James Lee Enzyme Kinetics Solution

BK10110302 V.PRASARNTH RAAJ VEERA RAO – BIOPROCESS

y = 0.033x + 0.0391

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35

S vs t Graph

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


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:

( ) ( )( ) ( )( )


( )( ) ( )( )

( )

( ) [( )( )]

( )( )

( )

( ) ( )( )

ub into :

( ) ( )( ) ( )( )

( ) ( )( )( ) ( )( )( )

Make as a subject,

( )( ( )( ))

( )( )( )

( ) ( )( )( )

( )(

( )( ))

( ) ( )( )( )

(

) ( ) ( )( )

ub into ,


d[p]

dt (

( )( )( )

( ) ( )( )

)

( )( )( )

( ) ( )( )

)

Since ( )

( )( )

( ) ( )( )

)

b) Since [ ] [ ]

d[p]

dt ( )( )

( )( )


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


V= d [P] = k5 [E0] [S]

d t k2 k4 + k4 [S] + [S]

k1 k3 k3

= Vm [S]

k2 k4 + k4 [S] + [S]

k1 k3 k3


Solution 2.4

a) Michaelis-Menten approach

The rate of product formation.

d[p]

dt ( )


Make E as the subject,

Since forward reaction = backward reaction.

( )

ubstitute into :

[( ) ( ) ( )] ( )

( ) ( )( ) ( )( ) ( )( )

Make as a subject:

( ) ( )( ) ( )( )


( )( ) ( )( )

( )

( ) [( )( )]

( )( )

( )

( ) ( )( )

ub into :

( ) ( )( ) ( )( )

( ) ( )( )( ) ( )( )( )

Make as a subject,

( )( ( )( ))

( )( )( )

( ) ( )( )( )

( )(

( )( ))

( ) ( )( )( )

(

) ( ) ( )( )

ub into ,


d[p]

dt (

( )( )( )

( ) ( )( )

)

( )( )( )

( ) ( )( )

)

Since ( )

( )( )

( ) ( )( )

)

b) Briggs-Haldane approach

←

→

←

→

←

→

→

The rate of product formation,

d(p)

dt ( )


Make as a subject,


Substrate consumption,

d( )

dt ( )( ) ( )( ) ( ) ( )

d( )

dt ( )( ) ( ) ( )

ubstitute into :

( )( ) ( ) ( )( ) ( )

( )( ( ) ) ( )( ) ( )

( )( ( ) ) ( ) ( )( ) ( )( ) ( )

( )( ( ) ( )) ( ) ( )( ) ( )

( ) ( ) ( )( ) ( )

( ( ) ( ))

ubstitute into

( )( ) ( )( )

( )( ) ( ) ( )( ) ( )

( ( ) ( )))( )

( )( ) ( ) ( ) ( ) ( )( ) ( ) ( )

( ( ) ( )))


( )( )( ( ) ( ))

( ) ( ) ( ) ( )( ) ( ) ( )

( )( ( ) ( ) ( ) )( ( ) ( )) ( ) ( )

( ) ( ) ( )

( ( ) ( ) ( ) )( ( ) ( ))

ubstitute into

d(p)

dt (

( ) ( )

( ( ) ( ) ( ) )( ( ) ( ))

v

d(p)

dt

v ( ) ( )

( ( ) ( ) ( ) )( ( ) ( ))


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


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 50 100 150 200 250 300 350

Lineweaver Burk Plot


y = -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

Non-Linear Regression Procedure


Solution 2.6

↔

→

Rate of product formation

v ( )

( )

Enzyme is preserved,

( )

( )

d( )

dt neg igib e

( )

( )( ) ( ) ( )

( )( ) ( )( )

( ) ( )( ) ( )

Substitute equation into

( ) ( ( ))( ) ( )

( ) ( ) ( )

( )

( )( ( ) ) ( )

( ) ( )

( ( ) )

Substitute into

v d(p)

dt (

( ) ( ( ) )

)

( )

( ( ) )

Assumptions:

[𝐸𝑂]small, 𝑑(𝐸𝑆)

𝑑𝑡

𝑣𝑚 𝑘 𝑘 𝐸𝑜

𝐾𝑚 𝑘 𝑘 𝑘


Dividing with the value of

[( )

( ( ) ) ]/

v v s

( ( ) )


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

FCSo Km + FCSo Cs - FKm Cs – FCs2 = 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


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 :

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))


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


Solution 2.9

Km = 0.03mol/L

rmax = 13mol/L min × 60 = 780mol/L hr

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 ) = rmax t

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 2.10

Km = 10g/L

rmax = 7g/L.min

a)

CSTR @ Steady State


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

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


b)

CSTR @ Steady State


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.

2 L

F=0.5L/min Cs0=50g/L

F=0.5L/min Cs1=?g/L


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 + [ ]

+ [ ]

[

[ ]

(

[ ]

)]}


[ES] =

[ ]

[ ]

[ [ ]

(

[ ]

)]

V = [ ]

=

[ ]

[ ]

[ [ ]

(

[ ]

)]

b) KSP = [ ][ ]

[ ] KPS =

[ ][ ]

[ ]

KSP = [ ][ ][ ]

[ ] KPS =

[ ][ ][ ]

[ ]

[ESP] = [ ][ ][ ]

[EPS] =

[ ][ ][ ]

Given: [ESP] = [EPS]

KS KSP = KP KPS

=

[ ]

=

[ ]

[ ]

[ ]

[ ]

[ ] [ ]

= [ ]

[ ]

[ ] [ ]

[ ]

[ ]

c)

Ks=Kps [ESP]=[EPS]

Kp=Ksp

[ ]

( )[ ]

( ) [ ] [ ] [ ]

[ ][ ]

[ ] [ ][ ] [ ]

𝑘𝑠𝑘𝑝 𝑘𝑝𝑠𝑘𝑠𝑝


[ ][ ]

[ ]( [ ]) [ ]

[ ]

( [ ])[ ]

[ ]

( [ ]) [ ]

Compare with

[ ]

[ ]

Hence, Vmax = [ ]

( [ ])

Km= [ ]

( [ ])

d) [ ]

[ ]

[ ]

*∫ [ ]

[ ]

∫

+

* [ ]

[ ]

+

[ [ ] [ ]]

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

[ ] [ ] n ([ ]

[ ])

[ ] [ ]

n ([ ] [ ])

n ([ ]

[ ])

([ ] [ ]

)


Y = mx+c

Y =

n ([ ]

[ ])

M=

X= ([ ] [ ]

)

C=

So we can plot a graph of

n ([ ]

[ ]) vs (

[ ] [ ]

)


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


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 } ---- 10


CE = (K4/K3) CEI


CEI = K8/K7CS CEIS2


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 }


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 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


y = 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 = 0.00489

Vmax = 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)


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 ) =

(E0)(S)

(ES) =

(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


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


Solution 2.18

[SO]1 = 0.1 mol/L [S0]2 = 0.3mol/L [E0] = 0.05 mol/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] + [ ]

[ ] (1+

[ ]

)

[E0] = [ES1] {1 +

[ ] (1+

[ ]

)}

[ES1] = [ ]

[ ] (

[ ]

)


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] + [ ][ ]

+ [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

[S2]. As [S1] increases, [ES1] also increases as in eq.4. [P2] also increases as in eq.2


Solution 2.19

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 2.20

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:

PEES

ESIIES

EISSEI

EIIE

ESSE

k

kk

kk

kk

kk

9

8,7

6,5

4,3

2,1

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,

SII

ISS

Kk

kK

k

k

Kk

kK

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 ESkrP (1)

The enzyme balance gives

][][][][][ 0 ESIEIESEE (2)

Divide (1) by (2),

][][][][

][

][

9

0 ESIEIESE

ESk

E

rP

(3)


Applied Law of mass action,

SK

SEES

ES

SE

K

KKs

]][[][

][

]][[

1

2 (4)

I

IK

IEEI

EI

IE

K

KK

]][[][

][

]][[

3

4 (5)

I

IK

IESESI

ESI

IES

k

kK

]][[][

][

]][[

7

8 (6)

Substitute (4), (5), (6) into (3),

IIS

SP

K

IES

K

IE

K

SEE

K

SEk

E

r

]][[]][[]][[][

]][[

][

9

0

ISIS

SP

KK

ISE

K

IE

K

SEE

K

SEk

E

r

]][][[]][[]][[][

]][[

][

9

0

Eliminate [E],

ISIS

SP

KK

IS

K

I

K

S

K

S

kE

r

]][[][][1

][

][ 90

Substitute 90 ][max

kErP

ISIS

S

P

P

KK

IS

K

I

K

S

K

S

r

r

]][[][][1

][

max

Multiply numerator and denominator by Ks,

II

SS

P

P

K

IS

K

IKSK

S

r

r

]][[][][

][

max


Rearranging,

II

SS

P

P

K

ISS

K

IKK

S

r

r

]][[][

][

][

max

)][

1]([)][

1(

][

max

II

SP

P

K

IS

K

IK

S

r

r

James Lee Enzyme Kinetics Solution

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