147965772-James-Lee-Enzyme-Kinetics-Solution.pdf

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


   ←     ←   →  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:

        


             []      ub into:

                 

Make



           as a subject,

ub  into,

                                      


Since

  

][]

  ddt[p]                          



           

Since [

       ddt[p]      


(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


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


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,

                           


Since

  

  ddt[p]                          



           

Briggs-Haldane approach

  ←    ←     ←   →  The rate of product formation,

  dp dt    Since the enzyme is preferred,

   

Make as a subject,

     


Substrate consumption,

d      dt d     dt ubstitute  into:               ( )         (   )

ubstitute  into            (  )       (  )


 ( )      ( )                 ( )         ubstitute  into             dp     dt   ( ) v    v   dp  dt   ( )


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 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


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


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


 ↔   →  Rate of product formation

v     



Enzyme is preserved,

         d negigibe dt                                           v  dp   dt           Substitute equation

Substitute

Assumptions: [ ]small,

 

  



into

into

 

         






   

Dividing

with the value of

 ]/ s  v   v  [




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

– –

–

–


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


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


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


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


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.


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

[

[

(

(

)]

)]}


[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]



  []  [] []    []  [] [][]  [] [][]   []


[][]  [] []  [] [] []  [] [] []  [] 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 [][]  [] 


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


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


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


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


 ↔   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


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


[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


–

–

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


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


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)


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)

147965772-James-Lee-Enzyme-Kinetics-Solution.pdf

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