 44,982x - 75,045

-500

0

500

1000

1500

2000

2500

0 5 10 15 20 25 30 35

CI/r

CI

Cs CI rP Cs/r KMI CI/r KI KS

4,7 0 0,0434 108,2949 54142,77 0 -75,04 54142,77

4,7 7,57 0,0285 164,9123 82451,44 265,614 265,4586 80165,39

4,7 30,3 0,0133 353,3835 176687 2278,195 1287,854 172625,6

10,8 0 0,0713 151,4727 75725,53 0 -75,04 75725,53

10,8 7,58 0,0512 210,9375 105458 148,0469 265,9084 102535,1

10,8 30,3 0,0266 406,015 202996,7 1139,098 1287,854 198330,5

30,3 0 0,1111 272,7273 136333,3 0 -75,04 136333,3

30,3 7,58 0,0909 333,3333 166636,4 83,38834 265,9084 162017,9

30,3 30,3 0,0581 521,5146 260727 521,5146 1287,854 254733,8

Tabel Perhitungan:

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