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:
THANKS FOR YOUR ATTENTION
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