Kinetika Reaksi Enzimatis

NUR ISTIANAH,ST,MT,M.Eng

Kinetika Reaksi Enzimatis

ALLPPT.com _ Free PowerPoint Templates, Diagrams and Charts

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

22

Nur Istianah-THP-FTP-UB -2014

ENZYME REACTION KINETICS

Most enzymes catalyse reactions and follow Michaelis–Ment en kinetics.

For a single enzyme and single substrate, the rate equation i

s:

23

Nur Istianah-THP-FTP-UB -2014

24

Nur Istianah-THP-FTP-UB -2014

For batch reaction, there is no inlet or outlet stream

where V is the volume of batch reactor which is constant vol

ume:

25

Nur Istianah-THP-FTP-UB -2014

S

S S = K

_M

S

S m

S s

C K

C r

dt r dC

 



^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 me mproduksi enzim dengan konsentrasi su bstrat awal sebesar 12x10 ^-3 M dan konve rsi bioreaktor 90%. Berapakah waktu yan g diperlukan untuk produksi enzim jika d ata kinetik reaktor diberikan sebagai beri kut:

• r _max = 2.5 mmol/m ³ .det; K _m = 8.9x10 ^-3 M

Solution

• Konsentrasi substrat akhir,

• Cs = (1-X

_A

)C

_s0

= (1-0.9)( 12x10

^-3

M) = 1.2x10

^-3

M

• r

_max

= 2.5 mmol/m

³

det= 2.5 x10

^-6

mol/L.det

 

 

jam t

t

C C C

K C

t r

_{s s}

s s m

5 . 3 det

12517

x10 2

. 1 x10

x10 12 2

. 1

x10 ln 12

8.9x10 x10

5 . 2

1 1 ln

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 s

ebesar 0.1 M dan di akhir fermentasi di

peroleh konsentrasi gula sebesar 0.025

M. Fermentasi berlangsung selama 4 ja

m 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 .

0 1 . 025 0

. 0

1 . ln 0

4 5x10 1

1 ln





 

 



  



 

 



  



jam M

r r

C C C

K C

r t

_{s s}

s s m

1 3

- max

. 017 .

0

025 .

0 5x10

) 025 .

0 ( 02 . 0





 

 

jam M

r r

C K

C r r

S m

S

dt V dX V

r X

X

F (

₀

 ) 

_x



F, Cs0

F, Cs V

on Accumulati G

Output -

Input  eneration 

the ratio of biomass ra te of generation to bi omass concentration, r_x/X, that is the specifi c growth rate; μ

 X 

r

_x



 D

 0 dt dX

_s

Steady state:

V D

F  

 1

0

 0

No cell in inlet:

X











 D

X DX

r X

V X F

x

.

)

(

₀

-3. Continuous stirred-tank – cont.

Monod rate:

s s

s

C K

C

 

^max





s s

s

C K

D C

 

  

^max

D C

_s

DK

^s

 



max

max max

1 1

1



 ^



s s

C K

D

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

C V K

C C C

F

s s

s s

s





 





 



^max

0

)

( 

C X K

C C C

D

s s

s s

s





 





 



^max

0

)

( 

Cs Cs

X Y X



 

0

0

 

 





 



 D

C DK Y

X

X

_s ^s

max 0

0



Case study

• Contoh soal 4:

• Kinetika Monod ( K s = 3 g/L) telah diperti

mbangkan dalam operasi CSTR untuk me

mproduksi yeast dengan konsentrasi subst

rat awal 50 g/L. Jika laju pertumbuhan yea

st spesifik maksimal sebesar 0.5 jam

^-1

, ber

apakah laju pertumbuhan yeast fermentas

i dan dilution rate untuk konversi reaksi se

besar 90% ? asumsikan tidak ada yeast di

awal fermentasi.

Solution

Cs = (1-X

_A

)Cs

₀

= (1-0.9)( 50 g/L) = 5 g/L

1 1

max

3125 .

0

g/L 5

g/L 3

g/L 5

5 . 0









 



 



jam D

x D jam

Cs K

D Cs

s





 

Case study

• Contoh soal 5:

• Hitunglah laju dilusi dari contoh soal 4 j

ika terdapat konsentrasi yeast di awal d

an akhir fermentasi sebesar 0.1 g/L dan

35 g/L!

Solution

• Pada kondisi ini berlaku persamaan berikut:

 X 

r

_x

r

x

X X

D ( 

₀

) 

S K

S

s



 

^max



) (

) (

0 0

X X

D X

X X

D r

^x

 

 



1 0

max

3134 .

1 0 . 0 35

) 35 ( 3125 .

0

) (





 



 

jam D

X X

S X K

S

D

^s



Case study

• Contoh soal 6:

• Berapakah yield fermentor pada contoh soal 5?

– Solution:

% 56 . 77 7756

. 5 0

50

1 . 0 35

0

0



 

 



 

Y

S S

X

Y X

Solution

• Pada kondisi ini berlaku persamaan berikut:

 X 

r

_x

r

x

X X

D ( 

₀

) 

S K

S

s



 

^max



) (

) (

0 0

X X

D X

X X

D r

^x

 

 



1 0

max

3134 .

1 0 . 0 35

) 35 ( 3125 .

0

) (





 



 

jam D

X X

S X K

S

D

^s



Case study 7

• A chemostat study was performed with yeast. The med ium flow rate was varied and the steady state concentr ation of cells and glucose in ther fermenter were meas ured and recorded. The inlet concentration of glucose was set at 100 g/L. The volume of the fermenter conte nts 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:

max max

1 1

1



 ^



s s

C K

Chemostat: D

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 3 1

. 1

max

max









46 . 2 2

. 3

max



 Ks

Ks



Persamaan laju reaksi

s s

s s

s

C X r C

C K

X r C

X r

 

 



46 .

2

78 .

0

 max



NUR ISTIANAH,ST,MT,M.Eng

DUAL SUBTRATE REACTION

ALLPPT.com _ Free PowerPoint Templates, Diagrams and Charts

Mechanisms of Single Enzyme with Dual Substrates

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

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

K =equilibrium or d issociation constant.

The total enzyme concentration

27

Nur Istianah-THP-FTP-UB -2014

The intermediates, complexes of ES1 and ES2, are defined b ased on equilibrium constants

The initial and total enzyme concentrations are defined base d on measurable components given below:

The free enzyme can also be defined based on the followin

g equation:

28

Nur Istianah-THP-FTP-UB -2014

29

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

Nur Istianah-THP-FTP-UB -2014

37

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/m

³

r

_max

= 100 mol/m

³

min

Chemostat (continuosly stirred-tank reactor) runs with various flow rates were carried out. If the inlet substrate cocentration is 300 mol/m

³

and the flow rate is 100 cm

³

/min, what is the steady-state substrate concentration of the outlet? The reactor volume is 300 cm

³

. Assume that the enzyme concentration in the reactor is constant so that the same kinetic parameters can be used.

Contoh Soal

Nur Istianah-THP-FTP-UB -2014

CSTR equation

…….1

…….2

…….3

…….4

Subtitusi pers 2 dan 3 :

SOLUTION

3. The enzym, cathepsin, hydrozes L-glutamyl – L – tyrosine to carboben zoxy – L – glutamic acid and L-tyrosine. It has been found (Frantz and St ephenson, J. Biol. Chem., 169, 359, 1947) that the glutamic acid formed i n the hydrolysis, inhibits (competitively) the progress of the reaction by f orming 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, K_I. (contr ibuted 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/r_max

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 C_I dan dan C_I/r sehingga didapatkan K_I= 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

C_I

Cs C_I r_P Cs/r K_MI C_I/r K_I K_S

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:

THANKS FOR YOUR ATTENTION

The best person is one giving something useful always