James Lee Enzyme Kinetics Solution

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

(

(

)

(

(

)

)( )

)( )

(

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)( )( )

(

)( )( )

as a subject, (

)(

(

( )( ))

(

) (

(

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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

(

(

)

(

(

)

)( )

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(

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(

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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

(

)

(

)

)( )

(

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(

)

(

( )( )

(

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)( )

(

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(

( )(

( ))

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( )(

( )( (

(

(

)

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( ))

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