y = 0.033x + 0.0391 y = 0.033x + 0.0391 0 0 0.2 0.2 0.4 0.4 0.6 0.6 0.8 0.8 1 1 1.2 1.2 0 0 55 1100 1155 2020 2255 3300 3355
S vs t Graph
S vs t Graph
Total volume = 44 + 5 + 1= 50ml Total volume = 44 + 5 + 1= 50ml From the graph the equation obtained From the graph the equation obtained y=0.033× + 0.03a 1y=0.033× + 0.03a 1 m= 0.033
m= 0.033
ɱɱ
mol / ml . min mol / ml . min a)a)
Activity of the β
Activity of the β
- glucosidase- glucosidase 0.033 x 50= 1.65 mumol / min 0.033 x 50= 1.65 mumol / min i) i) = 1.65mumol/min= 1.65mumol/min 0.1 0.1 mg/ml x 0.1mlmg/ml x 0.1ml = 165 units/mg protein = 165 units/mg protein ii)ii) = 1.65 mumol/min= 1.65 mumol/min 1ml of enzyme 1ml of enzyme
= 1.65 units/ml of enzyme = 1.65 units/ml of enzyme b)
b) Initial rate of reaction 0.033 mumol/ mL.minInitial rate of reaction 0.033 mumol/ mL.min Solution 2.1
←←
←←
→
→
Michaelis-Mente
Michaelis-Menten n approachapproach The rate of product formation. The rate of product formation.
dd[[pp]]
dtdt
Since the enzyme is preferred, Since the enzyme is preferred,
Make E as the subject, Make E as the subject,
Since forward reaction = backward
Since forward reaction = backward reaction.reaction.
ubstitute into:
ubstitute into:
[[
]
]
MakeMake
as a subject:as a subject:
Solution 2.2 Solution 2.2
a) a)
[[
]]
ub into:
ub into:
MakeMake
as a subject, as a subject,
ub into,
ub into,
d[p]
dt
Since
Since [
][
]
d[p]
dt
b)(a) E+S k 1 (ES)1
(ES)1 k 3 (ES)2
(ES)2 k 2 E+P
V=
[]
= k 5[ES]2[E0] = [E] + [ES] + [ES]2
[E] = [E0]-[ES]-[ES]2 k 2 = [E] [S] k 1 [ES]1 k 2 [ES]1= [Eo] [S]
–
[ES]1[S] k 1 [ES]1( k 2 + [S] ) = [E0] [S]–
[ES]2[S] k 1 [ES]1= [E0] [S]–
[ES]2[S] k 2 + [S] k 1 k 4 = [ES]1 k 3 [ES]2 k 4 [ES]2= [E0] [S]–
[ES]2[S] k 3 k 2 + [S] k 1 [ES]2 ( k 2k 4 + k 4[S] ) = [E0] [S]–
[ES]2 [S] k 1k 3 k 3 [ES]2( k 2 k 4 + k 4[S] + [S] )= [E0] [S] k 1k 3 k 3 [ES]2= [E0] [S] k 2k 4 + k 4[S] + [S] k 1k 3 k 3 Solution 2.3V= d [P] = k 5[E0] [S] d t k 2k 4 + k 4[S] + [S] k 1k 3 k 3 = Vm[S] k 2k 4 + k 4[S] + [S] k 1k 3 k 3
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:
Solution 2.4 a)
[
]
ub into:
Make
as a subject,
ub into,
d[p]
dt
Since
Briggs-Haldane approach
←
←
←
→
The rate of product formation,
dp
dt
Since the enzyme is preferred,
Make
as a subject,
Substrate consumption,
d
dt
d
dt
ubstitute into:
(
)
(
)
ubstitute into
(
)
(
)
(
)
(
)
(
)
ubstitute into
dp
dt
(
)
v
dp
dt
v
(
)
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 Solution 2.5
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 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
Type of Plot 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 1/s 1/v 312.5 9.009009 204.0816 6.756757 161.2903 6.993007 125 6.024096 105.2632 5
y = 3.3133x + 0.0191 0 0.01 0.02 0.03 0.04 0.05 0.06 0 0.002 0.004 0.006 0.008 0.01
Langmuir Plot
y = 0.0172x + 3.6342 0 1 2 3 4 5 6 7 8 9 10 0 50 100 150 200 250 300 350y = -0.0043x + 0.2645 0 0.05 0.1 0.15 0.2 0.25 0 5 10 15 20 25 30 35 40
Eadie-Hofstee Plot
0 0.05 0.1 0.15 0.2 0.25 0 0.002 0.004 0.006 0.008 0.01
↔
→
Rate of product formation
v
Enzyme is preserved,
d
dt negigibe
Substitute equation
into
Substitute
into
v dp
dt
Solution 2.6 Assumptions: []small,
Dividing
with the value of
[
]/
v
v
s
Solution 2.7
a) FCs0 - FCs + rSv = V
For Batch reactor F=0rSv = 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) =VmaxCsV FCSo Km+ FCSo Cs - FKm Cs–
FCs2=VmaxCsV (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/m3Data : 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 : y = 6.3852x + 59.571 0 20 40 60 80 100 120 0 2 4 6 8
(Cso-Cs)/ln(Cso/Cs)
(Cso-Cs)/ln(Cso/Cs) Linear ((Cso-Cs)/ln(Cso/Cs))a) Km =0.01 mol/L Cso = 3.4 x 10-4mol/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 10-4mol/L Solution 2.8Km= 0.03mol/L
rmax = 13mol/L min × 60 = 780mol/L hr
CSTR a) V = ? CSTR @ Stead State FCs0 - FCs + rSv = 0 F (Cs0 - Cs ) =
r
pv
10 (10–
0.5)=
V
V = 0.129 literb) Plug - Flow @ Stead State
Kmln
+
(Cs0 - Cs ) = rmaxt 0.03 ln
+
(10 - 0.5 ) = 780t 9.95899 = 780t t = 0.0123hr t = V/F = 0.0123 V = 0.0123 × 10 = 0.123liter F=10L/Hr Cs=0.5mol/L F=10L/Hr Cs=10mol/L Solution 2.9Km= 10g/L rmax = 7g/L.min CSTR @ Steady State FCs0 - FCs + rSv = 0 F (Cs0 - Cs ) =
r
pv
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 1 L 1 L F=0.5L/min Cs0=50g/L F=0.5L/min Cs2=? g/L F=0.5L/min Cs1=?g/L Solution 2.10 a)b) CSTR @ Steady State FCs0 - FCs + rSv = 0 F (Cs0 - Cs ) =
r
pv
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/LSince 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.
2 L F=0.5L/min
Cs0=50g/L
F=0.5L/min Cs1=?g/L
a) k 1 [E] [S] = k 2 [ES] [ES] = k 2 k 1[E] [S] k 1[E] [S] k 2 k 3[E] [P] = k 4[EP] [EP] = k 3 k 4[E] [P] k 5 [ES][P] = k 6[ESP] [ESP] = k 5[ES] [P] k 6 k 7[EP] [S] = k 8[ EPS ] [EPS] = k 7[EP] [ S ] K8 = k 7k 3[ S ] k 8 k 4[E] [P] From, [ESP] = k 5[P] k 2 k 6k 1[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 +[]
+[]
[
[]
(
[]
)]} Solution 2.11[ES] =
[]
[]
[
[]
[]
]
V =[]
=
[]
[]
[
[]
[]
]
b) KSP =[][]
[]
KPS =[][]
[]
KSP =[][][]
[]
KPS =[][][]
[]
[ESP] =[][][]
[EPS] =[][][]
Given: [ESP] = [EPS] KS KSP = KP KPS
=
[]
=[]
[]
[]
[]
[]
[]
=[]
[]
[]
[]
[]
[]
c)
Ks=Kps [ESP]=[EPS] Kp=Ksp []
[
]
[]
[]
[]
[][][]
[
][]
[]
[]
[]
[
][]
[]
[][]
[
[]
]
[] []
Compare with
[]
[]
Hence, Vmax =
[
]
[]
Km=
[]
[]
d)
[]
[]
[]
*∫
[]
[]
∫
+
*
[][]
+
[
[][]]
[][]
[
][
]
[
][]
[]
[
]
[
][]
n[]
[
]
[
][]
n[][
]
n
[
[]
]
[
][]
Y = mx+c Y =
n
[
[]
]
M=
X=
[
][]
C=
So we can plot a graph of
Rate: rp = k 9CES +k 10CEIS1 +k 10CEIS2 ---- 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= k 2/k 1 ---- 5 CECI/ CEI= K4/K3 ---- 7 CESCI/CEIS1 = K6/K5 ---- 6 CEICS/ CEIS2= K8/K7 ---- 8 By rearranging Equation 5, CE= (k 2/k 1) CsCES From Equation 2, CEo= [(k 2/k 1)CE+ 1] CES CES= CEo/[( k 2/k 1)CS+1] ---- 9 By rearranging Equation 6, CES= [(K6/K5)CI] CEIS1 From Equation 3, CEo= CEIS+CES+ (k 2/k 1) Cs CES = {CEIS1+ [1 + (k 2/k 1) Cs]( K6/K5)CI}CEIS1 = {1 + [1 + (K2/K1) Cs]( K6/K5)CI} CEIS1 CEIS1= CEo/ {1 + [1 + (k 2/k 1) Cs]( K6/K5)CI} ---- 10 By rearranging Equation 7, CE= (K4/K3) CEI By rearranging Equation 8, CEI= K8/K7CSCEIS2 Solution 2.14
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 = k 9CES +k 10CEIS1 +k 10CEIS2,
By substituting Equation 9, 10 & 11 into Equation 1, Therefore,
= k CEo/[ /k )C +1] + k10 CEo/ {1 + [1 + /k ) C ] /K )C } + k10 CEo/ {1 + [1 + /K )C
a) Based on the graphs
The y-intercept in Lineweaver
–
Burk plot is almost the same. Y-intercept => 3.8266; 3.6342Whereas 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 = -0.005In Line weaver
–
Burk Plot and Langmuir Plot both indicatesit’s a competitive inhibitor
Data : Lineweaver 1/s 1/Vo 1/Vi 312.5 9.009009 16.94915 204.0816 6.756757 14.08451 161.2903 6.993007 10.98901 125 6.024096 9.009009 105.2632 5 8 Langmuir s s/Vo S/Vi 0.0032 0.028829 0.054237 0.0049 0.033108 0.069014 0.0062 0.043357 0.068132 0.008 0.048193 0.072072 0.0095 0.0475 0.076 Solution 2.15y = 0.0172x + 3.6342 y = 0.0439x + 3.8266 0 2 4 6 8 10 12 14 16 18 20 0 100 200 300 400
Lineweaver-Burk Plot
1/Vo 1/Vi Linear (1/Vo) Linear (1/Vi) b) Y-intercept = 1/Vmax = 0.00489Vmax = 1/0.00489 = 204.5 mol /L.min Km/Vmax = 2.9883 Km=2.9883*204.5 =611mol/L y = 3.3133x + 0.0191 y = 2.9883x + 0.0489 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0 0.002 0.004 0.006 0.008 0.01
Langmuir Plot
s/Vo S/Vi Linear (s/Vo) Linear (S/Vi)(a) E + S
↔
ES + S
↔
→
k5 E + P(E0) = (E) +(ES) + (ESS)
(E) = (E0)
–
(ES)–
(ESS) ---
V =
= k 5(ES) ---
=
= (ES)(S)/ (k 4/ k 3) K2/ k 1= (E)(S) / (ES) K2/k 1(ES) = (E0)(S)–
(ES)(S) (ES)((k 2/k 1) + (S)) = (E0)(S)–
(ES)(S)2 /
(ES)( (k 2/k 1) + (S)(
) ) =
(E0)(S)–
(ES)(S)2 (ES) ( (k 2/k 1) + (S)(
) + (S)2 ) =
(E0)(S) (ES) =
(E0)(S) / (k 2/k 1) + (S)(
) + (S)2---
3→
V =
=
k 5(E0) (S) / (k 2/k 1) + (S)(
) + (S)2 =
Vm(S) / (k 2/k 1) + (S)(
) + (S)2(b) At low substrate concentration, 1/ Vm= 3.1209 Vm = 0.3204 Km/Vm= 106.07 Km/ 0.3204 = 106.7 K1 m = 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 Solution 2.16
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.minb) 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 Solution 2.17
[SO]1= 0.1 mol/L [S0]2= 0.3mol/L [E0] = 0.05 mol/L
[][]
[]
[][]
[]
[][]
[]
[][]
[]
[][]
[]
[][]
[]
V1=[]
= k 5[ES1] =3.5 [ES1] ---
V2=[]
= k 6[ES2] =2.8 [ES2] ---
[E0] = [E] + [ES1] + [ES2]
[E0] = [E] + [ES1] +
[][]
[E0] = [ES1] + [E] (1+
[]
) [E0] = [ES1] +[]
[]
(1+[]
) [E0] = [ES1] {1 +
[]
(1+[]
)} [ES1] =[]
[]
[]
Solution 2.18Vmt = [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] +
[][]
+ [ES2] [E0] = [E] (1+[]
)+ [ES2] [E0] =[]
[]
(1+[]
)+ [ES2] [E0] = [E2] [
[]
(1+
[]
)+ 1] Vmt = [S1]0–
[S2] + KMln[]
[]
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
Data : s s/v 6.7 22.33333 3.5 14 1.7 10.625 1/Vm = 2.3722 Vmax = 0.4215 mumol/L.min Km/Vm = 6.2429 Km = 6.2429(0.4215) =2.63mumol/L y = 2.3722x + 6.2429 0 5 10 15 20 25 0 1 2 3 4 5 6 7 8
Langmuir Plot
Solution 2.19Since 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: P E ES ESI I ES EIS S EI EI I E ES S E k k k k k k k k k 9 8 , 7 6 , 5 4 , 3 2 , 1 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,
SI I IS S K k k K k k K k k K k k 7 8 3 4 5 6 1 2
If the slower reaction, the product formation step, determines the rate of reaction according to Michaelis-Menten assumption, the rate can be expressed as:
] [
9 ES
k
r P (1)
The enzyme balance gives
] [ ] [ ] [ ] [ ]
[ E 0 E ES EI ESI (2)
Divide (1) by (2), ] [ ] [ ] [ ] [ ] [ ] [ 9
0 E ES EI ESI
ES k E r P (3) Solution 2.20
a. Write the kinetic model.
Applied Law of mass action, S K S E ES ES S E K K Ks [ ] [ ][ ] ] [ ] ][ [ 1 2 (4) I I K I E EI EI I E K K K [ ] [ ][ ] ] [ ] ][ [ 3 4 (5) I I K I ES ESI ESI I ES k k K [ ] [ ][ ] ] [ ] ][ [ 7 8 (6) Substitute (4), (5), (6) into (3), I I S S P K I ES K I E K S E E K S E k E r ] ][ [ ] ][ [ ] ][ [ ] [ ] ][ [ ] [ 9 0 I S I S S P K K I S E K I E K S E E K S E k E r ] ][ ][ [ ] ][ [ ] ][ [ ] [ ] ][ [ ] [ 9 0 Eliminate [E], I S I S S P K K I S K I K S K S k E r ] ][ [ ] [ ] [ 1 ] [ ] [ 0 9 Substitute [ 0] 9 max k E r P I S I S S P P K K I S K I K S K S r r ] ][ [ ] [ ] [ 1 ] [ max
Multiply numerator and denominator by Ks,
I I S S P P K I S K I K S K S r r ] ][ [ ] [ ] [ ] [ max
