147965772-James-Lee-Enzyme-Kinetics-Solution.pdf
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BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Total volume = 44 + 5 + 1= 50ml From the graph the equation obtained y=0.033× + 0.03a 1 m= 0.033
a)
ɱ
mol / ml . min
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
← ← → Michaelis-Menten Michaelis-Menten approach The rate of product formation.
ddt[p] 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:
Make
as a subject,
ub into,
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Since
][]
ddt[p]
Since [
ddt[p]
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
(a) E+S k 1
(ES)1
(ES)1 k 3 (ES)2 (ES)2 k 2
[] = k [ES]
V=
5
E+P
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] ) = [E 0] [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 2 k 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 1 k 3 k 3 [ES]2 = [E0] [S] k 2 k 4 + k 4 [S] + [S] k 1 k 3 k 3
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
V= d [P] = k 5 [E0] [S] dt k 2 k 4 + k 4 [S] + [S] k 1 k 3 k 3 =
Vm [S] k 2 k 4 + k 4 [S] + [S] k 1 k 3 k 3
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
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:
Make
as a subject,
ub into,
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Since
ddt[p]
Briggs-Haldane approach
← ← ← → The rate of product formation,
dp 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 dp dt ( ) v v dp dt ( )
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
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 0.0049
0.111 0.148
0.0062
0.143
0.008
0.166
0.0095
0.2
Type of Plot Langmuir Lineweaver-Burk Eadie-Hofstee Non-Linear Regression
Kinetic Parameters Vmax Km 0.2750 0.0047 0.0191 0.0057 0.2645 0.0043 0.2000 0.0032
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
y = 0.0172x + 3.6342
8 7 6 5 4 3 2 1 0 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
↔ → Rate of product formation
v
Enzyme is preserved,
d negigibe dt v dp dt Substitute equation
Substitute
Assumptions: [ ]small,
into
into
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Dividing
with the value of
]/ s v v [
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
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
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
+ (Cs - Cs ) = r t 0.03 ln + (10 - 0.5 ) = 780t Km ln
0
max
9.95899 = 780t t = 0.0123hr
t = V/F = 0.0123 V = 0.0123 × 10 = 0.123liter
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Km = 10g/L rmax = 7g/L.min
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 2
250+25Cs1-5Cs1-0.5Cs1 =7Cs1 2
0.5Cs1 -13Cs1-250=0 Cs1=38.86g/L
0.5 (38.86
–
Cs2) =
s s
(1)
(19.43-0.5 Cs2)(10+ Cs2)=7 Cs2 2
194.3+19.43Cs2-5Cs2-0.5Cs2 =7Cs2 2
0.5Cs2 -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 2
250+25Cs1-5Cs1-0.5Cs1 =14Cs1 2
0.5Cs1 -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
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 7 k 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] [
(
)]
[][] + [] + [E ] = [ES] {1 + [E0] = [ES] +
0
[
[
(
(
)]
)]}
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
[ES] =
V=
[] [] [[] [] ]
[] = [] [][] [] [ ]
b)
[][] [] [][][] K = [] [][][] [ESP] =
[][] [] [][][] K = [] [][][] [EPS] =
KSP =
KPS =
SP
Given:
PS
[ESP] = [EPS] KS KSP = KP KPS
=
[] = [] [] [] [] [] [] [] = [] [] [] [] [] c)
Ks=Kps Kp=Ksp
[ESP]=[EPS]
[] [] [] [] [] [][] [] [][] []
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
[][] [] [] [] [] [] [] [] [] [] Compare with
[] []
[] [] [] Km= [] Hence, Vmax =
[] [][]
d)
*∫ [] [] ∫ + [ ] * [] + [[] []] [][][][]
[][][][] [] [] n[[]] [][] n[][] [] [ ][] n []
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS Y = mx+c
n [] [] M= [ ][] X= C= Y=
So we can plot a graph of
n [] vs [][] []
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
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) Cs CES 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) C s ]( K6/K5)CI } CEIS1
CEIS1 = CEo/ {1 + [1 + (k 2/k 1) 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 + (K 4/K3)CI ]( K8/K7)CS } CEIS2
CEIS2 = CEo / {1 + [1 + (K 4/K3)CI ]( K8/K7)CS }
From Equation 1, since r p = k 9CES +k 10CEIS1 +k 10CEIS2, By substituting Equation 9, 10 & 11 into Equation 1, Therefore,
---- 11
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
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 = -0.016
;
X = -0.005
X=
–
In Line weaver Burk Plot and Langmuir Plot both indicates
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
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
Langmuir
it’s a competitive inhibitor
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Lineweaver-Burk Plot 20 18
y = 0.0439x + 3.8266
16 14 1/Vo
12
1/Vi
10
y = 0.0172x + 3.6342
8
Linear (1/Vo)
6
Linear (1/Vi)
4 2 0 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
↔ ES + S ↔
(a) E + S
→
k5
E+P
(E0 ) = (E) +(ES) + (ESS)
– –
(E) = (E0 ) (ES) (ESS) --------
= k (ES) ------ = V=
5
= (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 /
(E )(S) – (ES)(S) (ES) ( (k /k ) + (S)( ) + (S) ) = (E )(S) (ES) = (E )(S) / (k /k ) + (S)( ) + (S) -------- (ES)( (k 2/k 1) + (S)( ) ) = 2
2
1
0
3→
0
2
2
0
2
1
= k (E ) (S) / (k /k ) + (S)( ) + (S) = V (S) / (k /k ) + (S)( ) + (S)
V=
m
5
0
2
2
1
1
(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
2
2
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
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
[SO]1 = 0.1 mol/L
[S0]2 = 0.3mol/L
[][] [] [][] [] [][] [] [][] [] [][] []
[][] [] V1 =
[] = k [ES ] 5
=3.5 [ES1] --V2 =
1
[] = k [ES ] 6
=2.8 [ES2] ---
2
[E0] = [E] + [ES 1] + [ES 2]
[][] [] ) [E ] = [ES ] + [E] (1+ [] (1+ [] ) [E ] = [ES ] + [] (1+ [] )} [E ] = [ES ] {1 + [] [] [] [ES ] = [] [E0] = [E] + [ES 1] + 0
1
0
1
0
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 + [S 1] = 0.1 +0.0714 ln 0.1 - 0.0714 ln [S 1] 3.5 [ES1] t + [S 1] + 0.0644 = -0.0714 ln [S 1]
[] [] [] [] [S ] =
ln[S1] = 1
---
[E0] = [E] + [ES 1] + [ES 2]
[][] + [ES ] [] [E ] = [E] (1+ )+ [ES ] [] (1+ [] )+ [ES ] [E ] = [] (1+ [] )+ 1] [E ] = [E ] [ [] [] Vmt = [S ] – [S ] + K ln [] 2.8[ES ] t = 0.3 – [S ] + 0.2207ln [] [E0] = [E] +
2
0
2
0
2
0
2
1 0
2
M
1
2
–
2.8[ES1] t + [S 2] = 0.3 + 0.2207ln 0.3 0.2207ln [S2]
–
–
2.8[ES1] t + [S 2] 0.0343 = 0.2207ln [S2]
[] [] – – [] [] – [S ] = e – ln[S2] =
2
---
As [S1] increases, [ES 1] also increases as in eq.3. [P 1] also increases as in eq.1. This also occurs in [S2]. As [S 1] increases, [ES1] also increases as in eq.4. [P 2] also increases as in eq.2
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Data : s
s/v
6.7
22.33333
3.5
14
1.7
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
Since the Michaelis constant K M 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 k 9
ES E P
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 2 k 1
K S
K I
k 4 k 3
k 6
k 5
K IS
K SI
k 8 k 7
If the slower reaction, the product formation step, determines the rate of reaction according to Michaelis-Menten assumption, the rate can be expressed as: r P
k 9 [ ES ]
(1)
[ E 0 ] [ E ] [ ES ] [ EI ] [ ESI ]
(2)
The enzyme balance gives
Divide (1) by (2), r P
[ E 0 ]
k 9 [ ES ]
[ E ] [ ES ] [ EI ] [ ESI ]
(3)
BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS
Applied Law of mass action, Ks
K 2
[ E ][S ]
[ ES ]
K 1
K I
K I
K 4
[ E ][ I ]
[ EI ]
K 3
k 8
[ ES ][ I ]
k 7
[ ES ]
[ EI ]
[ ESI ]
[ E ][S ] K S
[ E ][ I ] K I
[ ESI ]
[ ES ][ I ] K I
Substitute (4), (5), (6) into (3),
[ E ][ S ]
k 9
r P
[ E 0 ]
[ E ]
[ E ][ S ] K S
K S
[ E 0 ]
[ E ]
[ E ][ S ] K S
K I
[ ES ][ I ] K I
[ E ][ S ]
k 9
r P
[ E ][ I ]
K S
[ E ][ I ]
K I
[ E ][ S ][ I ] K S K I
Eliminate [E],
[ S ] r P
[ E 0 ]k 9
Substitute r P max
K S
1
[ S ] K S
[ I ] K I
[ S ][ I ] K S K I
[ E 0 ]k 9
[ S ] r P r P
max
K S
1
[ S ] K S
[ I ] K I
[ S ][ I ] K S K I
Multiply numerator and denominator by K s, r P r P
max
K S
[ S ]
[ S ] K S [ I ] K I
[ S ][ I ] K I
(4)
(5)
(6)
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