BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.1
Total volume = 44 + 5 + 1= 50ml
From the graph the equation obtained
y=0.033× + 0.03a 1
m= 0.033 ɱ mol / ml . min
a) Activity of the β- glucosidase
0.033 x 50= 1.65 mumol / min
i)
= 1.65mumol/min
0.1 mg/ml x 0.1ml
= 165 units/mg protein
ii)
= 1.65 mumol/min
1ml of enzyme
= 1.65 units/ml of enzyme
b) Initial rate of reaction 0.033 mumol/ mL.min
S vs t Graph
1.2
y = 0.033x + 0.0391
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
35
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.2
→
←
→
←
→
a) Michaelis-Menten approach
The rate of product formation.
d[p]
dt
(
)
Since the enzyme is preferred,
Make E as the subject,
Since forward reaction = backward reaction.
( )
ubstitute
into
:
[(
(
Make
)
(
) (
)( )
)
(
(
)] ( )
)( )
(
)( )
as a subject:
(
)
(
)( )
(
)( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
(
)( )
(
)( )
(
(
)
[(
)
(
(
(
ub
into
(
(
)
(
(
)
)( )
)( )
(
)( )
)( )( )
(
)( )( )
as a subject,
(
)(
(
( )( ))
(
)
(
(
(
into
,
)
(
)( )( )
)( )( )
)(
)
(
ub
)( )
)
:
(
Make
)
)( )]
( )( ))
)( )( )
( )
( )( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
d[p]
dt
(
(
(
Since
(
)( )( )
)
( )
( )( )
)( )( )
)
( )
( )( )
)
( )( )
)
( )
( )( )
b) Since [ ]
[ ]
d[p]
dt
( )( )
( )( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.3
(a) E+S k1
(ES)1
(ES)1 k3 (ES)2
(ES)2 k2 E+P
V=
[ ]
= k5 [ES] 2
[E0] = [E] + [ES] + [ES]2
[E] = [E0]-[ES]-[ES]2
k2 = [E] [S]
k1 [ES]1
k2 [ES]1 = [Eo] [S] – [ES]1 [S]
k1
[ES]1 ( k2 + [S] ) = [E0] [S] – [ES]2 [S]
k1
[ES]1 = [E0] [S] – [ES]2 [S]
k2 + [S]
k1
k4 = [ES]1
k3
[ES]2
k4 [ES]2 = [E0] [S] – [ES]2 [S]
k3
k2 + [S]
k1
[ES]2 ( k2 k4 + k4 [S] ) = [E0] [S] – [ES]2 [S]
k1k3
k3
[ES]2 ( k2 k4 + k4 [S] + [S] )= [E0] [S]
k1 k3
k3
[ES]2 = [E0] [S]
k2 k4 + k4 [S] + [S]
k1 k3
k3
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
V= d [P] = k5 [E0] [S]
dt
k2 k4 + k4 [S] + [S]
k1 k3
k3
=
Vm [S]
k2 k4 + k4 [S] + [S]
k1 k3
k3
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.4
a) Michaelis-Menten approach
The rate of product formation.
d[p]
dt
(
)
Since the enzyme is preferred,
Make E as the subject,
Since forward reaction = backward reaction.
( )
ubstitute
into
:
[(
(
Make
)
(
) (
)( )
)
(
(
)] ( )
)( )
(
)( )
as a subject:
(
)
(
)( )
(
)( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
(
)( )
(
)( )
(
(
)
[(
)
(
(
(
ub
into
(
(
)
(
(
)
)( )
)( )
(
)( )
)( )( )
(
)( )( )
as a subject,
(
)(
(
( )( ))
(
)
(
(
(
into
,
)
(
)( )( )
)( )( )
)(
)
(
ub
)( )
)
:
(
Make
)
)( )]
( )( ))
)( )( )
( )
( )( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
d[p]
dt
(
(
(
Since
(
)( )( )
)
( )
( )( )
)( )( )
)
( )
( )( )
)
( )( )
)
( )
( )( )
b) Briggs-Haldane approach
→
←
→
←
→
←
→
The rate of product formation,
d(p)
dt
Since the enzyme is preferred,
Make as a subject,
(
)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Substrate consumption,
)
d(
( )( )
dt
)
d(
(
dt
ubstitute
into
(
)(
(
)(
)(
)( )
( )
)
)
( )
( )
( )
(
ubstitute
(
)
(
)
)( )
(
)
(
)
(
( )( )
(
)
)( )
(
)
(
( )(
( ))
( )
)
)
( )
( )(
)
( )
)
(
)
)
( )(
( )(
(
(
(
)
)
)
( ))
into
(
(
(
)( )
:
(
(
(
)(
)(
)
)
)(
)
( )
(
)( )
( )(
(
( ) ( )
( )
( ) ( )(
(
( )
)
(
)
( ))
)
( ))
( )
)( )
(
)
)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
(
(
)(
)(
( )
( ) ( )
( ) ( )
(
ubstitute
)(
)
( ))
( ) ( )(
( )
)(
)
( )
( )
( ))
(
)
( ) ( )
( ) ( )
(
( ) ( )
( )
)(
( )
( ))
into
d(p)
dt
(
( ) ( )
(
( ) ( )
( )
)(
( )
( ))
v
d(p)
dt
v
(
( ) ( )
( ) ( )
( )
)(
( )
( ))
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.5
Lineweaver- Burk Plot
x-intercept= - 1
km
y-intercept= 1/ V more
Equation obtained y= 0.0172 x + 3.6342
y-intercept = 3.6342= 1/ V max
V max = 0.275
x-intercept , y= 0
0.0172x + 3.6342=0
0.0172x = -3.6342
x= -211.291
x= -1/km
km = 1/211.291
= 0.00473
Longmuir Plot
Equation obtained y= 3.3133x + 0.0191
1/Vm = m = 3.3133
Vm=0.302
y-intercept= km/Vm = 0.0191
Km = 0,0191x 0.302
= 0.00577
Eadie-Hofstee Plot
Equation obtained y= -0.0043x + 0.2645
-Km = m = -3.3133
Km=0.302
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
y-intercept= Vm = 0.2645
Non-Linear Regression Procedure
From the graph, Vm=0.2
½ Vmax = 0.1,
Km=0.0032
Data for Graph plot :
Langmuir Plot
s
s/v
0.0032 0.028829
0.0049 0.033108
0.0062 0.043357
0.008 0.048193
0.0095
0.0475
Lineweaver-Burk Plot
1/s
1/v
312.5
9.009009
204.0816 6.756757
161.2903 6.993007
125
6.024096
105.2632
5
Eadie-Hofstee Plot
v/s
v
34.6875
0.111
30.20408 0.148
23.06452 0.143
20.75
0.166
21.05263
0.2
Non-Linear Regression Plot
S
v
0.0032
0.111
0.0049
0.148
0.0062
0.143
0.008
0.166
0.0095
0.2
Kinetic Parameters
Vmax
Km
Langmuir
0.2750
0.0047
Lineweaver-Burk
0.0191
0.0057
Eadie-Hofstee
0.2645
0.0043
Non-Linear Regression 0.2000
0.0032
Type of Plot
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Langmuir Plot
0.06
y = 3.3133x + 0.0191
0.05
0.04
0.03
0.02
0.01
0
0
0.002
0.004
0.006
0.008
0.01
Lineweaver Burk Plot
10
9
8
7
6
5
4
3
2
1
0
y = 0.0172x + 3.6342
0
50
100
150
200
250
300
350
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Eadie-Hofstee Plot
0.25
0.2
0.15
y = -0.0043x + 0.2645
0.1
0.05
0
0
5
10
15
20
25
30
35
40
Non-Linear Regression Procedure
0.25
0.2
0.15
0.1
0.05
0
0
0.002
0.004
0.006
0.008
0.01
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.6
↔
→
Rate of product formation
( )
v
(
)
Enzyme is preserved,
d( )
dt
(
(
)
(
)
Assumptions:
[𝐸𝑂 ]small,
neg igib e
)
( )( )
(
)(
)
(
)
(
)
(
)
(
)(
(
)
(
(
(
into
))( )
)
( )
(
(
)
)
( ) )
( )
( )
(
( )
d(p)
dt
(
(
( )
)
into
( )
(
)
( )( )
(
)
Substitute
v
(
( )( )
Substitute equation
(
)
𝑑(𝐸𝑆)
𝑑𝑡
)
𝑣𝑚
( )
( )
)
)
𝐾𝑚
𝑘 𝑘 𝐸𝑜
𝑘
𝑘
𝑘
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Dividing
[(
with the value of
( )
( )
v
)
]/
v s
(
( )
)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.7
a) FCs0 - FCs + rSv = V
For Batch reactor F=0
rSv = V
[ ]
=
[ ]
[
=
]
[
]
= 60mol/m3.min
b) Equation obtained y = 6.3852x + 59.571
m = Vmax
= 6.3852
y- intercept = - Km = 59.571
Km = - 59.571
c) FCs0 - FCs + rSv = 0
FCs0 - FCs = - rSv = rpv
FCs0 - FCs =
V
F = 0.0001m3/min
V = 0.0003m3
( FCSo - FCs ) (Km + Cs)
= Vmax CsV
2
FCSo Km + FCSo Cs - FKm Cs – FCs = Vmax CsV
(0.0001 (300)(200) + 0.0001(300)Cs – 0.001(200)Cs – 0.001Cs2 = 100 (0.0003)Cs )
6 + 0.03Cs – 0.02Cs – 0.001Cs2 = 0.03Cs
0.0001Cs2 + 0.02Cs – 6 = 0
Cs=165mol/m3
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Data :
Cs t t/ln(Cso/Cs) (Cso-Cs)/ln(Cso/Cs)
1 1
0.175322
52.42135
5 5
1.221197
72.0506
10 10
2.940141
85.26409
20 20
7.385387
103.3954
Graph :
(Cso-Cs)/ln(Cso/Cs)
120
y = 6.3852x + 59.571
100
80
(Cso-Cs)/ln(Cso/Cs)
60
Linear ((CsoCs)/ln(Cso/Cs))
40
20
0
0
2
4
6
8
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.8
a) Km =0.01 mol/L
Cso = 3.4 x 10 -4 mol/L
Cs = 0.9 x 3.4 x 10-4
= 3.06 x 10-4 mol/L
t= 5minutes
=
[ ]
[ ]
( 3.4x 10-4 – 3.06 x 10-4) = Vmax (3.06 x 10-4)
S
0.01 + (3.06 x 10-4)
6.8 x 10-6 = Vmax ( 0.03)
Vmax = 2.27 x 10-4 mol/L-min
b) 6.8 x 10-6 x 15 = 1.02 x 10-4 mol/L
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.9
Km = 0.03mol/L
rmax = 13mol/L min × 60 = 780mol/L hr
F=10L/Hr
Cs=10mol/L
F=10L/Hr
Cs=0.5mol/L
CSTR
a) V = ?
CSTR @ Stead State
FCs0 - FCs + rSv = 0
F (Cs0 - Cs ) = rpv
(
10 (10 – 0.5) =
)
V
V = 0.129 liter
b) Plug - Flow @ Stead State
Km ln
+ (Cs0 - Cs )
0.03 ln
+ (10 - 0.5 ) = 780t
= rmax t
9.95899 = 780t
t = 0.0123hr
t = V/F = 0.0123
V = 0.0123 × 10
= 0.123liter
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.10
Km = 10g/L
rmax = 7g/L.min
a)
F=0.5L/min
Cs0=50g/L
1L
F=0.5L/min
Cs1=?g/L
1L
F=0.5L/min
Cs2=? g/L
CSTR @ Steady State
FCs0 - FCs + rSv = 0
F (Cs0 - Cs ) = rpv
0.5 (50 – Cs1) =
(
s
)
s
(1)
(25-0.5Cs1)(10+ Cs1)=7Cs1
250+25Cs1-5Cs1-0.5Cs12=7Cs1
0.5Cs12-13Cs1-250=0
Cs1=38.86g/L
0.5 (38.86 – Cs2) =
(
s
)
s
(1)
(19.43-0.5 Cs2)(10+ Cs2)=7 Cs2
194.3+19.43Cs2-5Cs2-0.5Cs22=7Cs2
0.5Cs22-7.43Cs2-194.3 =0
Cs2=28.49g/L
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
b)
F=0.5L/min
Cs0=50g/L
2L
F=0.5L/min
Cs1=?g/L
CSTR @ Steady State
FCs0 - FCs + rSv = 0
F (Cs0 - Cs ) = rpv
0.5 (50 – Cs1) =
(
s
)
s
(2)
(25-0.5Cs1)(10+ Cs1)=14Cs1
250+25Cs1-5Cs1-0.5Cs12=14Cs1
0.5Cs12-6Cs1-250=0
Cs1=29.15g/L
Since in the Cs in two reactor system is less than Cs in one reactor system, therefore two reactor
system is more efficient than one reactor system as it indicates more substrates have been
consumed to form products.
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.11
a) k1 [E] [S] = k2 [ES]
[ES] = k2
k1 [E] [S]
k1 [E] [S]
k2
k3 [E] [P] = k4 [EP]
[EP] = k3
k4 [E] [P]
k5 [ES][P] = k6 [ESP]
[ESP] = k5 [ES] [P]
k6
k7 [EP] [S] = k8 [ EPS ]
[EPS] = k7 [EP] [ S ]
K8
= k7 k3 [ S ]
k8 k4 [E] [P]
From,
[ESP] = k5 [P] k2
k6k1 [E] [S]
[E0] = [E] + [ES] + [EP] + [ESP] + [EPS]
[E0] = [ES] + [ESP] + [E] + [EP] + [EPS]
[E0] = [ES] + [ESP] + [E] + [EP] +
[
][ ]
[ ]
[E0] = [ES] + [ESP] + [E] + [EP] + (
[ ][ ]
[E0] = [ES] + [ESP] + [E] +
[ ]
[E0] = [ES] + [ESP] + [E] [
[E0] = [ES] +
[
[E0] = [ES] {1 +
][ ]
+
[
][
[ ]
+
[
]
]
[
[ ]
+(
[ ]
(
[ ]
[
[ ]
(
)
)
)]
[ ]
(
[ ]
)]}
)]
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
[ES] =
V=
[ ]
b)
Given:
[ ]
=
[
]
[ ]
KSP =
[
KSP =
[ESP] =
[
[ ]
[
]
[ ]
(
[ ]
[
(
)]
[ ]
)]
KPS =
[
[ ][ ][ ]
[
]
KPS =
[ ][ ][ ]
[
]
[ ][ ][ ]
[EPS] =
[ ][ ][ ]
[
][ ]
]
][ ]
]
[
[ESP] = [EPS]
KS KSP = KP KPS
=
[ ]
=
=
[
[ ]
[
[ ]
[ ]
[
]
[
]
[ ]
[ ]
[ ]
]
]
[ ]
[ ]
c)
Ks=Kps
Kp=Ksp
𝑘𝑠
𝑘𝑝
𝑘𝑝𝑠
𝑘𝑠𝑝
[ESP]=[EPS]
[ ]
(
( )
[
[ ]
[ ][ ]
)[
[ ]
][ ]
]
[ ]
[ ]
[ ]
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
[
][ ]
[ ])
[ ](
[ ]
[ ]
[ ]
[ ])
[ ]
[ ]
[ ])
(
(
Compare with
[ ]
[ ]
Hence, Vmax = (
Km=
d)
[
]
[ ])
[ ]
(
[ ])
[ ]
[ ]
[ ]
*∫
[ ]
*
[
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
n(
[ ]
)
[ ]
+
[ ]
+
[ ] [ ]]
[ ] [ ]
[ ]
∫
[ ]
[ ]
n(
[ ]
)
[ ]
n([ ] [ ])
(
[ ]
[ ]
)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Y = mx+c
[ ]
)
]
n ([
Y=
M=
[
X= (
] [ ]
)
C=
So we can plot a graph of
[ ]
)
]
n ([
[
vs (
] [ ]
)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.14
Rate:
rp = k9CES +k10CEIS1 +k10CEIS2
---- 1
Enzyme balance:
CEo = CE + CES
---- 2
CEo = CEIS1 + CES + CE
---- 3
CEo = CEIS2 + CEI + CE
---- 4
The equilibrium reaction equations are as follows:
CE Cs / CES = k2/k1
---- 5
CECI / CEI = K4/K3
---- 7
CESCI /CEIS1 = K6/K5
---- 6
CEICS / CEIS2 = K8/K7
---- 8
By rearranging Equation 5,
CE = (k2/k1) Cs CES
From Equation 2,
CEo = [(k2/k1)CE + 1] CES
 CES = CEo /[( k2/k1)CS +1]
---- 9
By rearranging Equation 6,
CES = [(K6/K5)CI ] CEIS1
From Equation 3,
CEo = CEIS +CES + (k2/k1) Cs CES
= {CEIS1 + [1 + (k2/k1) Cs]( K6/K5)CI }CEIS1
= {1 + [1 + (K2/K1) Cs ]( K6/K5)CI } CEIS1
 CEIS1 = CEo/ {1 + [1 + (k2/k1) Cs ]( K6/K5)CI }
By rearranging Equation 7,
CE = (K4/K3) CEI
By rearranging Equation 8,
CEI = K8/K7CS CEIS2
---- 10
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
From Equation 4,
CEo = CEIS2 + CEI + [(K4/K3)CI]CEI
= CEIS2 + [1 + (K4/K3)CI ]CEI
= CEIS2 + [1 + (K4/K3)CI ]( K8/K7)CS CEIS2
CEo = {1 + [1 + (K4/K3)CI ]( K8/K7)CS } CEIS2
 CEIS2 = CEo / {1 + [1 + (K4/K3)CI ]( K8/K7)CS }
---- 11
From Equation 1, since rp = k9CES +k10CEIS1 +k10CEIS2,
By substituting Equation 9, 10 & 11 into Equation 1,
Therefore,
rp = k9 CEo /[( k2/k1)CS +1] + k10 CEo/ {1 + [1 + (k2/k1) Cs ]( K6/K5)CI } + k10 CEo / {1 + [1 + (K4/K3)CI
]( K8/K7)CS }
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.15
a) Based on the graphs
The y-intercept in Lineweaver – Burk plot is almost the same.
Y-intercept => 3.8266; 3.6342
Whereas in Langmuir Plot
Two equations obtained
Y = 2.9883x + 0.0489
Y = 3.3133x + 0.0191
When y=0
X=
;
X=
X = -0.016
;
X = -0.005
In Line weaver – Burk Plot and Langmuir Plot both indicates it’s a competitive inhibitor
Data :
Lineweaver
1/s
1/Vo
312.5
9.009009
204.0816 6.756757
161.2903 6.993007
125
6.024096
105.2632
5
Langmuir
s
s/Vo
0.0032 0.028829
0.0049 0.033108
0.0062 0.043357
0.008
0.048193
0.0095
0.0475
1/Vi
16.94915
14.08451
10.98901
9.009009
8
S/Vi
0.054237
0.069014
0.068132
0.072072
0.076
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Lineweaver-Burk Plot
20
18
16
14
12
10
8
6
4
2
0
y = 0.0439x + 3.8266
1/Vo
1/Vi
y = 0.0172x + 3.6342
Linear (1/Vo)
Linear (1/Vi)
0
100
200
300
400
Langmuir Plot
0.09
0.08
y = 2.9883x + 0.0489
0.07
0.06
s/Vo
y = 3.3133x + 0.0191
0.05
S/Vi
0.04
Linear (s/Vo)
0.03
Linear (S/Vi)
0.02
0.01
0
0
0.002
0.004
0.006
b) Y-intercept = 1/Vmax = 0.00489
Vmax = 1/0.00489 = 204.5 mol /L.min
Km/Vmax = 2.9883
Km=2.9883*204.5
=611mol/L
0.008
0.01
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.16
(a) E + S
↔
ES + S
↔
→k5 E + P
(E0 ) = (E) +(ES) + (ESS)
(E) = (E0 ) – (ES) – (ESS) -------V=
=
(
( )
= k5 (ES) ------
(
)( )
)
(
) = (ES)(S)/ (k4 / k3)
K2 / k1 = (E)(S) / (ES)
K2/k1 (ES) = (E0)(S) – (ES)(S)
(ES)((k2/k1) + (S)) = (E0)(S) – (ES)(S)2 /
(ES)( (k2/k1) + (S)( ) ) =
(E0)(S) – (ES)(S)2
(ES) ( (k2/k1) + (S)( ) + (S)2 ) =
(ES) =
(E0)(S)
(E0)(S) / (k2/k1) + (S)( ) + (S)2 --------
3→
V=
=
( )
=
k5 (E0) (S) / (k2/k1) + (S)( ) + (S)2
Vm(S) / (k2/k1) + (S)( ) + (S)2
(b) At low substrate concentration,
1/ Vm = 3.1209
Vm = 0.3204
Km/Vm = 106.07
Km / 0.3204 = 106.7
K1m = 33.98
At high substrate concentration,
1/ Vm = 3.0574
Vm = 0.3271
1/ K1. Vm = 0.0032
1/ Km(3.0574) = 0.0032
Km = 102mol/L
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.17
V= 5L
Cso = 100 mmol/L
F = 1 L/hr
Cs = 10m mol/L
a) F (s0 – FCs = rp V
1(100-10) = rp (5)
Rp = 18 m mol/ L.min
b) Find rp for each F and s
Equation obtained y= 0.0391x + 0.1641
M= 1/Vmax = 0.0391
Vmax= 25.57 m mol/L.min
Km/ Vm = 0.1641
Km= 4.197
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.18
[SO]1 = 0.1 mol/L
[S0]2 = 0.3mol/L
[ ][ ]
[
]
[ ][
]
]
[
[ ][
]
[
[ ][
]
]
]
[
[ ][
[
]
]
[ ][ ]
[
]
V1 =
[
]
= k5[ES1]
=3.5 [ES1] --V2 =
[
]
= k6[ES2]
=2.8 [ES2] ---
[E0] = [E] + [ES1] + [ES2]
[ ][
[E0] = [E] + [ES1] +
[E0] = [ES1] + [E] (1+
[E0] = [ES1] +
[E0] = [ES1] {1 +
[
[ES1] =
[
[
]
[
]
[
]
[
]
[
]
)
)
(1+
(1+
]
]
(
]
[
[
]
]
)
)}
[E0] = 0.05 mol/L
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Vmt = [S1]0 – [S2] + K – ln
[ ]
[ ]
3.5 [ES1] t = 0.1 – [S1] +0.0714 ln [
]
3.5 [ES1] t + [S1] = 0.1 +0.0714 ln 0.1 - 0.0714 ln [S1]
3.5 [ES1] t + [S1] + 0.0644 = -0.0714 ln [S1]
ln[S1] =
[
]
[
]
[S1] =
[
]
[
]
---
[E0] = [E] + [ES1] + [ES2]
[E0] = [E] +
[ ][
[E0] = [E] (1+
[E0] =
[
]
[
[E0] = [E2] [
[
]
+ [ES2]
[
]
]
(1+
]
)+ [ES2]
[
(1+
]
[
Vmt = [S1]0 – [S2] + KMln
)+ [ES2]
]
[
)+ 1]
]
[
]
2.8[ES1] t = 0.3 – [S2] + 0.2207ln [
]
2.8[ES1] t + [S2] = 0.3 + 0.2207ln 0.3 – 0.2207ln [S2]
2.8[ES1] t + [S2] – 0.0343 = – 0.2207ln [S2]
ln[S2] =
[
[S2] = e
[
]
[
]–
[
]–
–
]
–
---
As [S1] increases, [ES1] also increases as in eq.3. [P1] also increases as in eq.1. This also occurs in
[S2]. As [S1] increases, [ES1] also increases as in eq.4. [P2] also increases as in eq.2
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.19
Data :
s
6.7
3.5
1.7
s/v
22.33333
14
10.625
Langmuir Plot
25
y = 2.3722x + 6.2429
20
15
10
5
0
0
1
2
1/Vm = 2.3722
Vmax = 0.4215 mumol/L.min
Km/Vm = 6.2429
Km = 6.2429(0.4215)
=2.63mumol/L
3
4
5
6
7
8
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Solution 2.20
a. Write the kinetic model.
Since the Michaelis constant KM is not affected by the presence of the inhibitor (which has shown on
the given table); then this enzyme reaction is noncompetitive inhibition reaction.
Kinetic Model:
k 1, k 2
E  S 
 ES
k 3, k 4
E  I 
 EI
k 5, k 6
EI  S 
 EIS
k 7 ,k 8
ES  I 
 ESI
k9
ES 
EP
b. Derive the rate equation. State the assumptions.
Assumptions:


The dissociation constant for the first equilibrium reaction is the same as that of the third
equilibrium reaction.
The dissociation constant for the second equilibrium reaction is the same as that of the
fourth equilibrium reaction.
The two equilibrium reactions,
k
k2
 K S  6  K IS
k1
k5
k
k4
 K I  8  K SI
k3
k7
If the slower reaction, the product formation step, determines the rate of reaction according to
Michaelis-Menten assumption, the rate can be expressed as:
rP  k 9 [ ES ]
(1)
The enzyme balance gives
[ E0 ]  [ E ]  [ ES ]  [ EI ]  [ ESI ]
(2)
k 9 [ ES ]
rP

[ E0 ] [ E ]  [ ES ]  [ EI ]  [ ESI ]
(3)
Divide (1) by (2),
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Applied Law of mass action,
Ks 
K 2 [ E ][S ]
[ E ][S ]

 [ ES ] 
K1
[ ES ]
KS
(4)
KI 
K 4 [ E ][ I ]
[ E ][ I ]

 [ EI ] 
K3
[ EI ]
KI
(5)
KI 
k8 [ ES ][ I ]
[ ES ][ I ]

 [ ESI ] 
k7
[ ESI ]
KI
(6)
Substitute (4), (5), (6) into (3),
[ E ][ S ]
KS
rP

[ E ][ S ] [ E ][ I ] [ ES ][ I ]
[ E0 ]
[E] 


KS
KI
KI
k9
[ E ][ S ]
KS
rP

[ E ][ S ] [ E ][ I ] [ E ][ S ][ I ]
[ E0 ]
[E] 


KS
KI
KS KI
k9
Eliminate [E],
[S ]
KS
rP

[ S ] [ I ] [ S ][ I ]
[ E0 ]k 9
1


KS KI KS KI
Substitute rPmax  [ E0 ]k 9
rP
rPmax
[S ]
KS

[ S ] [ I ] [ S ][ I ]
1


KS KI KS KI
Multiply numerator and denominator by Ks,
rP

rPmax
[S ]
K [ I ] [ S ][ I ]
K S  [S ]  S

KI
KI
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Rearranging,
rP

rPmax
[S ]
K [I ]
[ S ][ I ]
KS  S
 [S ] 
KI
KI
rP

rPmax
[S ]
K S (1 
[I ]
[I ]
)  [ S ](1 
)
KI
KI